\documentclass[12pt]{amsart}
\title[Lecture 29]{Complex analysis: Lecture 29}
\author{}

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\begin{document}
\maketitle

\PP{29.0}{What is in these notes}
These notes contain some remarkable applications of the results we
have obtained so far
\footnote{This material is optional. We are going to
use the infinite product expansion of $\ds\frac{\sin(z)}{z}$ in
connection with Euler's gamma function later (Lecture 31, 32).
You will have to take it on faith, in case you choose not to
read these notes.}. Namely,\\

\begin{enumerate}
\item Cauchy's residue theorem (Lecture 26, \S 26.4).\\
\item Weierstrass' theorem on uniform convergence (Lecture 22, \S 22.6).\\
\item Our important inequality (Lecture 12, \S 12.7).\\
\end{enumerate}

We are going to use these to prove various expressions involving
trigonometric functions, almost all of them due to Euler. For instance:
\[
\cot(z) = \frac{1}{z} + \sum_{n=1}^{\infty}\left(\frac{1}{z-n\pi}+\frac{1}{z+n\pi}\right).
\]
\[
\cosec(z) = \frac{1}{z} + \sum_{n=1}^{\infty}
(-1)^n\left(\frac{1}{z-n\pi}+\frac{1}{z+n\pi}\right).
\]
\[
\frac{\sin(z)}{z} = \prod_{n=1}^{\infty} \left( 1-\frac{z^2}{n^2\pi^2}\right).
\]

These identities can be viewed as approximating trigonometric
functions by rational functions. The first two can also be thought
of as ``partial fraction decomposition", except now there
are infinitely many terms.\\

While these
and many other identities were obtained by Euler for computational purposes,
he used the last one to obtain the explicit value of $\ds\sum_{n=1}^{\infty}
\frac{1}{n^2}$. (You can read more about it on wikipedia - look for
{\em Basel problem}.)\\

\noindent {\em Euler's solution to the Basel problem.} Assuming the
validity of $\ds\frac{\sin(z)}{z} = \prod_{n=1}^{\infty} 
\left( 1-\frac{z^2}{n^2\pi^2}\right)$, we are going to
compare the coefficient of $z^2$ on both sides.\\
\[
\frac{\sin(z)}{z} = 1 - \frac{z^2}{3!} + \frac{z^4}{5!} - \cdots 
\ \ \text{from the Taylor series expansion of $\sin(z)$.}
\]
So, the left--hand side gives us $\ds-\frac{1}{6}$.
The right--hand side gives us an infinite sum, namely
$\ds - \sum_{n=1}^{\infty} \frac{1}{n^2\pi^2}$. Hence:
\[
-\frac{1}{6} = -\sum_{n=1}^{\infty}\frac{1}{n^2\pi^2} \Rightarrow
\boxed{
\sum_{n=1}^{\infty}\frac{1}{n^2} = \frac{\pi^2}{6}
}
\]
\ \\

\noindent I encourage you to read more about Euler's work on infinite series
in:\\

\begin{center} 
V.S. Varadarajan. {\em Euler and his work on infinite series.}
Bulletin of the American Mathematical Society (2007) volume 44, no. 4,
515-539.
\end{center}

You can download this paper from the following link:\\

\url{https://people.math.osu.edu/gautam.42/S20/Varadarajan-Euler.pdf}\\

\PP{29.1}{Generalized partial fractions}
Certain type of meromorphic functions can be written as an infinite sum
of rational fractions. This result is due to Mittag-Lefler
\footnote{G\"{o}sta Mittag-Leffler (1846-1927)} who proposed and solved
the problem of contructing meromorphic functions from the knowledge
of their behaviour near poles (known as Mittag-Leffler theorem).
The result given in this section is based on his paper from 1880,
and is a special case of Mittag-Leffler theorem.\\

Let $f:\C\dashrightarrow\C$ be a meromorphic function (see Lecture
26, \S 26.5, \S 26.6). Let $A\subset\C$
be the set of poles of $f$ in the complex plane. Let us
assume that the poles are arranged, $A=\{a_1,a_2,\ldots\}$ 
in a way that
$\ds 0<|a_1|\leq |a_2|\leq \cdots$ I am assuming that $0$
is not a pole of $f$\footnote{This is not a serious assumption
for the result, since we can just replace $f(z)$ by
$\ds f(z)-\frac{b}{z}$, if $f(z)$ did have a pole of order $1$
at $z=0$ with residue $b$. See Example \S 29.2 below}. \\

\noindent {\bf Assumption 1.}
Assume that all the poles of $f$ are of order $1$ (poles of order $1$
are also called {\em simple poles}), and let
$\ds\Res{z=a_k}{f(z)}=b_k$ for every $k\geq 1$.\\

\noindent {\bf Assumption 2.}
For every $m\in\Z_{\geq 1}$, it is possible to pick $R_m\in\R_{>0}$,
so that
\begin{enumerate} 
\item $\ds R_1<R_2<\ldots\ $ and $R_m\to\infty$ as $m\to\infty$.
\item $|a_j|\neq R_m$ (for every $j,m\geq 1$). (Meaning, the poles
of $f$ do not lie on the circle $C_m$ of radius $R_m$, centered at $0$).
\item There is a constant $M\in\R_{>0}$ such that
$\ds |f(z)|<M$ for every $z$ lying on $C_m$ (for every $m\geq 1$).
Just to clarify, $M$ is independent of $m\geq 1$. The same constant is
supposed to work for all circles.\\

\end{enumerate}

Then, for every $w\in\C\setminus A$:
\[
\boxed{
f(w) = f(0) + \sum_{k=1}^{\infty} b_k\left(\frac{1}{w-a_k} + \frac{1}{a_k}\right)
}
\]

\begin{proof}
Let $\Omega = \C\setminus A$, so that $f:\Omega\to\C$ is holomorphic.
For each $m\in\Z_{\geq 1}$, let $C_m$ be the counterclockwise
oriented circle of radius $R_m$, centered at $0$.\\

Let $N_m\in\Z_{\geq 1}$ denote the positive integer so that
\[
a_1,a_2,\ldots,a_{N_m} \in \text{Interior}(C_m) \qquad\text{and}\qquad
a_n\in\text{Exterior}(C), \text{ for every } n>N_m.
\]
The way the poles are arranged gives us $N_1\leq N_2\leq \ldots$\\

Consider a point $w\in\Omega$. Using Cauchy's integral formula, we
have (recall that each $a_k$ is a pole of order $1$, with residue
$b_k$):

\[
\frac{1}{2\pi\iota}\int_{C_m} \frac{f(z)}{z-w}\, dz
= f(w) + \sum_{k=1}^{N_m} \frac{b_k}{b_k-w}\ .
\]

Replacing $\ds\frac{1}{z-w} = \frac{1}{z}\left(1+\frac{w}{z-w}\right)$,
we get (again by Cauchy's integral formula):

\[
\frac{1}{2\pi\iota}\int_{C_m}\frac{f(z)}{z-w}\, dz = 
\frac{1}{2\pi\iota}\int_{C_m}\frac{f(z)}{z}\, dz + \frac{w}{2\pi\iota}
\int_{C_m}\frac{f(z)}{z(z-w)}\, dz 
\]
\[
= f(0) + \sum_{k=1}^{N_m}
\frac{b_k}{a_k} + \frac{w}{2\pi\iota}
\int_{C_m}\frac{f(z)}{z(z-w)}\, dz\ .
\]

Combining the two expressions, we get:

\[
f(w) - \left( f(0) + \sum_{k=1}^{N_m}b_k\left(\frac{1}{w-a_k}+\frac{1}{a_k}\right)
\right) = \frac{w}{2\pi\iota}\int_{C_m}\frac{f(z)}{z(z-w)}\, dz.
\]

Assume that $m$ is large enough so that $|w|<R_m$. Then we get,
using $|f(z)|<M$ for every $z$ on $C_m$, and $|z-w|\geq R_m-|w|$:

\[
\left|\frac{w}{2\pi\iota}\int_{C_m}\frac{f(z)}{z(z-w)}\, dz\right|
< \frac{|w|}{2\pi} \frac{M}{R_m(R_m-|w|)} 2\pi R_m = \frac{M|w|}{R_m-|w|}.
\]

By our assumption $R_m\to\infty$ as $m\to\infty$. So this integral goes
to $0$, as $m\to\infty$. Hence we obtain:

\[
f(w) = f(0) + \sum_{k=1}^{\infty}b_k\left(\frac{1}{w-a_k}+\frac{1}{a_k}\right).
\]

Note that the same argument also gives the uniform convergence of this infinite
sum, as follows. Let $K\subset\C$, a compact set be given such that $K\subset\Omega$.
Additionally, let $\varepsilon>0$ be given.
Let $a\in\R_{>0}$ be such that $|w|<a$ for every $w\in K$ (exists since $K$
is bounded).\\

Choose $m$ large enough that $\ds\frac{Ma}{R_m-a}<\varepsilon$. Take
$N = N_m$. Then, for every $w\in K$, we get:
\[
\left|\sum_{k=N+1}^{\infty} b_k\left(\frac{1}{w-a_k}+\frac{1}{a_k}\right)\right|
= \left|\frac{w}{2\pi\iota} \int_{C_m} \frac{f(z)}{z(z-w)}\, dz\right|
< \frac{Ma}{R_m-a} < \varepsilon.
\]
\end{proof}
\ \\

\PP{29.2}{Example} Let us derive the expression of $\cosec(z)$
as given in \S 29.0. The poles of $\cosec(z)$ are at $\pi\Z$. By l'h\^{o}pital
rule ($n\in\Z$):
\[
\lim_{z\to n\pi} (z-n\pi)\cosec(z) = \lim_{z\to n\pi} \frac{z-n\pi}{\sin(z)}
=\lim_{z\to n\pi} \frac{1}{\cos(z)} = \frac{1}{\cos(n\pi)} = (-1)^n.
\] 

So, all the poles are simple, and $\ds\Res{z=n\pi}{\cosec(z)}=(-1)^n$.\\

Consider the function $\ds f(z) = \cosec(z)-\frac{1}{z}$, so that it
does not have a pole at $0$. It remains to check that Assumption 2
of \S 29.1 holds. Once that is done, the result obtained in
\S 29.1 will give us:
\[
\boxed{
\cosec(z) = \frac{1}{z} + \sum_{n\in\Z_{\neq 0}}
(-1)^n \left(\frac{1}{z-n\pi} + \frac{1}{n\pi}\right)
}
\]
(Compare with the expansion given in \S 29.0 above, where the terms
with $n\in\Z_{\geq 1}$ and $-n$ are combined.)\\

\noindent {\em Verification of Assumption 2.} For each $m\in\Z_{\geq 1}$,
let $\ds R_m = \left(m+\frac{1}{2}\right)\pi$. Then,
\begin{itemize}
\item $R_1<R_2<\ldots\ $ and $R_m\to\infty$ as $m\to\infty$.
\item $\ds \cosec(z)-\frac{1}{z}$ does not have poles on the circle $C_m$ of radius $R_m$,
centered at $0$.
\end{itemize}

To find the uniform bound $M$, it would be sufficient to consider
$|\cosec(z)|$ alone, since $\ds \left|\cosec(z)-\frac{1}{z}\right|
\leq |\cosec(z)| + \frac{1}{|z|} < |\cosec(z)|+1$ (for $z$ lying
on any of $C_m$, $|z|=R_m>1$).\\

Let us break the circle $C_m$ into two parts (these are not disjoint,
only $C_m = P_1\cup P_2$ is needed).
\[
P_1 = \left\{z\in C_m : |\Im(z)|\geq \frac{\pi}{8}\right\}\qquad\text{and}\qquad
P_2 = \left\{z\in C_m : \left|z\pm \left(m+\frac{1}{2}\right)\pi\right|\leq 
\frac{\pi}{4}\right\}
\]
See Figure \ref{fig:cosec} below.\\

\begin{figure}[h]
\includegraphics[width=3in]{cosec.pdf}
\caption{$P_1$ consists of part of $C_m$ lying above $\ds y=\frac{\pi}{8}$
or lying below $y=-\frac{\pi}{8}$. $P_2$ consists of part of $C_m$
lying in the shaded discs, both of radii $\ds\frac{\pi}{4}$.}
\label{fig:cosec}
\end{figure}

For $z\in\C$ such that $\ds|\Im(z)|\geq \frac{\pi}{8}$, we have
\[
|\cosec(z)| = \frac{2}{|e^{\iota z}-e^{-\iota z}|} \leq
\frac{2}{||e^{\iota z}| - |e^{-\iota z}||} \leq \frac{2}{e^{|\Im(z)|}
-e^{-|\Im(z)|}} < \frac{2}{e^{\pi/8} - 1}.
\]
This gives us a bound (independent of $m$) on the part $P_1$ of $C_m$.\\

For the part $P_2$, note that $|\cosec(z)| = |\cosec(z+k\pi)|$ for any
$k\in\Z$. So, if $M_1$ is the absolute maximum of $|\cosec(z)|$ on
the closed disc $\left\{z\in\C : \left|z-\frac{\pi}{2}\right|\leq
\frac{\pi}{4}\right\}$, then (again independent of $m$), we have
$|\cosec(z)|\leq M_1$ for every $z$ lying in the part $P_2$ of $C_m$.\\

Now, take $M>\operatorname{Max}(M_1, 2/(e^{\pi/8}-1))$.\\

\PP{29.3}{Weierstrass' infinite product expansion}
A result parallel to that of \S 29.1 was obtained by Weierstrass
in 1876. Here we focus on zeroes of a holomorphic function,
instead of poles of a meromorphic function.\\

Let $f:\C\to\C$ be an entire holomorphic function. Assume
that $f$ vanishes at (and only at) points of a set 
$A\subset \C$, which (as in
\S 29.1 above) we arrange $A=\{a_1,a_2,\ldots\}$ so as
to have $0<|a_1|\leq |a_2|\leq \ldots$ Again, we are assuming
that $f(0)\neq 0$, but it is not a serious assumption.
(see Footnote 3 on page 2).\\

\noindent {\bf Assumption 1.} The order of vanishing of
$f$ at each $a_k$ is $1$.\\

\noindent {\bf Assumption 2.} Same as Assumption 2 of
\S 29.1, but for the function $\ds\frac{f'(z)}{f(z)}$.\\

Then, for every $z\in\C$:
\[
\boxed{
f(z) = f(0)e^{\frac{f'(0)}{f(0)}z}\prod_{n=1}^{\infty}
\left(
\left(1-\frac{z}{a_n}\right) e^{\frac{z}{a_n}}
\right)
}
\]

\begin{rem}
The notation $\ds\prod_{k=1}^{\infty} u_k(z)$ if for {\em infinite product}.
Explicitly, we consider the following sequence of functions:
\[
F_1(z) = u_1(z),\ \ F_2(z)=u_1(z)\cdot u_2(z),\ \ \cdots\ \ 
F_n(z) = u_1(z)\cdot u_2(z) \cdots u_n(z),\ \ \cdots
\]
Then, $\ds \prod_{k=1}^{\infty} u_k(z)$ is the limit
$\ds\lim_{n\to\infty} F_n(z)$. In order to ensure that the limit 
behaves well, we will have to verify that the sequence of
functions $\ds\{F_n(z)\}_{n=1}^{\infty}$ converges uniformly
(Weierstrass' theorem on uniform convergence \S 22.6).\\
\end{rem}

\begin{proof}
The proof in fact follows from the result obtained in
\S 29.1, applied to the function $\ds g(z) = \frac{f'(z)}{f(z)}$.
Note that $g(z)$ has simple poles at the points of $A$,
of residue $1$ (see Problem Set 7, problem 5:
$f(a)=0$ implies $\ds\Res{z=a}{\frac{f'(z)}{f(z)}}=$
order of vanishing of $f$ at $a$).\\

Thus we get:
\[
\frac{f'(z)}{f(z)} = \frac{f'(0)}{f(0)} + \sum_{n=1}^{\infty}
\left(
\frac{1}{z-a_n} + \frac{1}{a_n}
\right)
\]

We view this as a differential equation for $f(z)$, and try
to write $f(z)$ as its solution.\\

The following calculation is heuristic, but very instructive.
Note that $\ds\frac{f'(z)}{f(z)} = \frac{d}{dz}(\log(f(z)))$
(it is therefore called {\em logarithmic derivative}). So, taking
an antiderivative on both sides of the differential equation,
we get
(at least formally - i.e, without worrying about logarithms):
\[
\log(f(z)) = C+\frac{f'(0)}{f(0)}z+\sum_{n=1}^{\infty}
\left(
\log(z-a_n)+\frac{z}{a_n}
\right)\ ,
\]
where $C$ is a constant. Set $z=0$ to get (again, we are
not worrying about the convergence issues):
\[
\log(f(0)) = C + \sum_{n=1}^{\infty} \log(-a_n).
\]
Substituting it back in $\log(f(z))$ written above, we get:
\begin{align*}
\log(f(z)) &= \log(f(0)) + \frac{f'(0)}{f(0)}z+\sum_{n=1}^{\infty}
\left(
\log(z-a_n)-\log(-a_n)+\frac{z}{a_n}
\right)\\
&= \log(f(0)) + \frac{f'(0)}{f(0)}z+\sum_{n=1}^{\infty}
\left(
\log\left(1-\frac{z}{a_n}\right)+\frac{z}{a_n}
\right).
\end{align*}

Now take exponential to get the

\[
f(z) = f(0)e^{\frac{f'(0)}{f(0)}z}\prod_{n=1}^{\infty}
\left(\left(
1-\frac{z}{a_n}\right)
e^{\frac{z}{a_n}}
\right).
\]

Now that the final result is obtained, we can proceed as follows
(not relying on the heuristic calculation). (1) Prove that the infinite product
given on the right--hand side of the equation written 
above, converges uniformly on $\C$
\footnote{I am not going to write a proof here. In Lecture 32, we will discuss
the uniform convergence of an infinite product defining the
gamma function. A proof will be given then.}. (2) (easy)
Check that it solves the same
differential equation as $f(z)$, with the same initial value at $z=0$.
Hence, it equals $f(z)$.
\end{proof}
\ \\

\PP{29.4}{Example} Consider the function $\ds f(z) = \frac{\sin(z)}{z}$.
It is an entire holomorphic function, with zeroes of order $1$
at $z=n\pi$, $n\in\Z_{\neq 0}$. 
We have $f(0)=1$ (by l'h\^{o}pital) and $\ds\frac{f'(z)}{f(z)} = \cot(z)-\frac{1}{z}$,
whose value at $z=0$ is again computed using l'h\^{o}pital:
\[
\lim_{z\to 0} \frac{z\cos(z)-\sin(z)}{z\sin(z)} = \lim_{z\to 0}
\frac{-z\sin(z)}{z\cos(z)+\sin(z)} = \lim_{z\to 0}
\frac{-\sin(z)-z\cos(z)}{2\cos(z)-z\sin(z)} = 0.
\]
Leaving aside the
verification of the assumptions imposed in \S 29.3, the result
is:
\[
\boxed{
\frac{\sin(z)}{z} = \prod_{n\in\Z_{\neq 0}} \left(\left(
1-\frac{z}{n\pi}\right) e^{\frac{z}{n\pi}}\right)
}
\]
(Compare with the infinite product expansion written in \S 29.0 above,
where, again we have grouped the terms $n$ and $-n$ together).\\

Verification of Assumption 2 from \S 29.3 is absolutely similar to
the one for $\cosec(z)$ given in \S 29.2, except this time we
are going to have to find a bound on $\ds\frac{f'(z)}{f(z)} = 
\cot(z) - \frac{1}{z}$. The argument given
there works verbatim and I will not repeat it here.
\end{document}

