for all $m>N_{1}$. Since $|t_{2}|>1$, the inequality above implies that, for
any $m>N_{1}$,

$3|t_{1}|^{m+1}+|t_{1}|^{m+3}<|t_{2}|^{m+1}(|t_{2}|^{2}-1)=|t_{2}|^{m+3}-|t_{2}|^{m+1}.$

Adding both sides by $2|t_{2}|^{m+1}-|t_{1}|^{m+1}-|t_{1}|^{m+3}$, we may
infer that

$\displaystyle 3|t_{1}|^{m+1}+|t_{1}|^{m+3}$
$\displaystyle<|t_{2}|^{m+3}-|t_{2}|^{m+1}$ $\displaystyle\Updownarrow$
$\displaystyle 2|t_{2}|^{m+1}+2|t_{1}|^{m+1}$
$\displaystyle<|t_{2}|^{m+3}-|t_{1}|^{m+1}+|t_{2}|^{m+1}-|t_{1}|^{m+3}$
$\displaystyle\Updownarrow$ $\displaystyle 2(|t_{2}|^{m+1}+|t_{1}|^{m+1})$
$\displaystyle<\left[|t_{2}|^{m+2}-|t_{1}|^{m+2}\right]\cdot\left[|t_{2}|+|t_{1}|\right]$
$\displaystyle\Updownarrow$
$\displaystyle\frac{|t_{2}|^{m+1}+|t_{1}|^{m+1}}{|t_{2}|^{m+2}-|t_{1}|^{m+2}}$
$\displaystyle<\frac{|t_{2}|+|t_{1}|}{2}.$

It follows from (0.53) that $q(z)$ has its only one zero inside the circle
$|z|=\frac{|t_{2}|+|t_{1}|}{2}$, for all $m>N_{1}$.

Since $|t_{1}|<|t_{2}|$, we may infer that
$\frac{|t_{1}|+|t_{2}|}{2}<|t_{2}|$. Then

(0.55) $\frac{|t_{1}|+|t_{2}|}{2}=r|t_{2}|$

for some $0<r<1$. Recall from (0.54) that $|t_{2}|^{2}-1>0$. Since $0<r<1$,
there exists $N_{2}\in\mathbb{N}$ such that

$r^{m+1}<\frac{|t_{2}|^{2}-1}{4|t_{2}|+4|t_{1}|}\frac{1}{|t_{2}|},$

for all $m>N_{2}$. Observe that

(0.56)
$\frac{|t_{2}|^{2}-1}{2}=\frac{|t_{2}|^{2}+1-2}{2}=\frac{|t_{2}|^{2}+|t_{1}|\cdot|t_{2}|}{2}-1=\frac{|t_{2}|+|t_{1}|}{2}|t_{2}|-1.$

Then for any $m>N_{2}$,

$\displaystyle r^{m+2}$
$\displaystyle<\frac{|t_{2}|^{2}-1}{4|t_{2}|+4|t_{1}|}\frac{1}{|t_{2}|}$
$\displaystyle\Updownarrow$ $\displaystyle r^{m+2}$
$\displaystyle<\frac{1}{2(|t_{2}|+|t_{1}|)}\left[\frac{|t_{2}|^{2}-1}{2}\right]\frac{1}{|t_{2}|}$
$\displaystyle\Updownarrow$ $\displaystyle(|t_{2}|+|t_{1}|)r^{m+2}$
$\displaystyle<\frac{1}{2}\left[\left(\frac{|t_{2}|+|t_{1}|}{2}\right)|t_{2}|-1\right]\frac{1}{|t_{2}|}$

Multiplying both sides by $|t_{2}|^{m+2}$, we obtain, for any $m>N_{2}$,

$(|t_{2}|+|t_{1}|)r^{m+2}|t_{2}|^{m+2}<\frac{1}{2}\left[\left(\frac{|t_{2}|+|t_{1}|}{2}\right)|t_{2}|^{m+2}-|t_{2}|^{m+1}\right].$

Equation (0.55) induces that, for any $m>N_{2}$,

(0.57)
$(|t_{2}|+|t_{1}|)\left(\frac{|t_{1}|+|t_{2}|}{2}\right)^{m+2}<\frac{1}{2}\left[\left(\frac{|t_{2}|+|t_{1}|}{2}\right)|t_{2}|^{m+2}-|t_{2}|^{m+1}\right].$

Since $|t_{1}|<1<|t_{2}|$, there exists $N_{3}\in\mathbb{N}$ such that

$|t_{1}|^{m+2}\left(\frac{|t_{2}|+|t_{1}|}{2}\right)+|t_{1}|^{m+1}<\frac{|t_{2}|^{2}-1}{4},$

for all $m>N_{3}$. By Equation (0.56),

(0.58)
$|t_{1}|^{m+2}\left(\frac{|t_{2}|+|t_{1}|}{2}\right)+|t_{1}|^{m+1}<\frac{|t_{2}|^{2}-1}{4}=\frac{1}{2}\left[\left(\frac{|t_{2}|+|t_{1}|}{2}\right)|t_{2}|-1\right],$

for all $m>N_{3}$. Define

$N=\max(N_{1},N_{2},N_{3}).$

Let $m>N$, be arbitrary. Inequality (0.58) yields that

$\displaystyle\left(\frac{|t_{2}|+|t_{1}|}{2}\right)|t_{2}|^{m+2}-|t_{2}|^{m+1}-\left[|t_{1}|^{m+2}\left(\frac{|t_{2}|+|t_{1}|}{2}\right)+|t_{1}|^{m+1}\right]$
$\displaystyle>$
$\displaystyle\left(\frac{|t_{2}|+|t_{1}|}{2}\right)|t_{2}|^{m+2}-|t_{2}|^{m+1}-\frac{1}{2}\left[\left(\frac{|t_{2}|+|t_{1}|}{2}\right)|t_{2}|-1\right]$
$\displaystyle>$
$\displaystyle\left(\frac{|t_{2}|+|t_{1}|}{2}\right)|t_{2}|^{m+2}-|t_{2}|^{m+1}-\frac{|t_{2}|^{m+1}}{2}\left[\left(\frac{|t_{2}|+|t_{1}|}{2}\right)|t_{2}|-1\right]$
$\displaystyle=$
$\displaystyle\frac{1}{2}\left[\left(\frac{|t_{2}|+|t_{1}|}{2}\right)|t_{2}|^{m+2}-|t_{2}|^{m+1}\right]$

Then by Inequality (0.57),

$\displaystyle(|t_{2}|+|t_{1}|)\left(\frac{|t_{1}|+|t_{2}|}{2}\right)^{m+2}$
$\displaystyle<$
$\displaystyle\left(\frac{|t_{2}|+|t_{1}|}{2}\right)|t_{2}|^{m+2}-|t_{2}|^{m+1}-\left[|t_{1}|^{m+2}\left(\frac{|t_{2}|+|t_{1}|}{2}\right)+|t_{1}|^{m+1}\right].$

Applying triangle inequality, we may infer that

$\displaystyle\left(\frac{|t_{2}|+|t_{1}|}{2}\right)|t_{2}|^{m+2}-|t_{2}|^{m+1}-\left[|t_{1}|^{m+2}\left(\frac{|t_{2}|+|t_{1}|}{2}\right)+|t_{1}|^{m+1}\right]$
$\displaystyle\leq$
$\displaystyle\left|-\frac{|t_{2}|+|t_{1}|}{2}t_{2}^{m+2}\right|-\left|t_{2}^{m+1}-t_{1}^{m+1}+\frac{|t_{2}|+|t_{1}|}{2}t_{1}^{m+2}\right|$
$\displaystyle\leq$
$\displaystyle\left|-\frac{|t_{2}|+|t_{1}|}{2}t_{2}^{m+2}+\frac{|t_{2}|+|t_{1}|}{2}t_{1}^{m+2}+t_{2}^{m+1}-t_{1}^{m+1}\right|$
$\displaystyle=$
$\displaystyle\left|\frac{|t_{2}|+|t_{1}|}{2}(t_{1}^{m+2}-t_{2}^{m+2})+t_{2}^{m+1}-t_{1}^{m+1}\right|.$

Consequently,

$(|t_{2}|+|t_{1}|)\left(\frac{|t_{1}|+|t_{2}|}{2}\right)^{m+2}<\left|\frac{|t_{2}|+|t_{1}|}{2}(t_{1}^{m+2}-t_{2}^{m+2})+t_{2}^{m+1}-t_{1}^{m+1}\right|.$

Since $t_{1}t_{2}=-1$, it holds that $|t_{2}-t_{1}|=|t_{2}|+|t_{1}|$. Then

$(|t_{2}-t_{1}|)\left(\frac{|t_{1}|+|t_{2}|}{2}\right)^{m+2}<\left|\frac{|t_{2}|+|t_{1}|}{2}(t_{1}^{m+2}-t_{2}^{m+2})+t_{2}^{m+1}-t_{1}^{m+1}\right|.$

Thus, by Rouché’s Theorem (Theorem 15), we conclude that

$\displaystyle p(z)=$ $\displaystyle
t_{2}^{m+1}-t_{1}^{m+1}+(t_{1}^{m+2}-t_{2}^{m+2})z+(t_{2}-t_{1})(-1)^{m+1}z^{m+2}$
$\displaystyle q(z)=$ $\displaystyle-
t_{2}^{m+1}+t_{1}^{m+1}-(t_{1}^{m+2}-t_{2}^{m+2})z$

have exactly one zero inside the circle $|z|=\frac{|t_{1}|+|t_{2}|}{2}$. Since
$|t_{1}t_{2}|=1$, it holds that

$\left|\frac{1}{t_{2}}\right|=|t_{1}|=\frac{|t_{1}|+|t_{1}|}{2}<\frac{|t_{1}|+|t_{2}|}{2}.$

Notice that $\frac{1}{t_{2}}$ is a zero of $p(z)$. Hence $\frac{1}{t_{2}}$ is
the only zero of $p(z)$ inside the circle $|z|=\frac{|t_{1}|+|t_{2}|}{2}$. If
we substitute $\frac{1}{t_{2}}$ for $z$, we may infer that $\frac{1}{t_{2}}$
is not a zero of $H_{m}(z)$. Then $H_{m}(z)$ does not have a zero inside the
circle $|z|=\frac{|t_{1}|+|t_{2}|}{2}$, since, otherwise, $p(z)$ would have
two zeros inside the circle. Now let $m>N$ be an arbitrary, and $z^{\prime}$
be arbitrary root of $P_{m}(z)$. Applying Corollary 32, we obtain

$\displaystyle|z^{\prime}|\geq$
$\displaystyle\frac{\sqrt{|ac|}}{|c|}\frac{|t_{1}|+|t_{2}|}{2}$
$\displaystyle=$
$\displaystyle\frac{\sqrt{|ac|}}{|c|}\frac{1}{2}\left(\frac{1}{|t_{1}|}+\frac{1}{|t_{2}|}\right)$
$\displaystyle=$
$\displaystyle\frac{\sqrt{|ac|}}{|c|}\frac{1}{2}\left(\frac{2\sqrt{|ac|}}{\sqrt{b^{2}-4ac}-|b|}+\frac{2\sqrt{|ac|}}{\sqrt{b^{2}-4ac}+|b|}\right)$
$\displaystyle=$
$\displaystyle\frac{1}{2}\left(\frac{2|a|}{\sqrt{b^{2}-4ac}-|b|}+\frac{2|a|}{\sqrt{b^{2}-4ac}+|b|}\right)$
$\displaystyle=$
$\displaystyle\frac{1}{2}\left(\frac{1}{|\alpha|}+\frac{1}{|\beta|}\right).$

It concludes that

$\mathcal{Z}(P_{m})\cap
B\left(\frac{1}{2}\left(\frac{1}{|\alpha|}+\frac{1}{|\beta|}\right),0\right)=\emptyset,$

for all $m\geq N$. ∎

We are ready to prove the main theorem of this section and paper.

######  Theorem 46.

Let $a,b,c\in\mathbb{R}\backslash\\{0\\}$ where $b^{2}-4ac>0$. Suppose
$\\{P_{m}(z)\\}_{m\geq 0}$ is a sequence of functions of $z$ generated as
follows:

$\sum_{m=0}^{\infty}P_{m}(z)t^{m}=\frac{1}{(at^{2}+bt+c)(1-tz)}.$

Then

$\lim\mathcal{Z}(P_{m})=\left\\{z\in\mathbb{C}:|z|=\frac{1}{|t_{1}|}\right\\}$

where $t_{1}$ is the smallest (in modulus) zero of $at^{2}+bt+c$.

###### Proof.

If we denote $t_{2}$ for the other zero of $at^{2}+bt+c$, Proposition 27 and
Lemma 30 imply that

(0.59)
$\begin{split}&{a(t_{2}-t_{1})(t_{1}z-1)(t_{2}z-1)t_{1}^{m+1}t_{2}^{m+1}}P_{m}(z)\\\
=&t_{2}^{m+1}-t_{1}^{m+1}+(t_{1}^{m+2}-t_{2}^{m+2})z+(t_{2}-t_{1})t_{1}^{m+1}t_{2}^{m+1}z^{m+2}.\end{split}$

Let us define, for each $m=0,1,2,3,\dots$,

$Q_{m}(z):=t_{2}^{m+1}-t_{1}^{m+1}+(t_{1}^{m+2}-t_{2}^{m+2})z+(t_{2}-t_{1})t_{1}^{m+1}t_{2}^{m+1}z^{m+2}.$

We are to prove that $|t_{1}|\neq|t_{2}|$. Suppose, by contradiction, that
$|t_{1}|=|t_{2}|$. Since $t_{1},t_{2}$ are the zeros of $at^{2}+bt+c$, by
Vieta’s Formula, $t_{1}t_{2}=\frac{c}{a}$. Then $|t_{1}|=|t_{2}|$ implies that

$\left|\frac{c}{a}\right|=|t_{1}t_{2}|=|t_{1}||t_{2}|=|t_{1}|^{2}.$

This induces that $|t_{1}|=|t_{2}|=\sqrt{\left|\frac{c}{a}\right|}$. For the
case $ac<0$, we may infer that

$\left\\{\begin{array}[]{cc}t_{1}=&\sqrt{\left|\frac{c}{a}\right|}\\\
t_{2}=&-\sqrt{\left|\frac{c}{a}\right|}\end{array}\right.\text{ or
}\left\\{\begin{array}[]{cc}t_{1}=&-\sqrt{\left|\frac{c}{a}\right|}\\\
t_{2}=&\sqrt{\left|\frac{c}{a}\right|}\end{array}\right..$

Notice that, by Vieta’s Formula,

$-\frac{b}{a}=t_{1}+t_{2}=0.$

This implies that $b=0$, which contradicts against our condition $b\neq 0$.
Thus our contrary supposition is false. So it holds that $|t_{1}|\neq|t_{2}|$.
For the case $ac>0$, we may infer that

$t_{1}=t_{2}=-\sqrt{\frac{c}{a}},\text{ or }t_{1}=t_{2}=\sqrt{\frac{c}{a}}.$

By Vieta’s Formula, it holds that

$-\frac{b}{a}=t_{1}+t_{2}=\pm 2\sqrt{\frac{c}{a}}.$

By squaring both sides, we may infer that

$\frac{b^{2}}{a^{2}}=4\frac{c}{a}\Leftrightarrow b^{2}-4ac=0.$

It contradicts against our condition $b^{2}-4ac>0$. Thus our contrary
supposition is false. So it holds that $|t_{1}|\neq|t_{2}|$. Here we have, in
both cases $ac<0$ and $ac>0$, obtain $|t_{1}|\neq|t_{2}|$. Given that
$|t_{1}|\leq|t_{2}|$, it follows that we may conclude that $|t_{1}|<|t_{2}|$.

Let us define the entire functions

$\displaystyle\alpha_{1}(z):=$ $\displaystyle t_{2}-t_{2}^{2}z,$
$\displaystyle\alpha_{2}(z):=$ $\displaystyle-t_{1}+t_{1}^{2}z,$
$\displaystyle\alpha_{3}(z):=$ $\displaystyle
t_{1}t_{2}^{2}z^{2}-t_{1}^{2}t_{2}z^{2},$ $\displaystyle\beta_{1}(z):=$
$\displaystyle t_{2},$ $\displaystyle\beta_{2}(z):=$ $\displaystyle t_{1},$
$\displaystyle\beta_{3}(z):=$ $\displaystyle t_{1}t_{2}z.$

Observe that

$\displaystyle\sum_{k=1}^{3}\alpha_{k}(z)\beta_{k}(z)^{m}$ $\displaystyle=$
$\displaystyle(t_{2}-t_{2}^{2}z)t_{2}^{m}+(-t_{1}+t_{1}^{2}z)t_{1}^{m}+(t_{1}t_{2}^{2}z^{2}-t_{1}^{2}t_{2}z^{2})t_{1}^{m}t_{2}^{m}z^{m}$
$\displaystyle=$ $\displaystyle
t_{2}^{m+1}-t_{2}^{m+2}z-t_{1}^{m+1}+t_{1}^{m+2}z+t_{1}^{m+1}t_{2}^{m+2}z^{m+2}-t_{1}^{m+2}t_{2}^{m+1}z^{m+2}$
$\displaystyle=$ $\displaystyle
t_{2}^{m+1}-t_{1}^{m+1}+(t_{1}^{m+2}-t_{2}^{m+2})z+(t_{2}-t_{1})t_{1}^{m+1}t_{2}^{m+1}z^{m+2}$
$\displaystyle=$ $\displaystyle Q_{m}(z).$

Now our goal is to apply Theorem 42 to prove that
$\underline{\lim}\,\mathcal{Z}(Q_{m})=\overline{\lim}\,\mathcal{Z}(Q_{m})$ by
first checking the first condition of this theorem. Let us consider
$\beta_{1}$ and $\beta_{2}$. Let $w\in\mathbb{C}$ arbitrary such that $|w|=1$.
Observe that $|\beta_{1}(z)|=|t_{2}|$ and $|w\beta_{2}(z)|=|t_{1}|$ for any
$z\in\mathbb{C}$. Then $|\beta_{1}(z)|\neq|w\beta_{2}(z)|$, for all
$z\in\mathbb{C}$, and it implies that $\beta_{1}(z)\neq w\beta_{2}(z)$, for
all $z\in\mathbb{C}$.

Now let us consider $\beta_{1}$ and $\beta_{3}$. Let $w\in\mathbb{C}$ be
arbitrary such that $|w|=1$. With $a,c\neq 0$, Vieta’s Formula induces that
$t_{1},t_{2}\neq 0$. So $\frac{1}{t_{2}}$ exists. Observe that
$\left|\beta_{1}\left(\frac{1}{t_{2}}\right)\right|=|t_{2}|$, and that
$\left|w\beta_{3}\left(\frac{1}{t_{2}}\right)\right|=|t_{1}|$. Hence
$\beta_{1}\left(\frac{1}{t_{2}}\right)\neq
w\beta_{3}\left(\frac{1}{t_{2}}\right)$. Similarly, we may deduce that
$\beta_{2}\left(\frac{1}{t_{1}}\right)\neq
w\beta_{3}\left(\frac{1}{t_{1}}\right)$ for any $w$ with $|w|=1$. Since the
conditions of Theorem 42 are met, we apply this theorem and conclude
$\underline{\lim}\,\mathcal{Z}(Q_{m})=\overline{\lim}\,\mathcal{Z}(Q_{m})$.

The next goal is to prove that

$\lim\mathcal{Z}(Q_{m})=\left\\{z\in\mathbb{C}:|z|=\frac{1}{|t_{1}|}\right\\}\cup\left\\{\frac{1}{t_{2}}\right\\},$

by showing each set is a subset and a superset of the other.

Let $z^{*}\in\lim\mathcal{Z}(Q_{m})$ be arbitrary. Suppose, by contradiction,
that
$z^{*}\notin\left\\{z\in\mathbb{C};|z|=\frac{1}{|t_{1}|}\right\\}\cup\left\\{\frac{1}{t_{2}}\right\\}$.
Then $|z^{*}|\neq\frac{1}{|t_{1}|}$, and there are 2 cases:
$|z^{*}|<\frac{1}{|t_{1}|}$ or $|z^{*}|>\frac{1}{|t_{1}|}$.

If $|z^{*}|<\frac{1}{|t_{1}|}$, observe that
$|\beta_{3}(z^{*})|=|z^{*}|\cdot|t_{1}|\cdot|t_{2}|<|t_{2}|=|\beta_{1}(z^{*})|$,
and that $|\beta_{2}(z^{*})|=|t_{1}|<|t_{2}|=|\beta_{1}(z^{*})|$. Thus
$|\beta_{1}(z^{*})|>|\beta_{k}(z^{*})|$ for $k=2,3$. Notice that, since
$z^{*}\neq\frac{1}{t_{2}}$,

$|\alpha_{1}(z^{*})|=|t_{2}-t_{2}^{2}z^{*}|\neq 0.$

This, by Theorem 42, implies that $z^{*}\notin\lim\mathcal{Z}(Q_{m})$, which
contradicts our assumption.

If $|z^{*}|>\frac{1}{|t_{1}|}$, observe that
$|\beta_{3}(z^{*})|=|z^{*}|\cdot|t_{1}|\cdot|t_{2}|>|t_{2}|=|\beta_{1}(z^{*})|$.
Notice that
$|\beta_{2}(z^{*})|=|t_{1}|<|t_{2}|=|\beta_{1}(z^{*})|<|\beta_{3}(z^{*})|$.
Thus $|\beta_{3}(z^{*})|>|\beta_{k}(z^{*})|$ for $k=1,2$. Notice that

$|\alpha_{3}(z^{*})|=|t_{1}t_{2}^{2}z^{*2}-t_{1}^{2}t_{2}z^{*2}|=|t_{1}t_{2}|\cdot|t_{2}-t_{1}|\cdot|z^{*}|^{2}>0.$

This, by Theorem 42, implies that $z^{*}\notin\lim\mathcal{Z}(Q_{m})$, which
contradicts our assumption.

Hence in both cases $|z^{*}|<\frac{1}{|t_{1}|}$ or
$|z^{*}|>\frac{1}{|t_{1}|}$, we encounter contradictions, which induces that
$z^{*}\in\left\\{z\in\mathbb{C};|z|=\frac{1}{|t_{1}|}\right\\}\cup\left\\{\frac{1}{t_{2}}\right\\}$.
Thus
$\lim\mathcal{Z}(Q_{m})\subseteq\left\\{z\in\mathbb{C};|z|=\frac{1}{|t_{1}|}\right\\}\cup\left\\{\frac{1}{t_{2}}\right\\}$.

Now let
$z^{*}\in\left\\{z\in\mathbb{C};|z|=\frac{1}{|t_{1}|}\right\\}\cup\left\\{\frac{1}{t_{2}}\right\\}$
be arbitrary. First let us consider the case
$z^{*}\in\left\\{z\in\mathbb{C};|z|=\frac{1}{|t_{1}|}\right\\}$. Observe that

$|\beta_{1}(z^{*})|=|t_{2}|\geq|t_{1}|=|\beta_{2}(z^{*})|,$

and that

$|\beta_{3}(z^{*})|=|t_{1}|\cdot|t_{2}|\cdot|z^{*}|=|t_{2}|=|\beta_{1}(z^{*})|\geq|\beta_{2}(z^{*})|.$

Notice that

$|\alpha_{3}(z^{*})|=|t_{1}t_{2}^{2}z^{*2}-t_{1}^{2}t_{2}z^{*2}|=|t_{1}t_{2}|\cdot|t_{2}-t_{1}|\cdot|z^{*}|^{2}>0$

and

$|\alpha_{1}(z^{*})|=|t_{2}-t_{2}^{2}z^{*}|\geq|t_{2}|^{2}|z^{*}|-|t_{2}|=|t_{2}|\frac{|t_{2}|}{|t_{1}|}-|t_{2}|>|t_{2}|-|t_{2}|=0.$

Then Theorem 42 implies that $z^{*}\in\lim\mathcal{Z}(Q_{m})$. Now let us
consider the case $z^{*}=\frac{1}{t_{2}}$. Observe that

$|\beta_{3}(z^{*})|=|z^{*}|\cdot|t_{1}|\cdot|t_{2}|=|t_{1}|<|t_{2}|=|\beta_{1}(z^{*})|,$

and that

$|\beta_{2}(z^{*})|=|t_{1}|<|t_{2}|=|\beta_{1}(z^{*})|.$

Thus $|\beta_{1}(z^{*})|>|\beta_{k}(z^{*})|$ for $k=2,3$. Notice that, since
$z^{*}=\frac{1}{t_{2}}$,

$|\alpha_{1}(z^{*})|=|t_{2}-t_{2}^{2}z^{*}|=0.$

This, by Theorem 42, implies that $z^{*}\in\lim\mathcal{Z}(Q_{m})$.

It concludes that

$\lim\mathcal{Z}(Q_{m})\supseteq\left\\{z\in\mathbb{C};|z|=\frac{1}{|t_{1}|}\right\\}\cup\left\\{\frac{1}{t_{2}}\right\\}.$

Here we have shown that $\lim\mathcal{Z}(Q_{m})$ and
$\left\\{z\in\mathbb{C};|z|=\frac{1}{|t_{1}|}\right\\}\cup\left\\{\frac{1}{t_{2}}\right\\}$
are subset and superset of each other. Hence

$\lim\mathcal{Z}(Q_{m})=\left\\{z\in\mathbb{C};|z|=\frac{1}{|t_{1}|}\right\\}\cup\left\\{\frac{1}{t_{2}}\right\\}.$

The next step is to show

  1. 1.

$\overline{\lim}\,\mathcal{Z}(P_{m})\subseteq\overline{\lim}\,\left.\mathcal{Z}(Q_{m})\middle\backslash\left\\{\frac{1}{t_{2}}\right\\}\right.$,
and

  2. 2.

$\left.\underline{\lim}\,\mathcal{Z}(Q_{m})\middle\backslash\left\\{\frac{1}{t_{2}}\right\\}\right.\subseteq\underline{\lim}\,\mathcal{Z}(P_{m})$.

Let $z^{*}\in\overline{\lim}\,\mathcal{Z}(P_{m})$ be arbitrary. Choose $r>0$
and $N\in\mathbb{N}$ arbitrarily. Given
$z^{*}\in\overline{\lim}\,\mathcal{Z}(P_{m})$, there exists $n>N$ such that

$B(r,z^{*})\cap\mathcal{Z}(P_{n})\neq\emptyset.$

In view of Equation (0.59), we may deduce that
$\mathcal{Z}(P_{m})\subseteq\mathcal{Z}(Q_{m})$, for any $m\in\mathbb{N}$.
Then

$B(r,z^{*})\cap\mathcal{Z}(P_{n})\subseteq B(r,z^{*})\cap\mathcal{Z}(Q_{n}),$

which implies that

$B(r,z^{*})\cap\mathcal{Z}(Q_{n})\neq\emptyset.$

This concludes that

$\overline{\lim}\,\mathcal{Z}(P_{m})\subseteq\overline{\lim}\,\mathcal{Z}(Q_{m}).$

We need to show that
$\frac{1}{t_{2}}\notin\overline{\lim}\,\mathcal{Z}(P_{m})$. There are two
cases: $ac>0$ and $ac<0$. Let us consider the case $ac>0$. In view of
$|t_{1}|<|t_{2}|$, we may infer that $\frac{1}{t_{2}}\in
B\left(\frac{1}{|t_{1}|},0\right)$. Then there exists $r>0$ such that
$B\left(r,\frac{1}{t_{2}}\right)\subseteq B\left(\frac{1}{|t_{1}|},0\right)$.
By Theorem 36, we may infer that
$B\left(\frac{1}{t_{2}},r\right)\cap\mathcal{Z}(P_{m})=\emptyset$ for any
$m\geq 0$. This implies that

$\frac{1}{t_{2}}\notin\overline{\lim}\,\mathcal{Z}(P_{m}).$

Now let us consider the case $ac<0$. By Lemma 45, there exists
$N\in\mathbb{N}$ such that

$\mathcal{Z}(P_{m})\cap
B\left(\frac{1}{2}\left(\frac{1}{|t_{1}|}+\frac{1}{|t_{2}|}\right),0\right)=\emptyset,$

for any $m>N$. Since $|t_{1}|<|t_{2}|$, we may infer that

$\left|\frac{1}{t_{2}}\right|=\frac{1}{2}\left(\frac{1}{|t_{2}|}+\frac{1}{|t_{2}|}\right)<\frac{1}{2}\left(\frac{1}{|t_{1}|}+\frac{1}{|t_{2}|}\right).$

Then there exists $r>0$ such that $B\left(r,\frac{1}{t_{2}}\right)\subseteq
B\left(\frac{1}{2}\left(\frac{1}{|t_{1}|}+\frac{1}{|t_{2}|}\right),0\right)$.
It follows that

$\mathcal{Z}(P_{m})\cap B\left(r,\frac{1}{t_{2}}\right)=\emptyset,$

for all $m\geq N$. Thus
$\frac{1}{t_{2}}\notin\overline{\lim}\,\mathcal{Z}(P_{m})$. Here we have, in
both cases $ac>0$ and $ac<0$, that
$\frac{1}{t_{2}}\notin\overline{\lim}\,\mathcal{Z}(P_{m})$. Consequently, we
obtain

$\overline{\lim}\,\mathcal{Z}(P_{m})\subseteq\left.\overline{\lim}\,\mathcal{Z}(Q_{m})\middle\backslash\left\\{\frac{1}{t_{2}}\right\\}\right..$

Let
$z^{*}\in\left.\underline{\lim}\,\mathcal{Z}(Q_{m})\middle\backslash\left\\{\frac{1}{t_{2}},\frac{1}{t_{1}}\right\\}\right.$
be arbitrary. Choose

(0.60)
$0<r<\min\left(\left|\frac{1}{t_{1}}-z^{*}\right|,\left|\frac{1}{t_{2}}-z^{*}\right|\right)$

arbitrarily. Since $z^{*}\in\underline{\lim}\,\mathcal{Z}(Q_{m})$, there
exists $N\in\mathbb{N}$ such that

$B(r,z^{*})\cap\mathcal{Z}(Q_{m})\neq\emptyset,$

for any $m>N$. Let $n>N$ be arbitrary. Then there exists $z_{n}\in
B(r,z^{*})\cap\mathcal{Z}(Q_{n})$. Since $z_{n}\in B(r,z^{*})$, Inequality
(0.60) induces that

$|z_{n}-z^{*}|<r<\left|\frac{1}{t_{1}}-z^{*}\right|,$

This implies that $z_{n}\neq 1/t_{1}$. In a similar way, we can deduce that
$z_{n}\neq 1/t_{2}$. Given that $z_{n}\in\mathcal{Z}(Q_{n})$, $z_{n}$ is a
root of $Q_{n}$. Since $z_{n}\neq 1/t_{1},1/t_{2}$, Equation (0.59) implies
that $z_{n}$ is a root of $P_{n}(z)$. Hence
$B(r,z^{*})\cap\mathcal{Z}(P_{n})\neq\emptyset.$ This implies that
$z^{*}\in\underline{\lim}\,\mathcal{Z}(P_{m})$. Thus

$\left.\left\\{z\in\mathbb{C};|z|=\frac{1}{|t_{1}|}\right\\}\middle\backslash\left\\{\frac{1}{t_{2}},\frac{1}{t_{1}}\right\\}\right.\subseteq\underline{\lim}\,\mathcal{Z}(P_{m}).$

By Proposition 44, we may infer that
$1/t_{1}\in\underline{\lim}\,\mathcal{Z}(P_{m})$. It follows that

$\left.\underline{\lim}\,\mathcal{Z}(Q_{m})\middle\backslash\left\\{\frac{1}{t_{2}}\right\\}\right.\subseteq\underline{\lim}\,\mathcal{Z}(P_{m}).$

Here we have

  1. 1.

$\overline{\lim}\,\mathcal{Z}(P_{m})\subseteq\left.\overline{\lim}\,\mathcal{Z}(Q_{m})\middle\backslash\left\\{\frac{1}{t_{2}}\right\\}\right.$,

  2. 2.

$\left.\underline{\lim}\,\mathcal{Z}(Q_{m})\middle\backslash\left\\{\frac{1}{t_{2}}\right\\}\right.\subseteq\underline{\lim}\,\mathcal{Z}(P_{m})$,

  3. 3.

$\underline{\lim}\,\mathcal{Z}(P_{m})\subseteq\overline{\lim}\,\mathcal{Z}(P_{m})$
by Proposition 40, and

  4. 4.

$\underline{\lim}\,\mathcal{Z}(Q_{m})=\overline{\lim}\,\mathcal{Z}(Q_{m})=\lim\mathcal{Z}(Q_{m})$.

Consequently, we obtain

$\left.\lim\mathcal{Z}(Q_{m})\middle\backslash\left\\{\frac{1}{t_{2}}\right\\}\right.\subseteq\underline{\lim}\,\mathcal{Z}(P_{m})\subseteq\overline{\lim}\,\mathcal{Z}(P_{m})\subseteq\left.\lim\mathcal{Z}(Q_{m})\middle\backslash\left\\{\frac{1}{t_{2}}\right\\}\right..$

This implies that

$\lim\mathcal{Z}(P_{m})=\left\\{z\in\mathbb{C};|z|=\frac{1}{|t_{1}|}\right\\}.\qed$

## CONCLUSIONS

For a long time, many mathematicians studied the pattern of zeros of
functions. This area of research is still active nowadays. In this paper, we
reviewed what has been studied in the area and made our own contribution to
this subject. In the development of our theorems, we use the classical
Rouché’s Theorem, Eneström-Kakeya’s Theorem, and then study the limiting
behaviors of the zeros of our sequence of polynomials.

Our main theorems were about the generating function

$\frac{1}{(at^{2}+bt+c)(1-zt)}$

and a circle centered at the origin with radius $\frac{1}{|\alpha|}$ where
$\alpha$ is a zero of $at^{2}+bt+c$ with smallest modulus. When we were
working with Rouché’s Theorem, we found that, when $ac<0$, our zeros lied
inside a closed disk, and when $ac>0$, the zeros lied outside an open ball.
Using Eneström-Kakeya Theorem, we prove that, when $ac>0$, the zeros are
outside of the closed disk. However, there is a problem in applying Eneström-
Kakeya Theorem for the case $ac<0$. For any $m\geq 0$, the ratio of the
coefficient of $z^{m-2}$ to the coefficient of $z^{m-1}$ is always greater
than $1/|\alpha|$.

In this paper we cover the case $at^{2}+bt+c$ having real zeros with different
modulus. In begining of Section ‘Zero Distribution of $\\{P_{m}(z)\\}_{m\geq
0}$ about a Circle’, we saw that zeros are both inside and outside of the
circle when $at^{2}+bt+c$ has complex zeros. With Rouché’s Theorem, we might
be able to obtain the ratio of the number of zeros inside to the number of
zeros outside. Or we might be able to count exact number of zeros inside and
outside of the circle.

In Section ‘Application of Limit Distribution of Zeros’, we only cover the
case $at^{2}+bt+c$ has real zeros. An intuition is saying we maybe able to
handle the case of imaginary zeros.

######  Conjecture 47.

Let $\\{P_{m}(z)\\}_{m\geq 0}$ be a sequence of functions of $z$ generated as
follows:

$\sum_{m=0}^{\infty}P_{m}(z)t^{m}=\frac{1}{(at^{2}+bt+c)(1-tz)}$

where $a,b,c\in\mathbb{R}\backslash\\{0\\}$ and $b^{2}-4ac\leq 0$. Then
$\lim\mathcal{Z}(P_{m})=\left\\{z\in\mathbb{C};|z|=\frac{1}{|t_{1}|}\right\\}$
where $t_{1}$ is the smallest (in modulus) zero of $at^{2}+bt+c$.

Computer experiments suggest that this conjecture is true. However, we have
not been able to apply Theorem 42 directly for the reason that we will now
mention. If we follow the procedure as in the proof of Theorem 46, we need to
apply Theorem 42 to the polynomial

$Q_{m}(z)=t_{2}^{m+1}-t_{1}^{m+1}+(t_{1}^{m+2}-t_{2}^{m+2})z+(t_{2}-t_{1})t_{1}^{m+1}t_{2}^{m+1}z^{m+2}$

by setting

$\displaystyle\alpha_{1}(z):=$ $\displaystyle t_{2}-t_{2}^{2}z$
$\displaystyle\alpha_{2}(z):=$ $\displaystyle-t_{1}+t_{1}^{2}z$
$\displaystyle\alpha_{3}(z):=$ $\displaystyle
t_{1}t_{2}^{2}z^{2}-t_{1}^{2}t_{2}z^{2}$ $\displaystyle\beta_{1}(z):=$
$\displaystyle t_{2}$ $\displaystyle\beta_{2}(z):=$ $\displaystyle t_{1}$
$\displaystyle\beta_{3}(z):=$ $\displaystyle t_{1}t_{2}z$

However, in this conjecture, we may deduce that
$|\beta_{1}(z)|=|t_{1}|=|t_{2}|=|\beta_{2}(z)|$, and this violates the
conditions to apply Theorem 42.

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