\documentclass[12pt]{amsart}
\title[Lecture 26]{Complex analysis: Lecture 26}
\author{}

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\newtheorem*{ex}{Example}

\begin{document}
\maketitle

\PP{26.0}{}
Recall that we defined $\ds\Res{z=\alpha}{f(z)}$ for an isolated singularity
$\alpha\in\C$ of a function $f$. Thus, $f$ is defined, and holomorphic 
on the punctured disc of some radius $R\in\R_{>0}$, $f:D^{\times}(\alpha;R)\to\C$.
(Recall: $D^{\times}(\alpha;R)=\{z\in\C : 0<|z-\alpha|<R\}$.)\\

Let $\gamma$ be a positively oriented contour such that (i)
$\alpha\in\text{Interior}(\gamma)$, and (ii) $\text{Interior}(\gamma)
\setminus\{\alpha\}$ and $\gamma$ itself are in the domain of $f$. Then,\\

\[
\boxed{
\Res{z=\alpha}{f(z)} = \frac{1}{2\pi\iota}\int_{\gamma} f(z)\, dz
}
\]
\ \\

We saw in Lecture 25, \S 25.7, that if $\alpha$ is a pole of $f$, we
can use Cauchy's integral formula to compute the residue. In any case
(whether $\alpha$ is a pole or essential singularity), we have:
$\ds\Res{z=\alpha}{f(z)} = $ coefficient of $(z-\alpha)^{-1}$
in the Laurent series expansion of $f$ near $\alpha$.\\

\PP{26.1}{The point at $\infty$}
There is nothing special here. The behaviour of a function $f(z)$
near $z=\infty$ is investigated by performing the change of variables
$\ds z=\frac{1}{w}$ and studying the point $w=0$ of $\ds f\left(\frac{1}{w}\right)$.
In more detail:\\

\begin{enumerate}
\item $z=\infty$ is a singularity of $f(z)$
if $w=0$ is a singularity of $\ds f\left(\frac{1}{w}\right)$.\\

\item Singularity at $z=\infty$ of $f(z)$ is isolated if
$w=0$ is an isolated singularity of $\ds f\left(\frac{1}{w}\right)$.\\

\item $z=\infty$ is a removable singularity, a pole, or an essential
singularity of $f(z)$ if so is $w=0$ of $\ds f\left(\frac{1}{w}\right)$.\\

\end{enumerate}

\noindent {\em Examples.} (1) $\ds e^z$ has an essential singularity
at $\infty$. (2) $\ds f(z) = \frac{z+1}{z^2-3}$ has a removable singularity
at $\infty$. (3) Let $P(z)$ be a polynomial of degree $n$. Then $P(z)$
has a pole of order $n$ at $\infty$ (see \S 26.7 for a converse to
this statement).\\

The following exercise is given here for you to get used to the
statements written above.\\

\noindent {\em Exercise.} Let $f:\Omega\to\C$ be a holomorphic function,
defined on an open set $\Omega$. Prove that $\infty$ is an isolated
singularity of $f$ if, and only if there is a number $R\in\R_{>0}$
such that $\{z\in\C : |z|>R\}\subset\Omega$. In this case,
verify the following assertions:\\
\begin{enumerate}
\item $f(z)$ has a removable singularity at $\infty$ if, and only
if $\ds\lim_{z\to\infty} f(z)$ exists.\\
\item $f(z)$ has a pole of order $N$ at $\infty$ if, and only
if $\ds\lim_{z\to\infty} z^{-N}f(z)$ exists, and is non--zero.\\
\end{enumerate}

\PP{26.2}{Residue at $\infty$}
Again, assume that $\infty$ is an isolated singularity of
$f$. Meaning, there is a number $R\in\R_{>0}$ so that
$f$ is defined (and holomorphic, as always) on 
$\{z\in\C : |z|>R\}$. Thus, all the singularities of $f$
are contained in the closed disc:
$\ds \ol{D(0;R)} = \{z\in\C : |z|\leq R\}$.\\

Let $\gamma$ be a {\em clockwise oriented} contour, large enough
so as to have all the singularities of $f$ in the interior
of $\gamma$. For instance, pick a number $\rho>R$ and let
$\gamma = -C_\rho$ be the {\em clockwise oriented} 
circle of radius $\rho$, centered at $0$.
We define:\\
\[
\boxed{
\Res{z=\infty}{f(z)} = \frac{1}{2\pi\iota} \int_{\gamma} f(z)\, dz
}
\]

\begin{rem}
In the definition of $\ds\Res{z=\infty}{f(z)}$, the contour used
is {\em negatively} oriented. The reason is - as one travels along
$\gamma$, the point at $\infty$ should be to the left. I always
use the notation $C$ for {\em counterclockwise oriented} circles,
that is why there is a minus sign in $\gamma=-C_\rho$.\\
\end{rem}

The relation between residue of $f(z)$ at $\infty$, and the residue
of $f(w^{-1})$ at $0$ is obtained as follows. Upon the change
of variables $w=z^{-1}$, we get $\ds dw = -z^{-2}dz = -w^2 dz$.
Thus $\ds dz = - \frac{dw}{w^2}$.\\

Moreover (easy check) as $z$ moves over $-C_{\rho}$ (clockwise circle
of radius $\rho$, centered at $0$), $w=z^{-1}$ moves over $C_{\rho^{-1}}$
(counterclockwise circle of radius $\rho^{-1}$, centered at $0$).
This gives:

\[
\int_{-C_{\rho}} f(z)dz = \int_{C_{\rho^{-1}}} f\left(w^{-1}\right)
\frac{(-1)dw}{w^2}.
\]

Hence we arrive at the following identity.
\[
\boxed{
\Res{z=\infty}{f(z)} = - \Res{w=0}{w^{-2} f\left(w^{-1}\right)}
}
\]

\begin{ex}
Let us take a rational function $\ds f(z) = \frac{3z^2+1}{(z-1)(2z-1)^2}$.
It only has poles at $z=1$ and $z=1/2$. Let us compute its residue at
$\infty$ in two different ways.\\
(i) Directly from the definition. Take $\rho=2$ (anything bigger than $1$
will suffice), and let $C_\rho$ be the counterclockwise circle of radius
$\rho$, centered at $0$. 
I will leave the following calculation to you (it actually
follows easily from a result that we proved in Lecture 19, \S 19.5, page 7).
\[
\frac{1}{2\pi\iota} \int_{C_\rho} f(z)\, dz = \frac{3}{4}.
\]
Hence $\ds\Res{z=\infty}{f(z)} = -\frac{3}{4}$.\\

\noindent (ii) Using the change of variables $z=w^{-1}$.
\[
f(w^{-1}) = \frac{3w^{-2}+1}{(w^{-1}-1)(2w^{-1}-1)^2} = \frac{w(3+w^2)}
{(1-w)(2-w)^2}.
\]
Now we use $\ds\Res{z=\infty}{f(z)} = -\Res{w=0}{w^{-2}f(w^{-1})}$ to
carry out the calculation. Let $C$ be a small enough counterclockwise
circle centered at $0$ (small enough = radius $<1$).
\[
\Res{z=\infty}{f(z)} = - \frac{1}{2\pi\iota}\int_C
\frac{3+w^2}{w(1-w)(2-w)^2}\, dw = 
-\left[
\frac{3+w^2}{(1-w)(2-w)^2}\right]_{\text{Set } w=0}
= -\frac{3}{4}.
\]

\end{ex}

\PP{26.3}{A word on the non--isolated case}
Recall that an example of a non--isolated singularity was given in
Lecture 24, \S 24.2: $\ds f(z) = \cosec\left(z^{-1}\right)$.
The set of singularities is $\left\{\frac{1}{n\pi} : n\in\Z_{\neq 0}\right\}
\cup \{0\}$, and $0$ is not isolated. Meaning, no matter how
small radius $r>0$ we pick, the disc $D(0;r)$ will contain
a singularity other than $0$ (in fact, infinitely many).\\

In general, $\alpha\in\C$ is a non--isolated singularity of $f$,
if there exists a sequence $\{\alpha_n\}_{n=0}^{\infty}$ of
complex numbers such that:
\begin{itemize}
\item Each $\alpha_n$ is a {\em non--removable} singularity of $f$
(it could be a pole, or an essential singularity).\\

\item $\ds\lim_{n\to\infty} \alpha_n = \alpha$.\\

\end{itemize}

Note that in this case $\alpha$ cannot be removable, or a pole - 
if it were there would exist some positive real number $r\in\R_{>0}$
such that $f$ is defined on the punctured disc $D^{\times}(\alpha;r)$
(see, for instance, Lecture 25, \S 25.4).
The existence of such $r$ will contradict the fact that $\alpha_n\to\alpha$
as $n\to \infty$.\\

Hence, {\em non--isolated singularities can only be essential}.\\

\PP{26.4}{Cauchy's residue theorem}
Let $f$ be a holomorphic function, and $\gamma$ be a positively
oriented contour. Assume that there are finitely many points
$\alpha_1,\alpha_2,\ldots,\alpha_n\in\text{Interior}(\gamma)$
such that $\text{Interior}(\gamma)\setminus\{\alpha_1,\alpha_2,
\ldots,\alpha_n\}$ and $\gamma$ itself are in the domain of $f$.\\

\begin{figure}[h]
\includegraphics[width=2in]{residue.pdf}
\caption{$f$ is a holomorphic function defined on an open set,
which contains $\gamma$ and $\text{Interior}(\gamma)\setminus
\{\alpha_1,\alpha_2,\ldots,\alpha_n\}$.}
\label{fig:residue}
\end{figure}

\begin{thm}
\[
\boxed{
\frac{1}{2\pi\iota}\int_{\gamma} f(z)\, dz = \sum_{j=1}^n \Res{z=\alpha_j}{f(z)}
}
\]
\end{thm}

\begin{proof}
The proof of this theorem is based on the principle of contour deformation
(Lecture 16, \S 16.1). Namely, let $C_j$ be a small enough, counterclockwise
oriented circle, around $\alpha_j$ such that, for $k\neq j$, $\alpha_k$
is not on, or in the interior of $C_j$ (see Figure \ref{fig:residue}
above). By the principle of contour deformation:
\[
\frac{1}{2\pi\iota} \int_{\gamma} f(z)\, dz = 
\sum_{j=1}^n \frac{1}{2\pi\iota} \int_{C_j} f(z)\, dz
= \sum_{j=1}^n \Res{z=\alpha_j}{f(z)}.
\]
The last equality is by definition of the residue.
\end{proof}
\ \\

\begin{ex}
Let $C$ be the counterclockwise oriented circle of radius $2$, centered
at $0$. Compute $\ds \int_C \frac{e^{\pi z}}{z(z^2+1)}\, dz$.\\

\begin{align*}
\int_C \frac{e^{\pi z}}{z(z^2+1)}\, dz &= 
2\pi\iota\left(\Res{z=0}{\frac{e^{\pi z}}{z(z^2+1)}} + 
\Res{z=\iota}{\frac{e^{\pi z}}{z(z^2+1)}} + \Res{z=-\iota}
{\frac{e^{\pi z}}{z(z^2+1)}}\right)\\
&= 2\pi\iota\left(
\left[\frac{e^{\pi z}}{z^2+1}\right]_{\text{Set }z=0}
+\left[\frac{e^{\pi z}}{z(z+\iota)}\right]_{\text{Set }z=\iota}
+\left[\frac{e^{\pi z}}{z(z-\iota)}\right]_{\text{Set }z=-\iota}
\right)\\
&= 2\pi\iota \left(
1 + \frac{-1}{\iota(2\iota)} + \frac{-1}{-\iota(-2\iota)}
\right)\\
&= 4\pi\iota.
\end{align*}

\end{ex}

\PP{26.5}{Meromorphic functions}
A function $f$ is called {\em meromorphic}
\footnote{Greek: holo = whole/entire, mero = part. This is to highlight
that $f$ is only defined on a {\em part} of $\Omega$.}
on an open set $\Omega\subset\C$, if there exists a subset $A\subset\Omega$
such that:
\begin{itemize}
\item $f : \Omega\setminus A \to \C$ is holomorphic.\\

\item Every $\alpha\in A$ is either a removable singularity,
or a pole of $f$ (that is, $f$ is not allowed to have
essential singularities in $\Omega$).\\
\end{itemize}

\begin{ex}
Every holomorphic function is also meromorphic, with $A=\emptyset$.
$\cosec(z)$ and $\cot(z)$ are meromorphic functions on $\C$,
with $A = \{n\pi : n\in\Z\}$.
$e^{1/z}$ is not meromorphic on $\C$ (though it is holomorphic
on $\C\setminus\{0\}$, it has an essential singularity at
$0$).\\

A rational function, for instance $\ds f(z) = \frac{z^3+z+3}{z^5-1}$,
is meromorphic on $\C$ (in the example, $f(z)$ has poles at the
fifth roots of unity).\\
\end{ex}

The following notation is sometimes used for meromorphic functions:
$f:\Omega\dashrightarrow \C$ to indicate that $f$ may not be defined
on a subset $A\subset\Omega$. I am going to assume that
every point of $A$ is a pole (by including removable singularities
in the domain of $f$). A pole is always isolated, so the set $A$ itself has
to be {\em discrete} (meaning, for every $\alpha\in A$, there
exists $r\in\R_{>0}$ such that the punctured disc around
$\alpha$ of radius $r$ does not intersect $A$: 
$D^{\times}(\alpha;R)\cap A = \emptyset$). This leads to
the following important property of meromorphic functions.

\begin{prop}
Let $K\subset\C$ be a closed and bounded (in other words, compact)
set, such that $K\subset\Omega$. Then $K\cap A$ is finite.
\end{prop}

\begin{proof}
This is an application of Bolzano--Weierstrass theorem\footnote{A proof 
of Bolzano--Weierstrass
theorem (over $\R$) is given in Optional Reading A, \S A.2. I will leave
it to you (if you are interested) to carry out the proof over $\C$ - 
hint: use the version over $\R$ for the real and imaginary parts.}
: an infinite collection of points in a compact set
always have a cluster point. More precisely, if $\{\alpha_n\}_{n=0}^{\infty}$
is contained in a compact set $K$, then there exists $\alpha\in K$
with the property that for every $r>0$, there are infinitely many
$\alpha_n's$ with $|\alpha-\alpha_n|<r$.\\

Assume, for the sake of obtaining a contradiction, that
$K\cap A$ is infinite. By Bolzano--Weierstrass theorem, we will have
a point $\alpha\in K\subset\Omega$, such that for every
$r>0$, $D^{\times}(\alpha;r)\cap A\not=\emptyset$. This $\alpha$
is then a {\em non--isolated}, hence essential singularity
of $f$ (see \S 26.3 above). This contradicts the assumption that $f$ is meromorphic
($f$ is not allowed to have essential singularities in $\Omega$).
\end{proof}
\ \\

\PP{26.6}{Application 1: sum of residues}
Let $f:\C\dashrightarrow\C$ be a meromorphic function. Assume that
$\infty$ is an isolated singularity of $f$.\\

This means that there is a number $R\in\R_{>0}$ so that $f$
is defined and holomorphic on the open set $\{z\in\C : |z|>R\}$
(see \S 26.1 above). Hence, all the singularities (which are
necessarily poles since $f$ is meromorphic) are contained in
the compact set $K=\{z\in\C : |z|\leq R\}$. Therefore, by
Proposition 26.5 above, there are only finitely many of them,
say $\alpha_1,\alpha_2,\ldots,\alpha_n$.\\

Cauchy's residue theorem (\S 26.4), and the definition of $\Res{z=\infty}{f(z)}$
(\S 26.2) imply:
\[
\boxed{
\sum_{j=1}^n \Res{z=\alpha_j}{f(z)} + \Res{z=\infty}{f(z)} = 0
}
\]

\PP{26.7}{Application 2: entire functions with a pole at $\infty$}
The following argument is a mild generalization of the
one we saw in the proof of Liouville's theorem (Lecture 18, \S 18.2,
page 3).

Let $f:\C\to\C$ be a holomorphic function defined on the entire complex
plane (such functions are called {\em entire functions}). Assume
that $f$ has a pole of order $N$ at $\infty$.\\

\noindent {\bf Claim.} $f$ is a polynomial of degree $N$.\\

\begin{proof}
Let us take the Taylor series expansion of $f$ near $0$:
$\ds f(z) = \sum_{k=0}^{\infty} a_k z^k$. As $f$ is entire, the
radius of convergence of this series is $\infty$, by
Theorem (23.4) of Lecture 23.\\

As $f$ has a pole of order $N$ at $\infty$, by the exercise given
in \S 26.1 above, $\ds\lim_{z\to\infty} z^{-N}f(z)$ exists
and is non--zero. In particular it implies, by definition of the
limit, that there is a number $R\in\R_{>0}$, and a constant 
$M\in\R_{>0}$ so that
\[
|z^{-N}f(z)| < M \text{ for every } z\in\C \text{ such that } |z|>R.
\]

The coefficients of the Taylor series of $f(z)$ can be computed
as follows. Take any $\rho>0$, and let $C_{\rho}$ be the counterclockwise
circle of radius $\rho$ centered at $0$. Then:
\[
a_k = \frac{1}{2\pi\iota} \int_{C_\rho} \frac{f(z)}{z^{k+1}}\, dz.
\]
(we can take $\rho>0$ to be as large as we want, since $f$ is 
defined on the entire complex plane).\\

Now, taking $\rho>R$, we can estimate $|a_k|$ for $k\geq N+1$, using
the bound of $|z^{-N}f(z)|$ written above, and our important inequality
(see Lecture 12, \S 12.7, page 9). For every $\ell\geq 0$:

\[
|a_{N+1+\ell}| = \left|
\frac{1}{2\pi\iota} \int_{C_\rho} \frac{f(z)}{z^{N+\ell+2}}\, dz
\right| < \frac{1}{2\pi} \frac{M}{\rho^{\ell+2}} 2\pi\rho = \frac{M}{\rho^{\ell+1}}
\]

As $\ds\frac{M}{\rho^{\ell+1}}\to 0$ as $\rho\to\infty$, we conclude
that $|a_{N+1+\ell}|=0$ for every $\ell\geq 0$. Hence,
$f(z) = a_0 + a_1z + \cdots + a_N z^N$ is a polynomial.\\

Note that $\ds\lim_{z\to\infty} z^{-N}f(z) = a_N$ which was assumed
to be non--zero. Hence degree of $f$ is precisely $N$.
\end{proof}
\ \\

\PP{26.8}{Meromorphic functions on $\C$ with a pole at $\infty$} \\

\begin{thm}
Let us assume that we have a meromorphic function $f:\C\dashrightarrow \C$
which has a pole at $\infty$ (in particular, $\infty$ is an isolated
singularity of $f$).\\

Then, $f(z)$ is a rational function.
\end{thm}

\begin{proof}
Using the argument given in \S 26.6 above, we know $f$ will only
have finitely many poles $\{\alpha_1,\alpha_2,\ldots,\alpha_n\}\subset\C$.
For any $1\leq j\leq n$, we have the Laurent series expansion
of $f$ near $\alpha_j$ (remember: it is a pole, say of order $N_j$):
\[
f(z) = \frac{d^{(j)}_{N_j}}{(z-\alpha_j)^{N_j}} + \cdots + \frac{d^{(j)}_1}{z-\alpha_j}
+\sum_{k=0}^{\infty} c^{(j)}_k (z-\alpha_j)^k\ .
\]
Let us define: 
\[ R_j(z) = 
\frac{d^{(j)}_{N_j}}{(z-\alpha_j)^{N_j}} + \cdots + \frac{d^{(j)}_1}{z-\alpha_j}.
\]
Then, each $R_j(z)$ is a rational function, with only pole at $\alpha_j$.
Moreoever, $\ds\lim_{z\to\infty} R_j(z)=0$.\\

Let us define $\ds g(z) = f(z) - \sum_{j=1}^n R_j(z)$.\\
 
\noindent {\em Claim.} $g(z)$ is holomorphic on the entire complex plane. \\

Let us assume this claim to be true and proceed with the proof of
the theorem. Note that if $f(z)$ has a pole of order $N$ at $\infty$,
then so does $g(z)$. This is because $\ds\lim_{z\to\infty} z^{-N}R_j(z)=0$,
for every $j$, and therefore $\ds\lim_{z\to\infty} z^{-N}g(z) = \lim_{N\to\infty}
z^{-N}f(z)$. Hence $g(z)$ is a polynomial of degree $N$, by \S 26.7 above.
We can now conclude that $\ds f(z) = g(z) + \sum_{j=1}^n R_j(z)$ is a rational function,
being a sum of a polynomial and finite number of rational functions.\\

\noindent {\em Proof of the claim.} For $w\in\C$, $w\not\in\{\alpha_1,
\alpha_2,\ldots,\alpha_n\}$, it is clear that $g$ is holomorphic at $w$
(since $f$ and $R_j's$ are). So, let us consider the situation near
$w=\alpha_j$ ($j\in\{1,2,\ldots,n\}$). Each $R_\ell(z)$, with $\ell\neq j$,
is holomorphic at $\alpha_j$, since the only singularity of $R_\ell$
is $\alpha_\ell$. Moreover, near $\alpha_j$, we have the following
expansion $\ds f(z)-R_j(z) = \sum_{k=0}^{\infty} c^{(j)}_k (z-\alpha_j)^k$,
which means $\alpha_j$ is a removable singularity of $f(z)-R_j(z)$.
Hence we conclude that
\[
g(z) = f(z)-R_j(z) - \sum_{\begin{subarray}{c} \ell=1 \\ \ell\neq j\end{subarray}}
^n R_{\ell}(z) \text{ is holomorphic at } \alpha_j\ .
\]
So $g(z)$ has no singularities in the entire complex plane. The
claim follows.
\end{proof}
\end{document}

