that in turn acts as vector subtraction (tail-to-tail). When coupled to a
system of coordinates, we saw that we must be able to describe vectors in
terms of their lengths along each basis vector — we call these lengths the
components of the vector. Later on, to be consistent between algebra and
geometry, the we found rules for scalar multiplication and the dot product.
Finally, using areas of parallelograms, we built the component form of the
cross product AND inserted the idea of handedness into our coordinate systems
by defining cross product relationships between our basis vectors. Table 1.1
lists the equations that are the most important (and general) throughout this
chapter. I included the equation references to bring you back to the
discussion where we derived them, just in case you need some context to
refresh your memory about what each equation means.

Sprinkled throughout the chapter are several proofs that I recommended that
you try to do. All of them are practically identical to what I have already
done in the chapter, although they may have more components or a couple more
steps. I sincerely suggest that you try to do a few of them just to get a feel
for how these proofs are done on your own — in physics and math, one of the
best ways to deeply understand derivations or proofs is by putting in the time
to do it out for yourself (although in my experience, a physics and math
educations leaves very little time for anything but perpetual confusion…).

I know that for many of you reading this chapter, my lack of numerical values
will be unsettling — I will only get more algebraic as the chapters progress.
I do this on purpose though. My first reason is that it is frankly easier once
you get used to it. I do remember that the phase transition between needing
numbers in math and exclusively using letters is not a smooth one. It will
take time to master — probably as much time as it did for you when you first
started using the symbol $\pi$ instead of $3.14159$ back in the day. The
advantage to only using letters or symbols is that you eventually need not
worry about how the intermediate numerical values affect the outcome of your
mathematics. By extension, the necessary variables will be left in your
physical models of the natural world giving a deeper insight into how the
universe works. Unfortunately, unless you already have a handle on the
physics, injecting numerical values at intermediate steps will obfuscate this
insight. But again, it takes time (a.k.a. practice) to get used to doing
everything algebraically. Hopefully this chapter and the following will serve
as an external perturbation to make your phase transition that much easier.

TABLE 1.1: A summary of the important and general equations derived for vector
operations.

Equation Description | Equation Formula | Text Reference
---|---|---
3-Dimensional Cartesian Vector | $\vec{v}=v_{x}\hat{x}+v_{y}\hat{y}+v_{z}\hat{z}$ | Eq. 1.3
$n$-Dimensional Cartesian Vector | $\vec{v}=v_{1}\hat{x}_{1}+v_{2}\hat{x}_{2}+\dots=\sum\limits_{j=1}^{n}v_{j}\hat{x}_{j}$ | Eq. 1.4
Angle between $xy$ Components | $\tan\theta_{xy}=\dfrac{v_{y}}{v_{x}}$ | Eq. 1.5
Length/Magnitude of Vector | $|\vec{v}|=v=\sqrt{\sum\limits_{j=1}^{n}v_{j}^{2}}$ | Eq. 1.8
Sum of Two Vectors | $\vec{u}+\vec{v}=\sum\limits_{j=1}^{n}(u_{j}+v_{j})\hat{x}_{j}$ | Eq. 1.13
Difference of Two Vectors | $\vec{u}-\vec{v}=\sum\limits_{j=1}^{n}(u_{j}-v_{j})\hat{x}_{j}$ | Eq. 1.14
Scalar Multiplication | $a\vec{v}=\sum\limits_{j=1}^{n}(av_{j})\hat{x}_{j}$ | Eq. 1.15
Unit vector in Direction of $\vec{v}$ | $\hat{v}=\dfrac{\vec{v}}{|\vec{v}|}=\dfrac{\vec{v}}{v}$ | Eq. 1.16
Dot Product | $\vec{u}\cdot\vec{v}=\sum\limits_{j=1}^{n}u_{j}v_{j}=uv\cos{\theta}$ | Eq. 1.21
Projection of $\vec{v}$ onto $\vec{u}$ | $\textrm{proj}_{\hat{u}}(\vec{v})=\dfrac{\vec{u}\cdot\vec{v}}{|\vec{u}|}=\dfrac{\vec{u}\cdot\vec{v}}{u}$ | Eq. 1.23
Length/Magnitude and Dot Product | $|\vec{v}|=v=\sqrt{\vec{v}\cdot\vec{v}}$ | Eq. 1.25
Magnitude of Cross Product | $|\vec{v}\times\vec{u}|=uv\sin\theta$ | Eq. 1.27
Right-Handed Cartesian Basis | $\begin{aligned} \hat{x}\times\hat{y}&=\hat{z}\\\ \hat{z}\times\hat{x}&=\hat{y}\\\ \hat{y}\times\hat{z}&=\hat{x}\end{aligned}$ | Eq. 1.36
3-Dimensional Cross Product | $\begin{aligned} \vec{u}\times\vec{v}&=\;\;\;\;\left(v_{y}u_{z}-v_{z}u_{y}\right)\hat{x}\\\ &\;\;\;\;+\left(v_{z}u_{x}-v_{x}u_{z}\right)\hat{y}\\\ &\;\;\;\;+\left(v_{x}u_{y}-v_{y}u_{x}\right)\hat{z}\end{aligned}$ | Eq. 1.37
Definition of Linear Operator | $\mathcal{O}(a\vec{u}+b\vec{v})=a\mathcal{O}(\vec{u})+b\mathcal{O}(\vec{v})$ | Eq. 1.39

##  Chapter 2 Complex Algebra

In this chapter, we will start a perhaps unfamiliar form of mathematics for
many of you, but it is crucial to not only our understanding of quantum
phenomena, but also our understanding of differential equations, Fourier
Analysis of signals, all kinds of waves, and various forms of data analysis.
By the end of this chapter, we want to be able to add complex numbers to our
set of “game pieces” that we have been developing, and then eventually get to
the point that we can create functions of complex numbers. This chapter should
leave you in a pretty good position to understand most, if not all, the
complex mathematics you will cover in your undergraduate physics curriculum.
This chapter should also set you up to begin to learn the calculus of complex-
valued functions later on in your mathematics career.

###  2.1 The Lie of Imaginary Numbers

Before we get going, there is a common misconception that I want to clear up.
There is no such distinction between real and imaginary numbers in the
colloquial sense; that is, there is no set of objects that are somehow
tangible that we call the real numbers, $\mathds{R}$, versus the somehow
intangible objects called the imaginary numbers, $\mathds{I}$. These sets of
objects are certainly distinct mathematically, but that is due to a rotation
rather than some metaphysical and mystical separation that seems to exist by
calling two things real and imaginary.

The reason why I want to address this is because the term “imaginary” has a
totally different connotation in normal life than it does in mathematics. To
be clear, at some point in the development of our algebra system, some
mathematicians like Rene Decartes did truly believe that imaginary numbers
really were not a thing, but a rather convenient way out of an otherwise
harder problem [ComplexNumberHistory]. However, mathematicians today
effectively only use the word as a label whose name bears no deep meaning. We
credit people like Gauss, Cauchy, Euler, and Riemann, among others, for
changing the way we think about these objects mathematically. Gauss showed
that imaginary numbers are truly just an extension of the more conventional
real numbers, and had even tried to get them rebranded as lateral numbers
instead. His quote on this subject is below [Gauss_Quote].

> That this subject [imaginary numbers] has hitherto been surrounded by
> mysterious obscurity, is to be attributed largely to an ill adapted
> notation. If, for example, $+1$, $-1$, and the square root of $-1$ had been
> called direct, inverse and lateral units, instead of positive, negative and
> imaginary (or even impossible), such an obscurity would have been out of the
> question.

The problem with taking the word “imaginary” too literally in physics makes
the interpretation of certain natural phenomena appear as fake or something
akin to pseudoscience. For example, when describing an electron’s spin, the
imaginary (lateral!) unit $i=\sqrt{-1}$ appears when talking about the
projection of the spin vector along the $y$-axis. If we were to naively see
the presence of $i$, we might be tempted to conclude that there is something
mystical about this part of our physical world. Or even worse, we might
conclude that we could never measure the $y$ component of the spin because it
is imaginary! But this interpretation is not true. Gauss’ idea of lateral
numbers can be used to more appropriately explain the appearance of $i$ in
electron spins. In this case, as you will learn later in your physics career,
we can only ever know precisely the projection of an electron’s spin along one
axis in space; we denote the forward direction with a $+$ sign and the
backward direction with a $-$ sign. However, we can measure the statistical
effects of the spin’s vector components in the other two dimensions in 3D
space. What this means is we have a total of 3 sets of distinct pairs of basis
vectors in this spin space, which is supposed to have physical meaning in all
of 3D space. Without going into too much linear algebra, we essentially need 4
components to represent the final two perpendicular axes in 3D space — but
this is impossible with only the real numbers! Real components can only ever
give you the magnitude and direction that a vector points along a particular
line, as we discussed in the chapter on Vectors. Hence, our very real
measurements of the natural world force us to extend the real numbers to
include their lateral counterparts in order to accurately describe electron
spin.

Without further ado, I will quit my (legitimate) grumbling, and proceed with
our introduction to the world of complex algebra.

###  2.2 Some Important Definitions

Before we move on with more algebra, we need to lay down some ground rules for
these things that I’m calling complex numbers. The first time they are
typically introduced (although not the first time they were ever contrived
[ComplexNumberHistory]) is with the standard defining equation

$\displaystyle x^{2}+1=0,$ (2.1)

where one solves for $x$ to find $x=\pm\sqrt{-1}$. Since we showed in Section
1.2.1 that for any real number111Remember that we use the symbol $\mathds{R}$
to represent the set of all reals and the symbol $\in$ to mean “is an element
of.” $a\in\mathds{R}$, $a^{2}>0$, then we find that there can be no real
number $x$ that satisfies $x^{2}+1=0$. Thus we define the imaginary unit as

$\displaystyle i=\sqrt{-1}.$ (2.2)

Likewise, we could easily define the real unit with the following equation
$x^{2}-1=0$, and so we obtain $x=\pm 1$. These ideas fit in with our
understanding that $1$ represents “unit” length along a number line. We will
define any imaginary number $\alpha$ as $\alpha=ai$, where $a\in\mathds{R}$.
Such an object would solve the quadratic $x^{2}+a^{2}=0$ for $x$. Likewise, a
real number $a\in\mathds{R}$ would solve the quadratic $x^{2}-a^{2}=0$ for
$x$.

When we move to more complicated quadratics, we have an equation that looks
something like $ax^{2}+bx+c=0$, whose solutions are given by the quadratic
formula

$\displaystyle x=-\frac{b}{2a}\pm\frac{1}{2a}\sqrt{b^{2}-4ac}.$ (2.3)

If we pay attention to the quantity called the discriminant, $D=b^{2}-4ac$, we
should take note there are exactly three cases for $D$ given by the ordering
property of the reals. They are given explicitly as

$\displaystyle D>0$ $\displaystyle\Rightarrow b^{2}>4ac,$ $\displaystyle D=0$
$\displaystyle\Rightarrow b^{2}=4ac,$ (2.4) $\displaystyle D<0$
$\displaystyle\Rightarrow b^{2}<4ac.$

The third case, $D<0$, implies that we will again have a negative number
inside of a square-root, and so this implies that there will be an imaginary
unit involved in some way. Written explicitly, when $D<0$, then it must be
true that $-D>0$. Then

$\displaystyle
x=-\frac{b}{2a}\pm\frac{1}{2a}\sqrt{D}=-\frac{b}{2a}\pm\frac{1}{2a}\sqrt{-(-D)}=-\frac{b}{2a}\pm\frac{i}{2a}\sqrt{-D}=-\frac{b}{2a}\pm\frac{i}{2a}\sqrt{4ac-b^{2}}$
(2.5)

What we have now is something that hopefully is a little jarring to you,
especially if you have never seen complex algebra before. We have an
expression that is somehow telling us to add a real number $-b/2a$ with the
imaginary number $i\sqrt{4ac-b^{2}}/2a$. But can we? There is mathematically a
pretty large distinction between real numbers and imaginary numbers; the
square (and any other nonzero even power) of a real number is always positive.
Meanwhile, the square (and any other nonzero even power) of an imaginary
number is always negative. We proved the former, whereas we had to define the
latter. So what gives?

The way we deal with this conundrum is actually by defining something new. It
turns out that if you were jarred before, you were right, because there is no
way to add a nonzero real number with a nonzero imaginary number and get
either a purely real or imaginary number out. The sets of objects are just too
different. All of their differences can pretty much be reduced to the fact
that

$\displaystyle 1^{2}=1\textrm{ and }i^{2}=-1,$ (2.6)

and so we get around this issue by saying each of these numbers is totally
distinct from one another, but they can be combined to form a greater set of
numbers; just as the basis vectors $\hat{x}$ and $\hat{y}$ are totally
distinct, but can be combined to form a plane. Actually, if we take the real
unit and the imaginary unit as basis vectors, we can create the complex plane,
denoted by $\mathds{C}$. Furthermore, we define a complex number $z$ as the
vector sum of a real part $z=\mathrm{Re}(z)$ and an imaginary part
$z=\mathrm{Im}(z)$ like the following

$\displaystyle z=\mathrm{Re}(z)+i\,\mathrm{Im}(z).$ (2.7)

It is important to see then that this definition constrains both the real part
and the imaginary part to be real numbers! Please read that sentence again —
it confuses a lot of people. Even though the imaginary part of a complex
number is called the imaginary part, it itself is real. It represents the
projection (see Eq. 1.23) of the complex number in the direction of $i$,
whereas the real part is the projection of the complex number in the direction
of $1$.

Using this vector-like interpretation of complex numbers, then it must be true
the so-called “real axis” and “imaginary axis” must together span a complex
plane, just like the $x$-axis and $y$-axis span the $xy$-plane. By convention,
we denote $x=\mathrm{Re}(z)$ and $y=\mathrm{Im}(z)$, so we can write any
complex number $z\in\mathds{C}$ as

$\displaystyle z\in\mathds{C}\textrm{ if and only if }z=x+iy,\textrm{ where
}i=\sqrt{-1}\textrm{ and }x,y\in\mathds{R}.$ (2.8)

Since the complex numbers form a plane, then we can more easily see what Gauss
was talking about when he claimed that the imaginary unit should be instead
named the lateral unit. The existence of this plane just means that the real
numbers are accompanied by another orthogonal axis that has had the misfortune
of having us silly humans call them “imaginary”. The lateral numbers are
present with or without us claiming they are figments of our imaginations —
they are an extension of the real numbers into the generalized complex plane.
With that said, I will continue to refer to them as imaginary numbers just so
you get accustomed to the vernacular. But please do not think any of this is
some kind of fantasy. If you are willing to accept the existence of the real
number line, it is clear there must exist an accompanying imaginary (lateral!)
number line.

###  2.3 Conjugates and Magnitudes

A question that we can now ask since we have defined complex numbers is
whether there is a way to construct the real and imaginary parts of any
complex number computationally. In other words, if we know know any real
number $x$ and any real number $y$, we can compute a complex number $z=x+iy$.
That’s old news. The new question is whether we can find another complex
number $z^{\ast}$ if we know already know a complex number $z$ such that we
could calculate $x=\mathrm{Re}(z)$ and $y=\mathrm{Im}(z)$. If we were to
assume that this were possible, then we would have to start by clarifying that
if $z^{\ast}$ were complex, then $z^{\ast}=u+iv$ for some combination of real
numbers $u$ and $v$. Thus,

$\displaystyle z$ $\displaystyle=x+iy$ $\displaystyle z^{\ast}$
$\displaystyle=u+iv$

As of right now, we have two separate equations with two unknowns. By
“separate”, I mean that as of right now, we don’t have an equation to relate
any of the variables. Hence, we are free to choose that
$x=\mathrm{Re}(z)=(z+z^{\ast})/2$. Thus,

$\displaystyle
x=\mathrm{Re}(z)=\frac{1}{2}\left(z+z^{\ast}\right)=\frac{1}{2}\left[(x+u)+i(y+v)\right]=\frac{x+u}{2}+i\,\frac{y+v}{2}$

Since $x$ is purely real, then that means that it cannot be imaginary. Thus
the coefficient attached to $i$ must vanish; in other words, $y+v=0\Rightarrow
v=-y$. This means that

$\displaystyle
x=\frac{x+u}{2}+i\,\frac{y+v}{2}=\frac{x+u}{2}+i0=\frac{x+u}{2}$

By multiplying both sides by 2 and subtracting over the remaining x we then
find $x=u$. This means we have an expression for $z^{\ast}$, given $z$:

$\displaystyle\textrm{if }z=x+iy\textrm{ then }z^{\ast}=x+i(-y)=x-iy.$ (2.9)

The complex number $z^{\ast}$ is a very helpful quantity — so helpful, in
fact, that it is given the name of the complex conjugate to $\mathbf{z}$.
Additionally, this complex conjugate is unique, as based on our rules for the
real numbers in Section 1.2.1, there exists only one $v=-y\in\mathds{R}$ if
$y\in\mathds{R}$222This uniqueness property justifies our phrasing of the
complex conjugate instead of a complex conjugate.. Notice that
$z^{\ast}=x-iy=x+(-i)y$. This means that whenever we want a complex conjugate
of $z=x+iy$, then all we have to do is replace every $i$ with $-i$. If it is
not immediately clear why this is true, it’s because we have defined complex
numbers in such a way that any number may be written as a real part plus $i$
times an imaginary part, and then we defined the complex conjugate in terms of
those arbitrary real and imaginary parts. This rule is particularly helpful
for when you have a rather nasty function of complex variables, but you need
the conjugate to compute something useful.

Now what would be useful to compute with a complex conjugate? For starters, we
know that we can find the real part of $z$ (or $z^{\ast}$) with the following
formula.

$\displaystyle\mathrm{Re}(z)=\frac{z+z^{\ast}}{2}.$ (2.10)

We can actually calculate the imaginary part of $z$ as well. I leave it to you
to verify that

$\displaystyle\mathrm{Im}(z)=\frac{z-z^{\ast}}{2i}.$ (2.11)

To show this, set $y=\mathrm{Im}(z)$ and use Eq. 2.9 to solve for $y$. If the
division by $i$ is weird for you, then instead use
$1/i=(1/i)(i/i)=i/(i^{2})=i/(-1)=-i$. The third part of Example 2.3 shows
another helpful tool that the complex conjugate provides.

Consider the complex numbers $z=x+iy$ and $w=u+iv$, where
$x,y,u,v\in\mathds{R}$. We seek the real and imaginary parts of $z\pm w$,
$zw$, and $z/w$. 1. We start with $z\pm w$. $\displaystyle z\pm w$
$\displaystyle=(x+iy)\pm(u+iv)$ $\displaystyle=(x\pm u)+i(y\pm v).$ (2.12)
Hence, $\mathrm{Re}(z\pm w)=x\pm u$ and $\mathrm{Im}(z\pm w)=y\pm v$. 2. Next,
we compute $zw$. $\displaystyle zw$ $\displaystyle=(x+iy)(u+iv)$
$\displaystyle=xu+ixv+iyu+i^{2}yv$ $\displaystyle=(xu-yv)+i(xv+yu).$ (2.13)
Thus, $\mathrm{Re}(zw)=xu-yv$ and $\mathrm{Im}(zw)=xv+vu$. 3. Lastly, we seek
the quantity $z/w$ (this one is a little tricky). $\displaystyle\frac{z}{w}$
$\displaystyle=\frac{x+iy}{u+iv}$
$\displaystyle=\frac{x+iy}{u+iv}\,\frac{w^{\ast}}{w^{\ast}}$
$\displaystyle=\left(\frac{x+iy}{u+iv}\right)\left(\frac{u-iv}{u-iv}\right),\textrm{
Using Eq. \ref{eq: Complex Conjugate} for }w$
$\displaystyle=\frac{(x+iy)(u-iv)}{(u+iv)(u-iv)}$
$\displaystyle=\frac{(xu+yv)+i(xv-yu)}{u^{2}+v^{2}}$
$\displaystyle=\frac{xu+yv}{u^{2}+v^{2}}+i\,\frac{xv-yu}{u^{2}+v^{2}}.$ (2.14)
Therefore we have
$\displaystyle\mathrm{Re}\left(\frac{z}{w}\right)=\frac{xu+yv}{u^{2}+v^{2}}\textrm{
and }\mathrm{Im}\left(\frac{z}{w}\right)=\frac{xv-yu}{u^{2}+v^{2}}$ It is
important to note that since we can establish clear real and imaginary parts
for each of the arithmetic operations above — addition, subtraction,
multiplication, and division — these quantities are also complex numbers, by
definition. Thus, when we add, subtract, multiply, or divide complex numbers,
we will always get complex numbers back!

For a next calculation, let’s find the product of $z$ with its conjugate
$z^{\ast}$ which is often denoted as $z^{\ast}z$; we do so by direct
substitution.

$\displaystyle z^{\ast}z$ $\displaystyle=(x-iy)(x+iy)$
$\displaystyle=x^{2}-iyx+ixy-i^{2}y^{2}$
$\displaystyle=x^{2}-(-1)^{2}y^{2}+i(xy-yx)$ $\displaystyle=x^{2}+y^{2}+i0$
$\displaystyle=x^{2}+y^{2}.$

Outright, this formula may not look very impressive, so let’s try
$\sqrt{z^{\ast}z}$:

$\displaystyle\sqrt{z^{\ast}z}=\sqrt{x^{2}+y^{2}}=\sqrt{[\textrm{Re}(z)]^{2}+[\textrm{Im}(z)]^{2}}.$
(2.15)

Hopefully this catches your eye as the Pythagorean Theorem, or even more
importantly, the equation for the length of a two-dimensional vector Eq. 1.6,
where the $x$-component of the vector is just $x$ and the $y$-component is
just $y$! Using Eq. 2.15, called the complex modulus of $z$, we can calculate
a real modulus and and imaginary modulus if we consider only a purely real
number $x$ or purely imaginary number $\Upsilon=iy$.

$\displaystyle\sqrt{x^{\ast}x}$ $\displaystyle=\sqrt{x^{2}+0}=|x|,\textrm{ a
real number has no imaginary part}$
$\displaystyle\sqrt{\Upsilon^{\ast}\Upsilon}$
$\displaystyle=\sqrt{0+y^{2}}=|y|,\textrm{ an imaginary number has no real
part}$

where the absolute value bars are necessary because we implicitly took the
positive square root, and so our moduli should be positive for arbitrary
$x,y\in\mathds{R}$. But these expressions are just the magnitudes of the real
numbers $x$ and $y$. In other words, the complex modulus is the generalization
of magnitude — a.k.a. absolute value — in the complex plane! In other words,
we may write

$\displaystyle|z|=\sqrt{z^{\ast}z}=\sqrt{[\textrm{Re}(z)]^{2}+[\textrm{Im}(z)]^{2}},$
(2.16)

when talking about the magnitude of any complex number.

Before moving on, there are a couple of more things I want to talk about.
Firstly, what happens if we need to find the complex conjugate of a sum (or
difference)? Suppose we have two complex numbers $z$ and $w$, as we do in the
first part of Example 2.3, and we want to find $(z\pm w)^{\ast}$. We could do
so directly as shown below using Eq. 2.12,

$\displaystyle(z\pm w)^{\ast}=(x\pm u)+(-i)(y\pm
v)=(x-iy)\pm(u-iv)=z^{\ast}\pm w^{\ast}.$ (2.17)

Notice that we replaced the $+i$ in Eq. 2.12 with the $-i$ to find the
conjugate initially. Hence, when we have a sum (or difference) of two complex
numbers, then the conjugate of the sum (or difference) is just the sum (or
difference) of the conjugates. How about a product of two complex numbers?
Here we use Eq. 2.13,

$\displaystyle(zw)^{\ast}$ $\displaystyle=(xu-yv)+(-i)(xv+yu),$
$\displaystyle=(xu+i^{2}yv)-i(xv+yu),$ $\displaystyle=xu+(iy)(iv)-ixv-
iyu,\textrm{ combine like-terms in $x$ and $-iy$}$
$\displaystyle=x(u-iv)-iy(u-iv),$ $\displaystyle=(x-iy)(u-iv),$
$\displaystyle=z^{\ast}\,w^{\ast}.$ (2.18)

(Note derivation is a little tricky because I needed to remember that
$-1=i^{2}$ in the second step and then group one $i$ with $y$ and the other
with $v$ in the third step.) This relation shows that the conjugate of the
product is simply the product of the conjugates! At this point, however, we
have enough information to be certain that complex conjugation is NOT a linear
operation for complex numbers (see Eq. 1.39 for the definition of a linear
operator). To see this more clearly, we first extend our definition of a
linear operator to the complex plane

$\displaystyle\mathcal{O}(a\vec{z}+b\vec{w})=a\mathcal{O}(\vec{z})+b\mathcal{O}(\vec{w}),\;\;\textrm{for
all }a,b\in\mathds{C}.$ (2.19)

For right now, consider the complex vectors $\vec{z}$ and $\vec{w}$ as being
normal vectors whose components are complex-valued. The specific details on
these complex vectors are not totally necessary (we’ll save that for linear
algebra…) for right now. What is important here is that if we have to compute
$(a\vec{z}+b\vec{w})^{\ast}$, then we would have

$\displaystyle(a\vec{z}+b\vec{w})^{\ast}$
$\displaystyle=(a\vec{z})^{\ast}\pm(b\vec{w})^{\ast},\textrm{ by Eq. \ref{eq:
conjugate z pm w}}$ (2.20) $\displaystyle=a^{\ast}\,\vec{z}^{\ast}\pm
b^{\ast}\,\vec{w}^{\ast}.\textrm{ by Eq. \ref{eq: conjugate zw}}$ (2.21)

The only way this is equal to $a(\vec{z}^{\ast})+b(\vec{w}^{\ast})$ is if
$a=a^{\ast}$ and $b=b^{\ast}$, implying that $a,b\in\mathds{R}$. But since our
complex vectors have complex-valued components, and we have already shown that
multiplication between complex numbers produced complex numbers, it follows
that in general the condition that both $a$ and $b$ are real is NOT general.
Thus, in general, complex conjugation is not linear333This fact is
specifically exploited in quantum mechanics and quantum field theory when
dealing with time-reversal symmetry in real physical systems.!

We have shown that the conjugate of the sum is the sum of the conjugates (Eq.
2.17) and the conjugate of the product is the product of the conjugates (Eq.
2.18). Show now that the conjugate of the quotient is the quotient of the
conjugates. In other words, show that $(z/w)^{\ast}=z^{\ast}/w^{\ast}$. (Hint:
either start with $z^{\ast}/w^{\ast}$, substitute in $-i$ for the $+i$, and
then show that it is the conjugate of Eq. 2.14, OR start with the conjugate of
Eq. 2.14 and manipulate it algebraically into $z^{\ast}/w^{\ast}$. I think
both methods take similar amounts of algebra…)

With purely real numbers, it is true that $|a|=|-a|$ for all $a\in\mathds{R}$.
Thus, magnitudes are not unique; however, we can use a sign difference to
order our real numbers from least to greatest. This problem focuses on whether
we can do a similar thing in the complex plane — if we could, then it would be
true that only $z$ and $-z$ share the same magnitude, just like the reals. We
used $\sqrt{z^{\ast}z}$ as the definition of the magnitude of a complex
number. Show that the complex numbers cannot be ordered like the reals by
finding at least one other complex number that shares the same magnitude of
$z$ (and $-z$). For example, show that $|z|=|z^{\ast}|$ for all
$z=x+iy\in\mathds{C}$. (Hint: to avoid confusing yourself with all the $z$s
and asterisks, define $w=z^{\ast}$ and then find $|w|$ using Eq. 2.16.
Finally, compare your result with Eq. 2.15.)

###  2.4 Cartesian versus Polar Representations

The fact that the real and imaginary axes form a complex plane implies that we
should be able to draw any complex number in a geometric plane. Further, since
we have a Pythagorean Theorem-type relationship for the magnitude of the
complex number (and $1$ is totally distinct from $i$), we are able to infer
that the real and imaginary axes are orthogonal to one another — just like the
$x$-axis and the $y$-axis. Figure 2.1 shows a possible complex number $z=x+iy$
in the complex plane. Additionally, if we know the components $x$ and $y$,
then we could easily draw the conjugate of $z$, namely $z^{\ast}=x-iy$. It is
important to note that the points $z$ and $z^{\ast}$ are the complex numbers;
meanwhile, based on our knowledge of vectors, we could easily draw an arrow
from the origin to each complex number, where the length of each vector would
the the magnitude (or length) of the complex number. Now for some
trigonometry.

FIG. 2.1: If we consider $z=x+iy$, then we can draw it and its conjugate,
$z^{\ast}=x-iy$, in the complex plane.

From the definition of sines, cosines, and tangents (see Section 1.3.3), we
can use the geometrical right angle between the real and imaginary axes to
define the angle of the the complex number, or the so-called argument of a
complex number, denoted by $\mathrm{arg}(z)=\theta$. As is shown in Fig. 2.1,
$\theta$ is the angular elevation of the complex number above the $+x$-axis.
Since we know that $\mathrm{Im}(z^{\ast})=-\mathrm{Im}(z)$ while the real
parts are identical, we know that $\mathrm{arg}(z^{\ast})=-\mathrm{arg}(z)$.
Using these angles we can then determine

$\displaystyle\cos\theta$
$\displaystyle=\frac{\mathrm{adjacent}}{\mathrm{hypotenuse}}=\frac{x}{|z|}=\frac{x}{\sqrt{x^{2}+y^{2}}},$
(2.22) $\displaystyle\sin\theta$
$\displaystyle=\frac{\mathrm{opposite}}{\mathrm{hypotenuse}}=\frac{y}{|z|}=\frac{y}{\sqrt{x^{2}+y^{2}}}.$
(2.23)

By solving the second equalities for $x$ and $y$ and then substituting these
quantities in $z=x+iy$, we find

$\displaystyle z=|z|\cos\theta+i|z|\sin\theta=|z|(\cos\theta+i\sin\theta).$
(2.24)

Note that this picture is consistent with just substituting in $-i$ for $+i$
to obtain a conjugate, because

$\displaystyle
z^{\ast}=|z^{\ast}|[\cos\theta+(-i)\sin\theta]=|z|[\cos(-\theta)+i\sin(-\theta)].$
(2.25)

where we have used the result of Problem 2.3 to set $|z^{\ast}|=|z|$, and then
we used the even and odd symmetry of the sinusoids:
$\cos(-\theta)=\cos(\theta)$ and $\sin(-\theta)=-\sin\theta$.

Equation 2.24 shows that any complex number $z=x+iy$ can be represented as the
product of a radial part, and an angular part, as shown below

$\displaystyle z=\underbrace{|z|}_{\textrm{Radial
Part}}(\underbrace{\cos\theta+i\sin\theta}_{\textrm{Angular Part}}),$ (2.26)

as a complement to the real-and-imaginary-part representations. Again, for the
sake of completeness, the radial part is the modulus of the complex number
while the angular part is the argument of the complex number. Take note that
while the radial part is a real number like $\mathrm{Re}(z)$ and
$\mathrm{Im}(z)$, the angular part is still complex!

Thus, we have written any complex number in either a real-and-imaginary
representation and in a radial-and-angular representation. These terms,
although straight-to-the-point, are fairly clunky, so instead we name them the
Cartesian and Polar representations of a complex number, respectively. The
“pole” in this case is the distance (magnitude/length) the complex number is
from the origin. A much more useful (and elegant) way to write a complex
number $z=x+iy\in\mathds{C}$ and its conjugate in the polar representation is
actually with an imaginary exponential $\textrm{e}^{i\theta}$, given as

$\displaystyle z$ $\displaystyle=r\textrm{e}^{i\theta},$ (2.27) $\displaystyle
z^{\ast}$ $\displaystyle=r\textrm{e}^{-i\theta},$ (2.28)

where

$\displaystyle r$ $\displaystyle=|z|=\sqrt{x^{2}+y^{2}},$ (2.29)
$\displaystyle\theta$
$\displaystyle=\mathrm{arg}(z)=\arctan\left(\frac{y}{x}\right).$ (2.30)

Of course, to write such a thing, it would have to be true that

$\displaystyle\textrm{e}^{i\theta}=\cos\theta+i\sin\theta,$ (2.31)

where $\textrm{e}\approx 2.71828182846$ is the base of the natural logarithm.
Surprisingly, this result is true for all values of $\theta$! Remarkably, this
result is not too difficult to prove even though it took a supergenius like
Leonhard Euler444Pronounced oiler, unlike how they said it in The Imitation
Game, much to my chagrin… to first do it555Hence, it is usually called the
Euler identity. — it can be done with an introductory understanding of Taylor
Series in a Calculus II course — although it is a bit beyond the scope of this
chapter666Don’t worry! We will come back to it later on and YOU will prove it
in Problem 3.4.2 (I help you along though)..

It is initially a little strange to go from the Cartesian representation to
the polar representation, but ultimately using the polar representation is
more convenient than the Cartesian representation (otherwise physicists
wouldn’t bother with it!). So this problem is designed to have you practice
the conversion for a few important numbers in physics. (a) Show that
$i=\textrm{e}^{i\pi/2}$ by arguing $r(i)=1$ and $\mathrm{arg}(i)=\pi/2$. (b)
Show that, in the Cartesian representation,
$\sqrt{2}\,\textrm{e}^{-i\pi/4}=1-i$ (c) Consider the complex number
$z=\textrm{e}^{i\pi}$. What is $z+1$ in both Polar and Cartesian
representation?

Using the polar representation, it is possible for us to construct two of the
most widely used expressions in all of physics — we are going to rewrite sine
and cosine in terms of exponentials. To start consider the complex number of
unit magnitude given by Eq. 2.31. Then it must be true

$\displaystyle\mathrm{Re}\left(\textrm{e}^{i\theta}\right)=\cos\theta\in\mathds{R},$
(2.32)
$\displaystyle\mathrm{Im}\left(\textrm{e}^{i\theta}\right)=\sin\theta\in\mathds{R}.$
(2.33)

But by Eq. 2.10 and Eq. 2.11, we can write the real and imaginary parts of any
complex number as a superposition of it and its complex conjugate. Then,

$\displaystyle\cos\theta$
$\displaystyle=\frac{\textrm{e}^{i\theta}+\textrm{e}^{-i\theta}}{2},$ (2.34)
$\displaystyle\sin\theta$
$\displaystyle=\frac{\textrm{e}^{i\theta}-\textrm{e}^{-i\theta}}{2i}.$ (2.35)

Part of the reason why these relationships are so useful, is it allows us to
treat trigonometric functions — objects that are defined geometrically — as
exponential functions instead, which then allow us to employ a slew of useful
(and quick) multiplication, differentiation, and integration rules to
otherwise algebraically cumbersome functions. In physics, this polar
representation of trigonometric functions helps us describe certain geometries
and spaces in terms of essentially successive multiplications. As you can see
in the chapter on Fourier Analysis, being able to convert sines and cosines
into combinations of exponential functions makes otherwise impossible algebra
much simpler. As we continue through this book, I will highlight more
locations where these can be immediately implemented to make your life easier
because the sooner you begin to feel comfortable with Eqs. 2.34 and 2.35, the
faster you will begin to see through the mathematics of difficult subjects
like signals, optics, and quantum mechanics to understand the underlying
phenomena in a much more precise way. There is something potentially
unsettling by writing geometric formulas as exponential formulas; it seems to
imply there is a clear rotational aspect of multiplying two numbers, rather
than the simpler stretching-and-shrinking interpretation that was valid with
the real numbers. We study this more in the next section.

###  2.5 Multiplication = Dilation + Rotation

Let’s consider two complex numbers in polar representation, given by
$z=r\textrm{e}^{i\theta}$ and $w=\rho\textrm{e}^{i\phi}$. We will calculate
their product and difference in the polar representation.

$\displaystyle zw$
$\displaystyle=\left(r\textrm{e}^{i\theta}\right)\left(\rho\textrm{e}^{i\phi}\right)=r\rho\textrm{e}^{i\theta}\textrm{e}^{i\phi}=(r\rho)\,\textrm{e}^{i(\theta+\phi)},$
(2.36) $\displaystyle\frac{z}{w}$
$\displaystyle=\frac{r\textrm{e}^{i\theta}}{\rho\textrm{e}^{i\phi}}=\frac{r}{\rho}\,\textrm{e}^{i\theta}\textrm{e}^{-i\phi}=\left(\frac{r}{\rho}\right)\,\textrm{e}^{i(\theta-\phi)},$
(2.37)

Thus, by Eq. 2.36, multiplying a complex number by another is equivalent to
dilating the magnitude of the first by the second and rotating the first
complex number counter-clockwise by the second’s argument. Figure 2.2 shows
the geometry behind multiplication. Likewise, since division is the
multiplicative inverse, we should not be surprised by Eq. 2.37 which says that
dividing a complex number by another is equivalent to constricting777My word
choice for “anti-dilating.” the magnitude of the first by the second and
rotating the first complex number clockwise by the second’s argument.

FIG. 2.2: When we multiply the complex number $z=r\textrm{e}^{i\theta}$ by
$w=\rho\textrm{e}^{i\phi}$, the magnitude of $z$ is dilated by a factor of
$\rho$, while the angle of $z$ is rotated by $\mathrm{arg}(w)=\phi$. Thus
$|zw|=r\rho$ and $\mathrm{arg}(zw)=\theta+\phi$.

What is often the case, at least in physics, when we deal with complex
numbers, we are usually only taking about ones with unit magnitude — that is,
the complex modulus is equal to one. Any complex number of the form in Eq.
2.31 fits this description. The reason why these numbers are so important is
that they do not dilate the modulus of any complex number they are
multiplying. Instead, they exclusively rotate the complex number they are
multiplying. Figure 2.3 serves to help you get some intuition to how the
multiplication = rotation bit works with more concrete numerical examples like
$\pm 1$ and $\pm i$. Very often, physicists will talk of the phase of some
quantity (such as in electromagnetic theory, optics, or quantum mechanics),
and refer to the entire complex number $\textrm{e}^{i\theta}$ as this phase.
To be perfectly precise, the actual angle (argument) is the phase, NOT the
entire complex number. However, as we have seen by Eq. 2.36 where $r=1$,
multiplying by the entire number $z=\textrm{e}^{i\theta}$ only changes the
total phase of the product, since it rotates the original number by
$\mathrm{arg}(z)=\theta$. Thus, in terms of things we can experimentally
detect (and therefore know are truly there), we can only find the resultant
change of phase due to the multiplication of a complex number by another of
unit magnitude.

FIG. 2.3: In this figure, we multiply each of the following points along the
complex unit circle by the imaginary (lateral!) unit $i=\textrm{e}^{\pi i/2}$:
$(1,0)$, $(0,i)$, $(-1,0)$, and $(0,-i)$. Starting with $(1,0)$, we have
$1\cdot i=i=\textrm{e}^{\pi i/2}$ over the solid red arc. Hence, multiplying
$1$ by $i$, rotates $1$ by $\pi/2$ radians ($90^{\mathrm{o}}$). Next, we
multiply $i$ by $i$ in the dashed blue arc. But $i^{2}=-1=\textrm{e}^{\pi
i/2+\pi i/2}=\textrm{e}^{\pi i}$. Thus, again, multiplying by $i$ rotates our
starting complex number $i$ by $\pi/2$ radians. I leave you to confirm the
other two rotations for yourself.

Our ability to detect these so-called phase differences is actually rather
remarkable. For example, in both classical and quantum mechanics, magnetic
fields exert a torque on any charged object with angular momentum. This torque
causes the charged object to precess in a circle; in other words, the charged
object will behave similarly to how a toy top begins to wobble in circles
around its central axis before it falls over due to gravity. In this case, the
field that generates the wobble is the magnetic field instead of the
gravitational field. Anyway, in quantum mechanics, a typical experiment to
measure the phase difference goes something like this: generate a beam of
identical particles, split the beam into two parts, do something to one beam
and leave the other alone, then recombine the beams and see if anything
happens. So one such experiment deals with measuring the intensity of
particles with inherent angular momentum (spin) after splitting up the beam
and having one part travel through a magnetic field over some distance. The
amount those particles wobble due to the magnetic field then acts as the phase
difference between the particles in the beam. As you will eventually learn,
the intensity of the recombined beam is a function of the phase difference,
meaning we can experimentally vary either the strength of the magnetic field
or increase the distance the beam travels through it, and then change the
intensity of the resulting beam of particles! For an example of such an
experiment, check out [spin_interferometry].

Before proceeding, let’s take note of one special case of multiplication:
exponentiation. In other words, we can raise any real number $a\in\mathds{R}$
to the $n\in\mathds{R}$ power by successively multiplying $a$ by itself $n$
times. For example, if $a=2$ and $n=4$, then $a^{n}=2^{4}=2\cdot 2\cdot 2\cdot
2=16$. Likewise, if we have a negative exponent, then we divide successively,
while if $n$ has a noninteger fractional part, we take the appropriate root
$(2^{1/3}=\sqrt[3]{2})$. Using the polar representation of a complex number
$z=r\textrm{e}^{i\theta}$, we can generalize exponentiation rather
straightforwardly as

$\displaystyle
z^{n}=r^{n}\,(\textrm{e}^{i\theta})^{n}=r^{n}\,\textrm{e}^{in\theta}.$ (2.38)

By setting a new angle $\psi=n\theta$, and a new radial part as
$s=r^{n}\in\mathds{R}$, we have

$\displaystyle z^{n}=s\,\textrm{e}^{i\psi}.$ (2.39)

Making use of Euler’s Identity (Eq. 2.31), we see that

$\displaystyle z^{n}=s(\cos\psi+i\sin\psi)=r^{n}(\cos n\theta+i\sin n\theta).$
(2.40)

Since the quantity $z^{n}$ can be written in terms of a radial and angular
part, it must be complex-valued, as those parts yield real and imaginary
parts. Thus $z^{n}\in\mathds{C}$ which means our rules for complex numbers
lead to closure under exponentiation! In other words, we cannot ever possibly
end up with a non-complex-valued quantity purely by exponentiation888This is a
very good thing, mind you, for it essentially gives us motivation for starting
to see if more complicated algebraic and transcendental functions always
return complex numbers..

There is, meanwhile, a much more subtle identity that we may not have
initially noticed while showing $z^{n}\in\mathds{C}$. We made use of a
statement known as De Moivre’s Theorem that says for any integer $n$

$\displaystyle\left(\cos\theta+i\sin\theta\right)^{n}=\cos(n\theta)+i\sin(n\theta).$
(2.41)

Looking closely at Eq. 2.41, we see that this statement is actually incredibly
complicated. It manages to relate something as easy to compute as raising a
number to an integer power $n$ to the much more difficult trigonometric
functions, cosine and sine. But more importantly De Moivre’s Theorem puts the
exponent $n$ inside of the argument of the trig function! Usually, there is no
clear way to translate the argument of a trig function to anything outside of
the function — for example $\cos(2x)\neq 2\cos x\neq(\cos x)^{2}$ for every
value of $x$. Trigonometric functions just don’t behave nicely like this.
However, by making use of the complex plane, De Moivre’s Theorem gives us a
way of applying algebraically simply operations to compute otherwise very
difficult, if not outright impossible, quantities.

I do want to emphasize that De Moivre’s Theorem only applies for integer
exponents, whereas Eq. 2.40 applies for all possible values of $n$ (even
complex ones!). What gives? Where is there a difference? We look into this
question next.

####  2.5.1 Roots of Unity

Let’s consider the case in De Moivre’s Theorem where $n=1/2$. In this
particular case, then if De Moivre’s Theorem were to hold, then it would be
true (but it is not)

$\displaystyle\left(\cos\theta+i\sin\theta\right)^{1/2}=\cos\left(\frac{\theta}{2}\right)+i\sin\left(\frac{\theta}{2}\right).$

Let’s choose the easier case of $\theta=0$ and $\theta=2\pi$ since on the
left-hand side, we will have

$\displaystyle(\cos 0+i\sin 0)^{1/2}=(\cos 2\pi+i\sin 2\pi)^{1/2}=1^{1/2}.$

This follows since $\sin 0=\sin 2\pi=0$ and $\cos 0=\cos 2\pi=1$. There is no
problem yet. The issue arises when we look at the right-hand side to find

$\displaystyle\cos\left(\frac{0}{2}\right)+i\sin\left(\frac{0}{2}\right)=\cos
0$ $\displaystyle=1$
$\displaystyle\cos\left(\frac{2\pi}{2}\right)+i\sin\left(\frac{2\pi}{2}\right)=\cos\pi$
$\displaystyle=-1.$

(Remember that $\sin\pi=0$.) So, if De Moivre’s Theorem were to be trusted, we
would find that $1^{1/2}=1=-1$. In other words, we would find that our answer
implies that $1=-1$ and this contradicts our rules for real numbers.

This result is not too surprising from an Algebra II point-of-view since we
already know that both $(1)^{2}$ and $(-1)^{2}=1$, which we normally write
instead as $1^{1/2}=\pm 1$. But this means that the square-root function is
mutli-valued, and in general the $n^{\mathrm{th}}$-root is also multi-valued.
The problem with De Moivre’s Theorem is that it does not explicitly account
for this phenomenon. We could easily generalize it to account for the multi-
valuedness in exponentiation by rational exponents.

We start with the so-called Roots of Unity as they are both fundamental and
something that we have already developed the motivation for. Let’s start off
easy, and find the cubic roots of unity. We will proceed in the same way as we
did before with the square-roots of unity. Here we have

$\displaystyle\left(\cos\theta+i\sin\theta\right)^{1/3}=\cos\left(\frac{\theta}{3}\right)+i\sin\left(\frac{\theta}{3}\right).$

Like before, we choose $\theta=0$ and $\theta=2\pi$ because $\cos 0=\cos
2\pi=1$. There is actually another case we can consider, too:
$\theta=4\pi\Rightarrow\cos 4\pi=1$. Calculating each of these cases, we have

$\displaystyle\cos\left(\frac{0}{3}\right)+i\sin\left(\frac{0}{3}\right)=1$
$\displaystyle\cos\left(\frac{2\pi}{3}\right)+i\sin\left(\frac{2\pi}{3}\right)=-\frac{1}{2}+i\frac{\sqrt{3}}{2}$
$\displaystyle\cos\left(\frac{4\pi}{3}\right)+i\sin\left(\frac{4\pi}{3}\right)=-\frac{1}{2}-i\frac{\sqrt{3}}{2}$

And so the cubic-roots of unity we found are

$\displaystyle
1^{1/3}\in\left\\{1,-\frac{1}{2}+i\frac{\sqrt{3}}{2},-\frac{1}{2}-i\frac{\sqrt{3}}{2}\right\\}$

Hopefully, you aren’t satisfied with this derivation of the cubic-roots of
unity so far because I just arbitrarily decided to include the $4\pi$ part
when I didn’t include it for the square-roots. The reason why I was able to
include $4\pi$ for the cubic-roots and not for the square-roots is because of
the oscillatory behavior of sines and cosines — they repeat themselves every
$2\pi$ radians. So essentially, when looking for $\theta$ values that would
show the roots of unity are indeed multi-valued, I needed to watch out for two
criteria. First, I needed to make sure $\textrm{e}^{i\theta}=1$ since we are
taking about roots of unity, AND I needed to make sure that the right-hand
side did not repeat itself. Let’s analyze each criterion independently, and by
doing so, we will generalize the square and cubic cases to the
$n^{\mathrm{th}}$-root.

If we want $\textrm{e}^{i\theta}=1$, then we need $\cos\theta+i\sin\theta=1$.
But the value $1$ is totally real, hence the sine coefficient attached to the
$i$ must vanish for our $\theta$ values. This condition holds for
$\theta\in\\{0,\pm\pi,\pm 2\pi,\pm 3\pi,\dots\\}$. Next, we are dealing with
roots of unity, not roots of negative-unity. Thus, we can only allow for even
multiplies of $\pi$ so that $\cos(2\pi m)=1$, where $m\in\\{0,\pm 1,\pm 2,\pm
3,\dots\\}$. So this explains why I kept choosing $\theta=0,2\pi$ and then
$4\pi$. But why not the negatives, too? The answer to that comes again from
the even symmetry of the cosine function:

$\displaystyle\cos(-\theta)=\cos(\theta).$

In other words, the cosine functions ignore the overall negative sign inside
of their argument and produce the same result either way. Hence, we could
include the negative values, $-2\pi,-4\pi,$ et cetera, but we would ultimately
always recover the same set of possible $\theta$ values.

Now we move onto the second criterion which says that $\theta$ cannot make the
right-hand side repeat itself. We use the first criterion that we had
established where all of the roots of unity will have $\theta$ values that are
an even multiple of $2\pi$. Let’s call this multiple $m$ and write $1^{1/n}$
as

$\displaystyle 1^{1/n}=\cos\left(\frac{2\pi m}{n}\right)+i\sin\left(\frac{2\pi
m}{n}\right),\;\;m\in\\{0,1,2,\dots\\}.$

If we start with $m=0$, then we will always find $1^{1/n}=1$, as we expect,
since $1^{n}=1$. Then, as we continue to check through all of the possible
multiples we have, we find the first repeating value at $m=n$:

$\displaystyle 1^{1/n}\underbrace{=}_{m=n}\cos\left(\frac{2\pi
n}{n}\right)+i\sin\left(\frac{2\pi n}{n}\right)=\cos 2\pi=1$

But this returns the exact same value as $m=0$. Furthermore, any multiple of
$n$ will ALWAYS be the same as the $m=0$ case. Meanwhile, all of the possible
multiples up to $n$ are totally allowed because, in general,

$\displaystyle\frac{2\pi m}{n}\neq 2\pi,$

the only case where equality holds is when $m=n$. Thus, our
$n^{\mathrm{th}}$-roots of unity are given as

$\displaystyle 1^{1/n}=\cos\left(\frac{2\pi m}{n}\right)+i\sin\left(\frac{2\pi
m}{n}\right),\;\;m\in\\{0,1,\dots,n-1\\},$ (2.42)

in the Cartesian representation. In the polar representation, we pack the
sinusoids into the exponential function to find

$\displaystyle 1^{1/n}=\mathrm{exp}\left(\frac{2\pi
mi}{n}\right),\;\;m\in\\{0,1,\dots,n-1\\}.$ (2.43)

(Note that $\mathrm{exp}(x)=\textrm{e}^{x}$ for all $x$. I used the exp
representation because the equation would have looked gross if I used e
instead.)

FIG. 2.4: A few plots of the symmetry in the roots of unity. The orange points
are the complex numbers centered around their common origin. The vectors
attached to each point is to help with our geometrical interpretation of the
complex plane. The regular polygons within each unit circle represent the
internal symmetry within the set of the $n^{\mathrm{th}}$-roots of unity.

The $n^{\mathrm{th}}$-roots of unity are quite nice geometrically because they
chop up the unit circle ($2\pi$ radians) into regular $n$-gons by dividing the
full $2\pi$ radians into equally-spaced $2\pi/n$ intervals. A few of these are
shown in Fig. 2.4. What is more beautiful from a physical point-of-view is
that since the roots of unity form regular $n$-gons inscribed within a unit
circle, their sum must vanish. This can be directly applied to study
cylindrically (and polygonally) symmetrical systems in physics; for example,
point like-charges (masses) arranged in some polygonal shape whose net
electric (gravitational) fields must vanish at their geometrical center. In
more computational applications in physics, this property allows us to compute
discrete Fourier Transforms of signals which is a mathematical operation that
allows us to understand the signal in terms of its frequency-dependence
instead of its time-dependence999The Fourier Analysis chapter in this book
does not deal with the discrete transform, but the tools developed in the
chapter in the continuum can help you understand the intuition behind the
discrete version..

To show how the roots sum to zero explicitly, recall the square-roots of
unity: $1$ and $-1$. Together they form a line (lame), but their sum is
$1+(-1)=0$. We can also do the cubic-roots of unity that form an equilateral
triangle (less lame):

$\displaystyle
1-\frac{1}{2}+i\frac{\sqrt{3}}{2}-\frac{1}{2}-i\frac{\sqrt{3}}{2}=1-1+i0=0.$

Now let’s generalize (disclaimer: this will be one of the more abstract things
so far).

Using Eq. 2.42, find the Cartesian representation of the $4^{\mathrm{th}}$ and
$5^{\mathrm{th}}$ roots of unity and show that the sum of their roots is zero.

Consider the sum of the $n^{\mathrm{th}}$-roots of unity in polar
representation, written as $S_{n}$, and given as

$\displaystyle S_{n}=1+\textrm{e}^{2\pi i/n}+\dots+\textrm{e}^{2\pi
i(n-1)/n}=\sum_{m=0}^{n-1}\textrm{e}^{2\pi im/n}$

If we look closely at the summation, we will see that we actually have a
finite geometric series, defined generally as

$\displaystyle S_{n}=1+a+a^{2}+\dots+a^{n-1}=\sum_{m=0}^{n-1}a^{m},$ (2.44)

for some value $a$. In this case, $a=\textrm{e}^{2\pi i/n}$ and

$\displaystyle S_{n}=\sum_{m=0}^{n-1}\left(\textrm{e}^{2\pi i/n}\right)^{m}$

To determine what this finite sum is, we can employ a few
mathemagical101010Bad pun? tricks that I highly recommend you work out with
me. Knowing how they work WILL come in handy later on in your physics career,
at least when it comes to handling finite sums. If you work these tricks out
with me, then you are prone to remembering them in the future. Here we go.

We will calculate the quantity $1-S_{n}$ using the definition of $S_{n}$ given
in Eq. 2.44, and at the end we will substitute in $a=\textrm{e}^{2\pi i/n}$.

$\displaystyle 1-S_{n}$ $\displaystyle=1-\sum_{m=0}^{n-1}a^{m}$
$\displaystyle=1-(1+a+a^{2}+\dots a^{n-1})$
$\displaystyle=-(a+a^{2}+\dots+a^{n-1})$ $\displaystyle=-a(1+a+\dots+a^{n-2})$
$\displaystyle=-aS_{n-1}.$

In the last equality, I used the definition of the finite geometric series
again, except since the sum only goes to $n-2$, then that means the proper
subscript on $S$ is $n-1$ since $(n-1)-1=n-2$. In this form, we cannot proceed
because we needed $S_{n}$, NOT $S_{n-1}$. Thus we need to figure out a way to
relate the two different finite sums.

Let’s consider now $S_{n-1}=1+a+\dots+a^{n-2}$. If we look again at Eq. 2.44,
we will see that if we were to add $a^{n-1}$ to $S_{n-1}$, we would have

$\displaystyle S_{n-1}+a^{n-1}=1+a+\dots+a^{n-2}+a^{n-1}=1+a+\dots
a^{n-1}=S_{n}.$

Hence, $S_{n}-a^{n-1}=S_{n-1}$. Now we are going to substitute this expression
in for $S_{n-1}$ inside the last equality for $1-S_{n}$ and isolate $S_{n}$.

$\displaystyle 1-S_{n}=-aS_{n-1}=-a\left(S_{n}-a^{n-1}\right)=-aS_{n}+a^{n}.$

By moving the $S_{n}$ on the left-hand side to the right-hand side we have

$\displaystyle 1=S_{n}-aS_{n}+a^{n}\Rightarrow 1-a^{n}=(1-a)S_{n},$

and finally we have

$\displaystyle S_{n}=\frac{1-a^{n}}{1-a},$ (2.45)

which is the exact way to calculate a finite geometric series for any value of
$a\neq 1$ (the formula blows up at $a=1$ because if $a$ were 1 then we would
have a $S_{n}-S_{n}=0$ step in our derivation, thus we had to implicitly
assume that $a\neq 1$). Finally, we substitute in $a=\textrm{e}^{2\pi i/n}\neq
1$:

$\displaystyle S_{n}(\textrm{e}^{2\pi i/n})=\frac{1-(\textrm{e}^{2\pi
i/n})^{n}}{1-\textrm{e}^{2\pi i/n}}=\frac{1-\textrm{e}^{2\pi
in/n}}{1-\textrm{e}^{2\pi i/n}}=\frac{1-\textrm{e}^{2\pi
i}}{1-\textrm{e}^{2\pi i/n}}=\frac{1-1}{1-\textrm{e}^{2\pi i/n}}=0.$ (2.46)

And so it is true that the sum of all of the $n^{\mathrm{th}}$-roots of unity
will always be identically zero! Furthermore, since a geometric series is
built from multiplying the same object $a=\textrm{e}{2\pi i/n}$ by itself a
bunch of time, and we know that multiplication by a complex number of unit
magnitude is a pure rotation, then the complex number $\textrm{e}^{2\pi i/n}$
represents a symmetry present in the set of the $n^{\mathrm{th}}$-roots of
unity. The symmetry here is that our set of complex numbers are invariant
under a rotation of $2\pi/n$ radians — i.e. the number of points looks
identical if we rotate all of them by the same $2\pi/n$ angle! For example, if
we have the square-roots $\\{1,-1\\}$ and we rotate each number through the
complex plane by $2\pi/2=\pi$ radians, we have

$\displaystyle\\{1,-1\\}\xrightarrow{\textrm{rotate by }\pi\textrm{
radians}}\\{-1,1\\},$

as shown in Since the rotation $\textrm{e}^{\pi i}$ maps the set of square-
roots onto themselves, this rotation is the symmetry I was talking about that
is directly embedded within the set of square-roots of unity.

FIG. 2.5: The symmetry in the square-roots of unity is a rotation by $\pi$
radians since the set of roots is identical under this rotation.

Since the ambiguity in De Moivre’s Theorem is taken care of, we can return to
Eq. 2.40 to account for all of the different possible roots of $z$.
Specifically, we can talk about any root $1/n$, we have

$\displaystyle z^{1/n}$
$\displaystyle=\left(r\textrm{e}^{i\theta}\right)^{1/n}$
$\displaystyle=r^{1/n}\cdot 1^{1/n}\cdot\textrm{e}^{i\theta/n}$
$\displaystyle=r^{1/n}\,\textrm{e}^{2\pi
im/n}\,\textrm{e}^{i\theta/n},\;\;m\in\\{0,1,\dots,n-1\\}$
$\displaystyle=r^{1/n}\,\mathrm{exp}\left(i\frac{\theta+2\pi
m}{n}\right),\;\;m\in\\{0,1,\dots,n-1\\}$
$\displaystyle=r^{1/n}\left[\cos\left(\frac{\theta+2\pi
m}{n}\right)+i\sin\left(\frac{\theta+2\pi
m}{n}\right)\right],\;\;m\in\\{0,1,\dots,n-1\\}$ (2.47)

Then, by extension, a complex number $z$ to any rational power $p/q$ (for
example 4/3) can be written as

$\displaystyle z^{p/q}$
$\displaystyle=\left[r^{1/q}\,\mathrm{exp}\left(i\frac{\theta+2\pi
m}{q}\right)\right]^{p},\;\;m\in\\{0,1,\dots,q-1\\}$
$\displaystyle=r^{p/q}\,\mathrm{exp}\left(ip\frac{\theta+2\pi
m}{q}\right)^{p},\;\;m\in\\{0,1,\dots,q-1\\}$
$\displaystyle=r^{p/q}\left\\{\cos\left[\frac{p}{q}(\theta+2\pi
m)\right]+i\sin\left[\frac{p}{q}(\theta+2\pi
m)\right]\right\\},\;\;m\in\\{0,1,\dots,q-1\\},$ (2.48)

The power with raising a complex number to a rational exponent is that we can
then use this as a basis for raising a complex number to an irrational
exponent since we can always successively approximate an irrational number
with a rational number. For example, $\sqrt{2}\approx 1.414=1414/1000$. But
how about raising a complex number to a complex exponent? We will study this
and more in the next section.

###  2.6 Functions of a Complex Variable

In this section, we will study a few algebraic and transcendental functions of
a complex variable. Unfortunately, we will not have time to study these
functions beyond just domain and range, but rest assured that understanding
the inputs and outputs of complex functions is totally sufficient for a B.S.
in physics. As one studies either more math or more physics, ideas from
Complex Analysis become relevant if not actually crucial, whether they be in
the form of circuit analysis, fluid mechanics, or quantum field theory. I have
some references for anyone interested at this point in their studies
[complex_analysis_wikipedia_2018, beck_marchesi_pixton_sabalka_2017,
arnold_complexanalysis] (feel free to come back at a later date for them
though!).

I have used the term “map” before when talking about functions, and perhaps
you’ve heard others use that word in the same context. In more colloquial
settings, people use maps to get from one place to another, and in this
section the connection between a function and an everyday map will become
clearer. We will see that functions, particularly complex-valued functions,
serve to get from one place in the complex plane to another. To illustrate
this idea, let’s consider the following example where we see how regions of
the complex plane are connected through the function, or mapping,
$w(z)=z^{2}$.

Consider the complex-valued function $w(z)=z^{2}$. We want to study $w$ over a
unit square in the first quadrant of the complex plane. To start, we consider
$z=x+iy$ in the Cartesian representation. Then, $\displaystyle
w(z)=z^{2}=(x+iy)^{2}=x^{2}-y^{2}+2ixy.$ (2.49) If we define $w=u+iv$, then we
have $u=x^{2}-y^{2}$ and $v=2xy$ for the real and imaginary parts of $w$,
respectively. Now we define the boundary of the unit square for our input, or
domain, of interest in the complex plane $\displaystyle
z=\begin{cases}x+i0,&x\in[0,1]\\\ 1+iy,&y\in[0,1]\\\ x+1i,&x\in[0,1]\\\
0+iy,&y\in[0,1]\end{cases}$ If we substitute these coordinates for $z$ into
$w$, we have $\displaystyle w(z)=\begin{cases}x^{2},&x\in[0,1],\;y=0\\\
1-y^{2}+2iy,&x=1,\;y\in[0,1]\\\ x^{2}-1+2ix,&x\in[0,1],\;y=1\\\
-y^{2},&x=0,\;y\in[0,1]\end{cases}$ The “mapping”, in this case, comes in when
we rewrite $w(z)$ into its own set of real and imaginary parts, given by
$w=u+iv$. By looking at the four same regions, we have $\displaystyle
w(\textrm{unit cube})\in\begin{Bmatrix}u=x^{2},\;&v=0\\\ u=1-y^{2},\;&v=2y\\\
u=x^{2}-1,\;&v=2x\\\
u=-y^{2},\;&v=0\end{Bmatrix}=\begin{Bmatrix}u,&u\in[0,1]\\\
u=1-\frac{1}{4}v^{2},&v\in\left[0,2\right]\\\
u=\frac{1}{4}v^{2}-1,&v\in\left[0,2\right]\\\ u,&u\in[-1,0]\end{Bmatrix}$ It
is definitely difficult to visualize how the equation above defines a map, at
least not in this representation. Figure 2.6 shows how the equation above can
serve as a set of instructions of how to turn one plot, or regions of the
complex plane, is connected to another through the $w(z)=z^{2}$ function.

FIG. 2.6: In the left plot, we define the unit square in the $z$-plane. It is
through the $w(z)=z^{2}$ mapping that we connect the unit square to the
square-squared region in the $w$-plane in the right plot. The colored curves
represent sets of the complex plane that are connected through the
$w(z)=z^{2}$ “map.”

From Example 2.6, we can conclude a couple of things. First, complex-valued
functions unite different regions of the complex plane. And second,
specifically for this quadratic function, we chose to only consider the unit
square. We could have chosen a square of size 2, or $\pi$, or a gazillion.
Then $w$ would also increase in size by Eq. 2.49. I will leave it to you to
pick particular points on the square of side length $s$ and plug it into Eq.
2.49 to see how large the region in the $w$-plane becomes (hint: try the
points $(s,0),\;(s,s),$ and $(0,s)$). Since we can choose a square of any
size, we can artificially stretch the region in the $w$-plane to infinity as
we increase the domain in the $z$-plane. Notice though, that as long as our
domain is only in the first quadrant in the $z$-plane, we will only ever be in
the first or second quadrants in the $w$-plane. This region is sometimes
referred to as the upper-half plane. As you may see after doing Problem 2.6,
other regions will map to fill the lower-half plane, and by extension, we can
stretch regions out as well to fill the entire lower-half plane. Hence, the
function $w(z)=z^{2}$ maps the entire complex plane into the entire complex
plane! Actually, functions such as these are specifically named entire
functions since they are defined111111Technically, functions are entire only
if they converge everywhere in the plane. However, convergence is beyond the
scope of this chapter, so understanding entire functions as simply being
defined everywhere is good enough for right now.  everywhere in the plane.
Since we have begun to understand complex mappings in terms of complex domains
and ranges, we seek to study a few more fairly commonly-used complex functions
in physics.

Use Eq. 2.49 to show that the square defined by $\displaystyle
z=\begin{cases}x+i0,&x\in[-1,0],\;y=0\\\ -1+iy,&x=-1,\;y\in[0,1]\\\
x+1y,&x\in[-1,0],\;y=1\\\ 0+iy,&x=0,\;y\in[0,1]\end{cases}$ maps to the same
region in Fig. 2.6, except that it is reflected over the $u$-axis into the
lower-half plane.

####  2.6.1 Polynomials

For the first type of function, we consider functions of the form

$\displaystyle
w(z)=P_{n}(z)=\sum_{j=0}^{n}a_{j}z^{j}=a_{0}+a_{1}z+a_{2}z^{2}+\dots+a_{n}z^{n},$
(2.50)

where all of the constant coefficients $\\{a_{j}\\}$ are arbitrary complex
numbers, but each exponent is a nonnegative integer. This class of functions,
often denoted by the symbol $P_{n}$ for “polynomial of
$n^{\mathrm{th}}$-order,” is the complex generalization of the real
polynomials that you already know. This means that instead of only allowing
ourselves to input real numbers into a polynomial, we now are allowing
ourselves to insert a complex number $z=r\textrm{e}^{i\theta}$ as the argument
of the function (in other words, an element from the domain). With this value
of $z$ and by Eq. 2.40, we have

$\displaystyle
P_{n}\left(z=r\textrm{e}^{i\theta}\right)=\sum_{j=0}^{n}a_{j}r^{j}\textrm{e}^{ij\theta}=a_{0}+a_{1}r\textrm{e}^{i\theta}+a_{2}r^{2}\textrm{e}^{2i\theta}+\dots+a_{n}r^{n}\textrm{e}^{in\theta}.$
(2.51)

It is important to note that we are entirely allowed to compute every single
one of these terms using the polar representations of the coefficients, if we
had them given. Since each term is defined and computable, then we are
definitely allowed to add all of their corresponding real and imaginary parts.
Thus, for any $z\in\mathds{C}$ that we throw at this thing, we will always be
able to find its output as a complex number. Thus a general polynomial maps
the complex plane onto121212To be clear, there is a difference between the
words into and onto. A function from set $A$ to set $B$, $f:A\rightarrow B$,
is said to be onto if every element of $B$ is in the range of $f$. Pictorally,
this means that there are no parts of $B$ that are not connected by $f$ to
$A$. Otherwise, mathematicians use the word into. the complex plane. Our
conclusion is important because, just like with real numbers, complex
polynomials are actually relatively easy functions to handle, whether it be in
calculus, computational physics, mathematical modeling, et cetera, because we
can handle them term-wise. Other functions, (like the transcendental
functions) are not so nice.

Consider the complex quadratic, $\displaystyle
P_{2}(z)=i+\textrm{e}^{i\pi/4}z-2z^{2}.$ Use either the Cartesian or the polar
representation to evalutate $P_{2}(z)$ at the following points: $z_{1}=1$,
$z_{2}=-i\pi$, $z_{3}=1-i$. Also, write down the set of coefficients
$\\{a_{j}\\}$ (this part is NOT a trick. I just want to make sure you know
what the set of cofficients is).

####  2.6.2 Complex Exponentiation Revisted

The last time we dealt with complex exponentials was with deriving the
rational generalization to De Moivre’s Theorem in Eq. 2.48. We were explicit,
however, that the exponent itself had to be a rational number
$p/q\in\mathds{R}$, where $p$ and $q\neq 0$ are integers along the real axis.
Now we have to ask the question of whether it is possible to raise a complex
number $z$ to a constant complex exponent $\xi$. In other words, we want to
study the function $w(z)=z^{\xi}$.

Since $\xi\in\mathds{C}$, we choose to work with it in its Cartesian
representation. Let $\alpha=\mathrm{Re}(\xi)$ and $\beta=\mathrm{Im}(\xi)$
such that $\xi=\alpha+i\beta$. Thus

$\displaystyle w(z)=z^{\xi}=z^{\alpha+i\beta}=z^{\alpha}\,z^{i\beta}.$

Now, we will make the substitution of $z=r\textrm{e}^{i\theta}$ into the
equation above.

$\displaystyle w(z)$
$\displaystyle=\left(r\textrm{e}^{i\theta}\right)^{\alpha}\left(r\textrm{e}^{i\theta}\right)^{i\beta},$
$\displaystyle=\left(r^{\alpha}\textrm{e}^{i\theta\cdot\alpha}\right)\,\left(r^{i\beta}\,\textrm{e}^{i\theta\cdot
i\beta}\right),$
$\displaystyle=r^{\alpha}\textrm{e}^{i\alpha\theta}\,r^{i\beta}\textrm{e}^{-\beta\theta},$
$\displaystyle=r^{\alpha+i\beta}\textrm{e}^{-\beta\theta}\,\textrm{e}^{i\alpha\theta}.$

Now, before proceeding further, since $r$ is a real number, that means it lies
entirely along the real axis, and we can use our regular logarithms on it,
since those are defined over the positive real axis. Thus, we choose to write
$r=\textrm{e}^{\ln r}$, as the (real) functions $\textrm{e}^{x}$ and $\ln x$
are inverse functions such that $\textrm{e}^{\ln x}=x=\ln(\textrm{e}^{x})$.
When we make this substitution into $w$, we can derive our final result.

$\displaystyle w(z)$
$\displaystyle=\textrm{e}^{\ln(r)\cdot(\alpha+i\beta)}\textrm{e}^{-\beta\theta}\,\textrm{e}^{i\alpha\theta},$
$\displaystyle=\textrm{e}^{\alpha\ln(r)+i\beta\ln(r)}\textrm{e}^{-\beta\theta}\,\textrm{e}^{i\alpha\theta},$
$\displaystyle=\textrm{e}^{\alpha\ln(r)-\beta\theta}\,\textrm{e}^{i[\beta\ln(r)+\alpha\theta]},$
$\displaystyle=r^{\alpha}\,\textrm{e}^{-\beta\theta}\,\textrm{e}^{i[\beta\ln(r)+\alpha\theta]}.$
(2.52)

It is important to note that the long string of symbols given in Eq. 2.52 is
something that is entirely computable. What I mean by that is this expression
is written in the polar representation of a complex number, where

$\displaystyle|w(z)|$
$\displaystyle=r^{\alpha}\textrm{e}^{-\beta\theta}\in\mathds{R}^{+},$
$\displaystyle\mathrm{arg}[w(z)]$
$\displaystyle=\beta\ln(r)+\alpha\theta\in\mathds{R}.$

Hence Eq. 2.52 is already in the form of a complex number, regardless of which
values of $\alpha$ and $\beta$ we choose! To be clear, if we were to convert
back to a Cartesian representation then we would need to be careful with any
ambiguities in the rational exponents, as is given in Eq. 2.48. But otherwise,
we are free to map the complex plane $z$ onto131313Since there are many
possible roots for complex exponents, the $z$-plane actually gets mapped onto
the $w$-plane in multi-valued ways, which we won’t go into. These
peculiarities (translated to “headaches”in math-speak) are known as branch
cuts, and special note really should be paid to them from a mathematical
point-of-view. However, in physics, they are only noted if necessary.  the
complex plane $w$ via complex exponentiation.

Use the polar representation and Eq. 2.52 to show that
$i^{i}=\textrm{e}^{-\pi/2}$. This example shows that, while complex
exponentiation maps the complex plane onto the complex plane, exponentiation
does NOT necessarily map imaginary numbers into other imaginary numbers. This
is interesting because exponentiation will always map real numbers into other
real numbers.

####  2.6.3 Complex Logarithms

Since we have covered the case of exponentiation, it follows that we should
cover inverse exponentiation, that is, we need to discuss how to find the
logarithm of a complex number. To study this function, consider $w(z)=\ln(z)$,
where $\ln$ denotes “natural logarithm141414In many advanced mathematics, the
“common log”, denoted by $\log$, is used instead of $\ln$. However, the
“common log” of a number $x$ in physics is almost always given by
$\ln(x)/\ln(10)=\log_{10}(x)$. ”. Just as before, this problem is most easily
tackled in the polar representation.

$\displaystyle
w(z)=w(r\textrm{e}^{i\theta})=\ln(r\textrm{e}^{i\theta})=\ln(r)+\ln(\textrm{e}^{i\theta})=\ln(r)+i\theta.$
(2.53)

To be clear, in this derivation, we implicitly borrowed the logarithm of a
product rule from the real numbers, $\ln(ab)=\ln(a)+\ln(b)$, and we implicitly
enforce that the natural logarithm is the function-inverse of
$\textrm{e}^{x}$. It is very important to note, however, that ANY complex
number $\xi=r_{\xi}\textrm{e}^{i\theta_{\xi}}$ is invariant under a full
rotation about the origin — therefore $\xi$ invariant under a rotation of
$2\pi n$, where $n$ is an integer. This is equivalent to saying
$\xi\,\textrm{e}^{2\pi ni}=\xi\cdot 1=\xi$. However, the exponents are
additive, therefore

$\displaystyle\xi\textrm{e}^{2\pi
ni}=r_{\xi}\textrm{e}^{i\theta_{\xi}}\textrm{e}^{2\pi
ni}=r_{\xi}\textrm{e}^{i(\theta_{\xi}+2\pi n)}$

But if we apply this idea to $z=r\textrm{e}^{i\theta}$, then we have
$z=r\textrm{e}^{i(\theta+2\pi n)}$. Thus, when we take a logarithm, we have

$\displaystyle w(z)=\ln(r)+i(\theta+2\pi n)=\ln(r)+i\theta+2\pi ni.$ (2.54)

But wait. From Eq. 2.53, have

$\displaystyle w(z)=\ln(r)+i\theta=\ln(r)+i\theta+2\pi ni,$ (2.55)

which seems to imply that $2\pi ni=0$ if this equation is to be true. Since
$2\pi$ and $i$ are both nonzero quantities, this means that $n$ would have to
be zero! However, we said $n$ can be any integer, not just zero. What gives?
Did we accidentally stumble on a contradiction within complex algebra?

So the truth is that, with our current system of complex numbers, we actually
did find a contradiction, and it has to do with our interpretation of
multiplication as rotation and $\textrm{e}^{2\pi ni}=1$ in the polar
representation. But this contraction is not a totally new idea — it is similar
to the idea that rational exponents are multivalued. It turns out that in the
complex plane, the logarithm also accrues multiple values for the same input.
In this sense, the full logarithm is not a function since for any input there
are many outputs (so it does not pass the complex-version of the vertical line
test). How do we deal with this problem? We use the idea of branch cuts to
save the day, where a “branch cut” is a fancy name for specifying a particular
range of the logarithm over a particular domain. In fact, this approach is
identical to the one you learned in trigonometry, where we would only specify
the inverse cosine ($\arccos(x)=\cos^{-1}(x)$) or inverse sine
($\arcsin{}(x)=\sin^{-1}(x)$) on the interval $x\in[-1,1)$, since there are
(infinitely) many values of $x$ such that $\cos(x)=\pm 1$ or $\sin(x)=\pm 1$.
Although this is (sort of) passing-the-buck on the issue of having the
logarithm be mutlivalued, it DOES allow us to retain our interpretation of
multiplication as rotations in the complex plane — we just need to be careful
about which branch we’re talking about when using the logarithm.

So the problem with the logarithm is due to the multiplicity, or rotational
symmetry, surrounding the $2\pi n$ angle. Thus, it helps to define a principal
branch of the logarithm, so there really isn’t a lot of ambiguity surrounding
our equations. All we need to do is define the imaginary part of the logarithm
over a $2\pi$-interval to keep things consistent, since it is the imaginary
part in Eq. 2.53 that has the multiple values. The interval we choose may
initially seem odd, but actually is more intuitive in physics, and is given by
$\theta_{principal}\in[-\pi,\pi)$. Therefore, the principal logarithm is
defined as

$\displaystyle\mathrm{Ln}(z)=\mathrm{Ln}(r\textrm{e}^{i\theta})=\ln(r)+i\theta,\;\theta\in[-\pi,\pi).$
(2.56)

Here, the capital $\mathrm{L}$ signifies the special branch of the multivalued
logarithm given in Eq. 2.53. Notice though that there is a lowercase
$\mathrm{l}$ in the real part of Eq. 2.56. This is because $r$ is a positive
real number, and we can always take a normal, non-multivalued logarithm of any
positive real number. All of the multivaluedness, again, is only due to the
imaginary part.

You have probably been told for your whole life that it is impossible to take
a logarithm of a negative number. Unfortunately, that is just not true. Use
Eq. 2.53 defined for $\mathrm{arg}(z)\in[\pi,3\pi)$ to calculate $\ln(-1)$. To
check whether your answer makes sense, suppose you compute $\ln(-1)$, and let
us denote it by $w=\ln(-1)$. This would imply that $\textrm{e}^{w}=-1$, or
$\textrm{e}^{w}+1=0$. Do you recognize this equation? Next, compute $\ln(-a)$,
where $a$ is a positive real number. Hence, $-a$ represents any point along
the negative real axis with an absolute value of $a$. For example, if $a=9$,
then I would be asking you to evaluate $\ln(-9)$. Since the number $-a$
represents any negative real number, what can you conclude about $\ln(-a)$? Is
it ever a purely real number (Hint: No it isn’t.)? Is this why you’ve been
taught that you cannot take the logarithm of a negative number (Hint: Yes it
is.)?

####  2.6.4 Hyperbolic Trigonometry

The final class of functions that I want to discuss is within the set of
functions that most people refer to as hyperbolic trigonometry. The geometry
is actually remarkably simple to jump to from normal, or circular,
trigonometry. The latter is governed by the unit circle, defined by
$x^{2}+y^{2}=1$. We are instead going to be interested in the so-called unit
hyperbola, defined by $x^{2}-y^{2}=1$. Although the geometry may be
interesting, I am going to be honest with you: the only time I’ve ever
actually seen it be relevant is in orbital mechanics where trajectories
matter, or general relativity, where the geometry of spacetime is the
dynamical variable of interest (in other fields of physics, like Physics 1,
the dynamical variables are position and velocity as a functions of time).
With that said, the actual hyperbolic functions themselves are often
incredibly useful tools to use in physics. They, for example, are used to
describe how the electrostatic potential changes in space from a source. I
have included Fig. 2.7 to help you see what unit hyperbola looks like, and how
the hyperbolic trigonometric functions relate triangles to it.

FIG. 2.7: The unit hyperbola $x^{2}-y^{2}=1$ parameterized by the functions
$x=\cosh t$ and $y=\sinh t$. Think of the parameter $t$ as the time it takes
for the black point to move from the point $(x,y)=(1,0)$ to the point
$(x,y)=(x(t),y(t))$ along the unit hyperbola. FIG. 2.8: Plots of the
hyperbolic cosine (Eq. 2.57) and hyperbolic sine (Eq. 2.58) functions over a
few real numbers $x$.

The most commonly used hyperbolic functions are the hyperbolic cosine and the
hyperbolic sine functions, defined by

$\displaystyle\cosh{x}$
$\displaystyle=\frac{\textrm{e}^{x}+\textrm{e}^{-x}}{2},$ (2.57)
$\displaystyle\sinh{x}$
$\displaystyle=\frac{\textrm{e}^{x}-\textrm{e}^{-x}}{2},$ (2.58)

for any $x\in\mathds{R}$. Here the extra $\mathrm{h}$ that’s attached is the
hyperbolic part, and the $\cosh$ notation is pronounced cah-sh while the
$\sinh$ part is often pronounced sin-ch. I’m not 100% sure why there is an
added ch in the $\sinh$ pronounciation, but my best guess is that sin-h is
hard to say. Since I know that these real-valued functions will be brand new
to most of you reading, I included a plot of them in Fig. 2.8 so you can
visualize them. Notice that as $x\rightarrow+\infty$, the hyperbolic sine and
hyperbolic cosine become equal, whereas on the other side of the real axis,
the $\sinh$ and $\cosh$ approach values equal magnitudes but opposite signs.
Note that if we set $x=\cosh(t)$ and $y=\sinh(t)$, then

$\displaystyle x^{2}-y^{2}$ $\displaystyle=(\cosh t)^{2}-(\sinh t)^{2},$
$\displaystyle=\left(\frac{\textrm{e}^{t}+\textrm{e}^{-t}}{2}\right)^{2}-\left(\frac{\textrm{e}^{t}-\textrm{e}^{-t}}{2}\right)^{2},$
$\displaystyle=\frac{1}{4}(\textrm{e}^{2t}+\textrm{e}^{-2t}+2)-\frac{1}{4}(\textrm{e}^{2t}+\textrm{e}^{-2t}-2),$
$\displaystyle=\frac{1}{4}(\textrm{e}^{2t}+\textrm{e}^{-2t}-\textrm{e}^{2t}+\textrm{e}^{-2t}+2+2),$
$\displaystyle=1$

which shows that these definitions of the $\cosh$ and $\sinh$ appropriately
parameterize the unit hyperbola with the parameter $t$. To better understand
the parameterization, imagine $t$ as a unit of time, and $x(t)$ and $y(t)$ are
the $x,y$ positions of a point moving along the unit hyperbola.

It is also really important to notice that the $\cosh$ is an even function,
whereas the $\sinh$ is an odd function, in the sense that an even function
$f(x)$ has the property $f(-x)=f(x)$ while odd functions have the property
$f(-x)=-f(x)$. I will prove these properties for the $\cosh$ and $\sinh$ to
you below.

$\displaystyle\cosh(-x)$
$\displaystyle=\frac{\textrm{e}^{(-x)}+\textrm{e}^{-(-x)}}{2}=\frac{\textrm{e}^{-x}+\textrm{e}^{x}}{2}=\frac{\textrm{e}^{x}+\textrm{e}^{-x}}{2}=\cosh(x).$
$\displaystyle\sinh(-x)$
$\displaystyle=\frac{\textrm{e}^{(-x)}-\textrm{e}^{-(-x)}}{2}=\frac{\textrm{e}^{-x}-\textrm{e}^{x}}{2}=\frac{-(\textrm{e}^{x}-\textrm{e}^{-x})}{2}=-\sinh(x).$

I wanted to point these properties out to you because they actually
identically mimic those of the normal trigonometric functions, where the
cosine is even and the sine is odd! There are actually a slew of other
analogous properties between circular trigonometry and hyperbolic
trigonometry, but I will not address them here. From here I am going to shift
our attention to uniting circular trig with hyperbolic trig through the
complex plane.

We defined a hyperbolic cosine and a hyperbolic sine. Recall that the normal
(circular) tangent is the quotient of the sine and cosine. Use this quotient
to derive the hyperbolic tangent function over the real numbers $x$, denoted
by $\tanh{x}$ and pronounced tan-ch, and then sketch it so you see what it
looks like. Based on your expression for the $\tanh$, prove that it is an odd
function over the reals.

Okay, back to complex numbers. Consider the function $w(z)=\cosh{z}$. This
time, we will choose to work with $z$ in its Cartesian representation so that
$z=x+iy$. Then, using Eq. 2.57, we have

$\displaystyle w(z)$ $\displaystyle=\cosh(x+iy),$
$\displaystyle=\frac{\textrm{e}^{x+iy}+\textrm{e}^{-(x+iy)}}{2},$
$\displaystyle=\frac{\textrm{e}^{x}\textrm{e}^{iy}+\textrm{e}^{-x}\textrm{e}^{-iy}}{2},$
$\displaystyle=\frac{\textrm{e}^{x}\left(\cos y+i\sin
y\right)+\textrm{e}^{-x}\left(\cos y-i\sin y\right)}{2},$
$\displaystyle=\frac{\left(\textrm{e}^{x}+\textrm{e}^{-x}\right)\cos
y+i\left(\textrm{e}^{x}-\textrm{e}^{-x}\right)\sin y}{2},$
$\displaystyle=\frac{\textrm{e}^{x}+\textrm{e}^{-x}}{2}\,\cos
y+i\frac{\textrm{e}^{x}-\textrm{e}^{-x}}{2}\,\sin y,$
$\displaystyle=\cosh(x)\cos(y)+i\sinh(x)\sin(y).$ (2.59)

Thus, we have shown that the $\cosh$ of a complex argument mixes circular and
hyperbolic trigonometry. Furthermore, let’s assume that $z=iy$, meaning it is
a purely imaginary (lateral) number. Then

$\displaystyle\cosh(iy)=\cos(y).$ (2.60)

This relationship goes the other way as well. If we consider $\cos(iy)$ then
we have from Eq. 2.34

$\displaystyle\cos(iy)=\frac{\textrm{e}^{i(iy)}+\textrm{e}^{-i(iy)}}{2}=\frac{\textrm{e}^{-y}+\textrm{e}^{y}}{2}=\frac{\textrm{e}^{y}+\textrm{e}^{-y}}{2}=\cosh(y).$
(2.61)

Since the imaginary arguments can be used to convert between circular and
hyperbolic trig using these equations, we conclude that hyperbolic triangles
are really just a lateral/adjacent form of straight-line versions. This
conclusion could have actually been reached with the unit circle itself,
$x^{2}+y^{2}=1$, if we changed the $y$ part to an imaginary number
$y\rightarrow iy$, thus converting the unit circle to
$x^{2}+(iy)^{2}=x^{2}-y^{2}=1$, the unit hyperbola! Perhaps if we had not
given $i=\sqrt{-1}$ the name of “imaginary”, then we could have known more
about geometry, in general, immediately from the start.

I derived the relationships between the hyperbolic cosine and the circular
cosine. Using an almost identical treatment, derive the following equations:
$\displaystyle\sinh(x+iy)$ $\displaystyle=\sinh(x)\cos(y)+i\cosh(x)\sin(y),$
(2.62) $\displaystyle\sinh(iy)$ $\displaystyle=i\sin(y),$ (2.63)
$\displaystyle\sin(iy)$ $\displaystyle=i\sinh(y).$ (2.64)

###  2.7 Concluding Remarks

In this chapter, the fundamentals of complex algebra were introduced and used
to generalize a few commonly used real functions into the complex plane.
Knowing how to use complex numbers to study physical systems is particularly
helpful. For example, it is pretty common to treat all electromagnetic waves
as traveling complex exponentials given by
$\textrm{e}^{i(\vec{k}\cdot\vec{r}-\omega t)}$, where $\vec{k}$ is the
wavenumber of the wave, $\vec{r}$ is the position of the wavefront, $\omega$
is the frequency of the wave, and $t$ is the time. To be clear, the
electromagnetic waves are purely real objects meaning that all of their
physical properties would be quantities that we would associate with the real
line (they are not lateral quantities, whatever those may be…). So why are
these imaginary/lateral parts relevant in physics? It turns out that
exponentials have a plethora of useful properties in calculus and they are
much more manageable than the trigonometric waves defined by sines and
cosines. Thus, we usually only need to worry about the real part of the
electric and magnetic fields. Meanwhile, whenever the waves move through a
material, it turns out that the $\vec{k}$ can be identified as having both a
real part and a lateral part, where the latter actually plays a role in the
heat-loss of the wave in that medium (this property is called attenuation). In
other words, knowing how to split up complex numbers into real and imaginary
(lateral) parts and more can tell us very important things about how physical
objects interact with one another. And this case is one where we use complex
numbers to make our mathematical modeling simpler — it does not include
situations in physics where the whole complex plane is absolutely necessary,
such as anything where the word “quantum” is involved.

There is still a lot of topics within the topic of complex numbers at large,
but this chapter hopefully gives you a solid foundation in something that,
once you get the hang of it, will truly make that math you use in physics much
simpler, and by extension, your life much easier. Not to mention, complex
algebra at least makes things like numbers a little prettier to look at.

I have included a table of helpful equations (Table 2.1) for you to reference
for whenever you need them later on in your career. They also have a reference
to the text where they were described.

TABLE 2.1: A summary of the important and general equations in Complex
Algebra.

Equation Description | Equation Formula | Text Reference
---|---|---
Definition of Complex Numbers | $z=x+iy,\;\;x,y\in\mathds{R}$ |  Eq. 2.8
Definition of Complex Conjugate | $z=x-iy,\;\;x,y\in\mathds{R}$ |  Eq. 2.9
Real Part of a Complex Number | $\mathrm{Re}(z)=\dfrac{z+z^{\ast}}{2}\in\mathds{R}$ |  Eq. 2.10
Imaginary Part of a Complex Number | $\mathrm{Im}(z)=\dfrac{z-z^{\ast}}{2i}\in\mathds{R}$ |  Eq. 2.11
Addition of Complex Numbers | $z\pm w=(x\pm u)+i(y\pm v)$ |  Eq. 2.12
Product of Complex Numbers | $zw=(xu-yv)+i(xv+yu)$ |  Eq. 2.13
Quotient of Complex Numbers | $\dfrac{z}{w}=\dfrac{xu+yv}{u^{2}+v^{2}}+i\,\dfrac{xv-yu}{u^{2}+v^{2}}$ |  Eq. 2.14
Modulus of a Complex Number | $|z|=\sqrt{z^{\ast}z}=\sqrt{[\textrm{Re}(z)]^{2}+[\textrm{Im}(z)]^{2}}$ |  Eq. 2.16
Polar Representation | $z=|z|\textrm{e}^{i\,\mathrm{arg}(z)}$ |  Eq. 2.27
Argument of a Complex Number | $\mathrm{arg}(z)=\arctan\left[\dfrac{\mathrm{Im}(z)}{\mathrm{Re}(z)}\right]$ |  Eq. 2.30
Euler’s Identity | $\textrm{e}^{i\theta}=\cos\theta+i\,\sin\theta$ |  Eq. 2.31
Cosine Formula | $\cos\theta=\dfrac{\textrm{e}^{i\theta}+\textrm{e}^{-i\theta}}{2}$ |  Eq. 2.34
Sine Formula | $\sin\theta=\dfrac{\textrm{e}^{i\theta}-\textrm{e}^{-i\theta}}{2i}$ |  Eq. 2.35
Multiplication as Dilation and Rotation | $zw=(r\rho)\,\textrm{e}^{i(\theta+\phi)}$ |  Eq. 2.36
De Moivre’ Theorem (for Integer $n$) | $(\cos\theta+i\sin\theta)^{n}=\cos(n\theta)+i\sin(n\theta)$ |  Eq. 2.41
$n^{\mathrm{th}}$ Roots of Unity | $1^{1/n}=\exp\left(\dfrac{2\pi mi}{n}\right),\;\,m\in\\{0,1,\dots,n-1\\}$ |  Eq. 2.43
$n$-Term Geometric Series | $S_{n}=\sum\limits_{m=0}^{n-1}a^{m}=\dfrac{1-a^{n}}{1-a},\;a\neq 1$ |  Eq. 2.45
Sum of $n^{\mathrm{th}}$ Roots of Unity | $\sum\limits_{m=0}^{n-1}\left(\textrm{e}^{2\pi i/n}\right)^{m}=0$ |  Eq. 2.46
General Complex Exponentiation | $z^{\alpha+i\beta}=\left(r\textrm{e}^{i\theta}\right)^{\alpha+i\beta}=r^{n}\textrm{e}^{-\beta\theta}\,\textrm{e}^{i[\beta\ln(r)+\alpha\theta]}$ |  Eq. 2.52
Principal Logarithm | $\mathrm{Ln}(z)=\mathrm{Ln}(r\textrm{e}^{i\theta})=\ln(r)+i\theta,\;\;\theta\in[-\pi,\pi)$ |  Eq. 2.56

##  Chapter 3 Calculus of a Single Variable

Up until now, I have not assumed that you have any prior knowledge of
calculus, mostly because it is frankly unnecessary to understand vectors and
complex algebra. Thus, everything that we have covered is totally static. It,
in no way, describes anything that can possibly change at all. For example, if
the position vector of a particular object is known at a particular time, we
would thus far only have the tools to describe the object’s vector at that
time. Sure, we could resolve the vector into its components, we could talk
about the direction of the vector, we could talk about it being orthogonal to
other vectors, but we would have NO WAY of talking about how that vector is
changing at that time. Hence could not speak of how the object will move in
time.

You might be wondering why we could not just reference the object’s velocity
vector at that time, since surely the velocity will tell us about how the
position vector changes. As we will see, this is absolutely true. However,
none of the mathematical tools we’ve discussed so far allow us to describe
this change. Meanwhile, you can definitely intuit how position will change
because you have experienced changes in your everyday life. Our goal in this
chapter is to develop your natural intuition into something mathematically
precise and immutable — our goal is to develop calculus.

As a heads up, chapter is going to be dense. Like, degenerate matter dense111A
classical model of the degeneracy will probably suffice though.. We are going
to cover essentially the same amount of material that one would see as they
progressed from a Calculus I and II sequence of courses. Although we will
cover a lot of material, it will be condensed to a form that is sufficient for
implementation in physics. Truth be told, once you understand derivatives and
integrals from a geometric/changey perspective, then the number of dimensions
rarely makes a difference to a physicist. Sure, it might add a few extra terms
here and there, but the actual modeling of the natural world using rates of
change is what matters to us, and it is almost always easy, if not trivial, to
discuss the laws of physics using calculus. That is not to say that actually
solving problems in physics is trivial, it isn’t, but the nontriviality of the
problem solving in physics is partly due to the inclusion of our clunky form
of algebra with calculus. The other part is usually just because nature is
smarter than us.

I plan to introduce many calculus topics at are relevant for physics in ways
that are easy to visualize. I am not going to get too bogged down in the
details about which functions have which derivatives or antiderivatives,
although I will talk about a few important ones. Calculus really should not be
about the memorization of a bunch of special-case formulas; when it is, it is
usually very hard for students to see its value or applicability. Thus, I will
reference outside sources for specific formulas if they are not quick and easy
to derive (tables of derivatives and integrals are ubiquitous and freely
available on the Internet). Although the transition can be hard, once you are
able to see beyond the mess of special cases, using calculus to describe the
universe will be effortless, and you will probably wonder how you didn’t
already describe it is such a fluent way222Okay, so maybe this is a little
anecdotal and waxing poetic, but, hey, if using calculus were worse than what
we had before, why would we still be using it?.

###  3.1 Position, Velocity, and Acceleration

People usually have an almost instinctive understanding of how things move
(with the exception of feathers and bowling balls in free fall).
Unfortunately, it took an incredible amount of time for us to develop a means
of communicating our understanding with others in a precise way. We can thank
people like Galileo and Newton for figuring that last bit out for us. In
historical hindsight though, developing the crucial ideas of calculus are
pretty straightforward.

We are going to start by exploiting the position, velocity, and acceleration
of some object (maybe a car or volleyball or bug) and we will be careful to
remember that each is a vector. Furthermore, let us assume that we know the
position vectors at a few different times, and let us denote these vectors by
$\vec{r}(t)$. Using vector addition (recall tip-to-tail), we can then define a
vector that changes the position at one time $t$ to one at a later time
$t+\Delta t$. Here this $\Delta t$ represents an increment in time, for
example, it could represent a year, a century, a second, a femtosecond, a unit
of Planck time, et cetera. Physically, this increment in time would be
whatever time difference it takes to get the object from one position to the
next. This change vector at time $t$ is going to be denoted as
$\Delta\vec{r}(t)$ and connects the position at $t$, $\vec{r}(t)$, to the
position at $t+\Delta t$, $\Delta\vec{r}(t+\Delta t)$, as is shown in Fig.
3.1.

FIG. 3.1: If the position of an object is known at two times, perhaps at $t$
and $t+\Delta t$, it is possible to describe the change in position from
$\vec{r}(t)$ to $\vec{r}(t+\Delta t)$ as $\Delta\vec{r}(t)=\vec{r}(t+\Delta
t)-\vec{r}(t)$.

A question that follows from this set up would be something like, “Is it
possible to describe how quickly the object moved from $\vec{r}(t)$ to
$\vec{r}(t+\Delta t)$?” Intuitively, we might propose that the change in
position, or the position increment, $\Delta\vec{r}(t)$, is proportional to
the increment in time, $\Delta t$. For example, if we were running at a
specific pace for a longer period of time, we will travel farther. However,
time is scalar, whereas the change in position is definitely a vector by Fig.
3.1. Thus, we are left to define the velocity333For anyone who already knows
calculus, this is the average velocity. of the object at time $t$ as the
vector responsible for the change in the object’s position. This is denoted as
$\vec{v}(t)$ and can be written mathematically as

$\displaystyle\Delta\vec{r}(t)=\vec{r}(t+\Delta t)-\vec{r}(t)=\vec{v}(t)\Delta
t,$ (3.1)

or if we divide both sides by the scalar time increment, we have

$\displaystyle\vec{v}(t)=\frac{\Delta\vec{r}(t)}{\Delta
t}=\frac{\vec{r}(t+\Delta t)-\vec{r}(t)}{\Delta t},$ (3.2)

It is important to remember that at this point the $(t)$ things everywhere are
meant to symbolize that the letter immediately before is a function of $t$; it
does not mean multiply everything by $t$. For example, $\vec{v}(t)$ means that
the velocity is a function of time $t$ rather than multiply $\vec{v}$ and the
scalar $t$.

By looking at Eq. 3.2, we see that the equation has the form of a slope, in
that it is a function that has a change divided by a change, or more
colloquially, it has a rise-over-run type form. Furthermore, we can see where
the units of miles-per-hour or meters-per-second come from, since we multiply
the position vector measured in units of length by the scalar $1/\Delta t$
which has units of per-time. Also, using this definition, since the time
increment takes on the value of whatever time-recording device we have allows,
if there is no change in position, we conclude that there is zero velocity.
Likewise, if there is ever a time where the velocity is zero, then there will
be no resulting change in position. Alright, now that the relationship between
changes in position and velocity is established between two points in time,
what happens when we introduce more points? Or from a more scientific
perspective, what happens when we measure the position of an object at a
greater number of points in time?

FIG. 3.2: If the position of an object is known at three times, perhaps at
$t$, $t+\Delta t$, and $t+2\Delta t$, it is possible to describe the change in
the change in the position from $\vec{r}(t)$ to $\vec{r}(t+2\Delta t)$ as
$\Delta\Delta\vec{r}(t+\Delta t)=\Delta\vec{r}(t+\Delta t)-\Delta\vec{r}(t)$.
Since the change in position is a measure of velocity by Eq. 3.2, then this
figure shows the existence in the change in velocity, also known as the
acceleration.

For simplicity, let’s assume we know the position of a particle at three
points in time: $t$, $t+\Delta t$, and $t+2\Delta t$. Then, just as we did
before in Fig. 3.1, it is possible for us to connect the position at $t$ with
that at $t+\Delta t$, and we can now connect the position at $t+\Delta t$ with
that at $t+2\Delta t$. A picture of this scenario is shown in Fig. 3.2, and in
this figure, it is evident that there are really two changes in position:

$\displaystyle\Delta\vec{r}(t)$ $\displaystyle=\vec{r}(t+\Delta
t)-\vec{r}(t),$ $\displaystyle\Delta\vec{r}(t+\Delta t)$
$\displaystyle=\vec{r}(t+2\Delta t)-\vec{r}(t+\Delta t).$

Thus, we may ask “what is the change in the change in position?” Let us denote
this change of change in position as $\Delta\Delta\vec{r}(t+\Delta t)$, since
the change of change position occurs at $t+\Delta t$. Let’s denote the change
of change as $\Delta\Delta$, and evaluate it directly

$\displaystyle\Delta\Delta\vec{r}(t+\Delta t)=\Delta\vec{r}(t+\Delta
t)-\Delta\vec{r}(t)$

If we now assume that this change is proportional to some vector rate, just
like we did when we came up with the notion of velocity, and denote this rate
as $\vec{a}(t+\Delta t)$, then we can posit that

$\displaystyle\Delta\Delta\vec{r}(t+\Delta t)=\vec{a}(t+\Delta t)\Delta
t^{2},$

Here $\Delta t^{2}=(\Delta t)^{2}$. Solving for $\vec{a}(t+\Delta t)$
directly, we have

$\displaystyle\vec{a}(t+\Delta t)=\frac{\Delta\vec{r}(t+\Delta
t)-\Delta\vec{r}(t)}{\Delta t^{2}}=\frac{\frac{\Delta\vec{r}(t+\Delta
t)}{\Delta t}-\frac{\Delta\vec{r}(t)}{\Delta t}}{\Delta t}.$

In the second equality, I moved one of the $\Delta t$s into the numerator to
make two fractions because these fractions are defined explicitly in Eq. 3.2.
Using this definition, we have

$\displaystyle\vec{a}(t+\Delta t)=\frac{\vec{v}(t+\Delta t)-\vec{v}(t)}{\Delta
t},$ (3.3)

which we recognize as the definition of acceleration444Average acceleration to
be precise. Again, this equation is one that has the form of a slope. Further,
we would say that if there is a change in velocity, then there must have been
an acceleration to cause it. However, since the acceleration is related to the
change of change in position, we could also say that a “second-order” change
in position is due to the existence of an acceleration.

So what have we learned by measuring more points? Well, it appears that each
time we measure a new point, we can define a new rate associated with the
addition of a new position measurement. This is generally true for position
measurements. Actually, the names for a few of the even higher order changes
($3^{rd}-6^{th}$) are the jerk, snap, crackle, and pop
[thompson_systems_technology]. We also know that the little time increment
$\Delta t$ is some measurement, but its actual value depends on the time-
measuring device we have (for example, atomic clocks have smaller $\Delta t$s
than grandfather clocks). In principle, we could have exactly one position
measurement for every single time measurement we have, and let’s assume we
have $N$ total of these position measurements. Then we could write them down
in a table of sorts as

$\displaystyle\begin{array}[]{c|c|ccccc}\textbf{Measure:}&\textrm{Time
}t&t&t+\Delta t&t+2\Delta t&\dots&t+(N-1)\Delta t\\\
\hline\cr\textbf{Measure:}&\textrm{Position
}\vec{r}&\vec{r}(t)&\vec{r}(t+\Delta t)&\vec{r}(t+2\Delta
t)&\dots&\vec{r}(t+(N-1)\Delta t)\end{array}$

And so each time there is a change in position, we would presumably have a
corresponding velocity measurement, and additionally, each time we have a
change in velocity we would have an acceleration. So we would have two more
rows in that table

$\displaystyle\begin{array}[]{c|c|ccccc}\textbf{Measure:}&\textrm{Time
}t&t&t+\Delta t&t+2\Delta t&\dots&t+(N-1)\Delta t\\\
\hline\cr\textbf{Measure:}&\textrm{Position
}\vec{r}&\vec{r}(t)&\vec{r}(t+\Delta t)&\vec{r}(t+2\Delta
t)&\dots&\vec{r}(t+(N-1)\Delta t)\\\
\hline\cr\textbf{Calculate:}&\textrm{Velocity }\vec{v}&-&\vec{v}(t+\Delta
t)&\vec{v}(t+2\Delta t)&\dots&\vec{v}(t+(N-1)\Delta t)\\\
\hline\cr\textbf{Calculate:}&\textrm{Acceleration }\vec{a}&-&\vec{a}(t+\Delta
t)&\vec{a}(t+2\Delta t)&\dots&-\end{array}$

Here, the dashed marks appear since there are no changes initially for either
the position or velocity; thus there are no velocities or accelerations at
this time. To better help visualize this table, imagine that all of these
measurements came from an object’s trajectory as that shown in Fig. 3.3, where
the object starts at an initial position of $\vec{r}_{0}$ and moves to a final
position of $\vec{r}_{f}$. In the figure, the blue points represent the
positions as functions of time (remember that these are all vectors, hence the
ghostly vectors to $\vec{r}_{0}$ and $\vec{r}_{f}$), the red dashed vectors
represent the velocities as functions of time, and the green dotted vectors
represent the accelerations as functions of time.

FIG. 3.3: The position, velocity, and acceleration of particle as a function
of time $t$ and it meanders from an initial position $\vec{r}_{0}$ to a final
position $\vec{r}_{f}$. During the trip, it can have many different velocities
and accelerations, whose directions are represented by the red dashed arrows
and the green dotted arrows, respectively.

An important highlight from drawing out all of these position measurements is
that the velocity vectors can really be seen to connect all of the position
measurements together. More precisely, if we wanted to know the position of
the object at any time measurement, let’s say at the $j^{\mathrm{th}}$ time
increment, then we could find it by starting at $\vec{r}_{0}$ and then adding
all of the changes in position up until the $j^{\mathrm{th}}$ increment. In
other terms,

$\displaystyle\vec{r}(t+j\Delta t)$
$\displaystyle=\vec{r}_{0}+\Delta\vec{r}(t)+\Delta\vec{r}(t+\Delta
t)+\Delta\vec{r}(t+2\Delta t)+\dots\Delta\vec{r}(t+(j-1)\Delta t),$
$\displaystyle=\vec{r}_{0}+\vec{v}(t)\Delta t+\vec{v}(t+\Delta t)\Delta
t+\vec{v}(t+2\Delta t)\Delta t+\dots+\vec{v}(t+(j-1)\Delta t)\Delta t.$

All of the $\Delta t$s here show up from the implicit definition of the
velocity at a point. The equation above can be written much more succinctly
with summation notation, and so we have

$\displaystyle\vec{r}(t+j\Delta
t)=\vec{r}_{0}+\sum_{k=0}^{j-1}\vec{v}(t+k\Delta t)\Delta t.$ (3.4)

Although it is harder to visualize from Fig. 3.3 (but not impossible!), there
is actually an analogous summation relationship that exists to relate the
velocity at $t+j\Delta t$ to all of the intermediate accelerations:

$\displaystyle\vec{v}(t+j\Delta
t)=\vec{v}_{0}+\sum_{k=0}^{j-1}\vec{a}(t+k\Delta t)\Delta t.$ (3.5)

These equations have the form of the area of a rectangle, where $\Delta t$ is
the width of the rectangle, and the velocity or acceleration would be the
height of the rectangle. Furthermore, we actually add up all of the
intermediate areas to find something new.

In short, this is all that calculus is: either evaluating slopes or adding up
a bunch of boxes. If we want to describe how quickly something changes in
terms of another variable — for example, if we want to find how fast position
changes as a function of time — we would use a slope. If, on the other hand,
we wanted to describe the aggregate effects of many successive actions — for
example, if we want to find out where our final position is after moving with
a bunch of different velocities — we would add boxes. Even further, we can
undo the effect of taking a slope by adding a bunch of areas together; an
operational inversion analogous to how multiplying something by $4$ and then
dividing the result by $4$ returns the original something. And that is truly
all of calculus. There really is nothing else to conceptualize.

However, like most things, this conceptualization is easier said than done.
And that’s okay. As we progress through this chapter, we will study many more
applications of these ideas and refine them into some theorems. Fortunately
for us, the context of this chapter will largely be the natural world, and so
if the mathematics ever gets too messy, we can always find real-world
analogies to bolster our intuition and understand which message the math is
trying to convey.

###  3.2 Continuity

We have discussed how slopes and box-addition represent function-transitions
between position, velocity, and acceleration. But everything we did was built
on the assumption that we were taking measurements of time with some device
that only measured in increments of $\Delta t$. While this picture is accurate
from a numerical perspective, and definitely from an experimental point-of-
view, it is not necessarily physical and certainly not mathematical.

Physically, objects have positions at intermediate values between the points
we measure. We unfortunately are limited by how quickly we can measure
something’s position; however, this is no way implies that objects only move
in discrete ways. In principle, if we were able to “take more measurements”
then we could describe an object’s position, velocity, or acceleration at any
time $t$ that we wanted.

To better illustrate this idea, we assume that something only moves over a
fixed time interval, perhaps $t_{f}-t_{0}$ and that we have made a total of
$N$ measurements. Then the time increment would be

$\displaystyle\Delta t=\frac{t_{f}-t_{0}}{N}.$

Notice that this time increment gets smaller when the number of samples $N$
gets larger. If this is not clear, choose $t_{f}-t_{0}$ to be your favorite
number (physicists really like $1$) and then divide by successively larger
numbers — for example, $1/1=1$, $1/2=0.5$, $1/3=0.33\dots$, …, $1/100=0.01$, …
So as long as the $t_{f}-t_{0}$ is a fixed number, $\Delta t$ will get very
small as $N$ gets very large.

The small time increment is helpful, because it allows us to probe more time-
values. For example, if $t_{0}=1$, $t_{f}=2$, and $N=10$, then we can talk
about times like

$\displaystyle t_{0}+\Delta t$ $\displaystyle=1+1/10=1.1,$ $\displaystyle
t_{0}+2\Delta t$ $\displaystyle=1+2/10=1.2,$ $\displaystyle\vdots$
$\displaystyle t_{0}+(N-1)\Delta t$ $\displaystyle=1+9/10=1.9$

But we could never reach a time value like $1.11$, since our time increment is
larger than that extra $0.01$. In other words, our resolution is not good
enough to see a time increment of $0.01$. So what can we do? Instead of only
taking $10$ measurements, how about we take $100$? Then we could have

$\displaystyle t_{0}+\Delta t$ $\displaystyle=1+1/100=1.01,$ $\displaystyle
t_{0}+2\Delta t$ $\displaystyle=1+2/100=1.02,$ $\displaystyle\vdots$
$\displaystyle t_{0}+(N-1)\Delta t$ $\displaystyle=1+99/100=1.99$

By the same argument, we could jump down to a resolution of 0.001 by taking
1000 measurements to have

$\displaystyle t_{0}+\Delta t$ $\displaystyle=1+1/1000=1.001,$ $\displaystyle
t_{0}+2\Delta t$ $\displaystyle=1+2/1000=1.002,$ $\displaystyle\vdots$
$\displaystyle t_{0}+(N-1)\Delta t$ $\displaystyle=1+999/1000=1.999$

Hence, it is clear that by increasing the number of samples, we can know the
position over smaller time increments and therefore observe what happens at
those intermittent steps that we would have been incapable of observing
before. So now the question is: “how many samples do we need to get a time
increment to probe time up to any arbitrary level of precision $\epsilon$?”

To solve this problem, we just need the time increment to be less than this
arbitrary precision $\epsilon$ (before $\epsilon\in\\{0.1,0.01,0.001\\}$).
Thus, what we do is posit that $\Delta t<\epsilon$, or

$\displaystyle\frac{t_{f}-t_{0}}{N}<\epsilon\Rightarrow\frac{t_{f}-t_{0}}{\epsilon}<N,$
(3.6)

The second inequality555This inequality rule can be tricky for a lot of people
the first time they see it. Most people know that if you multiply an
inequality by a negative sign, then you must flip the inequality. The same is
true for division. For example, it is true that $1/2<1/1$, but it is not true
that $2<1$, when we invert the inequality. is our answer. I recommend you test
it for yourself.

The beautiful thing about this idea is that $\epsilon$ can literally be
anything, as long as it’s positive. That means we could increase it or
decrease it at will — this is the math equivalent of increasing or decreasing
the volume of your phone666Or whatever the cool sound-emitting technological
marvel you future people use., except there is no upper or lower bound! Thus,
we could, at least in principle, probe time measurements to any level of
precision we want. All we would need to do is take enough measurements.

In mathematics, we give this idea the name, limit of a sequence. In this case,
the sequence would be $\Delta t$, and the elements of the sequence would be
each $\Delta t$ with a incrementally higher $N$ value substituted in. In this
case, we would say that the limit of $\Delta t$ would be zero since we can
definitively get within a precision of $\epsilon$ of zero using the
inequalities above. More formally, we would say that the limit of some
sequence $a_{N}$, denoted by $L$, exists by the following criterion:

> The sequence $a_{N}$ is said to converge to the limit $L$ if there exists
> positive integers $n$ and $M$ such that $|a_{n}-L|<\epsilon$ for every
> $\epsilon>0$, whenever $n>M$.

When the limit does in fact exist, we say that

$\displaystyle\lim_{N\rightarrow\infty}a_{N}=L,$ (3.7)

where the calculation of how many measurements are required to reach a
precision of $\epsilon$ is given by the positive integer $M$ and the “taking
enough measurements” is taken into account when $N\rightarrow\infty$ in Eq.
3.7.

Knowing more and more precise time measurements is great, but what does it get
us? How can we be sure that there will truly be a position measurement for us
to take if we zoom in on the temporal axis with smaller and smaller time
increments? To the best of my knowledge, in physics, we assume that this is
always true for moving objects. But this would imply that as we more and more
precise time measurements, we could get more and more precise position
measurements. Or at least, if the position at one point in time is known, then
the position as $t+\Delta t$ should be nearby that at $t$. Otherwise, the
object would somehow jump/teleport to another location randomly. And that
would be weird/not anything we have ever observed, so in physics we assume
that we can always zoom in infinitely far for all spatial and temporal
variables ($N\rightarrow\infty$).

But how would we represent this mathematically? For this case, we will assume
that we have taken infinitely many temporal measurements so that we can zoom
in arbitrarily around some special point in time, let’s call it $t^{\prime}$.
If we have zoomed in to some arbitrary level of precision, now denoted by
$\delta t$777The lowercase delta is because our precision is tiny, so I
figured a little variable would be appropriate., then we can probe any time
values within what’s called an open ball of radius $\delta t$. The ball itself
is defined as the set of all $t$-values that are within the radius $\delta t$
of the central point $t^{\prime}$, and for those who are interested it is
written as

$\displaystyle\mathcal{B}(t^{\prime},\delta
t)=\\{t\in\mathds{R}:|t-t^{\prime}|<\delta t\\}.$

In a single dimension though, this ball is just the open interval
$t\in(t^{\prime}-\delta t,t^{\prime}+\delta t)$ along the real line. It is a
circular disk in two dimensions, and actually is a ball (sphere) in three
dimensions. In four and higher dimensions, it is harder to visualize888Yes,
this is a dare to prove me wrong..

Anyway, we want to be sure that an object doesn’t suddenly teleport elsewhere,
so we need the position of an object at time $t$ to be nearby the position at
time $t^{\prime}$. Therefore, if we can find a position ball whose radius
$|\delta\vec{r}|$ is somehow the result of the $\delta t$ precision we chose,
then we can guarantee that the position of the object at time $t$ will always
be within the vicinity of the position at time $t^{\prime}$. And most
importantly, means

$\displaystyle|\vec{r}(t)-\vec{r}(t^{\prime})|<|\delta\vec{r}|,$ (3.8)

where $|\delta\vec{r}|$ is a finite number, of again, arbitrary precision.

This whole idea that zooming in far on the independent variable consequently
leads to us freely zooming on the dependent variable is what is known as
continuity. To be more mathematically precise:

> A function $f:\mathds{R}\rightarrow\mathds{R}$ is said to be continuous at
> $x^{\prime}$ if there exists a $\delta>0$ for every $\epsilon>0$ such that
> for every $x$ where $|x-x^{\prime}|<\delta$ then implies
> $|f(x)-f(x^{\prime})|<\epsilon$.

When the above statement holds, we call the output $f(x^{\prime})$ the limit
of $f(x)$ as $x\rightarrow x^{\prime}$ which is written as

$\displaystyle\lim_{x\rightarrow x^{\prime}}f(x)=f(x^{\prime}).$ (3.9)

Sometimes, the limit is instead defined as $F$ so we can study functions where
$f(x^{\prime})$ need not be defined.

The definition above is what you may have seen/heard of before — it is the
$\epsilon$-$\delta$ definition of continuity — but it is strictly in one-
dimension, whereas actually everything else up until this point was in an
arbitrarily high number of spatial dimensions; so maybe we were taking about
an object only moving along a single axis (one-dimensional), or on a curve on
a surface (two-dimensional), or even in a trajectory through space (three-
dimensional). Here we condensed back to one-dimension with the
$\epsilon$-$\delta$ definition of continuity, because the mathematical
precision in higher dimensions is actually a little too nuanced to jump in
with immediately. Actually, to be completely honest, I don’t think the
definition, as it is written above, has ever come up in a physics class that
I’ve taken. However, the idea of being able to zoom in as far as we like does
come up constantly; that is why I spent so much time developing the intuition
behind the $\epsilon$-$\delta$ definition of continuity. Nevertheless, I do
want to unpack that definition a little, because it will be extremely relevant
for things like derivatives later on.

First things first, the statement $f:\mathds{R}\rightarrow\mathds{R}$
translates to “function $f$ that has real number inputs (first $\mathds{R}$)
and then outputs real numbers (second $\mathds{R}$).” This means we are only
dealing with the good-ol’ one-dimensional functions that you learned in high
school. The next thing that is listed is that this definition of continuity
only applies for functions at a single point $x^{\prime}$ in the domain. It
says nothing about the continuity of $f$ at nearby points. The actual
requirement for continuity is that there has to be an $\delta>0$, a.k.a. an
arbitrarily small number, to zoom in on $x$ nearby $x^{\prime}$ when we zoom
in on $f(x^{\prime})$ itself to a chosen precision of $\epsilon$. Therefore,
this definition posits that the existence of the $\epsilon$ precision
guarantees the existence of the $\delta$ precision in the independent
variable. This means that continuity allows us to continue taking smaller and
smaller independent-variable increments to measure smaller and smaller
dependent-variable increments, just like our intuition with an object’s
position informed us. So this definition, although I agree that it is pretty
intangible the first time you see it, incorporates all of what we’ve discussed
in the section so far, but it does it in a much more concise manner.

Example 3.2 shows how we might use the definition of continuity to show simple
functions are continuous. The truth is that it usually takes a different
algebra tricks to establish the continuity of individual functions, and
oftentimes these tricks are NOT obvious. If you ever take a course in
mathematical analysis, you will see it. Therefore, it usually helps to
establish the continuity of general classes of functions, like that of sums,
products, or compositions, because we can then argue that whatever messy
function we have is truly just a combination of continuous functions. Thus, I
am going to prove a few theorems to you regarding the continuity of sums,
products, and compositions.

Disclaimer: the proofs will be pretty abstract, and so if it is too hard to
follow at first, that’s okay. The fundamentally important continuity rules in
physics will be the numbered limit equations that follow each proof, but I
wouldn’t be able to live with myself if I just told you what they were without
justification. I mean, this is supposed to be science after all, right?

Definitions are great, but their utility really comes in applying them. In
this example, I will prove to you that constant and linear functions are
continuous. We first start with a constant function. Let’s assume our constant
function looks like $f(x)=A$ for all $x\in\mathds{R}$. If you don’t like the
$A$ here, replace it with the number $\pi$ and then, at the end, replace all
the $\pi$s with $A$s. Okay, so the proof goes something like this: Consider
the point $x^{\prime}\in\mathds{R}$ and the open ball
$\mathcal{B}(f(x^{\prime}),\epsilon)$ centered at $f(x^{\prime})$ with radius
$\epsilon>0$. More visually, this means we have zoomed in on the dependent
variable $f$ to a precision of $\epsilon$. We want to show that there exists a
$\delta>0$ such that $|x-x^{\prime}|<\delta$, no matter what $\epsilon$
precision we choose. Now consider the absolute difference
$|f(x)-f(x^{\prime})|$, where I will substitute in the values of $x$ and
$x^{\prime}$ into $f(x)=A$. $\displaystyle|f(x)-f(x^{\prime})|=|A-A|=0$ But by
definition, $\epsilon>0$. If we let $\delta=\epsilon>0$, then for every $x$
such that $|x-x^{\prime}|<\delta$, we have
$\displaystyle|f(x)-f(x^{\prime})|=|A-A|=0<\delta=\epsilon$ This shows that
$f$ is continuous at $x^{\prime}$. However, since this argument holds for all
$x^{\prime}\in\mathds{R}$, we conclude that constants functions are continuous
over all of the real numbers. For the next proof, we look at the linear
function $f(x)=bx$ for all $x\in\mathds{R}$, where $b\neq 0$ is a constant
coefficient. The proof is similar in set up to the one above, but its
execution will probably look very different to you the first time you see it.
Remember that the key is to show the existence of any $\delta>0$ for every
conceivable $\epsilon>0$. Consider the point $x^{\prime}\in\mathds{R}$ and the
open ball $\mathcal{B}(f(x^{\prime}),\epsilon)$ centered at $f(x^{\prime})$
with radius $\epsilon>0$. We will argue that for every $\epsilon>0$, we can
find a $\delta=\epsilon/|b|>0$ such that when $|x-x^{\prime}|<\delta$, we have
$|f(x)-f(x^{\prime})|<\epsilon$. $\displaystyle|f(x)-f(x^{\prime})|$
$\displaystyle=|bx-bx^{\prime}|$ $\displaystyle=|b(x-x^{\prime})|$
$\displaystyle=|b|\,\underbrace{|x-x^{\prime}|}_{<\delta}$
$\displaystyle<|b|\,\frac{\epsilon}{|b|}$ $\displaystyle=\epsilon$ Thus,
$|f(x)-f(x^{\prime})|<\epsilon$ when $|x-x^{\prime}|<\delta$, for all
$\delta>0$; thereby completing the proof that $f(x)=bx$ is continuous at
$x^{\prime}$. Further, since $x^{\prime}$ is arbitrary, again, this argument
holds for all of the real numbers. We conclude that linear functions are
continuous over all of the real numbers.

Before we proceed to this point, we need one very important theorem concerning
the absolute function: the so-called triangle inequality. It states simply
that for any $x,y\in\mathds{R}$, we have

$\displaystyle|x+y|\leq|x|+|y|.$ (3.10)

A proof of this statement is straightforward, and follows from the fact that
$x^{2}=(|x|)^{2}$ for all real numbers. It goes as something like this,

$\displaystyle(|x|+|y|)^{2}$ $\displaystyle=(|x|)^{2}+(|y|)^{2}+2|x|\,|y|$
$\displaystyle\geq(|x|)^{2}+(|y|)^{2}+2xy,\textrm{ since either $x$ or $y$
could possibly be negative}$ $\displaystyle=x^{2}+y^{2}+2xy$
$\displaystyle=(x+y)^{2}$ $\displaystyle=(|x+y|)^{2}.$

Thus, $(|x+y|)^{2}\leq(|x|+|y|)^{2}$. By taking the square root of both sides,
we have the triangle inequality; a theorem that gets its name because it
essentially says that if a triangle is made up of three sides, one side cannot
have a magnitude greater than the sum of the other two. This inequality will
help us establish that sums of continuous functions are continuous.

Now for continuity in summation. To prove this, consider the function
$h(x)=f(x)+g(x)$, where $f(x)$ and $g(x)$ are both continuous at $x^{\prime}$.
Since both $f$ and $g$ are continuous at $x^{\prime}$, then there must exist
$\delta$-precisions for both functions for every $\epsilon$-precision, namely
$\delta_{f}$ and $\delta_{g}$, respectively, such that

$\displaystyle|x-x^{\prime}|<\delta_{f}\textrm{ implies }|f(x)-f(x^{\prime})|$
$\displaystyle<\frac{\epsilon}{2},$
$\displaystyle|x-x^{\prime}|<\delta_{g}\textrm{ implies }|g(x)-g(x^{\prime})|$
$\displaystyle<\frac{\epsilon}{2},$

Then, by substitution into $h$, we have

$\displaystyle|h(x)-h(x^{\prime})|$
$\displaystyle=|f(x)+g(x)-f(x^{\prime})-g(x^{\prime})|$
$\displaystyle=|f(x)-f(x^{\prime})+g(x)-g(x^{\prime})|$
$\displaystyle\leq|f(x)-f(x^{\prime})|+|g(x)-g(x^{\prime})|,\textrm{ thanks to
the triangle inequality}$
$\displaystyle<\frac{\epsilon}{2}+\frac{\epsilon}{2},\textrm{ whenever
}|x-x^{\prime}|<\min\\{\delta_{f},\;\delta_{g}\\},$ $\displaystyle=\epsilon$

Thus, when we define $\delta=\min\\{\delta_{f},\;\delta_{g}\\}$, where the
$\min$ function outputs the smaller value of $\delta_{f}$ and $\delta_{g}$,
then we have that $|x-x^{\prime}|<\delta$ implies
$|h(x)-h(x^{\prime})|<\epsilon$ for every $\epsilon>0$. This means that when
$f$ and $g$ are continuous at $x^{\prime}$, so is the sum of the two. I want
to emphasize that this conclusion implies the following limit rule:

$\displaystyle\lim_{x\rightarrow
x^{\prime}}\left[f(x)+g(x)\right]=\left[\lim_{x\rightarrow
x^{\prime}}f(x)\right]+\left[\lim_{x\rightarrow x^{\prime}}g(x)\right].$
(3.11)

Hence, the limit of the sum is the sum of the limits. Furthermore, if (and
only if) both $f$ and $g$ are continuous over all of the real numbers, then so
is their sum $h$.

We next want to prove that the product of two functions that are continuous at
$x^{\prime}$ is also continuous at that point when two functions are
defined999The defined condition guarantees that I can write something like
$f(x^{\prime})$ and not have is accidentally blow up to infinity. at
$x^{\prime}$. Now let $h(x)=f(x)g(x)$, and let the definitions of $\delta_{f}$
and $\delta_{g}$ be modified a little from what is given above as

$\displaystyle|x-x^{\prime}|<\delta_{f}\textrm{ implies }|f(x)-f(x^{\prime})|$
$\displaystyle<\frac{\epsilon}{2(1+|g(x^{\prime})|)},$
$\displaystyle|x-x^{\prime}|<\delta_{g}\textrm{ implies }|g(x)-g(x^{\prime})|$
$\displaystyle<\frac{\epsilon}{2(1+|f(x^{\prime})|)}.$

These odd looking definitions will come in handy later on so that we will find
$|h(x)-h(x^{\prime})|<\epsilon$ and not some function of $\epsilon$. Then we
have by substitution,

$\displaystyle|h(x)-h(x^{\prime})|$
$\displaystyle=|f(x)g(x)-f(x^{\prime})g(x^{\prime})|,$
$\displaystyle=|f(x)g(x)+0-f(x^{\prime})g(x^{\prime})|,$
$\displaystyle=|f(x)g(x)-f(x^{\prime})g(x)+f(x^{\prime})g(x)-f(x^{\prime})g(x^{\prime})|,$
$\displaystyle=\left|\left[f(x)g(x)-f(x^{\prime})g(x)\right]+\left[f(x^{\prime})g(x)-f(x^{\prime})g(x^{\prime})\right]\right|,$
$\displaystyle\leq|f(x)-f(x^{\prime})|\,|g(x)|+|f(x^{\prime})|\,|g(x)-g(x^{\prime})|,\textrm{
via the triangle inequality}$
$\displaystyle<\frac{\epsilon}{2(1+|g(x^{\prime})|)}\,|g(x)|+|f(x^{\prime})|\,\frac{\epsilon}{2(1+|f(x^{\prime})|)},\textrm{
when }|x-x^{\prime}|<\min\\{\delta_{f},\;\delta_{g}\\}.$

Before continuing, we must figure out a way to deal with the $|g(x)|$ because
we do not know the specific value of $g$ at $x$ — we only know that
$g(x^{\prime})$ is defined and that $g$ is continuous at $x^{\prime}$. But
since $g$ is continuous at $x^{\prime}$, we are free to zoom in as far as we
like on $g(x^{\prime})$ and we are guaranteed to be able to find an
$x$-interval that corresponds to our $g$-precision. Thus, what we will do is
choose to limit our $g$-precision to be less than a set number, for example,
$1$. Then, by the continuity of $g$, there exists a $\delta_{1}$ such that

$\displaystyle|x-x^{\prime}|<\delta_{1}\textrm{ implies }|g(x)-g(x^{\prime})|$
$\displaystyle<1.$

Then we can say using the triangle inequality that

$\displaystyle|g(x)|=|g(x)+0|=|g(x)-g(x^{\prime})+g(x^{\prime})|\leq|g(x)-g(x^{\prime})|+|g(x^{\prime})|<1+|g(x^{\prime})|,$

where the last inequality holds for every $x$ such that
$|x-x^{\prime}|<\delta_{1}$. Okay, so now that we have these conditions in
place, and if we remember that $|f(x^{\prime})|\leq 1+|f(x^{\prime})|$, we can
define our $\delta$-precision such that
$\delta=\min\\{\delta_{f},\;\delta_{g},\;\delta_{1}\\}$. Then for every $x$
such that $|x-x^{\prime}|<\delta$, we have

$\displaystyle|h(x)-h(x^{\prime})|$
$\displaystyle<\frac{\epsilon}{2(1+|g(x^{\prime})|)}\,|g(x)|+|f(x^{\prime})|\,\frac{\epsilon}{2(1+|f(x^{\prime})|)},$
$\displaystyle\leq\frac{\epsilon\,(1+|g(x^{\prime})|)}{2(1+|g(x^{\prime})|)}+\frac{\epsilon\,(1+|f(x^{\prime})|)}{2(1+|f(x^{\prime})|)},$
$\displaystyle=\frac{\epsilon}{2}+\frac{\epsilon}{2}$
$\displaystyle=\epsilon.$

Since we were able to find an $\delta>0$ for every $\epsilon>0$ such
that$|x-x^{\prime}|<\delta$ implies/leads to $|h(x)-h(x^{\prime})|<\epsilon$,
then we can conclude that $h(x)=f(x)g(x)$ is continuous at $x^{\prime}$. This
time, the corresponding limit equation would look like

$\displaystyle\lim_{x\rightarrow
x^{\prime}}\left[f(x)g(x)\right]=\left[\lim_{x\rightarrow
x^{\prime}}f(x)\right]\,\left[\lim_{x\rightarrow x^{\prime}}g(x)\right],$
(3.12)

or the limit of a product is the product of limits. Just as was the case for
the the sum of continuous functions, if both $f$ and $g$ are continuous over
all of the reals, so is their product. However, if only one function is
continuous everywhere, then the product is NOT continuous everywhere — it,
too, is limited101010Nice pun, right? to only be continuous for the values
$x^{\prime}$ where both of its factors are.

The last continuity rule deals with compositions of functions. As a brief
review compositions are sometimes written as $g(f(x))$ or $g\circ f(x)$. They
are functions that look something like this: if $g(x)=x^{2}$ and $f(x)=4\sin
x$, then $g(f(x))=(4\sin x)^{2}$. Thus, the notation of $g(f(x))$ really just
means put whatever the output of $f$ is into $g$ as the input. Although this
class of functions may seem like more of a special case compared to sums or
products (it did to me the first time I learned of it), compositions of
functions appear everywhere in both calculus and physics. Thus, knowing how
they behave is imperative.

Our goal in particular is to show that compositions of continuous functions
are also continuous. To do this, we then need to show that a function
$h(x)=g(f(x))$ is continuous at the point $x^{\prime}$, given that $f$ is
continuous at $x^{\prime}$ and $g$ is continuous at $f(x^{\prime})$. To make
this proof a little clearer, define

$\displaystyle u=f(x),\;u^{\prime}=f(x^{\prime}).$

By the continuity of $g$ at $u^{\prime}$, we know that for every $\epsilon>0$,
there exists a $\delta_{u}>0$ such that

$\displaystyle|u-u^{\prime}|=|f(x)-f(x^{\prime})|<\delta_{u}\textrm{ implies
}|g(u)-g(u^{\prime})|<\epsilon.$

Then by the continuity of $f$ at $x^{\prime}$, we know that for every
$\epsilon_{f}$-precision, there exists a $\delta>0$ such that

$\displaystyle|x-x^{\prime}|<\delta\textrm{ implies
}|f(x)-f(x^{\prime})|=|u-u^{\prime}|<\epsilon_{f}.$

Therefore, since we want something that looks like

$\displaystyle|x-x^{\prime}|<\delta\textrm{ implies
}|h(x)-h(x^{\prime})|=|g(f(x))-g(f(x^{\prime}))|<\epsilon,$

then we choose $\epsilon_{f}=\delta_{u}$ such that our final statement reads
as

$\displaystyle|x-x^{\prime}|<\delta\textrm{ implies
}|f(x)-f(x^{\prime})|=|u-u^{\prime}|<\delta_{u}\textrm{ implies
}|g(u)-g(u^{\prime})|=|h(x)-h(x^{\prime})|<\epsilon.$

for every $\epsilon>0$. As long as both $f$ and $g$ are continuous everywhere,
then the composite function will be continuous everywhere. If there are any
points where $g$ is not continuous, then the composite function will not be
continuous at those points111111There is a huge caveat here: sometimes
functions that are not specifically defined at a point can still look
continuous at that point. These types of function discontinuities are called
removable discontinuities or holes.. The corresponding limit equation for this
composition property can be written as

$\displaystyle\lim_{x\rightarrow
x^{\prime}}\left[g(f(x))\right]=g\left(\lim_{x\rightarrow
x^{\prime}}f(x)\right),$ (3.13)

which can be summarized as: the limit of the composition is the composition of
the limits.

Again, these continuity rules (Eqs. 3.11, 3.12, and 3.13) are incredibly
useful tools to determine if any function is continuous, or at least
continuous at a specified point.

This problem is designed to show you how to use the continuity rules defined
by Eqs. 3.11, 3.12, and 3.13, given that any constant functions or linear
functions are continuous from Example 3.2. As a quick example of how to use
these continuity rules, by 3.11, it follows that the line $f(x)=bx+A$, where
$b$ and $A$ are constants, is continuous for every single real number input,
$x$. This is because both $bx$ and $A$ are continuous everywhere from Example
3.2, therefore their sum is also continuous everywhere. (a) For the first part
of the problem, use Eq. 3.12 to argue that $f(x)=bx^{2}$ is continuous
everywhere. (b) Then, using Eq. 3.11, argue that any parabola
$f(x)=ax^{2}+bx+c$ is continuous everywhere. (c) Now argue that any cubic
polynomial $P_{3}(x)=a_{3}x^{3}+a_{2}x^{2}+a_{1}x+a_{0}$ is continuous
everywhere, where every element of the set of coefficients
$\\{a_{n}\\}=\\{a_{0},a_{1},a_{2},a_{3}\\}$ is constant. (d) Finally, use the
same argument to show that any $n^{\mathrm{th}}$ order polynomial, denoted by
$P_{n}$, and written as
$P_{n}(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+\dots+a_{2}x^{2}+a_{1}x+a_{0}$ and
accompanied by the set of constant coefficients
$\\{a_{n}\\}=\\{a_{0},a_{1},a_{2}\dots,a_{n-1},a_{n}\\}$ is continuous
everywhere.

Although the ability to potentially take an infinite amount of measurements is
great, it is honestly more helpful to know, for example, how an object is
moving because we can then predict where it will be in the next instant in
time. In physics, equations like these are called equations of motion, and the