# Phase retrieval of entire functions and its implications for Gabor phase
retrieval

Matthias Wellershoff

###### Abstract

We consider the full or partial recovery of entire functions from magnitude
measurements on different subsets of the complex plane. This so-called _phase
retrieval for entire functions_ is inspired by its manifold connections to
other phase retrieval problems. A particular connection, illuminated in more
detail in this paper, is that to Gabor phase retrieval which can be made by
using the Bargmann transform and the Fock space.

By applying well-known techniques from complex analysis — in particular, the
famous Hadamard factorisation theorem — we develop some known and many new
results for the phase retrieval of entire functions. Among other things, we
provide a full classification of all (finite order) entire functions whose
magnitudes agree on two arbitrary lines in the complex plane as well as a full
classification of all entire functions of exponential type whose magnitudes
agree on infinitely many equidistant parallel lines.

Our results have interesting implications for Gabor phase retrieval such as
giving a full classification of all signals whose Gabor magnitudes agree on
two arbitrary lines in the time-frequency plane; or, yielding a machinery with
which to generate signals whose Gabor magnitudes agree on infinitely many
equidistant lines. The latter has already been harnessed to propose certain
counterexamples to sampled Gabor phase retrieval in the literature. We show
that one may also apply this machinery to generate so-called “universal
counterexamples”: signals which cannot be recovered (up to a global phase
constant) from magnitude measurements of their Gabor transforms on infinitely
many equidistant parallel lines — no matter how small the distance between
those lines.

Keywords Phase retrieval, finite order entire functions, Hadamard
factorisation theorem, Gabor transform, time-frequency analysis

Mathematics Subject Classification (2020) 30D20, 94A12

##  1 Introduction

In this paper, we consider the recovery of entire functions from magnitude-
only measurements. This problem has recently been considered in numerous
papers (e.g. [1, 5, 10, 13, 14, 18]) because it is connected to many different
phase retrieval problems. One particular instance of this is the work by
Mallat and Waldspurger on the recovery of analytic signals from Cauchy wavelet
transform magnitudes [13]. There, the authors use a factorisation theorem for
functions which are analytic on the upper half-plane to show that magnitude
information on two scales is sufficient for the recovery of analytic signals
from Cauchy wavelet transform magnitudes up to a global phase constant. This
is a notable result because it stood in contrast to the theory on Gabor phase
retrieval where it was not known whether magnitude information in two
frequency bins is sufficient for the recovery of signals. As the Gabor
transform is related to the Fock space of entire functions through its
relation with the Bargmann transform [4], we can approach the above gap in
knowledge from a complex-analytic perspective.

###  1.1 A short overview of existing research

The present paper extends earlier work by Mc Donald on phase retrieval of
entire functions from magnitude information on a single line in the complex
plane [14]. Mc Donald shows that such information is not sufficient for the
recovery of finite order entire functions (up to a global phase constant) and
uses the Hadamard factorisation theorem to give a full classification of all
the sources of non-uniqueness. We postpone a detailed description of the
result by Mc Donald to Section 3.

Another notable paper in connection with our work is [10]; not only does it
contain a result on the phase retrieval of certain analytic signals from
Cauchy wavelet transform magnitudes which predates the famous results by
Mallat and Waldspurger [13] mentioned above, it also includes some interesting
results on Gabor phase retrieval and the recovery of entire functions from
phaseless measurements which we redevelop here. In particular, we will
encounter the following uniqueness result.

######  Theorem 1 (Theorem 3.3 on p. 419 of [10]).

Let $\theta_{1},\theta_{2}\in[0,2\pi)$, with
$\theta_{1}-\theta_{2}\not\in\pi\mathbb{Q}$, and let
$f,g\in\mathcal{O}(\mathbb{C})$ be entire functions of finite order such that

$\left\lvert f(x\mathrm{e}^{\mathrm{i}\theta_{1}})\right\rvert=\left\lvert
g(x\mathrm{e}^{\mathrm{i}\theta_{1}})\right\rvert\mbox{ and }\left\lvert
f(x\mathrm{e}^{\mathrm{i}\theta_{2}})\right\rvert=\left\lvert
g(x\mathrm{e}^{\mathrm{i}\theta_{2}})\right\rvert,$

for $x\in\mathbb{R}$. Then, there exists an $\alpha\in\mathbb{R}$ such that
$f=\mathrm{e}^{\mathrm{i}\alpha}g$.

Hence, finite order entire functions whose magnitudes agree on two lines
intersecting at an angle $\theta\in\mathbb{R}\setminus\pi\mathbb{Q}$ must
agree up to a global phase constant. We will revisit this result in Section
4.1. The above result has the following noteworthy corollary.

######  Proposition 2 (Proposition 4.1 on p. 425 of [10]).

Let $\theta\in\mathbb{R}\setminus\pi\mathbb{Q}$ and let $f,g\in
L^{2}(\mathbb{R})$ such that

$\left\lvert\mathcal{G}f(x,0)\right\rvert=\left\lvert\mathcal{G}g(x,0)\right\rvert\mbox{
and
}\left\lvert\mathcal{G}f(x\cos\theta,x\sin\theta)\right\rvert=\left\lvert\mathcal{G}g(x\cos\theta,x\sin\theta)\right\rvert,$

for $x\in\mathbb{R}$. Then, there exists an $\alpha\in\mathbb{R}$ such that
$f=\mathrm{e}^{\mathrm{i}\alpha}g$.

Thus, two signals whose Gabor magnitudes agree on two lines in the time-
frequency plane intersecting at an angle
$\theta\in\mathbb{R}\setminus\pi\mathbb{Q}$ must agree up to global phase.
This is an extension of the classical uniqueness result for Gabor phase
retrieval which states that two signals whose Gabor magnitudes agree on the
entire time-frequency plane must agree up to global phase.

Let us finally highlight that the results which we present in this paper have
applications in the theory of sampled short-time Fourier transform phase
retrieval which has recently seen a stark increase in interest [1, 2, 3, 7, 8,
17, 18]. In particular, we were able to apply the results in this work to
generate the counterexamples presented in an earlier paper [3]: the genesis of
the counterexamples presented is not motivated there and, in fact, they are
the product of the machinery which we present here. In the meantime, our work
in [3] has been generalised to include window functions different than the
Gaussian [7]. We believe that all of these recent advances can be traced back
to the ideas which will be presented in the following.

###  1.2 Our contributions

As mentioned before, we extend results by Mc Donald on the recovery of entire
functions on a single line in the complex plane [14]. To be precise, we
consider two arbitrary lines in the complex plane and ask when one may recover
a finite order entire function from its magnitudes measured on those two
lines. In doing so, we redevelop Theorem 1 by Jaming on the unique recovery of
entire functions from magnitude measurements on two lines intersecting at an
“irrational angle” $\theta\in\mathbb{R}\setminus\pi\mathbb{Q}$. We moreover
consider lines intersecting at a “rational angle” $\theta\in\pi\mathbb{Q}$ as
well as parallel lines and show that in both cases the phase retrieval of
entire functions does not enjoy uniqueness. More precisely, we are able to
characterise all sources of non-uniqueness in these two cases. Let us
summarise our results in the following.

#### Lines intersecting at a rational angle

We prove the following result for lines intersecting at a rational angle.

######  Theorem 3 (Main theorem — I).

Let $\theta\in\pi\mathbb{Q}$ and let $n\in\mathbb{N}$ be the smallest integer
such that $\theta n\in\pi\mathbb{N}$. Let $f,g\in\mathcal{O}(\mathbb{C})$ be
non-zero entire functions of finite order. The following are equivalent:

(a)

$\left\lvert f\right\rvert$ and $\left\lvert g\right\rvert$ agree on two lines
intersecting at angle $\theta$ at the origin of the complex plane;

(b)

$f$ and $g$ may be factored as

$\displaystyle\begin{multlined}f(z)=r\mathrm{e}^{\mathrm{i}\phi}\exp\left(\sum_{\ell=1}^{q}c_{\ell}z^{\ell}\right)z^{m}\prod_{a\in\mathcal{X}}E\left(\frac{z}{a};p\right)^{m_{\mathcal{X}}(a)}\\\
\cdot\prod_{a\in\mathcal{Y}_{\mathrm{u}}}\prod_{k=1}^{n}E\left(\mathrm{e}^{2k\mathrm{i}\theta}\frac{z}{a};p\right)^{m_{\mathcal{Y}}(a)},\end{multlined}f(z)=r\mathrm{e}^{\mathrm{i}\phi}\exp\left(\sum_{\ell=1}^{q}c_{\ell}z^{\ell}\right)z^{m}\prod_{a\in\mathcal{X}}E\left(\frac{z}{a};p\right)^{m_{\mathcal{X}}(a)}\\\
\cdot\prod_{a\in\mathcal{Y}_{\mathrm{u}}}\prod_{k=1}^{n}E\left(\mathrm{e}^{2k\mathrm{i}\theta}\frac{z}{a};p\right)^{m_{\mathcal{Y}}(a)},$
$\displaystyle\begin{multlined}g(z)=r\mathrm{e}^{\mathrm{i}\psi}\exp\left(\sum_{\ell=1}^{q}c^{\prime}_{\ell}z^{\ell}\right)z^{m}\prod_{a\in\mathcal{X}}E\left(\frac{z}{a};p\right)^{m_{\mathcal{X}}(a)}\\\
\cdot\prod_{a\in\mathcal{Y}_{\mathrm{u}}}\prod_{k=1}^{n}E\left(\mathrm{e}^{2k\mathrm{i}\theta}\frac{z}{\overline{a}};p\right)^{m_{\mathcal{Y}}(a)},\end{multlined}g(z)=r\mathrm{e}^{\mathrm{i}\psi}\exp\left(\sum_{\ell=1}^{q}c^{\prime}_{\ell}z^{\ell}\right)z^{m}\prod_{a\in\mathcal{X}}E\left(\frac{z}{a};p\right)^{m_{\mathcal{X}}(a)}\\\
\cdot\prod_{a\in\mathcal{Y}_{\mathrm{u}}}\prod_{k=1}^{n}E\left(\mathrm{e}^{2k\mathrm{i}\theta}\frac{z}{\overline{a}};p\right)^{m_{\mathcal{Y}}(a)},$

where $r>0$, $\phi,\psi\in\mathbb{R}$, and where
$(c_{\ell})_{\ell=1}^{q},(c^{\prime}_{\ell})_{\ell=1}^{q}\in\mathbb{C}$ are
closely related.

We will explain the precise meaning of the notation used in the statement
above in Section 4. For now, we want to point the reader to what we believe to
be the most important take-away from this theorem. There are essentially two
reasons for which two entire functions $f$ and $g$ may agree on two
intersecting lines while not agreeing up to global phase. First, $f$ and $g$
might differ by a factor $\mathrm{e}^{p(z)}$, where $p$ is a special complex
polynomial. This is for instance the case when we consider the two lines
$\mathbb{R}$ and $\mathrm{i}\mathbb{R}$ along with the functions $f(z)=1$,
$g(z)=\exp(\mathrm{i}z^{2})$, for $z\in\mathbb{C}$. Secondly, $f$ and $g$
might have different root sets with rotation symmetry around the point of
intersection of the two lines. This is for instance the case when we consider
the two lines $\mathbb{R}$ and $\mathrm{i}\mathbb{R}$ along with the functions

$f(z)=1-\frac{z^{2}}{(1+\mathrm{i})^{2}},\quad
g(z)=1-\frac{z^{2}}{(1-\mathrm{i})^{2}},$

for $z\in\mathbb{C}$. Note that $f$ has the root set
$\mathcal{R}(f)=\\{1+\mathrm{i},-1-\mathrm{i}\\}$ and that $g$ has the root
set $\mathcal{R}(g)=\\{1-\mathrm{i},-1+\mathrm{i}\\}$. If we rotate
$1+\mathrm{i}$ by $2\theta=\pi$ (twice the angle of intersection) repeatedly,
we stay within the root set of $f$. The same is true for $g$. Theorem 3 states
that these two phenomena are the only sources of non-uniqueness for the phase
retrieval of entire functions with measurements taken on two intersecting
lines.

#### Parallel lines

We prove the following result for parallel lines.

######  Theorem 4 (Main theorem — II).

Let $\tau\in\mathbb{R}\setminus\\{0\\}$, and let
$f,g\in\mathcal{O}(\mathbb{C})$ be non-zero entire functions of finite order.
Then, the following are equivalent:

(a)

$\left\lvert f\right\rvert$ and $\left\lvert g\right\rvert$ agree on two
parallel lines with distance $\tau$ (parallel to $\mathbb{R}$);

(b)

$f$ and $g$ may be factored as

$\displaystyle\begin{multlined}f(z)=r\mathrm{e}^{\mathrm{i}\phi}\exp\left(\sum_{\ell=1}^{q}c_{\ell}z^{\ell}\right)z^{m}\prod_{a\in\mathcal{X}}E\left(\frac{z}{a};p\right)^{m_{\mathcal{X}}(a)}\\\
\cdot\prod_{a\in\mathcal{Y}_{\mathrm{u}}}\prod_{k\in\mathbb{Z}}E\left(\frac{z}{a+2k\mathrm{i}\tau};p\right)^{m_{\mathcal{Y}}(a)},\end{multlined}f(z)=r\mathrm{e}^{\mathrm{i}\phi}\exp\left(\sum_{\ell=1}^{q}c_{\ell}z^{\ell}\right)z^{m}\prod_{a\in\mathcal{X}}E\left(\frac{z}{a};p\right)^{m_{\mathcal{X}}(a)}\\\
\cdot\prod_{a\in\mathcal{Y}_{\mathrm{u}}}\prod_{k\in\mathbb{Z}}E\left(\frac{z}{a+2k\mathrm{i}\tau};p\right)^{m_{\mathcal{Y}}(a)},$
$\displaystyle\begin{multlined}g(z)=r\mathrm{e}^{\mathrm{i}\psi}\exp\left(\sum_{\ell=1}^{q}c_{\ell}^{\prime}z^{\ell}\right)z^{m}\prod_{a\in\mathcal{X}}E\left(\frac{z}{a};p\right)^{m_{\mathcal{X}}(a)}\\\
\cdot\prod_{a\in\mathcal{Y}_{\mathrm{u}}}\prod_{k\in\mathbb{Z}}E\left(\frac{z}{\overline{a}+2k\mathrm{i}\tau};p\right)^{m_{\mathcal{Y}}(a)},\end{multlined}g(z)=r\mathrm{e}^{\mathrm{i}\psi}\exp\left(\sum_{\ell=1}^{q}c_{\ell}^{\prime}z^{\ell}\right)z^{m}\prod_{a\in\mathcal{X}}E\left(\frac{z}{a};p\right)^{m_{\mathcal{X}}(a)}\\\
\cdot\prod_{a\in\mathcal{Y}_{\mathrm{u}}}\prod_{k\in\mathbb{Z}}E\left(\frac{z}{\overline{a}+2k\mathrm{i}\tau};p\right)^{m_{\mathcal{Y}}(a)},$

where $r>0$, $\phi,\psi\in\mathbb{R}$, and where
$(c_{\ell})_{\ell=1}^{q},(c^{\prime}_{\ell})_{\ell=1}^{q}\in\mathbb{C}$ are
closely related.

We note that two entire functions $f$ and $g$ which do not agree up to global
phase must have different root sets provided that their magnitudes agree on
two parallel lines in the time-frequency plane. Precisely, the above theorem
states that there is a certain symmetry in these root sets: indeed, if $a$ is
a root of $f$ which is not a root of $g$, then the same must be true for all
elements of $a+2\mathrm{i}\tau\mathbb{Z}$. In particular, the functions $f$
and $g$ described above must have infinitely many roots. We may consider the
parallel lines $\mathbb{R}$ and $\mathbb{R}+\mathrm{i}$ together with the
functions

$\displaystyle f(z)=\cosh\left(\frac{\pi
z}{2}\right)+\mathrm{i}\sinh\left(\frac{\pi z}{2}\right),$ $\displaystyle
g(z)=\cosh\left(\frac{\pi z}{2}\right)+\mathrm{i}\sinh\left(\frac{\pi
z}{2}\right),$

as an example. Note that the root set of $f$ is given by
$\mathcal{R}(f)=\\{\mathrm{i}/2+2k\mathrm{i}\,\mid\,k\in\mathbb{Z}\\}$ while
the root set of $g$ is given by
$\mathcal{R}(g)=\\{-\mathrm{i}/2+2k\mathrm{i}\,\mid\,k\in\mathbb{Z}\\}$.

#### Infinitely many equidistant parallel lines

We emphasise that the two theorems discussed above together with Theorem 1
fully characterise all entire functions of finite order whose magnitudes agree
on two arbitrary lines in the time-frequency plane. We also note that
considering two arbitrary lines in the time-frequency plane always leads to
symmetries in the root sets of the entire function under consideration: in
particular, intersecting lines correspond to rotation symmetries while
parallel lines correspond to translation symmetries. This seems to be the most
important insight stemming from our characterisation. One may use it to
consider three or more lines in the complex plane and try to understand how
the symmetries imposed by the additional lines change the characterisation. We
want to especially focus on infinitely many equidistant parallel lines. When
all the lines are equidistant, there are no additional symmetries in the root
set of $f$ and $g$ and we can prove the following result.

######  Theorem 5 (Main theorem — III).

Let $\tau\in\mathbb{R}\setminus\\{0\\}$ and let
$f,g\in\mathcal{O}(\mathbb{C})$ be non-zero entire functions of exponential
type. Then, the following are equivalent:

(a)

$\left\lvert f\right\rvert$ and $\left\lvert g\right\rvert$ agree on
infinitely many parallel lines with distance $\tau$ (parallel to
$\mathbb{R}$);

(b)

$f$ and $g$ may be factored as

$\displaystyle
f(z)=r\mathrm{e}^{\mathrm{i}\phi}\mathrm{e}^{cz}z^{m}\prod_{a\in\mathcal{X}}E\left(\frac{z}{a};1\right)^{m_{\mathcal{X}}(a)}\prod_{a\in\mathcal{Y}_{\mathrm{u}}}\prod_{k\in\mathbb{Z}}E\left(\frac{z}{a+2k\mathrm{i}\tau};1\right)^{m_{\mathcal{Y}}(a)},$
$\displaystyle
g(z)=r\mathrm{e}^{\mathrm{i}\psi}\mathrm{e}^{c^{\prime}z}z^{m}\prod_{a\in\mathcal{X}}E\left(\frac{z}{a};1\right)^{m_{\mathcal{X}}(a)}\prod_{a\in\mathcal{Y}_{\mathrm{u}}}\prod_{k\in\mathbb{Z}}E\left(\frac{z}{\overline{a}+2k\mathrm{i}\tau};1\right)^{m_{\mathcal{Y}}(a)},$

where $r>0$, $\phi,\psi\in\mathbb{R}$, and where $c,c^{\prime}\in\mathbb{C}$
are closely related.

We note that we may prove a similar result for general finite order entire
functions. However, the general result comes with a very complicated relation
for the polynomial coefficients
$(c_{\ell})_{\ell=1}^{q},(c^{\prime}_{\ell})_{\ell=1}^{q}\in\mathbb{C}$ which
seems useless in practice. We note moreover that the above result is the
design mechanism according to which we have created the counterexamples
presented in [2]. In connection with this, let us remark that the examples

$\displaystyle f(z)=\cosh\left(\frac{\pi
z}{2}\right)+\mathrm{i}\sinh\left(\frac{\pi z}{2}\right),$ $\displaystyle
g(z)=\cosh\left(\frac{\pi z}{2}\right)+\mathrm{i}\sinh\left(\frac{\pi
z}{2}\right),$

continue to agree in magnitude on the infinitely many equidistant parallel
lines $\mathbb{R}+\mathrm{i}\mathbb{Z}$.

#### Some implications for Gabor phase retrieval

Through the relation of the Gabor transform and the Bargmann transform, the
Theorems 1, 3 and 4 presented above give rise to a full characterisation of
all $L^{2}$-functions which do not agree up to global phase but whose Gabor
transform magnitudes agree on any two arbitrary lines in the time-frequency
plane. Moreover the third main theorem may be used to design $L^{2}$-functions
which do not agree up to global phase but whose Gabor transform magnitudes
agree on infinitely many parallel lines in the time-frequency plane. We will
explore all of these corollaries (and more) in greater detail in Section 5.

We finally want to remark that the techniques which we have used in the proofs
of the three main theorems may also be applied to generate “universal
counterexamples” for Gabor transform phase retrieval; specifically, we design
a function $f\in L^{2}(\mathbb{R})$ such that, for all $n\in\mathbb{N}$, there
exists a function $g_{n}\in L^{2}(\mathbb{R})$ such that $f$ and $g_{n}$ do
not agree up to global phase while

$\left\lvert\mathcal{G}f\right\rvert=\left\lvert\mathcal{G}g_{n}\right\rvert\mbox{
on }\mathbb{R}\times\frac{1}{n}\mathbb{Z}.$

This shows that creating finer and finer sampling sets must not necessarily
lead to uniqueness in sampled Gabor phase retrieval even when one of the
functions under consideration is fixed.

###  1.3 Notation

We denote the space of entire function by $\mathcal{O}(\mathbb{C})$ and,
therein (contained as a subspace), the space of polynomials of degree
$n\in\mathbb{N}_{0}$ with complex coefficients by $\mathbb{C}_{n}[z]$. For an
entire function $f$, we denote the set of roots by $\mathcal{R}(f)$ and the
set of non-zero roots by $\mathcal{R}_{\ast}(f)$. In both of these sets, we do
not record the multiplicity of the roots. Instead, we capture the multiplicity
using a multiset structure over the complex plane. For this purpose, let us
denote by $m_{f}:\mathcal{R}(f)\to\mathbb{N}$ the function mapping each root
of $f$ to its multiplicity. We can extend $m_{f}$ to all of $\mathbb{C}$ by
zero and call this extension $M_{f}:\mathbb{C}\to\mathbb{N}_{0}$. Remarkably,
the function $M_{f}$ captures all the information about the roots of $f$: the
complement of its null set corresponds to the set of roots of $f$ and its
action on the set of roots of $f$ contains all information about the
multiplicities. It is in this sense that the following definition should be
understood:

######  Definition 6 (Multiset).

A _multiset over $\mathbb{C}$_ is a function $M:\mathbb{C}\to\mathbb{N}_{0}$.

We finally denote $\mathbb{R}_{\ast}:=\mathbb{R}\setminus\\{0\\}$ as well as
$\mathbb{C}_{\ast}:=\mathbb{C}\setminus\\{0\\}$ following our notation for the
roots of entire functions.

###  1.4 Outline

In Section 2, we provide the necessary background on the theory of entire
functions and, in particular, the Hadamard factorisation theorem. Most of the
knowledge in this first section is summarised from the excellent book [16].

In Section 3, we present what we consider to be the basics of phase retrieval
for entire functions. Those include some simple and elegant statements for
setups in which measurements are taken on open subsets of $\mathbb{C}$. They
additionally include the aforementioned work by Mc Donald [14].

In Section 4, we present the core of our work. In particular, we provide full
classifications for entire functions of finite order whose magnitudes agree on
two intersecting lines or on two parallel lines in the complex plane.
Additionally, we use the classification of entire functions whose magnitudes
agree on two parallel lines in order to give a full classification of all
first order entire functions whose magnitudes agree on infinitely many
equidistant parallel lines.

In Section 5, we use the relationship between the Gabor transform, the
Bargmann transform and the Fock space of entire functions, in order to
classify all signals whose Gabor transforms agree on two intersecting lines or
two parallel lines in the time-frequency plane. Additionally, we present
theory which can be used to generate functions in $L^{2}(\mathbb{R})$ which do
not agree up to a global phase constant but whose Gabor transform magnitudes
agree on infinitely many equidistant parallel lines. We emphasise that this is
the exact machinery which was used to write our earlier work [2]. Finally, we
use Jensen’s formula from complex analysis in order to give an upper bound on
the number of roots that the Gabor transform can have in circles around the
origin.

In Section 6, we prove the existence of so-called “universal counterexamples”.
In particular, we use the theory on phase retrieval of entire functions with
magnitude information on infinitely many parallel lines in order to show that
there exists a function $f\in L^{2}(\mathbb{R})$ such that, for all
$n\in\mathbb{N}$, there exists a function $g_{n}\in L^{2}(\mathbb{R})$ which
does not agree with $f$ up to a global phase constant but which satisfies

$\left\lvert\mathcal{G}f\right\rvert=\left\lvert\mathcal{G}g_{n}\right\rvert\mbox{
on }\mathbb{R}\times\frac{1}{n}\mathbb{Z}.$

##  2 Hadamard’s factorisation theorem

It is well known that non-zero polynomials $p\in\mathbb{C}_{n}[z]$ may be
factored in terms of their roots $a_{1},\dots,a_{n}\in\mathbb{C}$. Indeed,
suppose that $p(0)\neq 0$ for simplicity. Then, it holds that

$p(z)=p(0)\left(1-\frac{z}{a_{1}}\right)\dots\left(1-\frac{z}{a_{n}}\right),\qquad
z\in\mathbb{C}.$

The same can be said for non-zero entire functions $\mathcal{O}(\mathbb{C})$
but the picture is a bit more complicated for them. The reasons for this are
broadly speaking twofold. First, there are many entire functions which do not
have any roots but are still distinct (e.g. $\exp(z)$ and $\exp(z^{2})$).
Secondly, it is, in general, not true that

$\prod_{a\in\mathcal{R}(f)}\left(1-\frac{z}{a}\right)^{m_{f}(a)}$

converges (think of $\cos(z)$). To deal with the latter problem, one works
with the so-called _primary factors_ ,

$E(z;0)=1-z,\qquad
E(z;p)=(1-z)\exp\left(\sum_{\ell=1}^{p}\frac{z^{\ell}}{\ell}\right),\qquad
z\in\mathbb{C},$

for $p\in\mathbb{N}$. The primary factors can be used to show that any non-
zero entire function can be expanded into a product of its roots:

$f(z)=\mathrm{e}^{g(z)}z^{m}\prod_{a\in\mathcal{R}_{\ast}(f)}E\left(\frac{z}{a};p(a)\right)^{m_{f}(a)},$

where $m=m_{f}(0)$, $g\in\mathcal{O}(\mathbb{C})$ and
$(p(a))_{a\in\mathcal{R}_{\ast}(f)}\in\mathbb{N}_{0}$ is some sequence of
integers. For our purposes this result (called _Weierstrass factorisation
theorem_) is not precise enough. It only contains little information on the
function $g$ and the integers $(p(a))_{a}$. Luckily enough, we always end up
having to work with entire functions of finite order.

######  Definition 7 (Finite order).

Let $f\in\mathcal{O}(\mathbb{C})$. We say that $f$ is of _finite order_ if
there exist $A,C,r>0$ such that

$\left\lvert f(z)\right\rvert\leq C\cdot\exp(\left\lvert
z\right\rvert^{A}),\qquad\left\lvert z\right\rvert>r.$

The infimum over all $A>0$ such that there exist $C,r>0$ for which the above
holds is called the _order of $f$_ and will be denoted by $\rho$.

Finite order functions have the nice property that their root sets are not too
dense.

######  Lemma 8.

Let $f\in\mathcal{O}(\mathbb{C})$ be non-zero and of finite order $\rho\geq
0$. Then, it holds that

$\sum_{a\in\mathcal{R}_{\ast}(f)}\frac{m_{f}(a)}{\left\lvert
a\right\rvert^{A}}$

converges for all $A>\rho$.

A beautiful proof of this lemma can be found on the pages 251 and 250 of [16].
It makes use of Jensen’s formula which relates the moduli of entire functions
to the moduli of their zeroes. Lemma 8 will be used multiple times throughout
this section. One of its most immediate uses is in the following corollary,
however.

######  Corollary 9 (Cf. [16]).

Let $f\in\mathcal{O}(\mathbb{C})$ be non-zero and of finite order $\rho\geq
0$. Then, there exists an integer $p\in\mathbb{N}_{0}$, with $p\leq\rho$, such
that

$\prod_{a\in\mathcal{R}_{\ast}(f)}E\left(\frac{z}{a};p\right)^{m_{f}(a)}$

converges to an entire function.

Interestingly, the above corollary allows for the use of the same integer $p$
in every factor of the product. It remains to refine the information on the
function $g$ in the Weierstrass factorisation theorem to obtain the famous
Hadamard factorisation theorem.

######  Theorem 10 (Hadamard factorisation theorem: cf. [16]).

Let $f\in\mathcal{O}(\mathbb{C})$ be non-zero and of finite order $\rho\geq
0$. Then, there exist $r>0$, $\phi\in\mathbb{R}$, $q\in\mathbb{N}_{0}$, with
$q\leq\rho$, $(c_{\ell})_{\ell=1}^{q}\in\mathbb{C}$, $m\in\mathbb{N}_{0}$ and
$p\in\mathbb{N}_{0}$, with $p\leq\rho$, such that

$f(z)=r\mathrm{e}^{\mathrm{i}\phi}\exp\left(\sum_{\ell=1}^{q}c_{\ell}z^{\ell}\right)z^{m}\prod_{a\in\mathcal{R}_{\ast}(f)}E\left(\frac{z}{a};p\right)^{m_{f}(a)},\qquad
z\in\mathbb{C}.$

We want to end this section by stating one more result of which we will make
use in the course of this paper. It links the moduli of the roots of certain
products to their order.

######  Lemma 11 (Cf. [16]).

Let $\\{a_{j}\\}_{j=1}^{\infty}\subset\mathbb{C}_{\ast}$ be a set of isolated
points with corresponding multiplicities
$\\{m_{j}\\}_{j=1}^{\infty}\subset\mathbb{N}$. Suppose that there exists an
$A>0$ such that

$\sum_{j=1}^{\infty}\frac{m_{j}}{\left\lvert a_{j}\right\rvert^{A}}$

converges and denote the infimum over all such $A>0$ by $\rho_{1}$. It holds
that

$f(z)=\prod_{j=1}^{\infty}E\left(\frac{z}{a_{j}};p\right)^{m_{j}},\qquad
z\in\mathbb{C},$ (1)

is a well-defined entire function of order $\rho_{1}$, for $p+1>\rho_{1}$.

######  Remark 12.

Typically, one calls the non-negative number $\rho_{1}\geq 0$ defined in the
lemma above the _exponent of convergence of $\\{a_{j}\\}_{j=1}^{\infty}$ with
multiplicities $\\{m_{j}\\}_{j=1}^{\infty}$_. In addition, one calls the
smallest integer $p_{1}\in\mathbb{N}_{0}$ such that the product in (1)
converges the _genus of the canonical product formed with
$\\{a_{j}\\}_{j=1}^{\infty}$ and multiplicities $\\{m_{j}\\}_{j=1}^{\infty}$_
and the corresponding product the _canonical product formed with
$\\{a_{j}\\}_{j=1}^{\infty}$ and multiplicities $\\{m_{j}\\}_{j=1}^{\infty}$_.
The attentive reader will have noticed that for an entire function of order
$\rho>0$ with roots $\\{a_{j}\\}_{j=1}^{\infty}$ with multiplicities
$\\{m_{j}\\}_{j=1}^{\infty}$, it holds that

$p_{1}\leq\rho_{1}\leq\rho.$

The integer $p$ whose existence is postulated by Hadamard’s factorisation
theorem can actually be chosen in different ways. Often, it is useful to
choose $p$ to be the genus of the canonical product formed with
$\mathcal{R}_{\ast}(f)$. In this way, the factorisation is unique. It is,
however, also possible to choose $p$ to be larger to the point where $p$ is
even larger than the order of $f$. In this case, one will need to adapt the
factorisation by chosing $q$ just as large, however.

##  3 Phase retrieval of entire functions: prior arts

In the following two sections, we want to investigate the recoverability of
$f\in\mathcal{O}(\mathbb{C})$ from the measurements

$\left\lvert f(z)\right\rvert,\qquad z\in\Omega,$ (2)

where $\Omega\subseteq\mathbb{C}$ is some (not necessarily open) subset of
$\mathbb{C}$. We call this the _phase retrieval problem for entire functions_.
It is inspired by its connection to the problem of phase retrieval from Gabor
and Cauchy wavelet transform measurements [4, 10, 13] which we will partially
explain in the Sections 5 and 6 of the present paper. Of course, the choice of
$\Omega$ dictates what features of $f$ can be recovered from the measurements
(2).

Let us start by highlighting the following two fundamental facts related to
the phase retrieval problem for entire functions. First, it is obvious that if
$\Omega\subseteq\Omega^{\prime}$ and one can recover a certain feature of $f$
from the measurements (2), then one can also recover that feature from
$\\{\left\lvert f(z)\right\rvert\,|\,z\in\Omega^{\prime}\\}$. Secondly, it is
also clear that no matter how large we make $\Omega$, we can never fully
recover $f$. Indeed, it holds that for all $\alpha\in\mathbb{R}$,

$\left\\{\left\lvert
f(z)\right\rvert\,\middle|\,z\in\mathbb{C}\right\\}=\left\\{\left\lvert
f(z)\mathrm{e}^{\mathrm{i}\alpha}\right\rvert\,\middle|\,z\in\mathbb{C}\right\\}.$

We will, however, say that $f$ may be uniquely recovered _up to global phase_
if it is possible to recover any element of the set
$\\{f\mathrm{e}^{\mathrm{i}\alpha}\,|\,\alpha\in\mathbb{R}\\}$ from the
measurements (2).

Some recoverability statements for the phase retrieval problem for entire
functions come with a proof that is in equal parts easy and elegant. We
suspect that these results have been known for a while but could not find a
reference for them. The first result is on the recovery of entire functions
from absolute value measurements on the entire complex plane.

######  Theorem 13.

Let $f\in\mathcal{O}(\mathbb{C})$. Then, $f$ may be uniquely recovered up to
global phase from the measurements

$\left\lvert f(z)\right\rvert,\qquad z\in\mathbb{C}.$

###### Proof.

If $f$ is zero, then the statement is true. Let us therefore consider
$f,g\in\mathcal{O}(\mathbb{C})$ non-zero such that

$\left\lvert f(z)\right\rvert=\left\lvert g(z)\right\rvert,\qquad
z\in\mathbb{C}.$

We will consider the function $h:=f/g$, which is holomorphic on
$\mathbb{C}\setminus\mathcal{R}(g)$. It holds that

$\left\lvert h(z)\right\rvert=1,\qquad z\in\mathbb{C}\setminus\mathcal{R}(g).$

In particular (as the roots of non-zero entire functions are isolated), we
know that for every $a\in\mathcal{R}(g)$, there exists a neighbourhood of $a$
on which $h$ is bounded. By Riemann’s theorem on removable singularities, it
follows that $h$ is holomorphically extendable over $a$. Therefore, $h$
extends to an entire function $H$. In addition,

$\left\lvert H(z)\right\rvert=1,\qquad z\in\mathbb{C},$

such that it follows from Liouville’s theorem that $H$ is constant. As $H$ has
unit absolute value, we know that $H=\mathrm{e}^{\mathrm{i}\alpha}$ for some
$\alpha\in\mathbb{R}$. Now, $h$ agrees with $H$ on
$\mathbb{C}\setminus\mathcal{R}(g)$ such that, we find

$f(z)=\mathrm{e}^{\mathrm{i}\alpha}g(z),\qquad
z\in\mathbb{C}\setminus\mathcal{R}(g).$

As both functions in the comparison above are entire (and the roots of non-
zero entire functions are isolated), it follows that
$f=\mathrm{e}^{\mathrm{i}\alpha}g$. ∎

A similar result may be proven starting from knowledge of $\left\lvert
f\right\rvert$ on some open set. To derive this, one only needs to substitute
the use of Liouville’s theorem with an argument involving the open mapping
theorem.

######  Theorem 14.

Let $\Omega\subset\mathbb{C}$ be open and $f\in\mathcal{O}(\mathbb{C})$. Then,
$f$ may be uniquely recovered up to global phase from the measurements

$\left\lvert f(z)\right\rvert,\qquad z\in\Omega.$

###### Proof.

If $f$ is zero, then the statement is true. Let us therefore consider
$f,g\in\mathcal{O}(\mathbb{C})$ non-zero such that

$\left\lvert f(z)\right\rvert=\left\lvert g(z)\right\rvert,\qquad z\in\Omega,$

We will consider the function $h:=f/g$, which is holomorphic on
$\mathbb{C}\setminus\mathcal{R}(g)$. It holds that

$\left\lvert h(z)\right\rvert=1,\qquad z\in\Omega\setminus\mathcal{R}(g),$

and thus that $h(\Omega\setminus\mathcal{R}(g))\subset\mathcal{S}^{0}$
($\hat{=}$ the complex unit circle). Let us suppose by contradiction that $h$
is non-constant. Then, it follows from the open mapping theorem that
$h(\Omega\setminus\mathcal{R}(g))$ is open. This contradicts the fact that
$h(\Omega\setminus\mathcal{R}(g))\subset\mathcal{S}^{0}$. Therefore, $h$ must
be constant. As $h$ has unit absolute value on
$\Omega\setminus\mathcal{R}(g)$, we know that
$h=\mathrm{e}^{\mathrm{i}\alpha}$, for some $\alpha\in\mathbb{R}$, on
$\Omega\setminus\mathcal{R}(g)$. Therefore,

$f(z)=\mathrm{e}^{\mathrm{i}\alpha}g(z),\qquad
z\in\Omega\setminus\mathcal{R}(g).$

As both functions in the comparison above are entire (and the zeroes of non-
zero entire functions are isolated), it follows that
$f=\mathrm{e}^{\mathrm{i}\alpha}g$. ∎

It follows immediately that we can recover $f$ up to global phase from the
measurements (2) on any set $\Omega\subset\mathbb{C}$ which contains an open
set. To complete our understanding of the phase retrieval problem of entire
functions, it is necessary to contemplate what happens in cases in which we
only know $\left\lvert f\right\rvert$ on sets which do not contain open sets.
In the following, we will gradually move towards this goal by considering
$\Omega$ to be different closed sets. We will start by considering the case in
which $\Omega$ is a single line in the complex plane.

###  3.1 Phase retrieval of entire functions from information on a single line

One can quickly convince oneself that knowledge of $\left\lvert f\right\rvert$
on a single line is not enough to uniquely recover $f$ up to global phase in
general. Indeed, consider $f,g\in\mathcal{O}(\mathbb{C})$ such that

$f(z)=1,\qquad g(z)=\mathrm{e}^{\mathrm{i}z},$

for $z\in\mathbb{C}$. Then, $\left\lvert f(x)\right\rvert=\left\lvert
g(x)\right\rvert$, for $x\in\mathbb{R}$. So $\left\lvert f\right\rvert$ and
$\left\lvert g\right\rvert$ agree on the real number line even though $f$ and
$g$ do not agree up to global phase. Similarly, we might consider
$f,g\in\mathcal{O}(\mathbb{C})$ such that

$f(z)=1+\mathrm{i}z,\qquad g(z)=1-\mathrm{i}z,$

for $z\in\mathbb{C}$. Again, we find $\left\lvert f(x)\right\rvert=\left\lvert
g(x)\right\rvert$, for $x\in\mathbb{R}$, in spite of $f$ and $g$ not agreeing
up to global phase.

######  Remark 15.

Note that given any single line $\Omega$ in the complex plane, there exists a
rigid motion $\mathfrak{r}:\mathbb{C}\to\mathbb{C}$ (a rotation followed by a
translation) such that $\mathfrak{r}(\Omega)=\mathbb{R}$. Note in addition
that rigid motions are entire functions, i.e.
$\mathfrak{r}\in\mathcal{O}(\mathbb{C})$. Now, if
$f,g\in\mathcal{O}(\mathbb{C})$ are two functions that do not agree up to
global phase and satisfy $\left\lvert f(x)\right\rvert=\left\lvert
g(x)\right\rvert$, for $x\in\mathbb{R}$ (we have demonstrated such functions
exist). Then,
$f\circ\mathfrak{r},g\circ\mathfrak{r}\in\mathcal{O}(\mathbb{C})$ (where
$\circ$ denotes function composition) do not agree up to global phase and
satisfy $\left\lvert f\circ\mathfrak{r}(z)\right\rvert=\left\lvert
g\circ\mathfrak{r}(z)\right\rvert$, for $z\in\Omega$. Therefore, we have that
_for all lines $\Omega$ in the complex plane, there exist entire functions
which do not agree up to global phase but whose magnitudes agree on $\Omega$._

In spite of the fact that entire functions are not recoverable from magnitude
information on a single line, one can fully characterise all finite order
functions $g\in\mathcal{O}(\mathbb{C})$ which satisfy

$\left\lvert g(x)\right\rvert=\left\lvert f(x)\right\rvert,\qquad
x\in\mathbb{R},$ (3)

for any given $f\in\mathcal{O}(\mathbb{C})$ of finite order. In this way, one
can show that the ambiguities exemplified by the two constructions at the
beginning of this subsection really present a complete picture of all possible
ambiguities. Of course, this characterisation carries over to general straight
lines in $\mathbb{C}$ by the use of rigid motions. The following
considerations are due to Mc Donald [14].

For an entire function $f\in\mathcal{O}(\mathbb{C})$, we define another entire
function via the involution

$f^{\ast}(z):=\overline{f(\overline{z})},\qquad z\in\mathbb{C}.$

If two entire functions satisfy equation (3), then it follows immediately that

$f^{\ast}(x)f(x)=\left\lvert f(x)\right\rvert^{2}=\left\lvert
g(x)\right\rvert^{2}=g^{\ast}(x)g(x),\qquad x\in\mathbb{R}.$

Since both $f^{\ast}f$ and $g^{\ast}g$ are entire, we conclude from the above
consideration that

$f^{\ast}f=g^{\ast}g$ (4)

because the sets of roots of non-zero functions are isolated. Comparing the
roots in the above equation yields the following important insight.

######  Lemma 16 (Root symmetry – I).

Let $f,g\in\mathcal{O}(\mathbb{C})$ be such that

$\left\lvert f(x)\right\rvert=\left\lvert g(x)\right\rvert,\qquad
x\in\mathbb{R}.$

Then, it holds that

$M_{f}(z)+M_{f}(\overline{z})=M_{g}(z)+M_{g}(\overline{z}),\qquad
z\in\mathbb{C},$

where we remind the reader that $M_{f},M_{g}:\mathbb{C}\to\mathbb{N}_{0}$
denote the multisets of roots of $f$ and $g$, respectively.

If $a\in\mathbb{C}$ is a root of $f$, then either $a$ is a root of $g$ or
$\overline{a}$ is a root of $g$. Hence, there is a mirror symmetry in the root
sets of $f$ and $g$ along the real number line. We can combine this
information with the Hadamard factorisation theorem to obtain a description of
all entire functions whose absolute values agree on a single line.

To rigorously describe this, it makes sense to split the roots of $f$ into two
sets: i. The roots of $f$ which are roots of $g$. ii. The roots of $f$ which
are not roots of $g$. We will do this by considering

$\mathcal{X}:=\mathcal{R}_{\ast}(f)\cap\mathcal{R}(g),\qquad\mathcal{Y}:=\left\\{a\in\mathcal{R}(f)\,\middle|\,m_{f}(a)>M_{g}(a)\right\\},$

together with the map $m_{\mathcal{X}}:{\mathcal{X}}\to\mathbb{N}$ given by

$m_{\mathcal{X}}(a):=\min\\{m_{f}(a),m_{g}(a)\\},\qquad a\in{\mathcal{X}},$

as well as the map $m_{\mathcal{Y}}:{\mathcal{Y}}\to\mathbb{N}$ given by

$m_{\mathcal{Y}}(a):=m_{f}(a)-M_{g}(a),\qquad a\in{\mathcal{Y}}.$

We will denote the trivial extensions of these maps to all of $\mathbb{C}$ by
$M_{\mathcal{X}}$ and $M_{\mathcal{Y}}$ as usual. It is not hard to see that

$\displaystyle M_{f}(z)$
$\displaystyle=\min\\{M_{f}(z),M_{g}(z)\\}+M_{f}(z)-\min\\{M_{f}(z),M_{g}(z)\\}$
$\displaystyle=M_{\mathcal{X}}(z)+\max\\{0,M_{f}(z)-M_{g}(z)\\}=M_{\mathcal{X}}(z)+M_{\mathcal{Y}}(z),$

for $z\in\mathbb{C}_{\ast}$. In addition, it follows from Lemma 16 that

$\displaystyle M_{g}(z)$
$\displaystyle=\min\\{M_{f}(z),M_{g}(z)\\}+M_{g}(z)-\min\\{M_{f}(z),M_{g}(z)\\}$
$\displaystyle=M_{\mathcal{X}}(z)+\max\\{0,M_{g}(z)-M_{f}(z)\\}$
$\displaystyle=M_{\mathcal{X}}(z)+\max\\{0,M_{f}(\overline{z})-M_{g}(\overline{z})\\}=M_{\mathcal{X}}(z)+M_{\mathcal{Y}}(\overline{z}),$

for $z\in\mathbb{C}_{\ast}$. It is now merely a technical exercise to prove
the following theorem:

######  Theorem 17 (Mc Donald decomposition; [14]).

Let $f,g\in\mathcal{O}(\mathbb{C})$ be non-zero and of finite order. The
following are equivalent:

(a)

For all $x\in\mathbb{R}$, $\left\lvert f(x)\right\rvert=\left\lvert
g(x)\right\rvert$.

(b)

There exist $r>0$, $\phi,\psi\in\mathbb{R}$, $q\in\mathbb{N}_{0}$,
$c_{\ell},c_{\ell}^{\prime}\in\mathbb{C}$, with
$\operatorname{Re}c_{\ell}=\operatorname{Re}c_{\ell}^{\prime}$, for
$\ell\in\\{1,\dots,q\\}$, $m\in\mathbb{N}_{0}$ and $p\in\mathbb{N}_{0}$ such
that

$\displaystyle
f(z)=r\mathrm{e}^{\mathrm{i}\phi}\exp\left(\sum_{\ell=1}^{q}c_{\ell}z^{\ell}\right)z^{m}\prod_{a\in\mathcal{X}}E\left(\frac{z}{a};p\right)^{m_{\mathcal{X}}(a)}\prod_{a\in\mathcal{Y}}E\left(\frac{z}{a};p\right)^{m_{\mathcal{Y}}(a)},$
$\displaystyle
g(z)=r\mathrm{e}^{\mathrm{i}\psi}\exp\left(\sum_{\ell=1}^{q}c_{\ell}^{\prime}z^{\ell}\right)z^{m}\prod_{a\in\mathcal{X}}E\left(\frac{z}{a};p\right)^{m_{\mathcal{X}}(a)}\prod_{a\in\mathcal{Y}}E\left(\frac{z}{\overline{a}};p\right)^{m_{\mathcal{Y}}(a)},$

for $z\in\mathbb{C}$.

###### Proof.

Let us first show that item (a) implies item (b). According to Hadamard’s
factorisation theorem and the considerations before, there exist
$r,r^{\prime}>0$, $\phi,\psi\in\mathbb{R}$, $q,q^{\prime}\in\mathbb{N}_{0}$,
$a_{\ell},b_{\ell}\in\mathbb{R}$, for $\ell\in\\{1,\dots,q\\}$,
$a_{\ell}^{\prime},b_{\ell}^{\prime}\in\mathbb{R}$, for
$\ell\in\\{1,\dots,q^{\prime}\\}$, $m\in\mathbb{N}_{0}$ and
$p,p^{\prime}\in\mathbb{N}_{0}$ such that

$\displaystyle
f(z)=r\mathrm{e}^{\mathrm{i}\phi}\exp\left(\sum_{\ell=1}^{q}(a_{\ell}+\mathrm{i}b_{\ell})z^{\ell}\right)z^{m}\prod_{a\in\mathcal{X}}E\left(\frac{z}{a};p\right)^{m_{\mathcal{X}}(a)}\prod_{a\in\mathcal{Y}}E\left(\frac{z}{a};p\right)^{m_{\mathcal{Y}}(a)},$
$\displaystyle
g(z)=r^{\prime}\mathrm{e}^{\mathrm{i}\psi}\exp\left(\sum_{\ell=1}^{q^{\prime}}(a_{\ell}^{\prime}+\mathrm{i}b_{\ell}^{\prime})z^{\ell}\right)z^{m}\prod_{a\in\mathcal{X}}E\left(\frac{z}{a};p^{\prime}\right)^{m_{\mathcal{X}}(a)}\prod_{a\in\mathcal{Y}}E\left(\frac{z}{\overline{a}};p^{\prime}\right)^{m_{\mathcal{Y}}(a)},$

for $z\in\mathbb{C}$. Note that we have used that Lemma 16 directly implies
that $m=M_{f}(0)=M_{g}(0)$. Further simplifications of the above formulae
follow from the considerations in Remark 12 according to which we may assume
that $p=p^{\prime}$ (by setting both of them equal to the maximum of the geni
of the canonical products formed with $\mathcal{R}_{\ast}(f)$ and with
$\mathcal{R}_{\ast}(g)$). Note that this implies that we may need to increase
the values of $q$ or $q^{\prime}$. We will also assume that $q=q^{\prime}$,
for simplicity. This is possible by setting the superfluous polynomial
coefficients to zero.

Now, let $x\in\mathbb{R}$ be arbitrary and compute

$\left\lvert
f(x)\right\rvert=r\exp\left(\sum_{\ell=1}^{q}a_{\ell}x^{\ell}\right)\left\lvert
x\right\rvert^{m}\prod_{a\in\mathcal{X}}\left\lvert
E\left(\frac{x}{a};p\right)\right\rvert^{m_{\mathcal{X}}(a)}\prod_{a\in\mathcal{Y}}\left\lvert
E\left(\frac{x}{a};p\right)\right\rvert^{m_{\mathcal{Y}}(a)}$

as well as

$\left\lvert
g(x)\right\rvert=r^{\prime}\exp\left(\sum_{\ell=1}^{q}a^{\prime}_{\ell}x^{\ell}\right)\left\lvert
x\right\rvert^{m}\prod_{a\in\mathcal{X}}\left\lvert
E\left(\frac{x}{a};p\right)\right\rvert^{m_{\mathcal{X}}(a)}\prod_{a\in\mathcal{Y}}\left\lvert
E\left(\frac{x}{\overline{a}};p\right)\right\rvert^{m_{\mathcal{Y}}(a)}.$

According to

$\left\lvert E\left(z;p\right)\right\rvert=\left\lvert
1-z\right\rvert\exp\left(\operatorname{Re}\sum_{\ell=1}^{p}\frac{z^{\ell}}{\ell}\right)=\left\lvert
1-\overline{z}\right\rvert\exp\left(\operatorname{Re}\sum_{\ell=1}^{p}\frac{\overline{z}^{\ell}}{\ell}\right)=\left\lvert
E\left(\overline{z};p\right)\right\rvert,$

for $z\in\mathbb{C}$, we have

$1=\frac{\left\lvert f(x)\right\rvert}{\left\lvert
g(x)\right\rvert}=\frac{r}{r^{\prime}}\exp\left(\sum_{\ell=1}^{q}(a_{\ell}-a_{\ell}^{\prime})x^{\ell}\right)$

and thus

$0=\log\frac{r}{r^{\prime}}+\sum_{\ell=1}^{q}(a_{\ell}-a_{\ell}^{\prime})x^{\ell}.$

As $x\in\mathbb{R}$ was arbitrary, we can compare coefficients. In this way,
we obtain $r=r^{\prime}$ as well as $a_{\ell}=a_{\ell}^{\prime}$, for
$\ell\in\\{1,\dots,q\\}$. Item (b) has been proven.

Next, we show that item (b) implies item (a). Consider $x\in\mathbb{R}$ and
note that it follows from item (b) that

$\left\lvert
f(x)\right\rvert=r\exp\left(\sum_{\ell=1}^{q}a_{\ell}x^{\ell}\right)\left\lvert
x\right\rvert^{m}\prod_{a\in\mathcal{X}}\left\lvert
E\left(\frac{x}{a};p\right)\right\rvert^{m_{\mathcal{X}}(a)}\prod_{a\in\mathcal{Y}}\left\lvert
E\left(\frac{x}{a};p\right)\right\rvert^{m_{\mathcal{Y}}(a)}$

as well as

$\left\lvert
g(x)\right\rvert=r\exp\left(\sum_{\ell=1}^{q}a_{\ell}x^{\ell}\right)\left\lvert
x\right\rvert^{m}\prod_{a\in\mathcal{X}}\left\lvert
E\left(\frac{x}{a};p\right)\right\rvert^{m_{\mathcal{X}}(a)}\prod_{a\in\mathcal{Y}}\left\lvert
E\left(\frac{x}{\overline{a}};p\right)\right\rvert^{m_{\mathcal{Y}}(a)}.$

Remember that $\left\lvert E(z;p)\right\rvert=\left\lvert
E(\overline{z};p)\right\rvert$, for $z\in\mathbb{C}$. Therefore, $\left\lvert
f(x)\right\rvert=\left\lvert g(x)\right\rvert$. ∎

##  4 Phase retrieval of entire functions: extensions

In the prior section, we had seen that magnitude measurements of entire
functions of finite order on any single line in the complex line are not
enough to recover the corresponding phases up to a global phase factor. We had
also seen that this is due to two ambiguities. First, magnitude information on
the line $\mathbb{R}$ is not enough to distinguish between different functions
of the form $\exp(iQ)$, where $Q\in\mathbb{R}_{n}[z]$ is a polynomial with
real-coefficients and complex argument. Secondly, and maybe more importantly,
magnitude information on a single line is not even enough to determine the
roots of the entire function which serves as the signal.

A natural question is whether it is possible to formulate a problem whose
solution is unique up to global phase by adding more restrictions on the
entire function under consideration. In this section, we want to partially
answer this question by determining what happens when we know the absolute
value of a finite order entire function on two arbitrary lines in the complex
plane. The results we derive can be extended to an arbitrary number of lines
in the complex plane in certain cases.

###  4.1 Intersecting lines

If we look at two arbitrary lines in the complex plane, there are two distinct
cases which can come up. In the first case, the lines intersect at some point.
In the second case, the lines never intersect and are, in consequence,
parallel. In this subsection, we want to focus on intersecting lines. We will
assume without loss of generality (remember Remark 15 on rigid motions) that
our lines meet at the origin of the complex plane. In this way, we can fully
focus on the angle $\theta$ in between the intersecting lines. Note that we
may assume (again without loss of generality) that one of the two lines
corresponds to the line $\mathbb{R}$ and that the other line intersects the
reals at an angle $\theta\in(0,\tfrac{\pi}{2}]$. Indeed, if this should not be
the case, we can rotate the entire complex plane about the origin until it is
the case. We will furthermore distinguish two cases. First, the case in which
$\theta\in(0,\tfrac{\pi}{2}]\setminus\pi\mathbb{Q}$ (we will call such angles
_irrational_). Secondly, the case in which
$\theta\in(0,\tfrac{\pi}{2}]\cap\pi\mathbb{Q}$ (we will call such angles
_rational_).

##### Irrational angles

We will start by considering general $\theta\in(0,\tfrac{\pi}{2}]$ and thus
leave considerations concerning the rationality of angles to the side for now.
If we have two non-zero functions $f,g\in\mathcal{O}(\mathbb{C})$ such that

$\left\lvert f(x)\right\rvert=\left\lvert
g(x)\right\rvert\qquad\mbox{and}\qquad\left\lvert
f(\mathrm{e}^{\mathrm{i}\theta}x)\right\rvert=\left\lvert
g(\mathrm{e}^{\mathrm{i}\theta}x)\right\rvert,$

for all $x\in\mathbb{R}$, then the symmetries in the root set described by
Lemma 16 are accompanied by additional symmmetries stemming from the equality
of the absolute values on $\mathrm{e}^{\mathrm{i}\theta}\mathbb{R}$.

######  Lemma 18 (Root symmetry — II).

Let $\theta\in(0,\tfrac{\pi}{2}]$ and let $f,g\in\mathcal{O}(\mathbb{C})$ be
such that

$\left\lvert f(\mathrm{e}^{\mathrm{i}\theta}x)\right\rvert=\left\lvert
g(\mathrm{e}^{\mathrm{i}\theta}x)\right\rvert,\qquad x\in\mathbb{R}.$

Then, it holds that

$M_{f}(\mathrm{e}^{\mathrm{i}\theta}z)+M_{f}(\mathrm{e}^{\mathrm{i}\theta}\overline{z})=M_{g}(\mathrm{e}^{\mathrm{i}\theta}z)+M_{g}(\mathrm{e}^{\mathrm{i}\theta}\overline{z}),\qquad
z\in\mathbb{C}.$

Viewed from a geometrical point of view, the above lemma states that there is
a mirror symmetry along the axis $\mathrm{e}^{\mathrm{i}\theta}\mathbb{R}$ in
the root sets of $f$ and $g$. This symmetry is especially striking when one
considers its implications for the multiset described by $M_{\mathcal{Y}}$. If
one does consider a root $a$ of $f$ which is not a root of $g$, then
$\overline{a}$ must be a root of $g$ according to the mirror symmetry along
the real line. If $\overline{a}$ is not a root of $f$, then
$\mathrm{e}^{2\mathrm{i}\theta}a$ must be a root of $f$ according to the
mirror symmetry along the line $\mathrm{e}^{\mathrm{i}\theta}\mathbb{R}$. In
this way, one observes a rotational symmetry in the roots. To make these
considerations precise, we remember the notations

$\mathcal{X}=\mathcal{R}_{\ast}(f)\cap\mathcal{R}(g),\qquad\mathcal{Y}=\left\\{a\in\mathcal{R}_{\ast}(f)\,\middle|\,m_{f}(a)>M_{g}(a)\right\\},$

as well as $m_{\mathcal{X}}:\mathcal{X}\to\mathbb{N}$,

$m_{\mathcal{X}}(a)=\min\\{m_{f}(a),m_{g}(a)\\},\qquad a\in\mathcal{X},$

and $m_{\mathcal{Y}}:\mathcal{Y}\to\mathbb{N}$,

$m_{\mathcal{Y}}(a)=m_{f}(a)-M_{g}(a),\qquad a\in\mathcal{Y}.$

$M_{\mathcal{X}}$ and $M_{\mathcal{Y}}$ were the multisets defined by
extending $m_{\mathcal{X}}$ and $m_{\mathcal{Y}}$ to all of $\mathbb{C}$ by
zero. One can then convince oneself that the Lemmata 16 and 18 imply that

$\displaystyle M_{\mathcal{Y}}\left(\mathrm{e}^{2\mathrm{i}\theta}z\right)$
$\displaystyle=\max\left\\{0,M_{f}\left(\mathrm{e}^{2\mathrm{i}\theta}z\right)-M_{g}\left(\mathrm{e}^{2\mathrm{i}\theta}z\right)\right\\}=\max\left\\{0,M_{g}\left(\overline{z}\right)-M_{f}\left(\overline{z}\right)\right\\}$
$\displaystyle=\max\left\\{0,M_{f}\left(z\right)-M_{g}\left(z\right)\right\\}=M_{\mathcal{Y}}\left(z\right),$

for $z\in\mathbb{C}$. It follows immediately that

$M_{\mathcal{Y}}\left(\mathrm{e}^{2k\mathrm{i}\theta}z\right)=M_{\mathcal{Y}}\left(z\right),\qquad
z\in\mathbb{C},$ (5)

for $k\in\mathbb{Z}$. The set $\mathcal{Y}$ is rotationally symmetric around
the origin with angle of rotation $2\theta$. Note that rational and irrational
angles create fundamentally different situations. For irrational angles, it
follows that $\mathcal{Y}$ must contain infinitely many elements whenever it
is non-empty: a contradiction for non-zero entire functions. For rational
angles, $\mathcal{Y}$ can have finite cardinality even when it is non-empty.

Let us first discuss irrational angles in greater detail. We want to start by
stating that the results we present in the following have first been
discovered and proven by Philippe Jaming in 2014 [10]. We were not aware of
this at the time we started our research on entire functions and do therefore
present these results as they occurred to us. In any case, credit for the
following result is due to Jaming.

From now on, we will assume that
$\theta\in(0,\tfrac{\pi}{2}]\setminus\pi\mathbb{Q}$. Let us also assume, by
contradiction, that $\mathcal{Y}$ is non-empty. It follows immediately from
equation (5) that if $a\in\mathcal{Y}$, then

$C_{a;\theta}:=\left\\{\mathrm{e}^{2k\mathrm{i}\theta}a\,\middle|\,k\in\mathbb{Z}\right\\}\subset\mathcal{Y}.$

Note that the countably infinite set $C_{a;\theta}$ is contained in the
compact subset

$\left\\{z\in\mathbb{C}\,\middle|\,\left\lvert z\right\rvert=\left\lvert
a\right\rvert\right\\}$

of the complex plane. Therefore, there must be a sequence in $C_{a;\theta}$
which converges. This is in contradiction to $f$ having an isolated set of
zeroes due to it being a non-zero entire function. We conclude that
$\mathcal{Y}=\emptyset$. Therefore, magnitude information on two lines
intersecting at an irrational angle is enough to uniquely determine the roots
of an entire function. It should come as no surprise that this allows us to
fully recover the entire function up to global phase from magnitude
information on two lines intersecting at an irrational angle.

######  Theorem 19 (Cf. Theorem 1).

Let $\theta\in(0,\tfrac{\pi}{2}]\setminus\pi\mathbb{Q}$, and let
$f,g\in\mathcal{O}(\mathbb{C})$ be of finite order. The following are
equivalent:

(a)

For all $x\in\mathbb{R}$, $\left\lvert f(x)\right\rvert=\left\lvert
g(x)\right\rvert$ and $\left\lvert
f(\mathrm{e}^{\mathrm{i}\alpha}x)\right\rvert=\left\lvert
g(\mathrm{e}^{\mathrm{i}\alpha}x)\right\rvert$.

(b)

There exists an $\alpha\in\mathbb{R}$ such that
$f=\mathrm{e}^{\mathrm{i}\alpha}g$.

###### Proof.

We can assume that $f$ and $g$ are non-zero. If one of them happens to be
zero, then the other one needs to be zero on the reals. Therefore, it does not
have isolated roots such that it is zero itself. We will only show that item
(a) implies item (b) and assume that the other direction is clear to the
reader. According to the Mc Donald decomposition and the considerations
before, there exist $r>0$, $\phi,\psi\in\mathbb{R}$, $q\in\mathbb{N}_{0}$,
$a_{\ell},b_{\ell},b^{\prime}_{\ell}\in\mathbb{R}$, for
$\ell\in\\{1,\dots,q\\}$, $m\in\mathbb{N}_{0}$ and $p\in\mathbb{N}_{0}$ such
that

$\displaystyle
f(z)=r\mathrm{e}^{\mathrm{i}\phi}\exp\left(\sum_{\ell=1}^{q}\left(a_{\ell}+\mathrm{i}b_{\ell}\right)z^{\ell}\right)z^{m}\prod_{a\in\mathcal{X}}E\left(\frac{z}{a},p\right)^{m_{\mathcal{X}}(a)},$
$\displaystyle
g(z)=r\mathrm{e}^{\mathrm{i}\psi}\exp\left(\sum_{\ell=1}^{q}\left(a_{\ell}+\mathrm{i}b^{\prime}_{\ell}\right)z^{\ell}\right)z^{m}\prod_{a\in\mathcal{X}}E\left(\frac{z}{a},p\right)^{m_{\mathcal{X}}(a)},$

for $z\in\mathbb{C}$.

We may therefore compute

$1=\frac{\left\lvert
f(\mathrm{e}^{\mathrm{i}\theta}x)\right\rvert}{\left\lvert
g(\mathrm{e}^{\mathrm{i}\theta}x)\right\rvert}=\exp\left(\sum_{\ell=1}^{q}(b_{\ell}^{\prime}-b_{\ell})\operatorname{Im}\left[\mathrm{e}^{\mathrm{i}\theta\ell}\right]x^{\ell}\right),\qquad
x\in\mathbb{R},$

such that

$0=\sum_{\ell=1}^{q}(b_{\ell}^{\prime}-b_{\ell})\sin(\theta\ell)x^{\ell},\qquad
x\in\mathbb{R}.$

After a comparison of coefficients and realising that
$\theta\ell\not\in\pi\mathbb{Z}$, for $\ell\in\\{1,\dots,q\\}$, we find
$b_{\ell}=b_{\ell}^{\prime}$, for $\ell\in\\{1,\dots,q\\}$. We conclude that
$f$ and $g$ differ by a global phase factor ($\alpha=\phi-\psi$). ∎

######  Remark 20.

Note that the theorem above deals with the special case where one of the two
intersecting lines is $\mathbb{R}$ and the other is
$\mathrm{e}^{\mathrm{i}\theta}\mathbb{R}$, with $\theta\in(0,\frac{\pi}{2}]$.
Note also that given any two lines $L_{1},L_{2}\subset\mathbb{C}$ that
intersect, we can find a rigid motion $\mathfrak{r}:\mathbb{C}\to\mathbb{C}$
such that $\mathfrak{r}(L_{1})=\mathbb{R}$ and
$\mathfrak{r}(L_{2})=\mathrm{e}^{\mathrm{i}\theta}\mathbb{R}$, with
$\theta\in(0,\frac{\pi}{2}]$, (after potentially relabelling $L_{1}$ and
$L_{2}$). With the help of this rigid motion, we can provide a result for any
two intersecting lines which meet at an irrational angle in the complex plane.

Finally, note that it is actually enough to assume that $\left\lvert
f(x)\right\rvert=\left\lvert g(x)\right\rvert$ and $\left\lvert
f(\mathrm{e}^{\mathrm{i}\theta}x)\right\rvert=\left\lvert
g(\mathrm{e}^{\mathrm{i}\theta}x)\right\rvert$, for $x\in X\subset\mathbb{R}$,
where $X$ is any set with accumulation point. Indeed, it follows from $f$ and
$g$ being entire functions (and the identity theorem), that the above
equations must extend to all of $\mathbb{R}$ in this case.

##### Rational angles

We will now turn our attention to rational angles
$\theta\in(0,\tfrac{\pi}{2}]\cap\pi\mathbb{Q}$. It is relatively easy to
construct $\Omega\subset\mathbb{C}$ consisting of two lines intersecting at a
rational angle ($\theta=\tfrac{\pi}{2}$) and $f,g\in\mathcal{O}(\mathbb{C})$
such that the magnitudes of $f$ and $g$ agree on $\Omega$ yet $f$ and $g$ are
not the same up to global phase. Indeed consider
$\Omega=\mathbb{R}\cup\mathrm{i}\mathbb{R}$ as well as

$f(z)=1-\frac{z^{2}}{(1+\mathrm{i})^{2}},\qquad
g(z)=1-\frac{z^{2}}{(1-\mathrm{i})^{2}},$

for $z\in\mathbb{C}$. Then, we have

$\left\lvert f(x)\right\rvert=\left\lvert
1-\frac{x^{2}}{(1+\mathrm{i})^{2}}\right\rvert=\left\lvert
1-\frac{x^{2}}{(1-\mathrm{i})^{2}}\right\rvert=\left\lvert
g(x)\right\rvert,\qquad x\in\mathbb{R},$

as well as

$\left\lvert f(\mathrm{i}x)\right\rvert=\left\lvert
1+\frac{x^{2}}{(1+\mathrm{i})^{2}}\right\rvert=\left\lvert
1+\frac{x^{2}}{(1-\mathrm{i})^{2}}\right\rvert=\left\lvert
g(\mathrm{i}x)\right\rvert,\qquad x\in\mathbb{R}.$

$\mathcal{R}_{\mathrm{u}}(f)$$\mathbb{R}$$\mathrm{i}\mathbb{R}$$\theta$$-\theta$$a$$\mathrm{e}^{2\mathrm{i}\theta}a$$a^{\prime}$$\mathrm{e}^{2\mathrm{i}\theta}a^{\prime}$
Figure 1: The splitting of $\mathcal{Y}$ according to the rotational symmetry
with angle $2\theta$. The dotted line indicates that no element of
$\mathrm{e}^{\mathrm{i}\theta}\mathbb{R}$ is included in $\mathcal{Y}$. Note
how for any $a,a^{\prime}\in\mathcal{Y}$, it holds that
$\mathrm{e}^{2k\mathrm{i}\theta}a,\mathrm{e}^{2k\mathrm{i}\theta}a^{\prime}\in\mathcal{Y}$,
for $k=1,\dots,n$.

Much like in the Mc Donald decomposition, we can classify all entire functions
of finite order whose magnitudes agree on two lines intersecting at a rational
angle. Remember from our considerations on irrational angles that two entire
functions whose magnitudes agree on two lines which meet at an angle $\theta$
will have rotation symmetries around the origin with angle $2\theta$ in their
root set. To describe this more precisely, we may consider the smallest
integer $n\in\mathbb{N}$ such that $n\theta\in\pi\mathbb{N}$ along with the
set

$\mathcal{Y}_{\mathrm{u}}:=\mathcal{Y}\cap\left\\{z\in\mathbb{C}\,\middle|\,\operatorname{\exists}r>0,~{}\phi\in[-\theta,\theta):z=r\mathrm{e}^{\mathrm{i}\phi}\right\\}$

which is illustrated in Figure 1. Looking at equation (5), it is not too hard
to see that

$\mathcal{Y}=\mathrm{e}^{2\mathrm{i}\theta(\mathbb{Z}/n\mathbb{Z})}\mathcal{Y}_{\mathrm{u}}.$

We may now show the following result.

######  Theorem 21 (Cf. Theorem 3).

Let $\theta\in\pi\mathbb{Q}\cap(0,\frac{\pi}{2}]$, let $n\in\mathbb{N}$ be the
smallest integer such that $\theta n\in\pi\mathbb{N}$, and let
$f,g\in\mathcal{O}(\mathbb{C})$ be non-zero and of finite order. The following
are equivalent:

(a)

For all $x\in\mathbb{R}$, $\left\lvert f(x)\right\rvert=\left\lvert
g(x)\right\rvert$ and $\left\lvert
f(\mathrm{e}^{\mathrm{i}\theta}x)\right\rvert=\left\lvert
g(\mathrm{e}^{\mathrm{i}\theta}x)\right\rvert$.

(b)

There exist $r>0$, $\phi,\psi\in\mathbb{R}$, $q\in\mathbb{N}_{0}$,
$a_{\ell},b_{\ell},b^{\prime}_{\ell}\in\mathbb{R}$, for
$\ell\in\\{1,\dots,q\\}$, $m\in\mathbb{N}_{0}$ and $p\in\mathbb{N}_{0}$ such
that

$\displaystyle b_{\ell}^{\prime}=b_{\ell},\qquad\mbox{for all
}\ell\in\\{1,\dots,q\\}\setminus n\mathbb{N},$

as well as

$\displaystyle\begin{multlined}f(z)=r\mathrm{e}^{\mathrm{i}\phi}\exp\left(\sum_{\ell=1}^{q}\left(a_{\ell}+\mathrm{i}b_{\ell}\right)z^{\ell}\right)z^{m}\prod_{a\in\mathcal{X}}E\left(\frac{z}{a};p\right)^{m_{\mathcal{X}}(a)}\\\
\cdot\prod_{a\in\mathcal{Y}_{\mathrm{u}}}\prod_{k=1}^{n}E\left(\mathrm{e}^{2k\mathrm{i}\theta}\frac{z}{a};p\right)^{m_{\mathcal{Y}}(a)},\end{multlined}f(z)=r\mathrm{e}^{\mathrm{i}\phi}\exp\left(\sum_{\ell=1}^{q}\left(a_{\ell}+\mathrm{i}b_{\ell}\right)z^{\ell}\right)z^{m}\prod_{a\in\mathcal{X}}E\left(\frac{z}{a};p\right)^{m_{\mathcal{X}}(a)}\\\
\cdot\prod_{a\in\mathcal{Y}_{\mathrm{u}}}\prod_{k=1}^{n}E\left(\mathrm{e}^{2k\mathrm{i}\theta}\frac{z}{a};p\right)^{m_{\mathcal{Y}}(a)},$
$\displaystyle\begin{multlined}g(z)=r\mathrm{e}^{\mathrm{i}\psi}\exp\left(\sum_{\ell=1}^{q}\left(a_{\ell}+\mathrm{i}b_{\ell}^{\prime}\right)z^{\ell}\right)z^{m}\prod_{a\in\mathcal{X}}E\left(\frac{z}{a};p\right)^{m_{\mathcal{X}}(a)}\\\
\cdot\prod_{a\in\mathcal{Y}_{\mathrm{u}}}\prod_{k=1}^{n}E\left(\mathrm{e}^{2k\mathrm{i}\theta}\frac{z}{\overline{a}};p\right)^{m_{\mathcal{Y}}(a)},\end{multlined}g(z)=r\mathrm{e}^{\mathrm{i}\psi}\exp\left(\sum_{\ell=1}^{q}\left(a_{\ell}+\mathrm{i}b_{\ell}^{\prime}\right)z^{\ell}\right)z^{m}\prod_{a\in\mathcal{X}}E\left(\frac{z}{a};p\right)^{m_{\mathcal{X}}(a)}\\\
\cdot\prod_{a\in\mathcal{Y}_{\mathrm{u}}}\prod_{k=1}^{n}E\left(\mathrm{e}^{2k\mathrm{i}\theta}\frac{z}{\overline{a}};p\right)^{m_{\mathcal{Y}}(a)},$

for $z\in\mathbb{C}$.

###### Proof.

We start by showing that item (a) implies item (b). According to the Mc Donald
decomposition and the considerations before the statement of Theorem 19, there
exist $r>0$, $\phi,\psi\in\mathbb{R}$, $q\in\mathbb{N}_{0}$,
$a_{\ell},b_{\ell},b^{\prime}_{\ell}\in\mathbb{R}$, for
$\ell\in\\{1,\dots,q\\}$, $m\in\mathbb{N}_{0}$ and $p\in\mathbb{N}_{0}$ such
that

$\displaystyle\begin{multlined}f(z)=r\mathrm{e}^{\mathrm{i}\phi}\exp\left(\sum_{\ell=1}^{q}\left(a_{\ell}+\mathrm{i}b_{\ell}\right)z^{\ell}\right)z^{m}\prod_{a\in\mathcal{X}}E\left(\frac{z}{a};p\right)^{m_{\mathcal{X}}(a)}\\\
\cdot\prod_{a\in\mathcal{Y}_{\mathrm{u}}}\prod_{k=1}^{n}E\left(\mathrm{e}^{2k\mathrm{i}\theta}\frac{z}{a};p\right)^{m_{\mathcal{Y}}(a)},\end{multlined}f(z)=r\mathrm{e}^{\mathrm{i}\phi}\exp\left(\sum_{\ell=1}^{q}\left(a_{\ell}+\mathrm{i}b_{\ell}\right)z^{\ell}\right)z^{m}\prod_{a\in\mathcal{X}}E\left(\frac{z}{a};p\right)^{m_{\mathcal{X}}(a)}\\\
\cdot\prod_{a\in\mathcal{Y}_{\mathrm{u}}}\prod_{k=1}^{n}E\left(\mathrm{e}^{2k\mathrm{i}\theta}\frac{z}{a};p\right)^{m_{\mathcal{Y}}(a)},$
$\displaystyle\begin{multlined}g(z)=r\mathrm{e}^{\mathrm{i}\psi}\exp\left(\sum_{\ell=1}^{q}\left(a_{\ell}+\mathrm{i}b_{\ell}^{\prime}\right)z^{\ell}\right)z^{m}\prod_{a\in\mathcal{X}}E\left(\frac{z}{a};p\right)^{m_{\mathcal{X}}(a)}\\\
\cdot\prod_{a\in\mathcal{Y}_{\mathrm{u}}}\prod_{k=1}^{n}E\left(\mathrm{e}^{2k\mathrm{i}\theta}\frac{z}{\overline{a}};p\right)^{m_{\mathcal{Y}}(a)},\end{multlined}g(z)=r\mathrm{e}^{\mathrm{i}\psi}\exp\left(\sum_{\ell=1}^{q}\left(a_{\ell}+\mathrm{i}b_{\ell}^{\prime}\right)z^{\ell}\right)z^{m}\prod_{a\in\mathcal{X}}E\left(\frac{z}{a};p\right)^{m_{\mathcal{X}}(a)}\\\
\cdot\prod_{a\in\mathcal{Y}_{\mathrm{u}}}\prod_{k=1}^{n}E\left(\mathrm{e}^{2k\mathrm{i}\theta}\frac{z}{\overline{a}};p\right)^{m_{\mathcal{Y}}(a)},$

for $z\in\mathbb{C}$.

Let $x\in\mathbb{R}$ be arbitrary and compute

$\left\lvert
f(\mathrm{e}^{\mathrm{i}\theta}x)\right\rvert=r\exp\left(\operatorname{Re}\sum_{\ell=1}^{q}\left(a_{\ell}+\mathrm{i}b_{\ell}\right)\mathrm{e}^{\mathrm{i}\theta\ell}x^{\ell}\right)\left\lvert
x\right\rvert^{m}\prod_{a\in\mathcal{X}}\left\lvert
E\left(\mathrm{e}^{\mathrm{i}\theta}\frac{x}{a};p\right)\right\rvert^{m_{\mathcal{X}}(a)}\\\
\cdot\prod_{a\in\mathcal{Y}_{\mathrm{u}}}\prod_{k=1}^{n}\left\lvert
E\left(\mathrm{e}^{(2k+1)\mathrm{i}\theta}\frac{x}{a};p\right)\right\rvert^{m_{\mathcal{Y}}(a)}$

as well as

$\left\lvert
g(\mathrm{e}^{\mathrm{i}\theta}x)\right\rvert=r\exp\left(\operatorname{Re}\sum_{\ell=1}^{q}\left(a_{\ell}+\mathrm{i}b_{\ell}^{\prime}\right)\mathrm{e}^{\mathrm{i}\theta\ell}x^{\ell}\right)\left\lvert
x\right\rvert^{m}\prod_{a\in\mathcal{X}}\left\lvert
E\left(\mathrm{e}^{\mathrm{i}\theta}\frac{x}{a};p\right)\right\rvert^{m_{\mathcal{X}}(a)}\\\
\cdot\prod_{a\in\mathcal{Y}_{\mathrm{u}}}\prod_{k=1}^{n}\left\lvert
E\left(\mathrm{e}^{(2k+1)\mathrm{i}\theta}\frac{x}{\overline{a}};p\right)\right\rvert^{m_{\mathcal{Y}}(a)}.$

After a change of index, the latter becomes

$\left\lvert
g(\mathrm{e}^{\mathrm{i}\theta}x)\right\rvert=r\exp\left(\operatorname{Re}\sum_{\ell=1}^{q}\left(a_{\ell}+\mathrm{i}b_{\ell}^{\prime}\right)\mathrm{e}^{\mathrm{i}\theta\ell}x^{\ell}\right)\left\lvert
x\right\rvert^{m}\prod_{a\in\mathcal{X}}\left\lvert
E\left(\mathrm{e}^{\mathrm{i}\theta}\frac{x}{a};p\right)\right\rvert^{m_{\mathcal{X}}(a)}\\\
\cdot\prod_{a\in\mathcal{Y}_{\mathrm{u}}}\prod_{k=1}^{n}\left\lvert
E\left(\mathrm{e}^{-(2k+1)\mathrm{i}\theta}\frac{x}{\overline{a}};p\right)\right\rvert^{m_{\mathcal{Y}}(a)}.$

We can now compute

$\displaystyle 1$ $\displaystyle=\frac{\left\lvert
f(\mathrm{e}^{\mathrm{i}\theta}x)\right\rvert}{\left\lvert
g(\mathrm{e}^{\mathrm{i}\theta}x)\right\rvert}$
$\displaystyle=\exp\left(\sum_{\ell=1}^{q}(b_{\ell}^{\prime}-b_{\ell})\sin(\theta\ell)x^{\ell}\right)\prod_{a\in\mathcal{Y}_{\mathrm{u}}}\prod_{k=1}^{n}\left\lvert\frac{E\left(\mathrm{e}^{(2k+1)\mathrm{i}\theta}\frac{x}{a};p\right)}{E\left(\mathrm{e}^{-(2k+1)\mathrm{i}\theta}\frac{x}{\overline{a}};p\right)}\right\rvert^{m_{\mathcal{Y}}(a)}$
$\displaystyle=\begin{multlined}\exp\left(\sum_{\ell=1}^{q}(b_{\ell}^{\prime}-b_{\ell})\sin(\theta\ell)x^{\ell}\right)\prod_{a\in\mathcal{Y}_{\mathrm{u}}}\prod_{k=1}^{n}\left\lvert\frac{\left(1-\mathrm{e}^{(2k+1)\mathrm{i}\theta}\frac{x}{a}\right)}{\left(1-\mathrm{e}^{-(2k+1)\mathrm{i}\theta}\frac{x}{\overline{a}}\right)}\right\rvert^{m_{\mathcal{Y}}(a)}\\\
\cdot\exp\left(m_{\mathcal{Y}}(a)\operatorname{Re}\sum_{\ell=1}^{p}\frac{1}{\ell}\left[\mathrm{e}^{(2k+1)\mathrm{i}\theta\ell}\frac{x^{\ell}}{a^{\ell}}-\mathrm{e}^{-(2k+1)\mathrm{i}\theta\ell}\frac{x^{\ell}}{\overline{a}^{\ell}}\right]\right)\end{multlined}\exp\left(\sum_{\ell=1}^{q}(b_{\ell}^{\prime}-b_{\ell})\sin(\theta\ell)x^{\ell}\right)\prod_{a\in\mathcal{Y}_{\mathrm{u}}}\prod_{k=1}^{n}\left\lvert\frac{\left(1-\mathrm{e}^{(2k+1)\mathrm{i}\theta}\frac{x}{a}\right)}{\left(1-\mathrm{e}^{-(2k+1)\mathrm{i}\theta}\frac{x}{\overline{a}}\right)}\right\rvert^{m_{\mathcal{Y}}(a)}\\\
\cdot\exp\left(m_{\mathcal{Y}}(a)\operatorname{Re}\sum_{\ell=1}^{p}\frac{1}{\ell}\left[\mathrm{e}^{(2k+1)\mathrm{i}\theta\ell}\frac{x^{\ell}}{a^{\ell}}-\mathrm{e}^{-(2k+1)\mathrm{i}\theta\ell}\frac{x^{\ell}}{\overline{a}^{\ell}}\right]\right)$
$\displaystyle=\begin{multlined}\exp\left(\sum_{\ell=1}^{q}(b_{\ell}^{\prime}-b_{\ell})\sin(\theta\ell)x^{\ell}\right)\\\
\cdot\prod_{a\in\mathcal{Y}_{\mathrm{u}}}\prod_{k=1}^{n}\exp\left(m_{\mathcal{Y}}(a)\sum_{\ell=1}^{p}\frac{x^{\ell}}{\ell}\operatorname{Re}\left[\mathrm{e}^{(2k+1)\mathrm{i}\theta\ell}a^{-\ell}-\mathrm{e}^{-(2k+1)\mathrm{i}\theta\ell}\overline{a}^{-\ell}\right]\right)\end{multlined}\exp\left(\sum_{\ell=1}^{q}(b_{\ell}^{\prime}-b_{\ell})\sin(\theta\ell)x^{\ell}\right)\\\
\cdot\prod_{a\in\mathcal{Y}_{\mathrm{u}}}\prod_{k=1}^{n}\exp\left(m_{\mathcal{Y}}(a)\sum_{\ell=1}^{p}\frac{x^{\ell}}{\ell}\operatorname{Re}\left[\mathrm{e}^{(2k+1)\mathrm{i}\theta\ell}a^{-\ell}-\mathrm{e}^{-(2k+1)\mathrm{i}\theta\ell}\overline{a}^{-\ell}\right]\right)$
$\displaystyle=\exp\left(\sum_{\ell=1}^{q}(b_{\ell}^{\prime}-b_{\ell})\sin(\theta\ell)x^{\ell}\right).$

Therefore,

$0=\sum_{\ell=1}^{q}(b_{\ell}^{\prime}-b_{\ell})\sin(\theta\ell)x^{\ell}.$

As $x\in\mathbb{R}$ was arbitrary, we can compare coefficients. In this way,
we obtain

$(b_{\ell}^{\prime}-b_{\ell})\sin(\theta\ell)=0,\qquad\ell\in\\{1,\dots,q\\}.$

For each $\ell\in\\{1,\dots,q\\}$ there are two cases: either
$\theta\ell\in\pi\mathbb{Z}$ or $\theta\ell\not\in\pi\mathbb{Z}$. In the
former case, our equation is trivial and $\ell\in\\{n,2n,\dots\\}$. In the
latter case, we conclude

$b_{\ell}^{\prime}=b_{\ell}.$

Finally, we can show that item (b) implies item (a). Note that the assumptions
in item (b) are stronger than the assumptions in item (b) of the Mc Donald
decomposition theorem. Therefore, it must hold that $\left\lvert
f(x)\right\rvert=\left\lvert g(x)\right\rvert$, for $x\in\mathbb{R}$. In
addition, we can follow the computations from earlier in this proof to see
that for $x\in\mathbb{R}$,

$\frac{\left\lvert f(\mathrm{e}^{\mathrm{i}\theta}x)\right\rvert}{\left\lvert
g(\mathrm{e}^{\mathrm{i}\theta}x)\right\rvert}=\exp\left(\sum_{\ell=1}^{q}(b_{\ell}^{\prime}-b_{\ell})\sin(\theta\ell)x^{\ell}\right)=1,\qquad
x\in\mathbb{R},$

since

$b_{\ell}^{\prime}=b_{\ell},\qquad\mbox{for all
}\ell\in\\{1,\dots,q\\}\setminus\\{n,2n,\dots\\}.$

We conclude that $\left\lvert
f(\mathrm{e}^{\mathrm{i}\theta}x)\right\rvert=\left\lvert
g(\mathrm{e}^{\mathrm{i}\theta}x)\right\rvert$, for $x\in\mathbb{R}$. ∎

######  Remark 22.

As mentioned before, we can use a rigid motion
$\mathfrak{r}:\mathbb{C}\to\mathbb{C}$ to transform the above result in such a
way that it holds for any two lines in the complex plane which intersect at a
rational angle. Additionally, it is enough to assume that $\left\lvert
f(x)\right\rvert=\left\lvert g(x)\right\rvert$ and $\left\lvert
f(\mathrm{e}^{\mathrm{i}\alpha}x)\right\rvert=\left\lvert
g(\mathrm{e}^{\mathrm{i}\alpha}x)\right\rvert$, for $x\in X\subset\mathbb{R}$,
where $X$ is any set with accumulation point.

###  4.2 Parallel lines

Having looked at the case of intersecting lines, we want to move on to
consider two (and more) parallel lines in the complex plane. In the first part
of this section, we will assume without loss of generality (and this is again
due to the holomorphy of rigid motions as discussed in Remark 15) that one of
those two parallel lines is given by the real numbers and the other of those
parallel lines is given by $\mathbb{R}+\mathrm{i}\tau$, for
$\tau\in\mathbb{R}_{\ast}=\mathbb{R}\setminus\\{0\\}$. Thereafter, we show
that we can use the same considerations to elucidate the case of infinitely
many parallel lines $\mathbb{R}+\mathrm{i}\tau\mathbb{Z}$.

##### Two parallel lines

We want to start by noting that it is not enough to know the magnitude of an
entire function of finite order on two distinct parallel lines in the complex
plane to identify the function up to global phase. Indeed, consider
$f,g\in\mathcal{O}(\mathbb{C})$ given by

$\displaystyle f(z)=\cosh\left(\frac{\pi
z}{2}\right)+\mathrm{i}\sinh\left(\frac{\pi z}{2}\right),$ $\displaystyle
g(z)=\cosh\left(\frac{\pi z}{2}\right)-\mathrm{i}\sinh\left(\frac{\pi
z}{2}\right),$

for $z\in\mathbb{C}$. Then, since the complex conjugate of any complex number
has the same magnitude as the number itself, it holds that

$\left\lvert f(x)\right\rvert=\left\lvert\cosh\left(\frac{\pi
x}{2}\right)+\mathrm{i}\sinh\left(\frac{\pi
x}{2}\right)\right\rvert=\left\lvert\cosh\left(\frac{\pi
x}{2}\right)-\mathrm{i}\sinh\left(\frac{\pi
x}{2}\right)\right\rvert=\left\lvert g(x)\right\rvert,$

for $x\in\mathbb{R}$. Additionally, we can use trigonometric identities to
convince ourself of the fact that

$\displaystyle\left\lvert f(x+\mathrm{i})\right\rvert$
$\displaystyle=\left\lvert\cosh\left(\frac{\pi
x}{2}+\frac{\pi\mathrm{i}}{2}\right)+\mathrm{i}\sinh\left(\frac{\pi
x}{2}+\frac{\pi\mathrm{i}}{2}\right)\right\rvert$
$\displaystyle=\begin{multlined}\Big{\lvert}\cosh\left(\frac{\pi
x}{2}\right)\left(\cos\left(\frac{\pi}{2}\right)-\sin\left(\frac{\pi}{2}\right)\right)\\\
+\mathrm{i}\sinh\left(\frac{\pi
x}{2}\right)\left(\cos\left(\frac{\pi}{2}\right)+\sin\left(\frac{\pi}{2}\right)\right)\Big{\rvert}\end{multlined}\Big{\lvert}\cosh\left(\frac{\pi
x}{2}\right)\left(\cos\left(\frac{\pi}{2}\right)-\sin\left(\frac{\pi}{2}\right)\right)\\\
+\mathrm{i}\sinh\left(\frac{\pi
x}{2}\right)\left(\cos\left(\frac{\pi}{2}\right)+\sin\left(\frac{\pi}{2}\right)\right)\Big{\rvert}$
$\displaystyle=\left\lvert\cosh\left(\frac{\pi
x}{2}\right)-\mathrm{i}\sinh\left(\frac{\pi
x}{2}\right)\right\rvert=\left\lvert\cosh\left(\frac{\pi
x}{2}\right)+\mathrm{i}\sinh\left(\frac{\pi x}{2}\right)\right\rvert$
$\displaystyle=\begin{multlined}\Big{\lvert}\cosh\left(\frac{\pi
x}{2}\right)\left(\cos\left(\frac{\pi}{2}\right)+\sin\left(\frac{\pi}{2}\right)\right)\\\
-\mathrm{i}\sinh\left(\frac{\pi
x}{2}\right)\left(\cos\left(\frac{\pi}{2}\right)-\sin\left(\frac{\pi}{2}\right)\right)\Big{\rvert}\end{multlined}\Big{\lvert}\cosh\left(\frac{\pi
x}{2}\right)\left(\cos\left(\frac{\pi}{2}\right)+\sin\left(\frac{\pi}{2}\right)\right)\\\
-\mathrm{i}\sinh\left(\frac{\pi
x}{2}\right)\left(\cos\left(\frac{\pi}{2}\right)-\sin\left(\frac{\pi}{2}\right)\right)\Big{\rvert}$
$\displaystyle=\left\lvert\cosh\left(\frac{\pi
x}{2}+\frac{\pi\mathrm{i}}{2}\right)-\mathrm{i}\sinh\left(\frac{\pi
x}{2}+\frac{\pi\mathrm{i}}{2}\right)\right\rvert=\left\lvert
g(x+\mathrm{i})\right\rvert,$

for $x\in\mathbb{R}$.

######  Remark 23.

Note that given any two distinct parallel lines
$L_{1},L_{2}\subset\mathbb{C}$, we may find a holomorphic map
$\mathfrak{h}:\mathbb{C}\to\mathbb{C}$ such that
$\mathfrak{h}(L_{1})=\mathbb{R}$ and
$\mathfrak{h}(L_{2})=\mathbb{R}+\mathrm{i}$. Therefore, the example given
above naturally translates to an example for any given two distinct parallel
lines (when we consider $f\circ\mathfrak{h}$ as well as $g\circ\mathfrak{h}$)
and we conclude that _given two parallel lines $L_{1}$ and $L_{2}$ in the
complex plane, there exist entire functions which do not agree up to global
phase and whose magnitudes agree on $L_{1}\cup L_{2}$._

As we have done twice at this point, we will continue by characterising the
non-uniqueness of phase retrieval in terms of the Hadamard factorisation of
finite order entire functions. Let us consider $\tau\in\mathbb{R}_{\ast}$ and
suppose that we have two non-zero functions $f,g\in\mathcal{O}(\mathbb{C})$
such that

$\left\lvert f(x)\right\rvert=\left\lvert
g(x)\right\rvert\qquad\mbox{and}\qquad\left\lvert
f(x+\mathrm{i}\tau)\right\rvert=\left\lvert g(x+\mathrm{i}\tau)\right\rvert,$

for all $x\in\mathbb{R}$. Then, the symmetries in the root set described by
Lemma 16 are accompanied by additional symmmetries stemming from the equality
of the absolute values on $\mathbb{R}+\mathrm{i}\tau$.

######  Lemma 24 (Root symmetry — III).

Let $\tau\in\mathbb{R}_{\ast}$ and let $f,g\in\mathcal{O}(\mathbb{C})$ be such
that

$\left\lvert f(x+\mathrm{i}\tau)\right\rvert=\left\lvert
g(x+\mathrm{i}\tau)\right\rvert,\qquad x\in\mathbb{R}.$

Then, it holds that

$M_{f}(z+\mathrm{i}\tau)+M_{f}(\overline{z}+\mathrm{i}\tau)=M_{g}(z+\mathrm{i}\tau)+M_{g}(\overline{z}+\mathrm{i}\tau),\qquad
z\in\mathbb{C}.$

Viewed from a geometrical point of view, the above lemma states that there is
a mirror symmetry along the axis $\mathbb{R}+\mathrm{i}\tau$ in the root sets
of $f$ and $g$. Again this symmetry becomes especially interesting when one
considers its implications for the multiset described by $M_{\mathcal{Y}}$. If
one does consider a root $a$ of $f$ which is not a root of $g$, then
$\overline{a}$ must be a root of $g$ according to the mirror symmetry along
the real line. If $\overline{a}$ is not a root of $f$, then
$a+2\mathrm{i}\tau$ must be a root of $f$ according to the mirror symmetry
along the line $\mathbb{R}+\mathrm{i}\tau$. In this way, one observes a
translational symmetry in the roots.

More precisely, one can convince oneself that the Lemmata 16 and 24 imply that

$\displaystyle M_{\mathcal{Y}}\left(z+2\mathrm{i}\tau\right)$
$\displaystyle=\max\left\\{0,M_{f}\left(z+2\mathrm{i}\tau\right)-M_{g}\left(z+2\mathrm{i}\tau\right)\right\\}$
$\displaystyle=\max\left\\{0,M_{g}\left(\overline{z}\right)-M_{f}\left(\overline{z}\right)\right\\}=\max\left\\{0,M_{f}\left(z\right)-M_{g}\left(z\right)\right\\}$
$\displaystyle=M_{\mathcal{Y}}\left(z\right),$

for $z\in\mathbb{C}$. It follows immediately that

$M_{\mathcal{Y}}\left(z+2k\mathrm{i}\tau\right)=M_{\mathcal{Y}}\left(z\right),\qquad
z\in\mathbb{C},$

for $k\in\mathbb{Z}$, and the set $\mathcal{Y}$ is symmetric with respect to
translation by $2\mathrm{i}\tau$. Note that it follows that $\mathcal{Y}$ must
contain infinitely many elements when it is non-empty. Unlike in the case of
irrational angles, this does not lead to a contradiction, however, since those
elements do not need to have an accumulation point. It is interesting to note
that $\mathcal{Y}\neq\emptyset$ implies that $f$ and $g$ must at least be of
order one by Lemma 11.

$\mathcal{R}_{\mathrm{u}}(f)$$\mathbb{R}$$\mathrm{i}\mathbb{R}$$\tau$$-\tau$$a$$a+2\mathrm{i}\tau$$a-2\mathrm{i}\tau$$a-4\mathrm{i}\tau$
Figure 2: Depiction of $\mathcal{Y}_{\mathrm{u}}$. Note that the dashed line
indicates that $\mathbb{R}+\mathrm{i}\tau$ is not contained in
$\mathcal{Y}_{\mathrm{u}}$.

We might introduce the set

$\mathcal{Y}_{\mathrm{u}}:=\mathcal{Y}\cap\left(\mathbb{R}+\mathrm{i}\left[-\tau,\tau\right)\right)$

visualised in Figure 2 to see that

$\mathcal{Y}=\mathcal{Y}_{\mathrm{u}}+2\mathrm{i}\tau\mathbb{Z}.$

We may now prove the following result.

######  Theorem 25 (Cf. Theorem 4).

Let $\tau\in\mathbb{R}_{\ast}$ and let $f,g\in\mathcal{O}(\mathbb{C})$ be non-
zero and of finite order. Then, the following are equivalent:

(a)

For all $x\in\mathbb{R}$, $\left\lvert f(x)\right\rvert=\left\lvert
g(x)\right\rvert$ and $\left\lvert f(x+\mathrm{i}\tau)\right\rvert=\left\lvert
g(x+\mathrm{i}\tau)\right\rvert$.

(b)

There exist $r>0$, $\phi,\psi\in\mathbb{R}$, $q\in\mathbb{N}_{0}$,
$a_{\ell},b_{\ell},b^{\prime}_{\ell}\in\mathbb{R}$, for
$\ell\in\\{1,\dots,q\\}$, $m\in\mathbb{N}_{0}$ and $p\in\mathbb{N}_{0}$ such
that

$\sum_{\ell=1}^{q}(b_{\ell}-b_{\ell}^{\prime})\operatorname{Im}\left[(x+\mathrm{i}\tau)^{\ell}\right]=\sum_{a\in\mathcal{Y}_{\mathrm{u}}}m_{\mathcal{Y}}(a)\sum_{k\in\mathbb{Z}}\Bigg{(}\log\left\lvert\frac{a+2(k-1)\mathrm{i}\tau}{a+2k\mathrm{i}\tau}\right\rvert\\\
+\sum_{\ell=1}^{p}\frac{1}{\ell}\operatorname{Re}\left[\left(\frac{x+\mathrm{i}\tau}{a+2k\mathrm{i}\tau}\right)^{\ell}-\left(\frac{x+\mathrm{i}\tau}{\overline{a}-2(k-1)\mathrm{i}\tau}\right)^{\ell}\right]\Bigg{)},$
(6)

for all $x\in\mathbb{R}$, as well as

$\displaystyle\begin{multlined}f(z)=r\mathrm{e}^{\mathrm{i}\phi}\exp\left(\sum_{\ell=1}^{q}\left(a_{\ell}+\mathrm{i}b_{\ell}\right)z^{\ell}\right)z^{m}\prod_{a\in\mathcal{X}}E\left(\frac{z}{a};p\right)^{m_{\mathcal{X}}(a)}\\\
\cdot\prod_{a\in\mathcal{Y}_{\mathrm{u}}}\prod_{k\in\mathbb{Z}}E\left(\frac{z}{a+2k\mathrm{i}\tau};p\right)^{m_{\mathcal{Y}}(a)},\end{multlined}f(z)=r\mathrm{e}^{\mathrm{i}\phi}\exp\left(\sum_{\ell=1}^{q}\left(a_{\ell}+\mathrm{i}b_{\ell}\right)z^{\ell}\right)z^{m}\prod_{a\in\mathcal{X}}E\left(\frac{z}{a};p\right)^{m_{\mathcal{X}}(a)}\\\
\cdot\prod_{a\in\mathcal{Y}_{\mathrm{u}}}\prod_{k\in\mathbb{Z}}E\left(\frac{z}{a+2k\mathrm{i}\tau};p\right)^{m_{\mathcal{Y}}(a)},$
$\displaystyle\begin{multlined}g(z)=r\mathrm{e}^{\mathrm{i}\psi}\exp\left(\sum_{\ell=1}^{q}\left(a_{\ell}+\mathrm{i}b_{\ell}^{\prime}\right)z^{\ell}\right)z^{m}\prod_{a\in\mathcal{X}}E\left(\frac{z}{a};p\right)^{m_{\mathcal{X}}(a)}\\\
\cdot\prod_{a\in\mathcal{Y}_{\mathrm{u}}}\prod_{k\in\mathbb{Z}}E\left(\frac{z}{\overline{a}+2k\mathrm{i}\tau};p\right)^{m_{\mathcal{Y}}(a)},\end{multlined}g(z)=r\mathrm{e}^{\mathrm{i}\psi}\exp\left(\sum_{\ell=1}^{q}\left(a_{\ell}+\mathrm{i}b_{\ell}^{\prime}\right)z^{\ell}\right)z^{m}\prod_{a\in\mathcal{X}}E\left(\frac{z}{a};p\right)^{m_{\mathcal{X}}(a)}\\\
\cdot\prod_{a\in\mathcal{Y}_{\mathrm{u}}}\prod_{k\in\mathbb{Z}}E\left(\frac{z}{\overline{a}+2k\mathrm{i}\tau};p\right)^{m_{\mathcal{Y}}(a)},$

for $z\in\mathbb{C}$.

###### Proof.

Let us first show that item (a) implies item (b). According to the Mc Donald
decomposition and the prior considerations there exist $r>0$,
$\phi,\psi\in\mathbb{R}$, $q\in\mathbb{N}_{0}$,
$a_{\ell},b_{\ell},b^{\prime}_{\ell}\in\mathbb{R}$, for
$\ell\in\\{1,\dots,q\\}$, $m\in\mathbb{N}_{0}$ and $p\in\mathbb{N}_{0}$ such
that

$\displaystyle\begin{multlined}f(z)=r\mathrm{e}^{\mathrm{i}\phi}\exp\left(\sum_{\ell=1}^{q}\left(a_{\ell}+\mathrm{i}b_{\ell}\right)z^{\ell}\right)z^{m}\prod_{a\in\mathcal{X}}E\left(\frac{z}{a};p\right)^{m_{\mathcal{X}}(a)}\\\
\cdot\prod_{a\in\mathcal{Y}_{\mathrm{u}}}\prod_{k\in\mathbb{Z}}E\left(\frac{z}{a+2k\mathrm{i}\tau};p\right)^{m_{\mathcal{Y}}(a)},\end{multlined}f(z)=r\mathrm{e}^{\mathrm{i}\phi}\exp\left(\sum_{\ell=1}^{q}\left(a_{\ell}+\mathrm{i}b_{\ell}\right)z^{\ell}\right)z^{m}\prod_{a\in\mathcal{X}}E\left(\frac{z}{a};p\right)^{m_{\mathcal{X}}(a)}\\\
\cdot\prod_{a\in\mathcal{Y}_{\mathrm{u}}}\prod_{k\in\mathbb{Z}}E\left(\frac{z}{a+2k\mathrm{i}\tau};p\right)^{m_{\mathcal{Y}}(a)},$
$\displaystyle\begin{multlined}g(z)=r\mathrm{e}^{\mathrm{i}\psi}\exp\left(\sum_{\ell=1}^{q}\left(a_{\ell}+\mathrm{i}b_{\ell}^{\prime}\right)z^{\ell}\right)z^{m}\prod_{a\in\mathcal{X}}E\left(\frac{z}{a};p\right)^{m_{\mathcal{X}}(a)}\\\
\cdot\prod_{a\in\mathcal{Y}_{\mathrm{u}}}\prod_{k\in\mathbb{Z}}E\left(\frac{z}{\overline{a}+2k\mathrm{i}\tau};p\right)^{m_{\mathcal{Y}}(a)},\end{multlined}g(z)=r\mathrm{e}^{\mathrm{i}\psi}\exp\left(\sum_{\ell=1}^{q}\left(a_{\ell}+\mathrm{i}b_{\ell}^{\prime}\right)z^{\ell}\right)z^{m}\prod_{a\in\mathcal{X}}E\left(\frac{z}{a};p\right)^{m_{\mathcal{X}}(a)}\\\
\cdot\prod_{a\in\mathcal{Y}_{\mathrm{u}}}\prod_{k\in\mathbb{Z}}E\left(\frac{z}{\overline{a}+2k\mathrm{i}\tau};p\right)^{m_{\mathcal{Y}}(a)},$

for $z\in\mathbb{C}$.

Let $x\in\mathbb{R}$ be arbitrary and compute

$\left\lvert
f(x+\mathrm{i}\tau)\right\rvert=r\exp\left(\operatorname{Re}\sum_{\ell=1}^{q}\left(a_{\ell}+\mathrm{i}b_{\ell}\right)(x+\mathrm{i}\tau)^{\ell}\right)\left\lvert
x+\mathrm{i}\tau\right\rvert^{m}\\\ \cdot\prod_{a\in\mathcal{X}}\left\lvert
E\left(\frac{x+\mathrm{i}\tau}{a};p\right)\right\rvert^{m_{\mathcal{X}}(a)}\prod_{a\in\mathcal{Y}_{\mathrm{u}}}\prod_{k\in\mathbb{Z}}\left\lvert
E\left(\frac{x+\mathrm{i}\tau}{a+2k\mathrm{i}\tau};p\right)\right\rvert^{m_{\mathcal{Y}}(a)},$

as well as

$\left\lvert
g(x+\mathrm{i}\tau)\right\rvert=r\exp\left(\operatorname{Re}\sum_{\ell=1}^{q}\left(a_{\ell}+\mathrm{i}b_{\ell}^{\prime}\right)(x+\mathrm{i}\tau)^{\ell}\right)\left\lvert
x+\mathrm{i}\tau\right\rvert^{m}\\\ \cdot\prod_{a\in\mathcal{X}}\left\lvert
E\left(\frac{x+\mathrm{i}\tau}{a};p\right)\right\rvert^{m_{\mathcal{X}}(a)}\prod_{a\in\mathcal{Y}_{\mathrm{u}}}\prod_{k\in\mathbb{Z}}\left\lvert
E\left(\frac{x+\mathrm{i}\tau}{\overline{a}+2k\mathrm{i}\tau};p\right)\right\rvert^{m_{\mathcal{Y}}(a)}.$

After a change of index, the latter becomes

$\left\lvert
g(x+\mathrm{i}\tau)\right\rvert=r\exp\left(\operatorname{Re}\sum_{\ell=1}^{q}\left(a_{\ell}+\mathrm{i}b_{\ell}^{\prime}\right)(x+\mathrm{i}\tau)^{\ell}\right)\left\lvert
x+\mathrm{i}\tau\right\rvert^{m}\\\ \cdot\prod_{a\in\mathcal{X}}\left\lvert
E\left(\frac{x+\mathrm{i}\tau}{a};p\right)\right\rvert^{m_{\mathcal{X}}(a)}\prod_{a\in\mathcal{Y}_{\mathrm{u}}}\prod_{k\in\mathbb{Z}}\left\lvert
E\left(\frac{x+\mathrm{i}\tau}{\overline{a}+2(k-1)\mathrm{i}\tau};p\right)\right\rvert^{m_{\mathcal{Y}}(a)}.$

We can now compute

$\displaystyle 1$ $\displaystyle=\frac{\left\lvert
f(x+\mathrm{i}\tau)\right\rvert}{\left\lvert g(x+\mathrm{i}\tau)\right\rvert}$
$\displaystyle=\exp\left(\sum_{\ell=1}^{q}(b_{\ell}^{\prime}-b_{\ell})\operatorname{Im}\left[(x+\mathrm{i}\tau)^{\ell}\right]\right)\prod_{a\in\mathcal{Y}_{\mathrm{u}}}\prod_{k\in\mathbb{Z}}\left\lvert\frac{E\left(\frac{x+\mathrm{i}\tau}{a+2k\mathrm{i}\tau};p\right)}{E\left(\frac{x+\mathrm{i}\tau}{\overline{a}-2(k-1)\mathrm{i}\tau};p\right)}\right\rvert^{m_{\mathcal{Y}}(a)}$
$\displaystyle=\begin{multlined}\exp\left(\sum_{\ell=1}^{q}(b_{\ell}^{\prime}-b_{\ell})\operatorname{Im}\left[(x+\mathrm{i}\tau)^{\ell}\right]\right)\prod_{a\in\mathcal{Y}_{\mathrm{u}}}\prod_{k\in\mathbb{Z}}\left\lvert\frac{1-\frac{x+\mathrm{i}\tau}{a+2k\mathrm{i}\tau}}{1-\frac{x+\mathrm{i}\tau}{\overline{a}-2(k-1)\mathrm{i}\tau}}\right\rvert^{m_{\mathcal{Y}}(a)}\\\
\cdot\exp\left(m_{\mathcal{Y}}(a)\sum_{\ell=1}^{p}\frac{1}{\ell}\operatorname{Re}\left[\left(\frac{x+\mathrm{i}\tau}{a+2k\mathrm{i}\tau}\right)^{\ell}-\left(\frac{x+\mathrm{i}\tau}{\overline{a}-2(k-1)\mathrm{i}\tau}\right)^{\ell}\right]\right)\end{multlined}\exp\left(\sum_{\ell=1}^{q}(b_{\ell}^{\prime}-b_{\ell})\operatorname{Im}\left[(x+\mathrm{i}\tau)^{\ell}\right]\right)\prod_{a\in\mathcal{Y}_{\mathrm{u}}}\prod_{k\in\mathbb{Z}}\left\lvert\frac{1-\frac{x+\mathrm{i}\tau}{a+2k\mathrm{i}\tau}}{1-\frac{x+\mathrm{i}\tau}{\overline{a}-2(k-1)\mathrm{i}\tau}}\right\rvert^{m_{\mathcal{Y}}(a)}\\\
\cdot\exp\left(m_{\mathcal{Y}}(a)\sum_{\ell=1}^{p}\frac{1}{\ell}\operatorname{Re}\left[\left(\frac{x+\mathrm{i}\tau}{a+2k\mathrm{i}\tau}\right)^{\ell}-\left(\frac{x+\mathrm{i}\tau}{\overline{a}-2(k-1)\mathrm{i}\tau}\right)^{\ell}\right]\right)$
$\displaystyle=\begin{multlined}\exp\left(\sum_{\ell=1}^{q}(b_{\ell}^{\prime}-b_{\ell})\operatorname{Im}\left[(x+\mathrm{i}\tau)^{\ell}\right]\right)\prod_{a\in\mathcal{Y}_{\mathrm{u}}}\prod_{k\in\mathbb{Z}}\left\lvert\frac{a+2(k-1)\mathrm{i}\tau}{a+2k\mathrm{i}\tau}\right\rvert^{m_{\mathcal{Y}}(a)}\\\
\cdot\exp\left(m_{\mathcal{Y}}(a)\sum_{\ell=1}^{p}\frac{1}{\ell}\operatorname{Re}\left[\left(\frac{x+\mathrm{i}\tau}{a+2k\mathrm{i}\tau}\right)^{\ell}-\left(\frac{x+\mathrm{i}\tau}{\overline{a}-2(k-1)\mathrm{i}\tau}\right)^{\ell}\right]\right).\end{multlined}\exp\left(\sum_{\ell=1}^{q}(b_{\ell}^{\prime}-b_{\ell})\operatorname{Im}\left[(x+\mathrm{i}\tau)^{\ell}\right]\right)\prod_{a\in\mathcal{Y}_{\mathrm{u}}}\prod_{k\in\mathbb{Z}}\left\lvert\frac{a+2(k-1)\mathrm{i}\tau}{a+2k\mathrm{i}\tau}\right\rvert^{m_{\mathcal{Y}}(a)}\\\
\cdot\exp\left(m_{\mathcal{Y}}(a)\sum_{\ell=1}^{p}\frac{1}{\ell}\operatorname{Re}\left[\left(\frac{x+\mathrm{i}\tau}{a+2k\mathrm{i}\tau}\right)^{\ell}-\left(\frac{x+\mathrm{i}\tau}{\overline{a}-2(k-1)\mathrm{i}\tau}\right)^{\ell}\right]\right).$

Therefore,

$\sum_{\ell=1}^{q}(b_{\ell}-b_{\ell}^{\prime})\operatorname{Im}\left[(x+\mathrm{i}\tau)^{\ell}\right]=\sum_{a\in\mathcal{Y}_{\mathrm{u}}}m_{\mathcal{Y}}(a)\sum_{k\in\mathbb{Z}}\Bigg{(}\log\left\lvert\frac{a+2(k-1)\mathrm{i}\tau}{a+2k\mathrm{i}\tau}\right\rvert\\\
+\sum_{\ell=1}^{p}\frac{1}{\ell}\operatorname{Re}\left[\left(\frac{x+\mathrm{i}\tau}{a+2k\mathrm{i}\tau}\right)^{\ell}-\left(\frac{x+\mathrm{i}\tau}{\overline{a}-2(k-1)\mathrm{i}\tau}\right)^{\ell}\right]\Bigg{)}.$

Finally, we show that item (b) implies item (a). Note that the decompositions
in item (b) satisfy the assumptions under item (b) of the Mc Donald
decomposition theorem. Therefore, it follows that $\left\lvert
f(x)\right\rvert=\left\lvert g(x)\right\rvert$, for $x\in\mathbb{R}$. We can
now follow the computations in the first part of this proof to find that for
$x\in\mathbb{R}$,

$\frac{\left\lvert f(x+\mathrm{i}\tau)\right\rvert}{\left\lvert
g(x+\mathrm{i}\tau)\right\rvert}=\exp\left(\sum_{\ell=1}^{q}(b_{\ell}^{\prime}-b_{\ell})\operatorname{Im}\left[(x+\mathrm{i}\tau)^{\ell}\right]\right)\\\
\cdot\prod_{a\in\mathcal{Y}_{\mathrm{u}}}\prod_{k\in\mathbb{Z}}\left\lvert\frac{a+2(k-1)\mathrm{i}\tau}{a+2k\mathrm{i}\tau}\right\rvert^{m_{\mathcal{Y}}(a)}\\\
\cdot\exp\left(m_{\mathcal{Y}}(a)\sum_{\ell=1}^{p}\frac{1}{\ell}\operatorname{Re}\left[\left(\frac{x+\mathrm{i}\tau}{a+2k\mathrm{i}\tau}\right)^{\ell}-\left(\frac{x+\mathrm{i}\tau}{\overline{a}-2(k-1)\mathrm{i}\tau}\right)^{\ell}\right]\right).$

Therefore, we have

$\log\frac{\left\lvert f(x+\mathrm{i}\tau)\right\rvert}{\left\lvert
g(x+\mathrm{i}\tau)\right\rvert}=\sum_{\ell=1}^{q}(b_{\ell}^{\prime}-b_{\ell})\operatorname{Im}\left[(x+\mathrm{i}\tau)^{\ell}\right]\\\
+\sum_{a\in\mathcal{Y}_{\mathrm{u}}}m_{\mathcal{Y}}(a)\sum_{k\in\mathbb{Z}}\Bigg{(}\log\left\lvert\frac{a+2(k-1)\mathrm{i}\tau}{a+2k\mathrm{i}\tau}\right\rvert\\\
+\sum_{\ell=1}^{p}\frac{1}{\ell}\operatorname{Re}\left[\left(\frac{x+\mathrm{i}\tau}{a+2k\mathrm{i}\tau}\right)^{\ell}-\left(\frac{x+\mathrm{i}\tau}{\overline{a}-2(k-1)\mathrm{i}\tau}\right)^{\ell}\right]\Bigg{)}=0.$

We conclude that $\left\lvert f(x+\mathrm{i}\tau)\right\rvert=\left\lvert
g(x+\mathrm{i}\tau)\right\rvert$. ∎

######  Remark 26.

Note that the theorem above deals with the special case where one of the two
parallel lines is $\mathbb{R}$ and the other is $\mathbb{R}+\mathrm{i}\tau$,
with $\tau\in\mathbb{R}_{\ast}$. Note also that given any two parallel lines
$L_{1},L_{2}\subset\mathbb{C}$, we can find a rigid motion
$\mathfrak{r}:\mathbb{C}\to\mathbb{C}$ such that
$\mathfrak{r}(L_{1})=\mathbb{R}$ and
$\mathfrak{r}(L_{2})=\mathbb{R}+\mathrm{i}\tau$, with
$\tau\in\mathbb{R}_{\ast}$. With the help of this rigid motion, we can provide
results for any two parallel lines in the complex plane.

Finally, as before, it is enough to assume that $\left\lvert
f(x)\right\rvert=\left\lvert g(x)\right\rvert$ as well as $\left\lvert
f(x+\mathrm{i}\tau)\right\rvert=\left\lvert g(x+\mathrm{i}\tau)\right\rvert$,
for $x\in X\subset\mathbb{R}$, where $X$ is any set with accumulation point.

The condition on the numbers $b_{\ell},b_{\ell}^{\prime}\in\mathbb{R}$, for
$\ell=1,\dots,q$, given in equation (6) seems of limited use in this
generality. The main difficulty in simplifying the relation between
$\\{b_{\ell}\\}_{\ell=1}^{q}$ and $\\{b_{\ell}^{\prime}\\}_{\ell=1}^{q}$ lies
in the fact that generally the sums involved in the relation are infinite and
therefore a straight-forward comparison of coefficients like the ones
performed in the proofs of Theorems 19 and 21 can only work when all the sums
one needs to reorder actually converge unconditionally. In the following
pages, we will illuminate one special case in which this turns out to be true
and thus the condition on the numbers $b_{\ell},b_{\ell}^{\prime}$ simplifies
significantly.

##### Infinitely many parallel lines

It is remarkable that the functions $f,g\in\mathcal{O}(\mathbb{C})$ given by

$\displaystyle f(z)=\cosh\left(\frac{\pi
z}{2}\right)+\mathrm{i}\sinh\left(\frac{\pi z}{2}\right),$ $\displaystyle
g(z)=\cosh\left(\frac{\pi z}{2}\right)-\mathrm{i}\sinh\left(\frac{\pi
z}{2}\right),$

for $z\in\mathbb{C}$, do have magnitudes which not only agree on $\mathbb{R}$
and $\mathbb{R}+\mathrm{i}$ but which agree on
$\mathbb{R}+\mathrm{i}\mathbb{Z}$. In other words, the magnitudes of $f$ and
$g$ agree on infinitely many parallel lines in the complex plane. To see this,
one can recycle the computation from before and convince oneself (by noting
that the terms involving a cosine must vanish when $n$ is odd and, likewise,
the terms involving sine must vanish when $n$ is even in the third step) that

$\displaystyle\left\lvert f(x+n\mathrm{i})\right\rvert$
$\displaystyle=\left\lvert\cosh\left(\frac{\pi x}{2}+\frac{\pi
n\mathrm{i}}{2}\right)+\mathrm{i}\sinh\left(\frac{\pi x}{2}+\frac{\pi
n\mathrm{i}}{2}\right)\right\rvert$
$\displaystyle=\begin{multlined}\Big{\lvert}\cosh\left(\frac{\pi
x}{2}\right)\left(\cos\left(\frac{\pi n}{2}\right)-\sin\left(\frac{\pi
n}{2}\right)\right)\\\ +\mathrm{i}\sinh\left(\frac{\pi
x}{2}\right)\left(\cos\left(\frac{\pi n}{2}\right)+\sin\left(\frac{\pi
n}{2}\right)\right)\Big{\rvert}\end{multlined}\Big{\lvert}\cosh\left(\frac{\pi
x}{2}\right)\left(\cos\left(\frac{\pi n}{2}\right)-\sin\left(\frac{\pi
n}{2}\right)\right)\\\ +\mathrm{i}\sinh\left(\frac{\pi
x}{2}\right)\left(\cos\left(\frac{\pi n}{2}\right)+\sin\left(\frac{\pi
n}{2}\right)\right)\Big{\rvert}$
$\displaystyle=\begin{multlined}\Big{\lvert}\cosh\left(\frac{\pi
x}{2}\right)\left(\cos\left(\frac{\pi n}{2}\right)+\sin\left(\frac{\pi
n}{2}\right)\right)\\\ -\mathrm{i}\sinh\left(\frac{\pi
x}{2}\right)\left(\cos\left(\frac{\pi n}{2}\right)-\sin\left(\frac{\pi
n}{2}\right)\right)\Big{\rvert}\end{multlined}\Big{\lvert}\cosh\left(\frac{\pi
x}{2}\right)\left(\cos\left(\frac{\pi n}{2}\right)+\sin\left(\frac{\pi
n}{2}\right)\right)\\\ -\mathrm{i}\sinh\left(\frac{\pi
x}{2}\right)\left(\cos\left(\frac{\pi n}{2}\right)-\sin\left(\frac{\pi
n}{2}\right)\right)\Big{\rvert}$
$\displaystyle=\left\lvert\cosh\left(\frac{\pi x}{2}+\frac{\pi
n\mathrm{i}}{2}\right)-\mathrm{i}\sinh\left(\frac{\pi x}{2}+\frac{\pi
n\mathrm{i}}{2}\right)\right\rvert=\left\lvert g(x+n\mathrm{i})\right\rvert,$

for $x\in\mathbb{R}$ and $n\in\mathbb{Z}$. This suggests that Theorem 25 might
be adaptable to the case in which we consider infinitely many parallel lines.
This hunch is further supported by the fact that additional equispaced
parallel lines will lead to symmetries with respect to translations which are
already implied by the symmetry stemming from the equations on $\mathbb{R}$
and $\mathbb{R}+\mathrm{i}\tau$. In other words, the magnitude of an entire
function on infinitely many equidistant parallel lines does not contain more
information on the roots of the function in question than the magnitude on two
parallel lines111Note that these two parallel lines have to be carefully
chosen to be two lines which have smallest distance between each other.. At
least for functions of order one, we can show that Theorem 25 is directly
adaptable to the case in which infinitely many lines are considered:

######  Theorem 27 (Cf. Theorem 5).

Let $\tau\in\mathbb{R}_{\ast}$ and let $f,g\in\mathcal{O}(\mathbb{C})$ be non-
zero and of order one. Then, the following are equivalent:

(a)

For all $x\in\mathbb{R}$ and $n\in\mathbb{Z}$, $\left\lvert
f(x+n\mathrm{i}\tau)\right\rvert=\left\lvert
g(x+n\mathrm{i}\tau)\right\rvert$.

(b)

There exist $r>0$, $\phi,\psi\in\mathbb{R}$,
$a^{\prime},b,b^{\prime}\in\mathbb{R}$ as well as $m\in\mathbb{N}_{0}$ such
that

$b=b^{\prime}+\frac{1}{\tau}\sum_{a\in\mathcal{Y}_{\mathrm{u}}}m_{\mathcal{Y}}(a)\sum_{k\in\mathbb{Z}}\left(\log\left\lvert\frac{a+2(k-1)\mathrm{i}\tau}{a+2k\mathrm{i}\tau}\right\rvert-2\tau\operatorname{Im}\frac{1}{a+2k\mathrm{i}\tau}\right)$

as well as

$\displaystyle\begin{multlined}f(z)=r\mathrm{e}^{\mathrm{i}\phi}\mathrm{e}^{(a^{\prime}+\mathrm{i}b)z}z^{m}\prod_{a\in\mathcal{X}}E\left(\frac{z}{a};1\right)^{m_{\mathcal{X}}(a)}\\\
\cdot\prod_{a\in\mathcal{Y}_{\mathrm{u}}}\prod_{k\in\mathbb{Z}}E\left(\frac{z}{a+2k\mathrm{i}\tau};1\right)^{m_{\mathcal{Y}}(a)},\end{multlined}f(z)=r\mathrm{e}^{\mathrm{i}\phi}\mathrm{e}^{(a^{\prime}+\mathrm{i}b)z}z^{m}\prod_{a\in\mathcal{X}}E\left(\frac{z}{a};1\right)^{m_{\mathcal{X}}(a)}\\\
\cdot\prod_{a\in\mathcal{Y}_{\mathrm{u}}}\prod_{k\in\mathbb{Z}}E\left(\frac{z}{a+2k\mathrm{i}\tau};1\right)^{m_{\mathcal{Y}}(a)},$
$\displaystyle\begin{multlined}g(z)=r\mathrm{e}^{\mathrm{i}\psi}\mathrm{e}^{(a^{\prime}+\mathrm{i}b^{\prime})z}z^{m}\prod_{a\in\mathcal{X}}E\left(\frac{z}{a};1\right)^{m_{\mathcal{X}}(a)}\\\
\cdot\prod_{a\in\mathcal{Y}_{\mathrm{u}}}\prod_{k\in\mathbb{Z}}E\left(\frac{z}{\overline{a}+2k\mathrm{i}\tau};1\right)^{m_{\mathcal{Y}}(a)},\end{multlined}g(z)=r\mathrm{e}^{\mathrm{i}\psi}\mathrm{e}^{(a^{\prime}+\mathrm{i}b^{\prime})z}z^{m}\prod_{a\in\mathcal{X}}E\left(\frac{z}{a};1\right)^{m_{\mathcal{X}}(a)}\\\
\cdot\prod_{a\in\mathcal{Y}_{\mathrm{u}}}\prod_{k\in\mathbb{Z}}E\left(\frac{z}{\overline{a}+2k\mathrm{i}\tau};1\right)^{m_{\mathcal{Y}}(a)},$

for $z\in\mathbb{C}$.

###### Proof.

Let us start by collecting three facts about the convergence of certain series
along with their proofs.

_Fact i_. Let $\ell\in\mathbb{Z}$ be arbitrary but fixed. Then,

$\sum_{a\in\mathcal{Y}_{\mathrm{u}}}m_{\mathcal{Y}}(a)\sum_{k\in\mathbb{Z}}\left\lvert\log\left\lvert\frac{a+2(k-\ell-1)\mathrm{i}\tau}{a+2(k-\ell)\mathrm{i}\tau}\right\rvert-2\tau\operatorname{Im}\frac{1}{a+2(k-\ell)\mathrm{i}\tau}\right\rvert<\infty.$

_Proof of Fact i_. Let us start by noting that

$\sum_{a\in\mathcal{Y}_{\mathrm{u}}}m_{\mathcal{Y}}(a)\sum_{k\in\mathbb{Z}}\left\lvert\log\left\lvert\frac{a+2(k-\ell-1)\mathrm{i}\tau}{a+2(k-\ell)\mathrm{i}\tau}\right\rvert-2\tau\operatorname{Im}\frac{1}{a+2(k-\ell)\mathrm{i}\tau}\right\rvert\\\
=\sum_{a\in\mathcal{Y}}m_{\mathcal{Y}}(a)\left\lvert\log\left\lvert\frac{a-2\mathrm{i}\tau}{a}\right\rvert-2\tau\operatorname{Im}\frac{1}{a}\right\rvert\\\
=\sum_{a\in\mathcal{Y}}m_{\mathcal{Y}}(a)\left\lvert\log\left\lvert
1-\frac{2\mathrm{i}\tau}{a}\right\rvert+\operatorname{Re}\frac{2\mathrm{i}\tau}{a}\right\rvert\\\
=\sum_{a\in\mathcal{Y}}m_{\mathcal{Y}}(a)\left\lvert\operatorname{Re}\left[\log\left(1-\frac{2\mathrm{i}\tau}{a}\right)+\frac{2\mathrm{i}\tau}{a}\right]\right\rvert\\\
\leq\sum_{a\in\mathcal{Y}}m_{\mathcal{Y}}(a)\left\lvert\log\left(1-\frac{2\mathrm{i}\tau}{a}\right)+\frac{2\mathrm{i}\tau}{a}\right\rvert.$

Since $f$ is an entire function, it follows that
$\mathcal{Y}\subset\mathcal{R}(f)$ is a set of isolated points. Therefore, we
have

$\sum_{\begin{subarray}{c}a\in\mathcal{Y}\\\ \left\lvert a\right\rvert\leq
4\left\lvert\tau\right\rvert\end{subarray}}m_{\mathcal{Y}}(a)\left\lvert\log\left(1-\frac{2\mathrm{i}\tau}{a}\right)+\frac{2\mathrm{i}\tau}{a}\right\rvert<\infty$

due to the sum being taken over a finite set. We may now use the Mercator
series of the logarithm and the fact that $f$ is of order one (see Lemma 8) to
see that

$\displaystyle\sum_{\begin{subarray}{c}a\in\mathcal{Y}\\\ \left\lvert
a\right\rvert>4\left\lvert\tau\right\rvert\end{subarray}}m_{\mathcal{Y}}(a)\left\lvert\log\left(1-\frac{2\mathrm{i}\tau}{a}\right)+\frac{2\mathrm{i}\tau}{a}\right\rvert$
$\displaystyle=\sum_{\begin{subarray}{c}a\in\mathcal{Y}\\\ \left\lvert
a\right\rvert>4\left\lvert\tau\right\rvert\end{subarray}}m_{\mathcal{Y}}(a)\left\lvert\sum_{j=2}^{\infty}\frac{1}{j}\left(\frac{2\mathrm{i}\tau}{a}\right)^{j}\right\rvert$
$\displaystyle\leq\frac{1}{2}\sum_{\begin{subarray}{c}a\in\mathcal{Y}\\\
\left\lvert
a\right\rvert>4\left\lvert\tau\right\rvert\end{subarray}}m_{\mathcal{Y}}(a)\sum_{j=2}^{\infty}\left\lvert\frac{2\mathrm{i}\tau}{a}\right\rvert^{j}$
$\displaystyle=\frac{1}{2}\sum_{\begin{subarray}{c}a\in\mathcal{Y}\\\
\left\lvert
a\right\rvert>4\left\lvert\tau\right\rvert\end{subarray}}m_{\mathcal{Y}}(a)\left\lvert\frac{2\mathrm{i}\tau}{a}\right\rvert^{2}\sum_{j=0}^{\infty}\left\lvert\frac{2\mathrm{i}\tau}{a}\right\rvert^{j}$
$\displaystyle\leq 2\tau^{2}\sum_{\begin{subarray}{c}a\in\mathcal{Y}\\\
\left\lvert
a\right\rvert>4\left\lvert\tau\right\rvert\end{subarray}}\frac{m_{\mathcal{Y}}(a)}{\left\lvert
a\right\rvert^{2}}\sum_{j=0}^{\infty}2^{-j}$
$\displaystyle=4\tau^{2}\sum_{\begin{subarray}{c}a\in\mathcal{Y}\\\
\left\lvert
a\right\rvert>4\left\lvert\tau\right\rvert\end{subarray}}\frac{m_{\mathcal{Y}}(a)}{\left\lvert
a\right\rvert^{2}}<\infty.$

_Fact ii._ Let $\ell^{\prime},\ell\in\mathbb{N}_{0}$, such that
$\ell^{\prime}\leq\ell$, be arbitrary but fixed. Then,

$\sum_{a\in\mathcal{Y}_{\mathrm{u}}}m_{\mathcal{Y}}(a)\sum_{k\in\mathbb{Z}}\left\lvert\frac{1}{a+2(k-\ell^{\prime})\mathrm{i}\tau}-\frac{1}{a+2(k-\ell)\mathrm{i}\tau}\right\rvert<\infty.$

_Proof of Fact ii_. We can estimate

$\sum_{a\in\mathcal{Y}_{\mathrm{u}}}m_{\mathcal{Y}}(a)\sum_{k\in\mathbb{Z}}\left\lvert\frac{1}{a+2(k-\ell^{\prime})\mathrm{i}\tau}-\frac{1}{a+2(k-\ell)\mathrm{i}\tau}\right\rvert\\\
=\sum_{a\in\mathcal{Y}}m_{\mathcal{Y}}(a)\left\lvert\frac{1}{a-2\ell^{\prime}\mathrm{i}\tau}-\frac{1}{a-2\ell\mathrm{i}\tau}\right\rvert\\\
=\sum_{a\in\mathcal{Y}}m_{\mathcal{Y}}(a)\left\lvert\frac{2(\ell^{\prime}-\ell)\mathrm{i}\tau}{(a-2\ell^{\prime}\mathrm{i}\tau)(a-2\ell\mathrm{i}\tau)}\right\rvert\\\
=2(\ell-\ell^{\prime})\left\lvert\tau\right\rvert\sum_{a\in\mathcal{Y}}\frac{m_{\mathcal{Y}}(a)}{\left\lvert
a-2\ell^{\prime}\mathrm{i}\tau\right\rvert\left\lvert
a-2\ell\mathrm{i}\tau\right\rvert}.$ (7)

We might again see that

$\sum_{\begin{subarray}{c}a\in\mathcal{Y}\\\ \left\lvert a\right\rvert\leq
4\ell\left\lvert\tau\right\rvert\end{subarray}}\frac{m_{\mathcal{Y}}(a)}{\left\lvert
a-2\ell^{\prime}\mathrm{i}\tau\right\rvert\left\lvert
a-2\ell\mathrm{i}\tau\right\rvert}<\infty$

since $f$ being entire implies that its roots are isolated and thus the above
sum is taken over a finite set. Additionally, we have (since $f$ is of order
one)

$\displaystyle\sum_{\begin{subarray}{c}a\in\mathcal{Y}\\\ \left\lvert
a\right\rvert>4\ell\left\lvert\tau\right\rvert\end{subarray}}\frac{m_{\mathcal{Y}}(a)}{\left\lvert
a-2\ell^{\prime}\mathrm{i}\tau\right\rvert\left\lvert
a-2\ell\mathrm{i}\tau\right\rvert}$
$\displaystyle\leq\sum_{\begin{subarray}{c}a\in\mathcal{Y}\\\ \left\lvert
a\right\rvert>4\ell\left\lvert\tau\right\rvert\end{subarray}}\frac{m_{\mathcal{Y}}(a)}{(\left\lvert
a\right\rvert-2\ell^{\prime}\left\lvert\tau\right\rvert)(\left\lvert
a\right\rvert-2\ell\left\lvert\tau\right\rvert)}$ $\displaystyle\leq
4\sum_{\begin{subarray}{c}a\in\mathcal{Y}\\\ \left\lvert
a\right\rvert>4\ell\left\lvert\tau\right\rvert\end{subarray}}\frac{m_{\mathcal{Y}}(a)}{\left\lvert
a\right\rvert^{2}}<\infty.$

_Fact iii_. Let $\ell^{\prime},\ell\in\mathbb{N}_{0}$, such that
$\ell^{\prime}\leq\ell$, be arbitrary but fixed. Then,

$\sum_{a\in\mathcal{Y}_{\mathrm{u}}}m_{\mathcal{Y}}(a)\sum_{k\in\mathbb{Z}}\left(\frac{1}{a+2(k-\ell^{\prime})\mathrm{i}\tau}-\frac{1}{a+2(k-\ell)\mathrm{i}\tau}\right)=0.$

_Proof of Fact iii._ Let us start by noting that

$\frac{1}{a+2(k-\ell^{\prime})\mathrm{i}\tau}-\frac{1}{a+2(k-\ell)\mathrm{i}\tau}\\\
=\sum_{j=\ell^{\prime}}^{\ell-1}\left(\frac{1}{a+2(k-j)\mathrm{i}\tau}-\frac{1}{a+2(k-j-1)\mathrm{i}\tau}\right),$

for $a\in\mathcal{Y}_{\mathrm{u}}$ and $k\in\mathbb{Z}$. By Fact ii (which
guarantees the unconditional convergence of the sums involved), we may write

$\sum_{a\in\mathcal{Y}_{\mathrm{u}}}m_{\mathcal{Y}}(a)\sum_{k\in\mathbb{Z}}\left(\frac{1}{a+2(k-\ell^{\prime})\mathrm{i}\tau}-\frac{1}{a+2(k-\ell)\mathrm{i}\tau}\right)\\\
=\sum_{j=\ell^{\prime}}^{\ell-1}\sum_{a\in\mathcal{Y}_{\mathrm{u}}}m_{\mathcal{Y}}(a)\sum_{k\in\mathbb{Z}}\left(\frac{1}{a+2(k-j)\mathrm{i}\tau}-\frac{1}{a+2(k-j-1)\mathrm{i}\tau}\right).$

We may now consider

$\sum_{k\in\mathbb{Z}}\left(\frac{1}{a+2(k-j)\mathrm{i}\tau}-\frac{1}{a+2(k-j-1)\mathrm{i}\tau}\right)\\\
=\lim_{N\to\infty}\sum_{k=-N}^{N}\left(\frac{1}{a+2(k-j)\mathrm{i}\tau}-\frac{1}{a+2(k-j-1)\mathrm{i}\tau}\right)\\\
=\lim_{N\to\infty}\left(\frac{1}{a+2(N-j)\mathrm{i}\tau}-\frac{1}{a-2(N+j+1)\mathrm{i}\tau}\right)=0.$

We can now turn to the proof of the Theorem. Let us first show that item (b)
implies item (a). If we assume that item (b) holds, then we know by the Mc
Donald decomposition theorem that $\left\lvert f\right\rvert=\left\lvert
g\right\rvert$ on the real numbers. We may now consider $x\in\mathbb{R}$ as
well as $n\in\mathbb{Z}_{\ast}$ and compute

$\left\lvert
f(x+n\mathrm{i}\tau)\right\rvert=r\mathrm{e}^{\operatorname{Re}\left((a^{\prime}+\mathrm{i}b)(x+n\mathrm{i}\tau)\right)}\left\lvert
x+n\mathrm{i}\tau\right\rvert^{m}\prod_{a\in\mathcal{X}}\left\lvert
E\left(\frac{x+n\mathrm{i}\tau}{a};1\right)\right\rvert^{m_{\mathcal{X}}(a)}\\\
\prod_{a\in\mathcal{Y}_{\mathrm{u}}}\prod_{k\in\mathbb{Z}}\left\lvert
E\left(\frac{x+n\mathrm{i}\tau}{a+2k\mathrm{i}\tau};1\right)\right\rvert^{m_{\mathcal{Y}}(a)}.$

as well as (by multiplying over $n-k$)

$\left\lvert
g(x+n\mathrm{i}\tau)\right\rvert=r\mathrm{e}^{\operatorname{Re}\left((a^{\prime}+\mathrm{i}b^{\prime})(x+n\mathrm{i}\tau)\right)}\left\lvert
x+n\mathrm{i}\tau\right\rvert^{m}\prod_{a\in\mathcal{X}}\left\lvert
E\left(\frac{x+n\mathrm{i}\tau}{a};1\right)\right\rvert^{m_{\mathcal{X}}(a)}\\\
\prod_{a\in\mathcal{Y}_{\mathrm{u}}}\prod_{k\in\mathbb{Z}}\left\lvert
E\left(\frac{x+n\mathrm{i}\tau}{\overline{a}+2(n-k)\mathrm{i}\tau};1\right)\right\rvert^{m_{\mathcal{Y}}(a)}.$

Therefore, we find that

$\displaystyle\frac{\left\lvert f(x+n\mathrm{i}\tau)\right\rvert}{\left\lvert
g(x+n\mathrm{i}\tau)\right\rvert}$
$\displaystyle=\mathrm{e}^{\operatorname{Re}\left(\mathrm{i}(b-b^{\prime})(x+n\mathrm{i}\tau)\right)}\prod_{a\in\mathcal{Y}_{\mathrm{u}}}\prod_{k\in\mathbb{Z}}\left\lvert\frac{E\left(\frac{x+n\mathrm{i}\tau}{a+2k\mathrm{i}\tau};1\right)}{E\left(\frac{x+n\mathrm{i}\tau}{\overline{a}+2(n-k)\mathrm{i}\tau};1\right)}\right\rvert^{m_{\mathcal{Y}}(a)}$
$\displaystyle=\begin{multlined}\mathrm{e}^{(b^{\prime}-b)\operatorname{Im}(x+n\mathrm{i}\tau)}\prod_{a\in\mathcal{Y}_{\mathrm{u}}}\prod_{k\in\mathbb{Z}}\left\lvert\frac{1-\frac{x+n\mathrm{i}\tau}{a+2k\mathrm{i}\tau}}{1-\frac{x+n\mathrm{i}\tau}{\overline{a}+2(n-k)\mathrm{i}\tau}}\right\rvert^{m_{\mathcal{Y}}(a)}\\\
\cdot\exp\left(m_{\mathcal{Y}}(a)\operatorname{Re}\left[\frac{x+n\mathrm{i}\tau}{a+2k\mathrm{i}\tau}-\frac{x+n\mathrm{i}\tau}{\overline{a}+2(n-k)\mathrm{i}\tau}\right]\right)\end{multlined}\mathrm{e}^{(b^{\prime}-b)\operatorname{Im}(x+n\mathrm{i}\tau)}\prod_{a\in\mathcal{Y}_{\mathrm{u}}}\prod_{k\in\mathbb{Z}}\left\lvert\frac{1-\frac{x+n\mathrm{i}\tau}{a+2k\mathrm{i}\tau}}{1-\frac{x+n\mathrm{i}\tau}{\overline{a}+2(n-k)\mathrm{i}\tau}}\right\rvert^{m_{\mathcal{Y}}(a)}\\\
\cdot\exp\left(m_{\mathcal{Y}}(a)\operatorname{Re}\left[\frac{x+n\mathrm{i}\tau}{a+2k\mathrm{i}\tau}-\frac{x+n\mathrm{i}\tau}{\overline{a}+2(n-k)\mathrm{i}\tau}\right]\right)$
$\displaystyle=\begin{multlined}\mathrm{e}^{n\tau(b^{\prime}-b)}\prod_{a\in\mathcal{Y}_{\mathrm{u}}}\prod_{k\in\mathbb{Z}}\left\lvert\frac{a+2(k-n)\mathrm{i}\tau}{a+2k\mathrm{i}\tau}\right\rvert^{m_{\mathcal{Y}}(a)}\\\
\cdot\exp\left(m_{\mathcal{Y}}(a)\operatorname{Re}\left[\frac{x+n\mathrm{i}\tau}{a+2k\mathrm{i}\tau}-\frac{x+n\mathrm{i}\tau}{\overline{a}+2(n-k)\mathrm{i}\tau}\right]\right),\end{multlined}\mathrm{e}^{n\tau(b^{\prime}-b)}\prod_{a\in\mathcal{Y}_{\mathrm{u}}}\prod_{k\in\mathbb{Z}}\left\lvert\frac{a+2(k-n)\mathrm{i}\tau}{a+2k\mathrm{i}\tau}\right\rvert^{m_{\mathcal{Y}}(a)}\\\
\cdot\exp\left(m_{\mathcal{Y}}(a)\operatorname{Re}\left[\frac{x+n\mathrm{i}\tau}{a+2k\mathrm{i}\tau}-\frac{x+n\mathrm{i}\tau}{\overline{a}+2(n-k)\mathrm{i}\tau}\right]\right),$

whenever this fraction is well-defined. From this computation, it is clear
that $\left\lvert f(x+n\mathrm{i}\tau)\right\rvert=\left\lvert
g(x+n\mathrm{i}\tau)\right\rvert$ if

$b-b^{\prime}=\frac{1}{n\tau}\sum_{a\in\mathcal{Y}_{\mathrm{u}}}m_{\mathcal{Y}}(a)\sum_{k\in\mathbb{Z}}\bigg{(}\log\left\lvert\frac{a+2(k-n)\mathrm{i}\tau}{a+2k\mathrm{i}\tau}\right\rvert\\\
+\operatorname{Re}\left[\frac{x+n\mathrm{i}\tau}{a+2k\mathrm{i}\tau}-\frac{x+n\mathrm{i}\tau}{\overline{a}+2(n-k)\mathrm{i}\tau}\right]\bigg{)}.$

It will turn out that this condition on the coefficients
$b,b^{\prime}\in\mathbb{R}$ is equivalent to the condition stated in the
theorem. Showing this is a lengthy technical exercise which must in particular
explain why the condition above is independent of $x\in\mathbb{R}$ and
$n\in\mathbb{Z}_{\ast}$.

Let us consider $n\in\mathbb{N}$ (the case in which $-n\in\mathbb{N}$ can be
dealt with analogously) as well as $x\in\mathbb{R}$. It holds that

$\log\left\lvert\frac{a+2(k-n)\mathrm{i}\tau}{a+2k\mathrm{i}\tau}\right\rvert=\sum_{\ell=1}^{n}\log\left\lvert\frac{a+2(k-\ell)\mathrm{i}\tau}{a+2(k-\ell+1)\mathrm{i}\tau}\right\rvert,$

for $a\in\mathcal{Y}_{\mathrm{u}}$ and $k\in\mathbb{Z}$, as well as

$\operatorname{Re}\left[\frac{x+n\mathrm{i}\tau}{a+2k\mathrm{i}\tau}-\frac{x+n\mathrm{i}\tau}{\overline{a}+2(n-k)\mathrm{i}\tau}\right]=x\operatorname{Re}\left[\frac{1}{a+2k\mathrm{i}\tau}-\frac{1}{a+2(k-n)\mathrm{i}\tau}\right]\\\
-n\tau\operatorname{Im}\left[\frac{1}{a+2k\mathrm{i}\tau}+\frac{1}{a+2(k-n)\mathrm{i}\tau}\right].$

We may now rewrite the coefficients of the sum relating $b$ and $b^{\prime}$:

$\log\left\lvert\frac{a+2(k-n)\mathrm{i}\tau}{a+2k\mathrm{i}\tau}\right\rvert+\operatorname{Re}\left[\frac{x+n\mathrm{i}\tau}{a+2k\mathrm{i}\tau}-\frac{x+n\mathrm{i}\tau}{\overline{a}+2(n-k)\mathrm{i}\tau}\right]\\\
=\sum_{\ell=1}^{n}\log\left\lvert\frac{a+2(k-\ell)\mathrm{i}\tau}{a+2(k-\ell+1)\mathrm{i}\tau}\right\rvert-n\tau\operatorname{Im}\left[\frac{1}{a+2k\mathrm{i}\tau}+\frac{1}{a+2(k-n)\mathrm{i}\tau}\right]\\\
+x\operatorname{Re}\left[\frac{1}{a+2k\mathrm{i}\tau}-\frac{1}{a+2(k-n)\mathrm{i}\tau}\right].$

We can elegantly add zero to the above expression and obtain

$\log\left\lvert\frac{a+2(k-n)\mathrm{i}\tau}{a+2k\mathrm{i}\tau}\right\rvert+\operatorname{Re}\left[\frac{x+n\mathrm{i}\tau}{a+2k\mathrm{i}\tau}-\frac{x+n\mathrm{i}\tau}{\overline{a}+2(n-k)\mathrm{i}\tau}\right]\\\
=\sum_{\ell=1}^{n}\left(\log\left\lvert\frac{a+2(k-\ell)\mathrm{i}\tau}{a+2(k-\ell+1)\mathrm{i}\tau}\right\rvert-2\tau\operatorname{Im}\frac{1}{a+2(k-\ell+1)\mathrm{i}\tau}\right)\\\
-\tau\sum_{\ell=1}^{n}\operatorname{Im}\left[\frac{1}{a+2k\mathrm{i}\tau}-\frac{1}{a+2(k-\ell+1)\mathrm{i}\tau}\right]\\\
+\tau\sum_{\ell=1}^{n}\operatorname{Im}\left[\frac{1}{a+2(k-\ell+1)\mathrm{i}\tau}-\frac{1}{a+2(k-n)\mathrm{i}\tau}\right]\\\
+x\operatorname{Re}\left[\frac{1}{a+2k\mathrm{i}\tau}-\frac{1}{a+2(k-n)\mathrm{i}\tau}\right].$

According to the Facts i and ii, it follows that the sum relating $b$ and
$b^{\prime}$ splits into $3n+1$ parts:

$\frac{1}{n\tau}\sum_{a\in\mathcal{Y}_{\mathrm{u}}}m_{\mathcal{Y}}(a)\sum_{k\in\mathbb{Z}}\bigg{(}\log\left\lvert\frac{a+2(k-n)\mathrm{i}\tau}{a+2k\mathrm{i}\tau}\right\rvert\\\
+\operatorname{Re}\left[\frac{x+n\mathrm{i}\tau}{a+2k\mathrm{i}\tau}-\frac{x+n\mathrm{i}\tau}{\overline{a}+2(n-k)\mathrm{i}\tau}\right]\bigg{)}\\\
=\frac{1}{n\tau}\sum_{\ell=1}^{n}\sum_{a\in\mathcal{Y}_{\mathrm{u}}}m_{\mathcal{Y}}(a)\sum_{k\in\mathbb{Z}}\bigg{(}\log\left\lvert\frac{a+2(k-\ell)\mathrm{i}\tau}{a+2(k-\ell+1)\mathrm{i}\tau}\right\rvert\\\
-2\tau\operatorname{Im}\frac{1}{a+2(k-\ell+1)\mathrm{i}\tau}\bigg{)}\\\
-\frac{1}{n}\sum_{\ell=1}^{n}\sum_{a\in\mathcal{Y}_{\mathrm{u}}}m_{\mathcal{Y}}(a)\sum_{k\in\mathbb{Z}}\operatorname{Im}\left[\frac{1}{a+2k\mathrm{i}\tau}-\frac{1}{a+2(k-\ell+1)\mathrm{i}\tau}\right]\\\
+\frac{1}{n}\sum_{\ell=1}^{n}\sum_{a\in\mathcal{Y}_{\mathrm{u}}}m_{\mathcal{Y}}(a)\sum_{k\in\mathbb{Z}}\operatorname{Im}\left[\frac{1}{a+2(k-\ell+1)\mathrm{i}\tau}-\frac{1}{a+2(k-n)\mathrm{i}\tau}\right]\\\
+\frac{x}{n\tau}\sum_{a\in\mathcal{Y}_{\mathrm{u}}}m_{\mathcal{Y}}(a)\sum_{k\in\mathbb{Z}}\operatorname{Re}\left[\frac{1}{a+2k\mathrm{i}\tau}-\frac{1}{a+2(k-n)\mathrm{i}\tau}\right].$

According to Fact iii, we find that $2n+1$ sums of these $3n+1$ sums vanish.
We conclude that

$\frac{1}{n\tau}\sum_{a\in\mathcal{Y}_{\mathrm{u}}}m_{\mathcal{Y}}(a)\sum_{k\in\mathbb{Z}}\bigg{(}\log\left\lvert\frac{a+2(k-n)\mathrm{i}\tau}{a+2k\mathrm{i}\tau}\right\rvert\\\
+\operatorname{Re}\left[\frac{x+n\mathrm{i}\tau}{a+2k\mathrm{i}\tau}-\frac{x+n\mathrm{i}\tau}{\overline{a}+2(n-k)\mathrm{i}\tau}\right]\bigg{)}\\\
=\frac{1}{n\tau}\sum_{\ell=1}^{n}\sum_{a\in\mathcal{Y}_{\mathrm{u}}}m_{\mathcal{Y}}(a)\sum_{k\in\mathbb{Z}}\bigg{(}\log\left\lvert\frac{a+2(k-\ell)\mathrm{i}\tau}{a+2(k-\ell+1)\mathrm{i}\tau}\right\rvert\\\
-2\tau\operatorname{Im}\frac{1}{a+2(k-\ell+1)\mathrm{i}\tau}\bigg{)}\\\
=\frac{1}{n\tau}\sum_{\ell=1}^{n}\sum_{a\in\mathcal{Y}_{\mathrm{u}}}m_{\mathcal{Y}}(a)\sum_{k\in\mathbb{Z}}\left(\log\left\lvert\frac{a+2(k-1)\mathrm{i}\tau}{a+2k\mathrm{i}\tau}\right\rvert-2\tau\operatorname{Im}\frac{1}{a+2k\mathrm{i}\tau}\right)\\\
=\frac{1}{\tau}\sum_{a\in\mathcal{Y}_{\mathrm{u}}}m_{\mathcal{Y}}(a)\sum_{k\in\mathbb{Z}}\left(\log\left\lvert\frac{a+2(k-1)\mathrm{i}\tau}{a+2k\mathrm{i}\tau}\right\rvert-2\tau\operatorname{Im}\frac{1}{a+2k\mathrm{i}\tau}\right),$

where we changed the index of summation in the second step. It follows (under
the assumptions in item (b)) that $\left\lvert
f(x+n\mathrm{i}\tau)\right\rvert=\left\lvert
g(x+n\mathrm{i}\tau)\right\rvert$.

All that is left to do is to show that item (a) implies item (b). Since we can
recycle some of our considerations from above this is now relatively simple.
Indeed, it follows from Theorem 25 that there must exist $r>0$,
$\phi,\psi\in\mathbb{R}$, $a^{\prime},b,b^{\prime}\in\mathbb{R}$, and
$m\in\mathbb{N}_{0}$ such that

$b-b^{\prime}=\frac{1}{\tau}\sum_{a\in\mathcal{Y}_{\mathrm{u}}}m_{\mathcal{Y}}(a)\sum_{k\in\mathbb{Z}}\bigg{(}\log\left\lvert\frac{a+2(k-1)\mathrm{i}\tau}{a+2k\mathrm{i}\tau}\right\rvert\\\
+\operatorname{Re}\left[\frac{x+\mathrm{i}\tau}{a+2k\mathrm{i}\tau}-\frac{x+\mathrm{i}\tau}{\overline{a}-2(k-1)\mathrm{i}\tau}\right]\bigg{)},$

for all $x\in\mathbb{R}$, as well as

$\displaystyle
f(z)=r\mathrm{e}^{\mathrm{i}\phi}\mathrm{e}^{(a^{\prime}+\mathrm{i}b)z}z^{m}\prod_{a\in\mathcal{X}}E\left(\frac{z}{a};1\right)^{m_{\mathcal{X}}(a)}\prod_{a\in\mathcal{Y}_{\mathrm{u}}}\prod_{k\in\mathbb{Z}}E\left(\frac{z}{a+2k\mathrm{i}\tau};1\right)^{m_{\mathcal{Y}}(a)},$
$\displaystyle
g(z)=r\mathrm{e}^{\mathrm{i}\psi}\mathrm{e}^{(a^{\prime}+\mathrm{i}b^{\prime})z}z^{m}\prod_{a\in\mathcal{X}}E\left(\frac{z}{a};1\right)^{m_{\mathcal{X}}(a)}\prod_{a\in\mathcal{Y}_{\mathrm{u}}}\prod_{k\in\mathbb{Z}}E\left(\frac{z}{\overline{a}+2k\mathrm{i}\tau};1\right)^{m_{\mathcal{Y}}(a)},$

for $z\in\mathbb{C}$. We have just shown that (when we showed that item (b)
implies item (a))

$\frac{1}{\tau}\sum_{a\in\mathcal{Y}_{\mathrm{u}}}m_{\mathcal{Y}}(a)\sum_{k\in\mathbb{Z}}\left(\log\left\lvert\frac{a+2(k-1)\mathrm{i}\tau}{a+2k\mathrm{i}\tau}\right\rvert+\operatorname{Re}\left[\frac{x+\mathrm{i}\tau}{a+2k\mathrm{i}\tau}-\frac{x+\mathrm{i}\tau}{\overline{a}-2(k-1)\mathrm{i}\tau}\right]\right)\\\
=\frac{1}{\tau}\sum_{a\in\mathcal{Y}_{\mathrm{u}}}m_{\mathcal{Y}}(a)\sum_{k\in\mathbb{Z}}\left(\log\left\lvert\frac{a+2(k-1)\mathrm{i}\tau}{a+2k\mathrm{i}\tau}\right\rvert-2\tau\operatorname{Im}\frac{1}{a+2k\mathrm{i}\tau}\right).$

Therefore, item (b) follows. ∎

We want to finish this section by noting that we can also prove a result for
entire functions of finite order $\rho>1$.

######  Lemma 28.

Let $\tau\in\mathbb{R}_{\ast}$ and let $f,g\in\mathcal{O}(\mathbb{C})$ be non-
zero and of finite order. Then, the following are equivalent:

(a)

For all $x\in\mathbb{R}$ and $n\in\mathbb{Z}$, $\left\lvert
f(x+n\mathrm{i}\tau)\right\rvert=\left\lvert
g(x+n\mathrm{i}\tau)\right\rvert$.

(b)

There exist $r>0$, $\phi,\psi\in\mathbb{R}$, $q\in\mathbb{N}_{0}$,
$a_{\ell},b_{\ell},b^{\prime}_{\ell}\in\mathbb{R}$, for
$\ell\in\\{1,\dots,q\\}$, $m\in\mathbb{N}_{0}$ and $p\in\mathbb{N}_{0}$ such
that

$\sum_{\ell=1}^{q}(b_{\ell}-b_{\ell}^{\prime})\operatorname{Im}\left[(x+n\mathrm{i}\tau)^{\ell}\right]=\sum_{a\in\mathcal{Y}_{\mathrm{u}}}m_{\mathcal{Y}}(a)\sum_{k\in\mathbb{Z}}\Bigg{(}\log\left\lvert\frac{a+2(k-n)\mathrm{i}\tau}{a+2k\mathrm{i}\tau}\right\rvert\\\
+\sum_{\ell=1}^{p}\frac{1}{\ell}\operatorname{Re}\left[\left(\frac{x+n\mathrm{i}\tau}{a+2k\mathrm{i}\tau}\right)^{\ell}-\left(\frac{x+n\mathrm{i}\tau}{\overline{a}+2(n-k)\mathrm{i}\tau}\right)^{\ell}\right]\Bigg{)},$

for all $x\in\mathbb{R}$ and all $n\in\mathbb{Z}$, as well as

$\displaystyle\begin{multlined}f(z)=r\mathrm{e}^{\mathrm{i}\phi}\exp\left(\sum_{\ell=1}^{q}\left(a_{\ell}+\mathrm{i}b_{\ell}\right)z^{\ell}\right)z^{m}\prod_{a\in\mathcal{X}}E\left(\frac{z}{a};p\right)^{m_{\mathcal{X}}(a)}\\\
\cdot\prod_{a\in\mathcal{Y}_{\mathrm{u}}}\prod_{k\in\mathbb{Z}}E\left(\frac{z}{a+2k\mathrm{i}\tau};p\right)^{m_{\mathcal{Y}}(a)},\end{multlined}f(z)=r\mathrm{e}^{\mathrm{i}\phi}\exp\left(\sum_{\ell=1}^{q}\left(a_{\ell}+\mathrm{i}b_{\ell}\right)z^{\ell}\right)z^{m}\prod_{a\in\mathcal{X}}E\left(\frac{z}{a};p\right)^{m_{\mathcal{X}}(a)}\\\
\cdot\prod_{a\in\mathcal{Y}_{\mathrm{u}}}\prod_{k\in\mathbb{Z}}E\left(\frac{z}{a+2k\mathrm{i}\tau};p\right)^{m_{\mathcal{Y}}(a)},$
$\displaystyle\begin{multlined}g(z)=r\mathrm{e}^{\mathrm{i}\psi}\exp\left(\sum_{\ell=1}^{q}\left(a_{\ell}+\mathrm{i}b_{\ell}^{\prime}\right)z^{\ell}\right)z^{m}\prod_{a\in\mathcal{X}}E\left(\frac{z}{a};p\right)^{m_{\mathcal{X}}(a)}\\\
\cdot\prod_{a\in\mathcal{Y}_{\mathrm{u}}}\prod_{k\in\mathbb{Z}}E\left(\frac{z}{\overline{a}+2k\mathrm{i}\tau};p\right)^{m_{\mathcal{Y}}(a)},\end{multlined}g(z)=r\mathrm{e}^{\mathrm{i}\psi}\exp\left(\sum_{\ell=1}^{q}\left(a_{\ell}+\mathrm{i}b_{\ell}^{\prime}\right)z^{\ell}\right)z^{m}\prod_{a\in\mathcal{X}}E\left(\frac{z}{a};p\right)^{m_{\mathcal{X}}(a)}\\\
\cdot\prod_{a\in\mathcal{Y}_{\mathrm{u}}}\prod_{k\in\mathbb{Z}}E\left(\frac{z}{\overline{a}+2k\mathrm{i}\tau};p\right)^{m_{\mathcal{Y}}(a)},$

for $z\in\mathbb{C}$.

###### Proof.

We will only show that item (a) implies item (b). The other direction is much
simpler and its proof is overwhelmingly similar to what has already been
presented in the proof of Theorem 25. Let us assume that item (a) holds. Then,
it follows from Theorem 25 that there must exist $r>0$,
$\phi,\psi\in\mathbb{R}$, $q\in\mathbb{N}_{0}$,
$a_{\ell},b_{\ell},b^{\prime}_{\ell}\in\mathbb{R}$, for
$\ell\in\\{1,\dots,q\\}$, $m\in\mathbb{N}_{0}$ and $p\in\mathbb{N}_{0}$ such
that

$\displaystyle\begin{multlined}f(z)=r\mathrm{e}^{\mathrm{i}\phi}\exp\left(\sum_{\ell=1}^{q}\left(a_{\ell}+\mathrm{i}b_{\ell}\right)z^{\ell}\right)z^{m}\prod_{a\in\mathcal{X}}E\left(\frac{z}{a};p\right)^{m_{\mathcal{X}}(a)}\\\
\cdot\prod_{a\in\mathcal{Y}_{\mathrm{u}}}\prod_{k\in\mathbb{Z}}E\left(\frac{z}{a+2k\mathrm{i}\tau};p\right)^{m_{\mathcal{Y}}(a)},\end{multlined}f(z)=r\mathrm{e}^{\mathrm{i}\phi}\exp\left(\sum_{\ell=1}^{q}\left(a_{\ell}+\mathrm{i}b_{\ell}\right)z^{\ell}\right)z^{m}\prod_{a\in\mathcal{X}}E\left(\frac{z}{a};p\right)^{m_{\mathcal{X}}(a)}\\\
\cdot\prod_{a\in\mathcal{Y}_{\mathrm{u}}}\prod_{k\in\mathbb{Z}}E\left(\frac{z}{a+2k\mathrm{i}\tau};p\right)^{m_{\mathcal{Y}}(a)},$
$\displaystyle\begin{multlined}g(z)=r\mathrm{e}^{\mathrm{i}\psi}\exp\left(\sum_{\ell=1}^{q}\left(a_{\ell}+\mathrm{i}b_{\ell}^{\prime}\right)z^{\ell}\right)z^{m}\prod_{a\in\mathcal{X}}E\left(\frac{z}{a};p\right)^{m_{\mathcal{X}}(a)}\\\
\cdot\prod_{a\in\mathcal{Y}_{\mathrm{u}}}\prod_{k\in\mathbb{Z}}E\left(\frac{z}{\overline{a}+2k\mathrm{i}\tau};p\right)^{m_{\mathcal{Y}}(a)},\end{multlined}g(z)=r\mathrm{e}^{\mathrm{i}\psi}\exp\left(\sum_{\ell=1}^{q}\left(a_{\ell}+\mathrm{i}b_{\ell}^{\prime}\right)z^{\ell}\right)z^{m}\prod_{a\in\mathcal{X}}E\left(\frac{z}{a};p\right)^{m_{\mathcal{X}}(a)}\\\
\cdot\prod_{a\in\mathcal{Y}_{\mathrm{u}}}\prod_{k\in\mathbb{Z}}E\left(\frac{z}{\overline{a}+2k\mathrm{i}\tau};p\right)^{m_{\mathcal{Y}}(a)},$

for $z\in\mathbb{C}$. We may now compute

$\left\lvert
f(x+n\mathrm{i}\tau)\right\rvert=r\exp\left(\operatorname{Re}\sum_{\ell=1}^{q}\left(a_{\ell}+\mathrm{i}b_{\ell}\right)(x+n\mathrm{i}\tau)^{\ell}\right)\left\lvert
x+n\mathrm{i}\tau\right\rvert^{m}\\\ \cdot\prod_{a\in\mathcal{X}}\left\lvert
E\left(\frac{x+n\mathrm{i}\tau}{a};p\right)\right\rvert^{m_{\mathcal{X}}(a)}\prod_{a\in\mathcal{Y}_{\mathrm{u}}}\prod_{k\in\mathbb{Z}}\left\lvert
E\left(\frac{x+n\mathrm{i}\tau}{a+2k\mathrm{i}\tau};p\right)\right\rvert^{m_{\mathcal{Y}}(a)}.$

By multiplying over $n-k$ instead of $k$ in the rightmost product of $g$, one
can obtain

$\left\lvert
g(x+n\mathrm{i}\tau)\right\rvert=r\exp\left(\operatorname{Re}\sum_{\ell=1}^{q}\left(a_{\ell}+\mathrm{i}b_{\ell}^{\prime}\right)(x+n\mathrm{i}\tau)^{\ell}\right)\left\lvert
x+n\mathrm{i}\tau\right\rvert^{m}\\\ \cdot\prod_{a\in\mathcal{X}}\left\lvert
E\left(\frac{x+n\mathrm{i}\tau}{a};p\right)\right\rvert^{m_{\mathcal{X}}(a)}\prod_{a\in\mathcal{Y}_{\mathrm{u}}}\prod_{k\in\mathbb{Z}}\left\lvert
E\left(\frac{x+n\mathrm{i}\tau}{\overline{a}+2(n-k)\mathrm{i}\tau};p\right)\right\rvert^{m_{\mathcal{Y}}(a)}.$

We may now compute the fraction (whenever it is well-defined)

$\frac{\left\lvert f(x+n\mathrm{i}\tau)\right\rvert}{\left\lvert
g(x+n\mathrm{i}\tau)\right\rvert}\\\
=\exp\left(\sum_{\ell=1}^{q}(b_{\ell}^{\prime}-b_{\ell})\operatorname{Im}\left[(x+n\mathrm{i}\tau)^{\ell}\right]\right)\prod_{a\in\mathcal{Y}_{\mathrm{u}}}\prod_{k\in\mathbb{Z}}\left\lvert\frac{E\left(\frac{x+n\mathrm{i}\tau}{a+2k\mathrm{i}\tau};p\right)}{E\left(\frac{x+n\mathrm{i}\tau}{\overline{a}+2(n-k)\mathrm{i}\tau};p\right)}\right\rvert^{m_{\mathcal{Y}}(a)}\\\
=\begin{multlined}\exp\left(\sum_{\ell=1}^{q}(b_{\ell}^{\prime}-b_{\ell})\operatorname{Im}\left[(x+n\mathrm{i}\tau)^{\ell}\right]\right)\prod_{a\in\mathcal{Y}_{\mathrm{u}}}\prod_{k\in\mathbb{Z}}\left\lvert\frac{1-\frac{x+n\mathrm{i}\tau}{a+2k\mathrm{i}\tau}}{1-\frac{x+n\mathrm{i}\tau}{\overline{a}+2(n-k)\mathrm{i}\tau}}\right\rvert^{m_{\mathcal{Y}}(a)}\\\
\cdot\exp\left(m_{\mathcal{Y}}(a)\sum_{\ell=1}^{p}\frac{1}{\ell}\operatorname{Re}\left[\left(\frac{x+n\mathrm{i}\tau}{a+2k\mathrm{i}\tau}\right)^{\ell}-\left(\frac{x+n\mathrm{i}\tau}{\overline{a}+2(n-k)\mathrm{i}\tau}\right)^{\ell}\right]\right)\end{multlined}\exp\left(\sum_{\ell=1}^{q}(b_{\ell}^{\prime}-b_{\ell})\operatorname{Im}\left[(x+n\mathrm{i}\tau)^{\ell}\right]\right)\prod_{a\in\mathcal{Y}_{\mathrm{u}}}\prod_{k\in\mathbb{Z}}\left\lvert\frac{1-\frac{x+n\mathrm{i}\tau}{a+2k\mathrm{i}\tau}}{1-\frac{x+n\mathrm{i}\tau}{\overline{a}+2(n-k)\mathrm{i}\tau}}\right\rvert^{m_{\mathcal{Y}}(a)}\\\
\cdot\exp\left(m_{\mathcal{Y}}(a)\sum_{\ell=1}^{p}\frac{1}{\ell}\operatorname{Re}\left[\left(\frac{x+n\mathrm{i}\tau}{a+2k\mathrm{i}\tau}\right)^{\ell}-\left(\frac{x+n\mathrm{i}\tau}{\overline{a}+2(n-k)\mathrm{i}\tau}\right)^{\ell}\right]\right)\\\
=\begin{multlined}\exp\left(\sum_{\ell=1}^{q}(b_{\ell}^{\prime}-b_{\ell})\operatorname{Im}\left[(x+n\mathrm{i}\tau)^{\ell}\right]\right)\prod_{a\in\mathcal{Y}_{\mathrm{u}}}\prod_{k\in\mathbb{Z}}\left\lvert\frac{a+2(k-n)\mathrm{i}\tau}{a+2k\mathrm{i}\tau}\right\rvert^{m_{\mathcal{Y}}(a)}\\\
\cdot\exp\left(m_{\mathcal{Y}}(a)\sum_{\ell=1}^{p}\frac{1}{\ell}\operatorname{Re}\left[\left(\frac{x+n\mathrm{i}\tau}{a+2k\mathrm{i}\tau}\right)^{\ell}-\left(\frac{x+n\mathrm{i}\tau}{\overline{a}+2(n-k)\mathrm{i}\tau}\right)^{\ell}\right]\right).\end{multlined}\exp\left(\sum_{\ell=1}^{q}(b_{\ell}^{\prime}-b_{\ell})\operatorname{Im}\left[(x+n\mathrm{i}\tau)^{\ell}\right]\right)\prod_{a\in\mathcal{Y}_{\mathrm{u}}}\prod_{k\in\mathbb{Z}}\left\lvert\frac{a+2(k-n)\mathrm{i}\tau}{a+2k\mathrm{i}\tau}\right\rvert^{m_{\mathcal{Y}}(a)}\\\
\cdot\exp\left(m_{\mathcal{Y}}(a)\sum_{\ell=1}^{p}\frac{1}{\ell}\operatorname{Re}\left[\left(\frac{x+n\mathrm{i}\tau}{a+2k\mathrm{i}\tau}\right)^{\ell}-\left(\frac{x+n\mathrm{i}\tau}{\overline{a}+2(n-k)\mathrm{i}\tau}\right)^{\ell}\right]\right).$

The relation between the coefficients $b_{\ell}$ and $b_{\ell}^{\prime}$ does
follow from considering the logarithm of the above equation. ∎

Unfortunately, the condition on the coefficients
$b_{\ell},b_{\ell}^{\prime}\in\mathbb{R}$, for $\ell\in\\{1,\dots,q\\}$, i.e.
the relation

$\sum_{\ell=1}^{q}(b_{\ell}-b_{\ell}^{\prime})\operatorname{Im}\left[(x+n\mathrm{i}\tau)^{\ell}\right]=\sum_{a\in\mathcal{Y}_{\mathrm{u}}}m_{\mathcal{Y}}(a)\sum_{k\in\mathbb{Z}}\Bigg{(}\log\left\lvert\frac{a+2(k-n)\mathrm{i}\tau}{a+2k\mathrm{i}\tau}\right\rvert\\\
+\sum_{\ell=1}^{p}\frac{1}{\ell}\operatorname{Re}\left[\left(\frac{x+n\mathrm{i}\tau}{a+2k\mathrm{i}\tau}\right)^{\ell}-\left(\frac{x+n\mathrm{i}\tau}{\overline{a}+2(n-k)\mathrm{i}\tau}\right)^{\ell}\right]\Bigg{)},$

for all $x\in\mathbb{R}$ and all $n\in\mathbb{Z}$, is not very clear unless
$\rho=1$. Specifically, the following problem arises. Suppose that we want to
construct two entire functions of finite order such that their magnitudes
agree on the infinite set of parallel lines
$\mathbb{R}+\mathrm{i}\tau\mathbb{Z}$. We can choose a set of isolated points
$\mathcal{Y}_{\mathrm{u}}\subset\mathbb{R}+\mathrm{i}[-\tau,\tau)$ along with
some multiplicities $m_{\mathcal{Y}}:\mathcal{Y}_{\mathrm{u}}\to\mathbb{N}$,
and an integer $p\in\mathbb{N}$ such that

$\sum_{a\in\mathcal{Y}_{\mathrm{u}}}\sum_{k\in\mathbb{Z}}\frac{m_{\mathcal{Y}}(a)}{\left\lvert
a+2k\mathrm{i}\tau\right\rvert^{p+2}}<\infty$

and define

$f(z):=\prod_{a\in\mathcal{Y}_{\mathrm{u}}}\prod_{k\in\mathbb{Z}}E\left(\frac{z}{a+2k\mathrm{i}\tau};p\right)^{m_{\mathcal{Y}}(a)}.$

We might as well choose another integer $q\in\mathbb{N}$ and some coefficients
$b_{\ell}$, for $\ell\in\\{1,\dots,q\\}$, and define

$g(z):=\exp\left(\mathrm{i}\sum_{\ell=1}^{q}b_{\ell}z^{\ell}\right)\prod_{a\in\mathcal{Y}_{\mathrm{u}}}\prod_{k\in\mathbb{Z}}E\left(\frac{z}{\overline{a}+2k\mathrm{i}\tau};p\right)^{m_{\mathcal{Y}}(a)}.$

It is however not clear if we can choose the coefficients such that

$\sum_{\ell=1}^{q}b_{\ell}\operatorname{Im}\left[(x+n\mathrm{i}\tau)^{\ell}\right]=\sum_{a\in\mathcal{Y}_{\mathrm{u}}}m_{\mathcal{Y}}(a)\sum_{k\in\mathbb{Z}}\Bigg{(}\log\left\lvert\frac{a+2k\mathrm{i}\tau}{a+2(k-n)\mathrm{i}\tau}\right\rvert\\\
+\sum_{\ell=1}^{p}\frac{1}{\ell}\operatorname{Re}\left[\left(\frac{x+n\mathrm{i}\tau}{\overline{a}+2(n-k)\mathrm{i}\tau}\right)^{\ell}-\left(\frac{x+n\mathrm{i}\tau}{a+2k\mathrm{i}\tau}\right)^{\ell}\right]\Bigg{)},$

for all $x\in\mathbb{R}$ and all $n\in\mathbb{Z}$. The above is a necessary
condition for $f$ and $g$ to agree in magnitude on the lines
$\mathbb{R}+\mathrm{i}\tau\mathbb{Z}$.

Of course, if we design the functions $f$ and $g$ such that their order is
one, then we can use the Theorem 27 in which the condition on the coefficients
$b,b^{\prime}\in\mathbb{R}$ is actually tractable. Let us consider the
following simple example: we let $\tau=1$,

$\mathcal{Y}_{\mathrm{u}}=\left\\{\frac{\mathrm{i}}{2}\right\\},$

$m_{\mathcal{Y}}(\mathrm{i}/2)=1$, as well as $b=\pi/2$ and define

$f(z)=\mathrm{e}^{\frac{\pi\mathrm{i}}{2}z}\prod_{k\in\mathbb{Z}}E\left(\frac{2z}{(4k+1)\mathrm{i}};1\right).$

We can now compute

$b^{\prime}=\frac{\pi}{2}-\sum_{k\in\mathbb{Z}}\left(\log\left\lvert\frac{4k-3}{4k+1}\right\rvert+\frac{4}{4k+1}\right)=-\frac{\pi}{2}$

and define

$g(z)=\mathrm{e}^{-\frac{\pi\mathrm{i}}{2}z}\prod_{k\in\mathbb{Z}}E\left(\frac{2z}{(4k+3)\mathrm{i}};1\right).$

It follows from Theorem 27 that

$\left\lvert f(x+n\mathrm{i})\right\rvert=\left\lvert
g(x+n\mathrm{i})\right\rvert,\qquad x\in\mathbb{R},~{}n\in\mathbb{Z}.$

Finally, one can actually show that

$\displaystyle f(z)=\cosh\left(\frac{\pi
z}{2}\right)+\mathrm{i}\sinh\left(\frac{\pi z}{2}\right),$ $\displaystyle
g(z)=\cosh\left(\frac{\pi z}{2}\right)-\mathrm{i}\sinh\left(\frac{\pi
z}{2}\right).$

Therefore, we recover the examples from the beginning of this paragraph in
this way.

Of course, we can now find two entire functions $f$ and $g$ of general order
$\rho$ whose magnitudes agree on the lines $\mathbb{R}+\mathrm{i}\mathbb{Z}$
but which do not agree up to global phase by picking
$h\in\mathcal{O}(\mathbb{C})$ of order $\rho$ and considering

$\displaystyle f(z)=h(z)\left(\cosh\left(\frac{\pi
z}{2}\right)+\mathrm{i}\sinh\left(\frac{\pi z}{2}\right)\right),$
$\displaystyle g(z)=h(z)\left(\cosh\left(\frac{\pi
z}{2}\right)-\mathrm{i}\sinh\left(\frac{\pi z}{2}\right)\right).$

But if we suppose that we have an arbitrary but fixed function
$f\in\mathcal{O}(\mathbb{C})$ of order $\rho>1$, then it is not clear whether
we can harness the lemma above to specifiy all $g\in\mathcal{O}(\mathbb{C})$
such that $\left\lvert f\right\rvert$ and $\left\lvert g\right\rvert$ agree on
$\mathbb{R}+\mathrm{i}\tau\mathbb{Z}$. Simplifying and clarifying the relation
on the coefficients $b_{\ell},b_{\ell}^{\prime}\in\mathbb{R}$, for
$\ell\in\\{1,\dots,q\\}$, might thus be a tempting topic for future research.

##  5 Phase retrieval from Gabor measurements

In this section, we will show how our work for entire functions relates to
Gabor phase retrieval through the intimate relationship of the Gabor
transform, the Bargmann transform and the Fock space of second order entire
functions. We do so in keeping with the ordering chosen in section 4 of
presenting results on intersecting lines in the time-frequency plane in the
first subsection and presenting results on parallel lines in the time-
frequency plane in the second subsection. In a third subsection, we will
discuss a restriction on the root set of the Gabor transform imposed by
Jensen’s formula from complex analysis.

As stated above, we want to make use of the relation between the Gabor
transform, the Bargmann transform and the Fock space. To do this, we start by
introducing the relevant notation in line with the excellent exposition in
[4]. First, let us consider the Gaussian $\phi(t):=\mathrm{e}^{-\pi t^{2}}$,
for $t\in\mathbb{R}$. We may define the Gabor transform of a signal $f\in
L^{2}(\mathbb{R})$ as

$\mathcal{G}f(x,\omega):=\int_{\mathbb{R}}f(t)\phi(t-x)\mathrm{e}^{-2\pi\mathrm{i}t\omega}\,\mathrm{d}t,\qquad
x,\omega\in\mathbb{R}.$

Similarly, we might define the Bargmann transform of $f\in L^{2}(\mathbb{R})$
as

$\mathcal{B}f(z):=\int_{\mathbb{R}}f(t)\mathrm{e}^{2\pi tz-\pi
t^{2}-\frac{\pi}{2}z^{2}}\,\mathrm{d}t,\qquad z\in\mathbb{C}.$

Thirdly, we introduce the Fock space $\mathcal{F}^{2}(\mathbb{C})$ as the
Hilbert space of entire functions $F$ such that

$\left\lVert F\right\rVert_{\mathcal{F}}^{2}=\int_{\mathbb{C}}\left\lvert
F(z)\right\rvert\mathrm{e}^{-\pi\left\lvert
z\right\rvert^{2}}\,\mathrm{d}z<\infty,$

equipped with the inner product

$(F,G)_{\mathcal{F}}:=\int_{\mathbb{C}}F(z)\overline{G(z)}\mathrm{e}^{-\pi\left\lvert
z\right\rvert^{2}}\,\mathrm{d}z,$

for $F,G\in\mathcal{F}^{2}(\mathbb{C})$. One can show that, after rescaling by
the factor $2^{1/4}$, the Bargmann transform is a unitary operator from
$L^{2}(\mathbb{R})$ onto $\mathcal{F}^{2}(\mathbb{C})$ (see Theorem 3.4.3 on
p. 56 of [4]). This implies that the Bargmann transform is bijective with
inverse