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\begin{document}
\title{Complex Methods\\Course P3}
\author{T.~W.~K\"{o}rner}
\maketitle
\begin{footnotesize}
\noindent
{\bf Small print}
The syllabus for the course is defined by
the Faculty Board Schedules (which are minimal for lecturing
and maximal for examining).
I should {\bf very much} appreciate being told
of any corrections or possible improvements
and might even part with a small reward to the
first finder of particular errors. This document
is written in \LaTeXe\ and should be accessible
from my home page.
My e-mail address is \verb+twk@dpmms.cam.ac.uk+.
\end{footnotesize}
\tableofcontents
\section{Complex differentiability is like real differentiability}%
\label{S Algebra}
The complex numbers
have algebraic properties which are very similar to
those of the real numbers (formally they are both fields)
except that there is no order on the complex numbers.
This similarity means that we can define differentiability
in the complex case in exactly the same way as we
did in the real case.
\begin{definition} A function $f:{\mathbb C}\rightarrow{\mathbb C}$
is differentiable at $z$ with derivative $f'(z)$ if
\[\left|\frac{f(z+h)-f(z)}{h}-f'(z)\right|\rightarrow 0\]
as $|h|\rightarrow 0$.
\end{definition}
Exactly the same proofs as in the real case produce
exactly the same elementary properties of differentiation.
\begin{lemma}\label{L Easy differentiation}
(i) The constant function given by $f(z)=c$ for all $z\in{\mathbb C}$
is everywhere differentiable with derivative $f'(z)=0$.
(ii) The function given by $f(z)=z$ for all $z\in{\mathbb C}$
is everywhere differentiable with $f'(z)=1$.
(iii) If $f,\ g:{\mathbb C}\rightarrow{\mathbb C}$ are
both differentiable at $z$, then so is $f+g$ with
$(f+g)'(z)=f'(z)+g'(z)$.
(iii) If $f,\ g:{\mathbb C}\rightarrow{\mathbb C}$ are
both differentiable at $z$, then so is their product $f\times g$ with
$(f\times g)'(z)=f'(z)g(z)+f(z)g'(z)$.
(iv) If $f:{\mathbb C}\rightarrow{\mathbb C}$ is nowhere zero
and $f$ is differentiable at $z$, then so is $1/f$ with
$(1/f)'(z)=-f'(z)/(f(z))^{2}$.
(v) If $f:{\mathbb C}\rightarrow{\mathbb C}$ is
differentiable at $z$ and $g:{\mathbb C}\rightarrow{\mathbb C}$
is differentiable at $f(z)$ then the composition
$g\circ f$ is differentiable at $z$ with
$(g\circ f)'(z)=f'(z)g'(f(z))$.
(vi) If $P(z)=\sum_{n=0}^{N}a_{n}z^{n}$, then $P$ is everywhere
differentiable with derivative given by
$P'(z)=\sum_{n=1}^{N}na_{n}z^{n-1}$.
\end{lemma}
The following extensive generalisation of
part~(iv) of Lemma~\ref{L Easy differentiation}
was proved
in Analysis~I (course~C5).
\begin{theorem}\label{T, power series analytic}
Let $a_{j}\in{\mathbb C}$ $[0\leq j]$. Then,
either $\sum_{n=0}^{\infty}a_{n}z^{n}$ converges for all
$z$ and we write $R=\infty$, or there exists a real number
$R\geq 0$ such that $\sum_{n=0}^{\infty}a_{n}z^{n}$
converges for all $|z|R$.
($R$ is called the \emph{radius of convergence}
of $\sum_{n=0}^{\infty}a_{n}z^{n}$.)
If we write $f(z)=\sum_{n=0}^{\infty}a_{n}z^{n}$
for $|z|0$
such that $z\in\Omega$ whenever $|z-w|<\delta$.
\end{definition}
Thus, wherever we are in an open set, we can move some
fixed distance (depending on the point chosen)
in any direction whilst remaining within
the set.
\begin{definition} Let $\Omega\subseteq{\mathbb C}$
be an open set. We say that a function $f:\Omega\rightarrow{\mathbb C}$
is analytic on $\Omega$ if $f$ is differentiable at
every point of $\Omega$.
\end{definition}
Since this is a non-rigorous treatment, it operates under
the assumption that everything is well behaved. One of the
surprises and one the great advantages of the rigorous
treatment given in course~C12 (Further Analysis) is that
it reveals that \emph{all analytic functions are well behaved
analytic functions}.
The rigorous treatment reveals that the next lemma is
true for all analytic functions although our `proof'
seems to require extra conditions.
\begin{lemma}\label{L, analytic to harmonic}
Suppose $\Omega\subseteq{\mathbb C}$ is open
and $f:\Omega\rightarrow{\mathbb C}$ is analytic. Then,
defining $u$ and $v$, as usual we have $u$
harmonic (that is, satisfying Laplace's equation
$\triangledown^{2}u=0$)
on $\tilde{\Omega}=\{(x,y)\in{\mathbb R}^{2}\,:\,x+iy\in\Omega\}$.
The same is true of $v$.
\end{lemma}
Lemma~\ref{L, analytic to harmonic} has an important converse.
\begin{lemma}\label{L, hamonic to analytic}
If $u$ is harmonic, then it is locally
the real part of an analytic function.
\end{lemma}
Formally, we can restate Lemma~\ref{L, hamonic to analytic}
as follows.
\begin{lemma} Let $D=\{z\in{\mathbb C}\,:\,|z-a|0,
\ 2\pi+\theta_{0}>\theta>\theta_{0}\}\]
and $\tilde{\Omega}=\{(x,y)\,:\,x+iy\in\Omega\}$. If
$u(x,y)=\log |x+iy|=\log (x^{2}+y^{2})^{1/2}$, then $u$
is the real part of an analytic function
defined by $\log z=\log r+i\theta$
where $r$ and $\theta$ form the unique solution
of $z=re^{i \theta}$ with $r>0$ and
$2\pi+\theta_{0}>\theta>\theta_{0}$.
(ii) If we define a real valued function $u$ on
$\tilde{\Gamma}={\mathbb R}^{2}\setminus\{{\boldsymbol 0}\}$
by $u(x,y)=\log |x+iy|$,
then $u$ is harmonic but we cannot find a real valued function $v$
on $\tilde{\Gamma}$ such that,
if we write $f(x+iy)=u(x,y)+iv(x,y)$, the
function $f:\Gamma={\mathbb C}\setminus\{0\}
\rightarrow{\mathbb C}$ is analytic.
\end{example}
\begin{definition}\label{D, logarithm}
Let
\[\Omega=\{z=re^{i\theta}\,:\,r>0,
\ 2\pi+\theta_{0}>\theta>\theta_{0}\}.\]
If we define
\[\log z=\log r+i\theta,\]
where $r$ and $\theta$ form the unique solution
of $z=re^{i \theta}$ with $r>0$ and
$2\pi+\theta_{0}>\theta>\theta_{0}$,
then $\log$ is called a branch of the logarithm function.
\end{definition}
\begin{lemma}\label{L, properties logarithm}
With the notation of Definition~\ref{D, logarithm},
we have the following results.
(i) $\log:\Omega\rightarrow{\mathbb C}$ is analytic.
(ii) $\log(\Omega)=\{w\,:\, 2\pi+\theta_{0}>\Im w>\theta_{0}\}
=\Lambda$, say.
(iii) $\exp(\log z)=z$ for all $z\in\Omega$.
(iv) If $\Im w\notin 2\pi{\mathbb Z}+\theta_{0}$, then
$\log(\exp w)=w$.
(v) $\log'(z)=z^{-1}$ for all $z\in\Omega$.
(vi) If $z_{1}$, $z_{2}$, $z_{1}z_{2}\in\Omega$,
then $\log z_{1}z_{2}=\log z_{1}+\log z_{2}+2n\pi i$ for
some $n\in{\mathbb Z}$.
\end{lemma}
\begin{lemma} There does not exist a continuous
function $L:{\mathbb C}\setminus\{0\}\rightarrow{\mathbb C}$
such that $\exp(L(z))=z$.
\end{lemma}
\begin{exercise}\label{Exercise, logarithm 1}
We use the notation of
Definition~\ref{D, logarithm}. Show that we cannot choose
$\theta_{0}$ so that $\log z_{1}z_{2}=\log z_{1}+\log z_{2}$
for all $z_{1}$, $z_{2}$, $z_{1}z_{2}\in\Omega$.
\end{exercise}
In Analysis~I, you saw that the easiest way to
define $x^{\alpha}$ when $x$ and $\alpha$ are real
and $x>0$ is to write $x^{\alpha}=\exp(\alpha \log x)$.
\begin{definition}\label{D, power}
We use the notation of
Definition~\ref{D, logarithm}. Suppose $\alpha\in{\mathbb C}$.
We define the map $z\mapsto z^{\alpha}$ on
$\Omega$ by $z^{\alpha}=\exp(\alpha\log z)$.
We call the resulting function a branch of $z^{\alpha}$.
\end{definition}
\begin{lemma} If we define $p_{\alpha}:\Omega\rightarrow{\mathbb C}$
by $p_{\alpha}(z)=z^{\alpha}$ as in Definition~\ref{D, power},
then $p_{\alpha}$ is analytic on $\Omega$. We have
\[p_{\alpha}'(z)=\alpha p_{\alpha-1}(z).\]
If $\alpha$ is real,
$r>0$ and $2\pi+\theta_{0}>\theta>\theta_{0}$ then
$p_{\alpha}(z)=r^{\alpha}\exp i\alpha\theta$
where $r^{\alpha}$ has its traditional meaning.
\end{lemma}
Except in the simplest circumstances, it is probably
best to deal with $z^{\alpha}$ by rewriting it
as $\exp(\alpha\log z)$.
If $0>\theta_{0}>-2\pi$, it is traditional to refer to
the function defined by
\[\log re^{i\theta}=\log r +i\theta\]
for $r>0$ and $2\pi+\theta_{0}>\theta>\theta_{0}$
as the principal branch of the logarithm
(with a similar convention for the associated powers).
This has the same effect and utility as my referring
to myself as the King of Siam.
\section{Conformal mapping} We start with our definition
of a conformal map.
\begin{definition} Let $\Omega$ and $\Gamma$ be open subsets
of ${\mathbb C}$. We say that $f:\Omega\rightarrow\Gamma$
is a conformal map if $f$ is bijective and analytic
and $f'$ never vanishes.
\end{definition}
In more advanced work it is shown that, if
$f$ is bijective and analytic, then $f'$ never
vanishes. The phrase `and $f'$ never vanishes' can
then be omitted from the definition.
\begin{lemma}\label{L conformal inverse}
Let $\Omega$ and $\Gamma$ be open subsets
of ${\mathbb C}$. If $f:\Omega\rightarrow\Gamma$
is conformal, then $f^{-1}:\Gamma\rightarrow\Omega$
is analytic. We have
\[(f^{-1})'(w)=\frac{1}{f'(f^{-1}(w))},\]
so $f^{-1}$ is also conformal.
\end{lemma}
\begin{exercise}\label{E, conformal equivalence}
We say that open subsets $\Omega$ and $\Gamma$
of ${\mathbb C}$ are conformally equivalent
if there exists a conformal map $f:\Omega\rightarrow\Gamma$.
Show that conformal equivalence is an equivalence relation.
\end{exercise}
The reader is warned that some mathematicians use
definitions of conformal mapping
which are not equivalent
to ours. (The most common change is to drop the
condition that $f$ is bijective but to continue
to insist that $f'$ is never zero.) Sometimes
people use conformal simply to mean angle
preserving, so you must be prepared to be asked
`Show that an analytic map with non-zero derivative
is conformal'.
So far as~1B examinations are concerned, we are chiefly
interested in the following conformal maps.
(i) $z\mapsto z+a$. Translation. Takes ${\mathbb C}$ to
${\mathbb C}$.
(ii) $z\mapsto e^{i\theta}z$ where $\theta$ is real.
Rotation. Takes ${\mathbb C}$ to ${\mathbb C}$.
(iii) $z\mapsto \lambda z$ where $\lambda$ is real and $\lambda>0$.
Dilation (scaling). Takes ${\mathbb C}$ to ${\mathbb C}$.
(iv) $z\mapsto z^{-1}$. Inversion in unit circle
followed by reflection in real axis. Takes circles and straight
lines `not through the origin' to circles
and circles and lines `through the origin'
to straight lines.
Takes ${\mathbb C}\setminus\{0\}$ to ${\mathbb C}\setminus\{0\}$.
[Note, in this course we are not interested in the
`point at infinity'.]
(v) $z\mapsto z^{\alpha}$ with $\alpha$ real and $\alpha>0$.
[N.B. You must specify a branch!]
Takes (appropriate) sectors
(of the form $\{re^{i\theta}\,:\,r>0,\ \theta_{1}>\theta>\theta_{2}\}$)
to sectors with base angle multiplied by $\alpha$.
(vi) $z\mapsto \exp z$. Takes (appropriate) planks
\[\{z\,:\, \theta_{1}>\Im z>\theta_{2}\}\]
to sectors. The map $z\mapsto\log z$
[N.B. You must specify a branch!] does the reverse.
We observe that maps of the type~(i) to~(iv) are M\"{o}bius
and together generate the M\"{o}bius group.
M\"{o}bius maps were extensively discussed in the
first year course `Algebra and Geometry'.
Observe also
that we do not really need maps of type~(v) explicitly,
since we can obtain them using maps of the type~(iii)
and~(vi).
The author strongly recommends constructing conformal
maps in a large number of simple steps, as the composition
of the simple maps given above, rather than trying to
do everything at once.
\begin{example}\label{Example, sector to disc}
Find a conformal map taking
\[\Omega=\{z\,:\,\Im z>0,\ \Re z>0,\ |z|<1\}\]
to the unit disc $D=\{z\,:\, |z|<1\}$.
Explain why the map $z\mapsto z^{4}$ does not work.
\end{example}
It should be noted that conformal mapping problems like
Example~\ref{Example, sector to disc} do not have
unique solutions since there are non-trivial conformal
maps of the disc into itself (for example rotation).
In the early days of aviation, conformal mappings
(of a very slightly more complicated kind) were
used to find the flow of air past the wings of aeroplanes.
The method depended on the following result.
\begin{lemma}\label{L conformal harmonic}
Let $\Omega$ and $\Gamma$ be open subsets
of ${\mathbb C}$ and $f:\Omega\rightarrow\Gamma$
a conformal map. Set
\[\tilde{\Omega}=\{(x,y)\,:\,x+iy\in\Omega\},
\ \tilde{\Gamma}=\{(x,y)\,:\,x+iy\in\Gamma\}\]
and let
${\displaystyle
\left(
\begin{matrix}u\\v\end{matrix}
\right)
:\tilde{\Omega}\rightarrow\tilde{\Gamma}}$
be the mapping given by $f(x+iy)=u(x,y)+iv(x,y)$.
Then, if $\phi:\tilde{\Gamma}\rightarrow{\mathbb R}$
is harmonic, so is $\psi:\tilde{\Omega}\rightarrow{\mathbb R}$
where $\psi(x,y)=\phi(u(x,y),v(x,y))$.
\end{lemma}
It must be admitted that the use of Lemma~\ref{L conformal harmonic}
and the general practice of conformal mapping at
1B level and substantially above it depends on the
fact that, for the kind of $\Omega$ and $\Gamma$ considered,
the conformal map $f:\Omega\rightarrow\Gamma$
does, indeed,
behave well near the boundaries. The reader is warned
that, should she ever attend an advanced pure course
on conformal maps or try to use theorems which merely
guarantee the existence of such a map $f$ without
actually giving an explicit construction, this
assumption can no longer be relied
on\footnote{\begin{verse}
In the midst of the word he was trying to say,\\
In the midst of his laughter and glee,\\
He had softly and silently vanished away --\\
For the Snark \emph{was} a Boojum, you see.
\end{verse}}.
\section{Contour integration and Cauchy's theorem}
It is natural to define the integral of a function
$F:{\mathbb R}\rightarrow{\mathbb C}$ by
\[\int_{a}^{b}F(t)\,dt=\int_{a}^{b}\Re F(t)\,dt
+i\int_{a}^{b}\Im F(t)\,dt.\]
In the course C9 (Analysis) it is shown that
this definition produces an integral with all the
properties we want. In addition, the following useful
lemma is proved.
\begin{lemma}\label{L; bound integral}
If $F:[a,b]\rightarrow{\mathbb C}$
is continuous then
\[\left|\int_{a}^{b}F(t)\,dt\right|
\leq (b-a)\sup_{t\in[a,b]}|F(t)|.\]
\end{lemma}
We summarise this result in a slogan
\begin{center}
{\bf modulus integral $\leq$ length $\times$ supremum}.
\end{center}
Next we wish to define the contour integral.
\[\int_{C}f(z)\,dz\]
where $C$ is a path in $\mathbb C$.
Roughly speaking
\[\int_{C}f(z)\,dz\approx\sum_{j=1}^{N}f(z_{j})(z_{j}-z_{j-1})\]
where the polygonal path joining $z_{0}$, $z_{1}$,
\dots $z_{N}$ is a `good approximation to $C$'.
We formalise this idea as follows.
(A function $g:{\mathbb R}\rightarrow{\mathbb C}$ is said
to be differentiable if $\Re g$ and $\Im g$ are.
We write $g'(t)=(\Re g)'(t)+i(\Im g)'(t)$.)
\begin{definition} If $\gamma:[a,b]\rightarrow{\mathbb C}$
is a sufficiently smooth\footnote{Continuously differentiable
will certainly do.} function describing the path $C$
and $f:{\mathbb C}\rightarrow{\mathbb C}$ is continuous,
we define
\[\int_{C}f(z)\,dz=\int_{a}^{b}f(\gamma(t))\,\gamma'(t)dt.\]
\end{definition}
If a path $C$ is made up of a path $C_{1}$ followed by a path $C_{2}$
followed by a path $C_{3}$ \dots followed by a path $C_{n}$
with each path satisfying the conditions of our definition,
then we take
\[\int_{C}f(z)\,dz=\sum_{r=1}^{n}\int_{C_{r}}f(z)\,dz.\]
It is, more or less, clear that our definitions are unambiguous
but a rigorous development would need to prove this.
If a contour begins and ends at the same point
we call it at closed contour. In this case,
some older texts use the
pleasant notation
\[\oint_{C}f(z)\,dz=\int_{C}f(z)\,dz.\]
Lemma~\ref{L; bound integral} now takes the following form.
\begin{lemma} Under the conditions above,
\[\left|\int_{C}f(z)\,dz\right|\leq \text{length $C$}\times
\sup_{z\in C}|f(z)|.\]
\end{lemma}
The next result is very important.
\begin{lemma}\label{L; simplest pole}
Suppose $a,\ w\in{\mathbb C}$ and $r\in{\mathbb R}$
with $r>0$.
Let $C$ be the path $w+r\exp i\theta$ described
as $\theta$ runs from $0$ to $2\pi$ (less formally, the
circle radius $r$ and centre $w$ described once anticlockwise).
Then
\begin{alignat*}{2}
\int_{C}\frac{1}{z-a}\,dz&=2\pi i&&\qquad\text{if $|a-w|r$}.
\end{alignat*}
\end{lemma}
Note that this illustrates an important point
\begin{center}
\begin{bf}
Change of contour is not change of variable.
\end{bf}
\end{center}
The next result has very little to do with the course
but I could not resist including it.
\begin{lemma}
If $C$ is a closed contour which does not cross over itself
and is described once anticlockwise then
\[\int_{C}z^{*}\,dz=2i\times\text{Area enclosed by $C$}.\]
\end{lemma}
We now come to our master theorem.
\begin{theorem}[Cauchy's theorem] Let $\Omega$ be an
open, simply connected (that is all in a single piece
and with no holes) set in ${\mathbb C}$ and
$f:\Omega\rightarrow{\mathbb C}$ be an analytic function.
Then
\[\int_{C}f(z)\,dz=0\]
\end{theorem}
\noindent{\bf Note} Observe that Lemma~\ref{L; simplest pole}
shows that the `no holes' condition can not be dropped.
Given a particular $\Omega$ it is usually trivial to check
that it has no holes but a rigorous development
of complex analysis for general $\Omega$ is somewhat delicate.
With our sturdy English common sense we have banished
the study of holes\footnote{Called homology by its
practitioners.} to the higher reaches of pure mathematics
but, in the US, some textbooks of mathematical methods for
engineers devote quite a lot of time to it.
To save ink in future, we shall call $\Omega$ a simply connected
domain if it is
an open simply connected (that is one piece without holes)
subset of ${\mathbb C}$. You should note that this notation is
not universal.
Here is a nice application of Cauchy's theorem which
foreshadows much of the course.
\begin{lemma} If $\lambda$ is real, then
\[\int_{-\infty}^{\infty}e^{-i\lambda x}e^{-x^{2}/2}\,dx
=(2\pi)^{1/2}e^{-\lambda^{2}/2}.\]
In particular,
\[\int_{-\infty}^{\infty}\cos(\lambda x)e^{-x^{2}/2}\,dx
=(2\pi)^{1/2}e^{-\lambda^{2}/2}.\]
\end{lemma}
We remind the reader that
`Change of contour is not change of variable'.
\section{Applications of Cauchy's theorem}
Cauchy's theorem has far reaching implications.
Our first result depends on the introduction of
a well understood singularity.
\begin{theorem}[Cauchy's formula]%\label{T; Cauchy formula}
Let $\Omega$
be a simply connected domain and $f:\Omega\rightarrow{\mathbb C}$
be analytic. If $C$ is a closed contour in $\Omega$
which does not cross over itself
and is described once anticlockwise,
and $a$ lies inside $C$,
then
\[\int_{C}\frac{f(z)}{z-a}\,dz=2\pi if(a).\]
\end{theorem}
Given a particular $C$ and a particular $a$,
it is usually trivial to check that $C$
does not cross over itself
and is described once anticlockwise, and
that $a$ lies inside $C$.
However a rigorous development
of these notions is somewhat delicate (to repeat
the refrain of our song). We shall say that a $C$
which does not cross over itself
and is described once anticlockwise
is a simple closed contour.
Differentiating under the integral (not hard to justify
with the ideas of Analysis II)
with respect to $a$ we get
the following result.
\begin{theorem}[Cauchy's formula, extended version]%
\label{T; extended Cauchy}
Let $\Omega$
be a simply connected domain and $f:\Omega\rightarrow{\mathbb C}$
be analytic. If $C$ is a simple closed contour in $\Omega$,
and $a$ lies inside $C$
then $f$ is $n$ times differentiable with
\[n!\int_{C}\frac{f(z)}{(z-a)^{n+1}}\,dz=2\pi i f^{(n)}(a).\]
\end{theorem}
It is worth emphasising part of the result just given.
\begin{theorem}
Let $\Omega$ be a open subset of ${\mathbb C}$.
If $f:{\Omega}\rightarrow{\mathbb C}$ is once
complex differentiable then it is infinitely complex
differentiable.
\end{theorem}
This is a truly remarkable result, it is surely worth
going to course~C12 simply to see it proved rigorously!
Another remarkable result is the following.
\begin{theorem}[Taylor's theorem]\label{T; Taylor}
Let $\Omega$ be a open subset of ${\mathbb C}$
and $f:{\Omega}\rightarrow{\mathbb C}$ be an analytic
function. Suppose the disc
\[D(b,\rho)=\{z\,:\,|z-b|<\rho\}\]
(with $\rho>0$) is a subset of $\Omega$.
Then we can find $a_{n}$ such that
\[f(z)=\sum_{n=0}^{\infty}a_{n}(z-b)^{n}\]
for all $z\in D(b,\rho)$.
If $00$
such that $D(w,\rho)\subseteq\Omega$ and $a_{n}\in{\mathbb C}$
such that
$f(z)=\sum_{n=0}^{\infty}a_{n}(z-w)^{n}$
for all $z\in D(w,\rho)$.)
\end{theorem}
\begin{exercise} Deduce Theorem~\ref{T; extended Cauchy},
in the case that $C$ is a circle, from Theorem~\ref{T; Taylor}
and results on power series.
\end{exercise}
Taylor's theorem for analytic functions has a striking
and useful generalisation.
\begin{theorem}[Laurent's expansion]\label{T; Laurent}
Let $\Omega$ be a open subset of ${\mathbb C}$
and let $b\in{\Omega}$.
Suppose that $f:{\Omega}\setminus\{b\}
\rightarrow{\mathbb C}$ is an analytic
function and the disc
\[D(w,\rho)=\{z\,:\,|z-b|<\rho\}\]
(with $\rho>0$) is a subset of $\Omega$.
Then we can find $a_{n}$ such that
\[f(z)=\sum_{n=-\infty}^{\infty}a_{n}(z-b)^{n}\]
for all $z\in D(b,\rho)$.
If $0N$, we say that
$w$ is a \emph{pole} or more specifically
that $w$ is a \emph{pole of order} $N$.
Sometimes a pole of order $1$ is called a
\emph{simple pole}\footnote{Complex analysts are much
attached to the word `simple'. Comment is unnecessary.}.
If there does not exist an $N$ with
$a_{-n}=0$ for all $n\geq N$, we call $w$
an \emph{essential singularity}.
\end{definition}
The next lemma is really just a commentary on
Definition~\ref{D; singularity}
\begin{lemma} We continue with the notation
of Theorem~\ref{T; Laurent} and Definition~\ref{D; singularity}.
(i) The point $b$ is a removable singularity if and
only if we can find
an analytic function
$\tilde{f}:\Omega\rightarrow{\mathbb C}$
such that $f(z)=\tilde{f}(z)$ for all $z\in\Omega\setminus\{b\}$.
(ii) The point $b$ is a pole of order exactly $N$ if and
only if we can find
an analytic function $h:\Omega\rightarrow{\mathbb C}$
with $h(b)\neq 0$
such that $f(z)=(z-b)^{-N}h(z)$
for all $z\in\Omega\setminus\{b\}$.
\end{lemma}
Thus the behaviour of an analytic function in
the neighbourhood of a removable singularity
or a pole is no more difficult to understand
than the behaviour of an analytic function
away from singularities.
We shall see in the next section that there are
particular reasons for wishing to calculate
the residue.
\begin{lemma} (i) If $f(z)=g(z)(z-a)^{-1}$ with $g$
analytic in a disc centre $a$ and, then, if $g(a)\neq 0$,
$f$ has simple pole at $a$ with residue
$g(a)$ and, if $g(a)=0$, $f$ has a removable singularity at $a$.
(i) If $f(z)=g(z)/h(z)$ with $g$ and $h$
analytic in a disc centre $a$ and $h(a)=0$, $h'(a)\neq 0$
then, if $g(a)\neq 0$, $f$ has simple pole at $a$ with residue
$g(a)/h'(a)$ and, if $g(a)=0$,
$f$ has a removable singularity at $a$.
\end{lemma}
If the pole is not simple, then power series expansion
is often, though not always, the simplest way of proceeding.
\begin{example} (The short question on paper I, 1998.)
Find the residues of the following functions at $z=0$,
using the principle branch of $\log$ in (iii).
\begin{center}
(i) $\cot z$,\ \ (ii) ${\displaystyle \frac{\sin z -z}{z^{4}}}$,
\ \ (iii) ${\displaystyle \frac{\log(\cos z)}{z(1-\cos z)}}$,\\
(iv) ${\displaystyle \frac{\cos z}{z^{2}}}$
\ \ \text{and}
\ \ \ (v) ${\displaystyle z^{3}\exp\left(\frac{1}{z}\right)}$.
\end{center}
\end{example}
\section{Calculus of residues}
The reason for devoting special attention to the
coefficient of $(z-w)^{-1}$ in the Laurent expansion
(recall that we called it the `residue') is revealed
in the next theorem\footnote{If you
invent a new and useful branch of mathematics,
then you too can
have all the theorems named after you.}.
\begin{theorem}[Cauchy's residue theorem]%
\label{T; Residue} Let
$\Omega$ be a simply connected
domain and $w_{1}$, $w_{2}$, \dots, $w_{n}$
distinct points in $\Omega$. Let
\[f:\Omega\setminus\{w_{1},\ w_{2},\ \dots ,\ w_{n}\}\rightarrow{\mathbb C}\]
be analytic and let the residue at $w_{j}$ be $\tau_{j}$
$[1\leq j\leq n]$. If $C$ is a simple closed contour
enclosing a region which contains
$\{w_{1},\ w_{2},\ \dots ,\ w_{n}\}$ [N.B. we do not
allow the $w_{j}$ to lie on C], then
\[\int_{C}f(z)\,dz=2\pi i\sum_{j=1}^{n}\tau_{j}.\]
\end{theorem}
\begin{exercise} Show that Theorem~\ref{T; extended Cauchy}
is a special case of Theorem~\ref{T; Residue}.
In the same spirit, deduce Theorem~\ref{T; extended Cauchy}
from Theorem~\ref{T; Residue} and the statement that
every analytic function satisfies Taylor's theorem
\[f(w+h)=\sum_{j=0}^{\infty}\frac{f^{(j)}(w)}{j!}h^{j}\]
for $|h|$ sufficiently small.
\end{exercise}
Here are some typical applications of Cauchy's residue theorem.
I have tried to place them in increasing order of complexity.
\begin{example} Show that
\[\int_{-\infty}^{\infty}\frac{1}{1+x^{4}}\,dx
=\frac{\pi}{2^{1/2}}.\]
Deduce the value of $\int_{0}^{\infty}\frac{1}{1+x^{4}}\,dx$.
\end{example}
\begin{example} Show that, if $\lambda$ is real,
\[\int_{-\infty}^{\infty}\frac{e^{i\lambda x}}{1+x^{2}}\,dx
=\pi e^{-|\lambda|}.\]
\end{example}
Our next integral requires a preliminary lemma.
\begin{lemma}[Jordan's lemma] Suppose
$f:{\mathbb C}\rightarrow{\mathbb C}$ is continuous
on the region $|z|>R_{0}$ and satisfies the inequality
$|f(z)|R_{0}$. If $C(r)$ is the contour
consisting of the semicircle $re^{i\theta}$ described
as $\theta$ runs from $0$ to $\pi$ then,
provided $\lambda$ is real and strictly positive,
\[\int_{C(r)}f(z)e^{i\lambda z}\,dz\rightarrow 0\]
as $r\rightarrow\infty$.
\end{lemma}
In the opinion of the writer, it is slightly
unsporting to use Jordan's lemma when simpler estimates
will do. It should also be noted that, if we \emph{genuinely}
need to use Jordan's lemma in the evaluation of a real
integral, then that integral may only exist for certain definitions
of the integral.
\begin{example}\label{Example; Hardy} Show that
\[\int_{0}^{\infty}\frac{\sin x}{x}\,dx=\frac{\pi}{2}.\]
\end{example}
\begin{exercise} If $t\in{\mathbb R}$, let us write
\[F(t)=\int_{0}^{\infty}\frac{\sin tx}{x}\,dx.\]
Show, using the result of Example~\ref{Example; Hardy}
and change of variable, that
\begin{alignat*}{2}
F(t)&=\frac{\pi}{2}&&\qquad\text{for $t>0$},\\
F(0)&=0,\\
F(t)&=-\frac{\pi}{2}&&\qquad\text{for $t<0$}.
\end{alignat*}
\end{exercise}
\begin{example} Show that, if $\alpha$ is real and $-1<\alpha<1$,
then
\[\int_{0}^{\infty}\frac{x^{\alpha}}{1+x^{2}}\,dx=
\frac{\pi}{2\cos(\alpha\pi/2)}.\]
What happens if $\alpha$ lies outside the range $(-1,1)$?
\end{example}
Our final example uses a slightly different idea.
\begin{example}
Show that, if $a$ is real and $a>0$,
\[\int_{0}^{\pi}\frac{a}{a^{2}+\sin^{2}\theta}\,d\theta
=\frac{\pi}{(1+a^{2})^{1/2}}.\]
\end{example}
There is a mixture of good and bad news about contour
integration.
(1) Most examples (particularly at 1B level) are based
on combining a limited number of tricks. If you are stuck,
try to identify parts of the problem which you have met before.
(2) The only way, for most people, to become fluent
in contour integration is to do lots of examples
yourself.
(3) Almost every book on complex analysis in your
college library\footnote{Often an architectural
gem and well worth visiting for its own sake.}
will contain a chapter with a large collection of
worked examples for you to take as model.
\section{Fourier transforms} Many systems
in nature, engineering and mathematics are
linear and allow us to build complex solutions
as linear combinations of simpler solutions.
Thus, for example, light and sound may be considered
as a mixture of simple, single frequency waves.
Mathematically we start by considering a single
frequency wave
\[e^{i\omega t},\]
we then consider a sum of a finite number of such
simple waves
\[\sum_{j=1}^{n}a_{j}e^{i\omega_{j} t}\]
with $a_{j}\in{\mathbb C}$ and are then driven to consider
the integral analogue
\[\int_{-\infty}^{\infty}F(\omega)e^{i\omega t}\,d\omega.\]
To emphasise the connection with Fourier series
(see the course Mathematical Methods,~C10) we use the
following definition.
\begin{definition} If $f:{\mathbb R}\rightarrow{\mathbb C}$
is reasonably well behaved, we define
\[\hat{f}(\lambda)
=\int_{-\infty}^{\infty}f(t)e^{-i\lambda t}\,dt,\]
and call the function
$\hat{f}:{\mathbb R}\rightarrow{\mathbb C}$
the \emph{Fourier transform}.
\end{definition}
This is a methods course, so we shall not go into what is meant
by good behaviour. However, the condition that $f$, $f'$ and $f''$
are continuous and
$t^{2}f(t),\ t^{2}f'(t),\ t^{2}f''(t)\rightarrow 0$
as $|t|\rightarrow\infty$ are amply sufficient
for our purpose (much less is required, but there
always has to be some control over behaviour towards
infinity).
The following results form part of the grammar of
Fourier transforms.
\begin{lemma} (i) If $a\in{\mathbb R}$, let us write
$f_{a}(t)=f(t-a)$. Then
\[\hat{f}_{a}(\lambda)=e^{-ia\lambda}\hat{f}(\lambda).\]
(Translation on one side gives phase change on other.)
(ii) If $K\in{\mathbb R}$ and $K>0$, let us write
$f_{K}(t)=f(Kt)$. Then
\[\hat{f}_{K}(\lambda)=K^{-1}\hat{f}(\lambda/K).\]
(Narrowing on one side gives broadening on the other.)
(iii) $\hat{f}(\lambda)^{*}=(f^{*})\hat{\ }(-\lambda)$.
(iv) $(\hat{f})'(\lambda)
=-i\hat{F}(\lambda)$ where $F(t)=tf(t)$.
(v) $(f')\hat{\ }(\lambda)=i\lambda\hat{f}(\lambda)$.
\end{lemma}
The next result is both elegant and important.
\begin{lemma}\label{L, Dual formula}
We have
\[\int_{-\infty}^{\infty}f(t)\hat{g}(t)\,dt
=\int_{-\infty}^{\infty}\hat{f}(\lambda)g(\lambda)\,d\lambda.\]
\end{lemma}
Taking $g(\lambda)=\exp(-(K^{-1}\lambda)^{2}/2)$
and allowing $K\rightarrow\infty$,
we obtain the key inversion formula.
\begin{theorem}[Inversion formula]
We have $f\hat{\ }\hat{\ }(t)=2\pi f(-t)$.
\end{theorem}
In other words,
\[f(t)=\frac{1}{2\pi}
\int_{-\infty}^{\infty}\hat{f}(\omega)e^{i\omega t}\,d\omega.\]
Thus we can break down any (well behaved) function
into its constituent frequencies and then reconstruct it.
The inversion formula gives a uniqueness result which
is often more useful than the inversion formula itself.
\begin{theorem}[Uniqueness] If $\hat{f}=\hat{g}$ then
$f=g$.
\end{theorem}
Combining the inversion formula with Lemma~\ref{L, Dual formula},
we get the following formula which is much loved by
1B~examiners and of considerable theoretical importance.
\begin{lemma}[Parseval's formula]\footnote{The opera has
an `f' and goes on for longer.} We have
\[\int_{-\infty}^{\infty}|f(t)|^{2}\,dt
=\frac{1}{2\pi}\int_{-\infty}^{\infty}|\hat{f}(\lambda)|^{2}\,d\lambda.\]
\end{lemma}
Fourier transforms are closely linked with the
important operation of convolution.
\begin{definition} If $f,\ g:{\mathbb R}\rightarrow{\mathbb C}$
are well behaved, we define their \emph{convolution}
$f*g:i{\mathbb R}\rightarrow{\mathbb C}$ by
\[f*g(t)=\int_{-\infty}^{\infty}f(t-s)g(s)\,ds.\]
\end{definition}
\begin{lemma} We have
$\widehat{f*g}(\lambda)=\hat{f}(\lambda)\hat{g}(\lambda)$.
\end{lemma}
For many mathematicians and engineers, Fourier transforms are
important because they convert convolution into multiplication
and convolution is important because it is transformed
by Fourier transforms into multiplication.
We shall see that convolutions occur naturally in the
study of differential equations.
It also occurs in probability theory where the sum $X+Y$
of two independent random variables $X$ and $Y$ with
probability densities $f_{X}$ and $f_{Y}$ is
$f_{X+Y}=f_{X}*f_{Y}$. In the next section
we outline the connection of convolution with signal
processing.
\section{Signals and such like} Suppose we have a black box
${\mathcal K}$. If we feed in a signal
$f:{\mathbb R}\rightarrow{\mathbb C}$ we will get out
a transformed signal
${\mathcal K}f:{\mathbb R}\rightarrow{\mathbb C}$.
Simple black boxes will have the following properties
(1) \emph{Time invariance} If ${\mathcal T}_{a}f(t)=f(t-a)$,
then ${\mathcal K}({\mathcal T}_{a}f)(t)=({\mathcal K}f)(t-a)$.
In other words, ${\mathcal K}{\mathcal T}_{a}
={\mathcal T}_{a}{\mathcal K}$.
(2) \emph{Causality} If $f(t)=0$ for $t<0$, then
$({\mathcal K}f)(t)=0$ for $t<0$. (The response
to a signal cannot precede the signal.)
(3) \emph{Stability} Roughly speaking, the black box
should consume rather than produce energy. Roughly
speaking, again, if there exists a $R$ such that
$f(t)=0$ for $|t|\geq R$, then we should have
$({\mathcal K}f)(t)\rightarrow 0$ as $t\rightarrow\infty$.
If conditions like this do not apply, both our mathematics
and our black box have a tendency to explode.
(Unstable systems may be investigated using a close
relative of the Fourier transform called the Laplace transform.)
(4) \emph{Linearity} In order for the methods of this course
to work, our black box must be linear, that is
\[{\mathcal K}(af+bg)=a{\mathcal K}(f)+b{\mathcal K}(g).\]
(Engineers sometimes spend a lot of effort converting
non-linear systems to linear for precisely this reason.)
As our first example of such a system, let us consider
the differential equation
\begin{equation*}
\tag*{$\bigstar$}
F''(t)+(a+b)F'(t)+ab F(t)=f(t)
\end{equation*}
(where $a,\ b>0$), subject to the boundary condition
$F(t),\ F'(t)\rightarrow 0$ as $t\rightarrow -\infty$.
We take ${\mathcal K}f=F$.
Before we can solve the system using Fourier transforms
we need a preliminary definition and lemma.
\begin{definition} The Heaviside function
$H:{\mathbb R}\rightarrow{\mathbb R}$ is given by
\begin{alignat*}{2}
H(t)&=0&&\qquad\text{for $t<0$},\\
H(t)&=1&&\qquad\text{for $t\geq 0$.}
\end{alignat*}
\end{definition}
\begin{lemma} Suppose that $\Re \alpha<0$. Then, if we set
$e_{\alpha}(t)=e^{\alpha t}H(t)$, we obtain
\[\hat{e}_{\alpha}(\lambda)=\frac{1}{i\lambda-\alpha}.\]
\end{lemma}
(Some applied mathematicians would leave out
the condition $\Re \alpha<0$ in the lemma
just given and most would write $\hat{H}(\lambda)=1/(i\lambda)$.
The study of Laplace transforms reveals why this reckless
behaviour does not lead to disaster.)
\begin{lemma}\label{Lemma, big star}
The solution $F={\mathcal K}f$ of
\begin{equation*}
\tag*{$\bigstar$}
F''(t)+(a+b)F'(t)+ab F(t)=f(t)
\end{equation*}
(where $a,\ b>0$), subject to the boundary condition
$F(t),\ F'(t)\rightarrow 0$ as $t\rightarrow -\infty$,
is given by
\[{\mathcal K}f=K\star f
\ \text{where}\ K(t)=\frac{e^{-bt}-e^{-at}}{a-b}H(t).\]
\end{lemma}
Observe that $K(t)=0$ for $t\leq 0$ and so, if $f(t)=0$
for $t\leq 0$, we have
\begin{alignat*}{2}
{\mathcal K}f(t)&=K\star f(t)=0\qquad\text{for $t\leq 0$},\\
{\mathcal K}f(t)&=K\star f(t)=\int_{0}^{t}f(s)K(t-s)\,ds
\qquad\text{for $t>0$}.
\end{alignat*}
Thus ${\mathcal K}$ is indeed causal.
There is another way of analysising black boxes. Let
$g_{n}$ be a sequence of functions such that
(i) $g_{n}(t)\geq 0$ for all $t$,
(ii) ${\displaystyle \int_{-\infty}^{\infty}g_{n}(t)\,dt=1}$,
(iii) $g_{n}(t)=0$ for $|t|>1/n$.
\noindent In some sense, the $g_{n}$ `converge' towards
the `idealised impulse function' $\delta$ whose
defining property runs as follows.
\begin{definition} If $f:{\mathbb R}\rightarrow{\mathbb R}$
is a well behaved function then
\[\int_{-\infty}^{\infty}f(t)\delta(t)\,dt=f(0).\]
\end{definition}
If the black box is well behaved we expect ${\mathcal K}f_{n}$
to converge to some function $E$. We write
\[{\mathcal K}\delta=E\]
and say that the response of the black box to the delta
function is the elementary solution $E$. Note that,
since our black box is causal, $K(t)=0$ for $t<0$.
If $f$ is a ordinary function, we define its translate
by some real number $a$ to be $f_{a}$ where
$f_{a}(t)=f(t-a)$. In the same way, we define the
translate by $a$ of the delta function by $a$ to be
$\delta_{a}$ where $\delta_{a}(t)=\delta(t-a)$ or,
more formally, by
\[\int_{-\infty}^{\infty}f(t)\delta_{a}(t)\,dt=
\int_{-\infty}^{\infty}f(t)\delta(t-a)\,dt=f(a).\]
Since our black box is time invariant, we have
\[{\mathcal K}\delta_{a}=E_{a}\]
and, since it is linear,
\[{\mathcal K}\sum_{j=1}^{n}\lambda_{j}\delta_{a_{j}}(t)=
\sum_{j=1}^{n}\lambda_{j}E_{a_{j}}(t).\]
In particular, if $F$ is a well behaved function,
\begin{align*}
{\mathcal K}\sum_{j=-MN}^{MN}N^{-1}F(j/N)\delta_{j/N}(t)&=
\sum_{j=-MN}^{MN}N^{-1}F(j/N)E_{j/N}(t)\\
&=\sum_{j=-MN}^{MN}N^{-1}F(j/N)E(t-j/N).
\end{align*}
Crossing our fingers and allowing $M$ and $N$ to tend
to infinity, we obtain
\[{\mathcal K}F(t)=\int_{-\infty}^{\infty}F(s)E(t-s)\,ds,\]
so
\[{\mathcal K}F=F*E.\]
Thus the response of the black box to a signal $F$
is obtained by convolving $F$ with the response of
the black box to the delta function. (This is why
the acoustics of concert halls are tested by letting off
starting pistols.) We now understand the importance
of convolution, delta functions and elementary solutions
in signal processing and the study of partial differential
equations.
To see what happens in our specific example, we use Fourier
transform methods find the elementary solution
of equation $\bigstar$.
\begin{lemma}\label{Lemma, Big elementary}
The solution $E={\mathcal K}\delta$ of
\begin{equation*}
\tag*{$\bigstar$}
E''(t)+(a+b)E'(t)+ab E(t)=\delta(t)
\end{equation*}
(where, $a,\ b>0$), subject to the boundary condition
$E(t),\ E'(t)\rightarrow 0$ as $t\rightarrow -\infty$,
is given by
\[E(t)=\frac{e^{-bt}-e^{-at}}{a-b}H(t).\]
\end{lemma}
Observe that Lemma~\ref{Lemma, Big elementary}
implies Lemma~\ref{Lemma, big star} and vice versa.
\section{Miscellany} The previous sections form a complete
course and I shall be happy simply to cover it.
If there is more time I will talk about some secondary
topics.
\begin{example} (Question~7, Paper II, 1996.)
(i) Find the poles of $\cot z$, and the residues at them.
Show that the first three terms of the Laurent expansion
of $\cot z$ in $0<|z|<\pi$ are
\[\frac{1}{z}-\frac{z}{3}-\frac{z^{3}}{45}.\]
(ii) Let $\Gamma_{N}$ be the rectangular contour with
vertices at $\pm(N+\frac{1}{2})\pm iN$, where $N$
is any positive integer. Show that, on this contour,
$|\cot\pi z|\leq\coth \pi N$. Hence, show that, for any
integer $r\geq 2$,
\[\int_{\Gamma_{N}}\frac{\cot\pi z}{z^{r}}\,dz
\rightarrow 0\ \text{as}\ N\rightarrow\infty.\]
Hence, using Part (a), show that
\[\sum_{n=1}^{\infty}\frac{1}{n^{2}}=\frac{\pi^{2}}{6},
\ \ \sum_{n=1}^{\infty}\frac{1}{n^{4}}=\frac{\pi^{4}}{90}.\]
\end{example}
The interested student can push matters a little further.
\begin{exercise} Show that, if $k$ is a strictly positive integer,
\[\sum_{n=1}^{\infty}\frac{1}{n^{2k}}=A_{k}\pi^{2k}\]
where $A_{k}$ is rational.
\end{exercise}
Over two centuries have passed since Euler obtained
the correct formula for $\sum_{n=1}^{\infty}\frac{1}{n^{2k}}$
but, apart from a recent result of Ap{\'e}ry to the
effect that $\sum_{n=1}^{\infty}\frac{1}{n^{3}}$ is
irrational, we know nothing about
$\sum_{n=1}^{\infty}\frac{1}{n^{2k+1}}$.
The next result is not part of the course but crops
up from time to time as a problem in the examinations.
It is also quite useful to have the general idea
of this result before embarking on the more
cautious modern treatment in more advanced courses like C12.
\begin{theorem}
Let $\Omega$
be a simply connected domain and $f:\Omega\rightarrow{\mathbb C}$
be analytic. Suppose that
$C$ is a simple closed contour in $\Omega$.
If $f$ has no zeros on $C$ and finitely many
zeros within $C$, then the change in argument of $f$
round $C$
\[[\arg f]_{C}=2\pi N\]
where $N$ is the number of zeros of $f$ within $C$,
multiple zeros being counted multiply.
\end{theorem}
Finally, I include a note on the derivative $\delta'$
of the $\delta$ function. Personally, I consider this
a bridge too far for this level, but some examiners
can not refrain from introducing it.
Suppose
$f$, $g:{\mathbb R}\rightarrow{\mathbb C}$ are well
behaved (in particular $f$, $f'$, $g$, $g'$ decrease
rapidly towards infinity). Then integration by
parts gives
\[\int_{-\infty}^{\infty}f(t)g'(t)\,dt
=-\int_{-\infty}^{\infty}f'(t)g(t)\,dt.\]
If $f$ and all its derivatives are well behaved (with rapid
decrease towards infinity) we are tempted to relax the conditions
on $g$. If $g=H$, the Heaviside function, this
leads to the \emph{formal} manipulations
\begin{align*}
\int_{-\infty}^{\infty}f(t)H'(t)\,dt
&=-\int_{-\infty}^{\infty}f'(t)H(t)\,dt\\
&=-\int_{0}^{\infty}f'(t)\,dt\\
&=f(0)
\end{align*}
and to the satisfactory conclusion that, in
some sense, $H'=\delta$.
If $g=\delta$ we get
\[\int_{-\infty}^{\infty}f(t)\delta'(t)\,dt
=-\int_{-\infty}^{\infty}f'(t)\delta(t)\,dt=-f'(0).\]
The minus sign confuses many students (and, possibly,
some of their elders and betters) but leaving out the
minus sign leads to all sorts of inconsistencies.
\section{Exercises} These exercises are divided into
three groups. I suggest that you work through the
questions in Part~A in order, getting as far as you
can in your alloted supervisions. Part~B consists
of questions for those who get through Part~A quickly.
They are not more difficult and will give you
further practice. Part~C consists of a few questions
which are a bit skew to the course but which
are quite interesting. The notation (Q$x$, Paper~X, 19AB)
tells you that the question is based on question~$x$
on Paper~X in 19AB, but I have often made slight changes.
\pagebreak[3]
\begin{center}
\bf{Part A}
\end{center}
\begin{question} Prove Lemma~\ref{L Easy differentiation}.
\end{question}
\begin{question} Prove Theorem~\ref{T Exponential}.
\end{question}
\begin{question} (This is Exercise~\ref{Exercise, logarithm 1}.)
We use the notation of
Definition~\ref{D, logarithm}. Show that we can not choose
$\theta_{0}$ so that $\log z_{1}z_{2}=\log z_{1}+\log z_{2}$
for all $z_{1}$, $z_{2}$, $z_{1}z_{2}\in\Omega$.
\end{question}
\begin{question} (i) Write out the standard properties of powers
$x^{\alpha}$ when $x$ and $\alpha$ are real and $x>0$.
(For example $(xy)^{\alpha}=x^{\alpha}y^{\alpha}$.)
Investigate the extent to which they remain true
in the complex case.
(ii) Show that $z^{1/3}$ has three possible branches
on ${\mathbb C}\setminus\{x\, :\, \text{$x$ real and $x\geq 0$}\}$.
Show that the same is true for $z^{2/3}$.
For each real $\alpha$ determine the number of branches
(possibly infinite) of $z^{\alpha}$.
\end{question}
\begin{question} (This is Exercise~\ref{E, conformal equivalence}.)
We say that open subsets $\Omega$ and $\Gamma$
of ${\mathbb C}$ are conformally equivalent
if there exists a conformal map $f:\Omega\rightarrow\Gamma$.
Show that conformal equivalence is an equivalence relation.
\end{question}
\begin{question} Let
\[\Lambda=\{w\,|\, 2\pi+\theta_{0}>\Im w>\theta_{0}\},
\ \Omega=\{z=re^{i\theta}\,:\,r>0,
\ 2\pi+\theta_{0}>\theta>\theta_{0}\}.\]
Let $f(z)=\exp z$ for $z\in\Lambda$. Show that
$f:\Lambda\rightarrow\Omega$ is conformal and use
Lemma~\ref{L conformal inverse} to establish
the existence of a function $\log$ with properties
given in Lemma~\ref{L, properties logarithm}.
\end{question}
\begin{question} (Q7(b), Paper~I, 1993)
For each of the following conformal
maps $f_{j}$ and simply connected
domains $D_{j}$ find the image of the
domain under the map (as usual $z=x+iy$).
\begin{alignat*}{4}
\text{(i)}&\ \ &f_{1}(z)&=1/(1+z),&&\qquad
D_{1}=\{x+iy\,:\, x^{2}+y^{2}<1,\ y>0\}\\
\text{(ii)}&\ \ &f_{2}(z)&=z^{2},&&\qquad
D_{2}=\{x+iy\,:\, x>0,\ y<0\}\\
\text{(iii)}&\ \ &f_{3}(z)&=\log z,&&\qquad
D_{3}=\{x+iy\,:\, y>0\}
\end{alignat*}
(You should make it clear which branch of $\log$ you
choose for $f_{3}$.)
Hence, or otherwise, show that
\[g(z)=\frac{1}{\pi}\log\left(
-\frac{1}{4}\left(\frac{1-z}{1+z}\right)^{2}
\right)\]
is a conformal map of $\{x+iy\,:\, x^{2}+y^{2}<1,\ y>0\}$
onto the infinite strip
\[\{x+iy\,:\, 00$.
Hence evaluate
\[\int_{0}^{\infty}\frac{\sin^{2}x}{1+x^{2}}\,dx.\]
\end{question}
\begin{question} (Q8(b), Paper~IV, 1994) Consider
the integral
\[I(a)=\int_{0}^{2\pi}\frac{d\theta}{(1+a\cos\theta)^{2}}\]
where $0<|a|<1$. By means of the substitution $z=e^{i\theta}$,
express $I(a)$ as an integral around the contour
$|z|=1$ and hence show that
\[I(a)=\frac{2\pi}{(1-a^{2})^{3/2}}.\]
\noindent[The examiner added that no credit would be given
for answers obtained by real methods.]
\end{question}
\begin{question} (Q16, Paper~II, 1997)
By integrating a branch of $(\log z)/(1+z^{4})$
about a suitable contour, show that
\[\int_{0}^{\infty}\frac{\log x}{1+x^{4}}\,dx
=-\frac{\pi^{2}}{8\sqrt{2}}\ ,\]
and evaluate
\[\int_{0}^{\infty}\frac{1}{1+x^{4}}\,dx.\]
\end{question}
\begin{question} (A golden oldie, last set as Q16, Paper~I, 1998)
Let
\[I(\alpha)=\int_{0}^{\infty}\frac{x^{\alpha}}{(x+1)^{3}}\,dx,\]
where $\alpha$ is real.
Use real methods to find the range of $\alpha$ for which the integral
converges. Use real methods to evaluate $I(0)$ and $I(1)$.
Now consider the integral of $z^{\alpha}/(z+1)^{3}$ around
a contour consisting of two circles of radius $R$ and $\epsilon$ and
straight lines on both sides of a cut along the positive
real axis. What restrictions must be placed on $\alpha$
for the contributions from the circles to become negligible
as $r\rightarrow\infty$ and $\epsilon\rightarrow 0$?
Under such conditions, show that
\[I(\alpha)=\frac{\pi\alpha(1-\alpha)}{2\sin\pi\alpha}.\]
Show that $I$ is continuous at $0$ and $1$.
\end{question}
\begin{question} (Q7(a), Paper~II, 1994) The
function $h(t)$ vanishes for $t<0$. The integral
\[P(t)=\int_{-\infty}^{t}h(t-\tau)f(\tau)\,d\tau\]
has the property that
\[\frac{df}{dt}=\int_{-\infty}^{t}h(t-\tau)P(\tau)\,d\tau\]
for all well behaved $f$.
Find $\hat{h}(\omega)^{2}$.
\end{question}
\begin{question} (Q7, Paper~II, 1998)
Suppose that
\[\hat{f}(\omega)=\frac{e^{i\omega}-e^{-i\omega}}{i\omega}\]
Find $f$ by the following two methods.
(i) By using formulae for such things as the Fourier
transforms of derivatives, translates
and Heaviside type functions
together with the uniqueness of Fourier transforms.
(ii) Directly from the inversion formula.
\noindent[\emph{Hint: You will need to distinguish between,
$t<-1$, $-11$.}]
\end{question}
\begin{question} (Q16, Paper~II, 1998) Use Fourier transform
methods to solve the following integral equation
for $f(t)$,
\begin{equation*}
f(t)+\int_{0}^{\infty}e^{-s}f(t-s)\,ds=
\begin{cases}
e^{-t}& \text{if $t\geq 0$,}\\
0& \text{if $t<0$.}
\end{cases}
\end{equation*}
Evaluate the convolution integral for your solution
and hence confirm that $f(t)$ solves the
integral equation in the form stated above.
\end{question}
\begin{question} (Q8(a), Paper~IV, 1994)
The analytic function $f(z)$ has $P$ poles and $Z$ zeros.
All the poles and zeros lie strictly within the smooth,
non-self-intersecting curve $C$. Using Cauchy's integral
formulas, show that, if all the poles and zeros are simple,
\[\int_{C}\frac{f'(z)}{f(z)}\,dz=2\pi(Z-P).\]
Explain how your result must be modified if the poles and zeros
are not simple and prove the modified result.
Restate your result in terms of the argument of $f$.
\end{question}
\begin{center}
\bf{Part B}
\end{center}
\begin{question}\label{Exercise, no Taylor}
Cauchy gave the following example of a well behaved
real function with no useful Taylor expansion
about $0$. It is important that you work through it
at some stage in your mathematical life.
Let $E:{\mathbb R}\rightarrow{\mathbb R}$ be defined
by $E(t)=\exp(-1/t^{2})$ for $t\neq 0$ and $E(0)$.
Use induction to show that $E$ is infinitely differentiable
with
\begin{align*}
E^{(n)}(t)&=Q_{n}(1/t)E(t)\qquad\text{for $t\neq 0$},\\
E^{(n)}(0)&=0.
\end{align*}
For which values of $t$ is it true that
\[E(t)=\sum_{n=0}^{\infty}\frac{E^{n}(0)t^{n}}{n!}?\]
Why does this not contradict Theorem~\ref{T; Taylor}?
\end{question}
\begin{question} (Q7, Paper~I, 1999) For each of the following
five functions, state the region of the complex plane
in which it is complex differentiable. State also the region in which
it has partial derivatives and the region in which
they satisfy the Cauchy-Riemann conditions.
\begin{align*}
f_{1}(z)&=|z|,\\
f_{2}(z)&=e^{-z},\\
f_{3}(z)&=z^{*},\\
f_{4}(z)&=(z-1)^{3},\\
f_{5}(z)&=|z|^{2}\\
\end{align*}
and $f_{6}$ given by
\begin{equation*}
f_{6}(x+iy)=
\begin{cases}
\frac{xy}{(x^{2}+y^{2})^{1/2}}& \text{if $x+iy\neq0$,}\\
0& \text{if $x+iy=0$.}
\end{cases}
\end{equation*}
\noindent[\emph{Note that, in one case, the region of
complex differentiability does not coincide with that
of the validity of the Cauchy-Riemann equations.}]
\end{question}
\begin{question} (Q17, Paper~IV, 1999) Write down the
Cauchy-Riemann equations for the real and imaginary parts
of the analytic function $w(z)=u(x,y)+iv(x,y)$, where $z=x+iy$.
Show that $\triangledown^{2}u=\triangledown^{2}v=0$
(i.e. that $u$ and $v$ are harmonic functions). Prove
also that the curves of constant $u$ in the $x,y$ plane
intersect those of constant $v$ orthogonally.
Find analytic functions $w_{1}(z)$ and $w_{2}(z)$ that are real
for real $z$ and for which the following functions are
their respective real parts:
(i)\ \ \ $u_{1}(x,y)=e^{x}\cos y$,
(ii)\ \ \ ${\displaystyle u_{2}(x,y)=
\frac{x(x^{2}+y^{2}+1)}{2(x^{2}+y^{2})}.}$
From your answer to (ii) find a non-zero harmonic function that
vanishes on the circle $x^{2}+y^{2}=1$ and on the line
$y=0$.
In case~(i), find the images in the $z$-plane of the circles
$|w_{1}|=\rho$, for constant $\rho$.
In case~(ii), find the images in the $w_{2}$-plane of the circles
$|z|=r$, for constant $r>1$.
\end{question}
\begin{question} (Q7(b), Paper~I, 1996)
In four of the following five cases
there exists a bijective analytic map $f:U\rightarrow V$.
In one case there is a topological reason why no
such map is possible. Find a suitable $f$ in the four cases
and briefly explain the fifth.
\begin{alignat*}{3}
&\text{(i)}&&\ U=\{z\in{\mathbb C}\,:\,\Im(z)>0\},&&
\qquad V=\{z\in{\mathbb C}\,:\,\Im(z)>0\}\\
&\text{(ii)}&&\ U=\{z\in{\mathbb C}\,:\,|z|<1\},&&
\qquad V=\{z\in{\mathbb C}\,:\,\Re(z)>0,\ \Im(z)>0\}\\
&\text{(iii)}&&\ U=\{z\in{\mathbb C}\,:\,2>|z|>1\},&&
\qquad V=\{z\in{\mathbb C}\,:\,|z|<1\}\\
&\text{(iv)}&&\ U={\mathbb C}\setminus\{z\in{\mathbb R}\,:\,z\geq 0\},&&
\qquad V={\mathbb C}\setminus\{z\in{\mathbb R}\,:\,|z|\geq 1\}\\
&\text{(v)}&&\ U=\{z\in{\mathbb C}\,:\,0<\Im(z)<1\},&&
\qquad V={\mathbb C}\setminus\{z\in{\mathbb R}\,:\,z\geq 0\}.
\end{alignat*}
\noindent[\emph{In case (iv) you may find it useful to consider
the effect of a translation followed by the map $z\mapsto 1/z$.}]
\end{question}
\begin{question} (Q7(a), Paper~I, 1995) Find the residue
at each of the poles of the function
\[f(z)=\frac{1}{z^{2}(1+z^{4})}\]
in the complex plane.
\end{question}
\begin{question} (Q7, Paper~II, 1999) You are asked to find the
Laurent expansion about $z=0$ for each of the following
three functions.
\begin{align*}
f_{1}(z)&=e^{1/z},\\
f_{2}(z)&=z^{-1/2},\\
f_{3}(z)&=\frac{\sinh z}{z^{3}}.
\end{align*}
In one case, you reply that you can not supply such
an expansion. Why?
In the other two cases, where there is a Laurent expansion,
state the nature of the singularity at $z=0$ and find its residue.
Show that there is a function $f$ analytic on ${\mathbb C}$
except, possibly, at finitely many points such that
\[f(z)=\sum_{n=1}^{\infty}z^{-n}\]
for $|z|>1$. Find any singularities of $f(z)$ in the region
$|z|\leq 1$ and find the residues at those singularities.
\end{question}
\begin{question} (Q7(a), Paper~II, 1995)
What are the poles and associated residues of $f(z)=(\cosh z)^{-1}$
in the complex $z$-plane?
By considering a rectangular contour, or otherwise, evaluate
the Fourier transform
\[\int_{-\infty}^{\infty}
\frac{e^{-ikx}}{\cosh x}\,dx.\]
\end{question}
\begin{question} Show that, for $a>b>0$, we have
\[I(a,b)=\int_{0}^{\infty}\frac{\cos x}{(x^{2}+a^{2})(x^{2}+b^{2})}\,dx
=\frac{\pi}{2(a^{2}-b^{2})}\left(\frac{1}{be^{b}}-\frac{1}{ae^{a}}\right).\]
Find $I(a,a)$ for $a>0$ and check that $I(a,b)\rightarrow I(a,a)$
as $b\rightarrow a$.
Find $I(a,b)$ for all real non-zero values of $a$ and $b$.
\end{question}
\begin{question} (Q7(b), Paper~II, 1993)
By integrating around an appropriate closed curve in the
complex plane cut along one half of the real axis,
show that
\[I(a)=\int_{0}^{\infty}\frac{x^{a-1}}{1+x+x^{2}}\,dx
=\frac{2\pi}{\sqrt{3}}
\cos\left(\frac{2\pi a+\pi}{6}\right)
\csc(\pi a)\]
if $00$,}\\
0& \text{if $t<0$.}
\end{cases}
\end{equation*}
Hence determine the function $y(t)$.
Check, by direct substitution, that $y$ is, indeed, the
required solution.
\noindent[\emph{The examiner said that you could use
the Fourier inversion theorem and Jordan's lemma without
proof. In spite of this invitation, I suggest you might
be better off using the uniqueness of Fourier transforms.}]
\end{question}
\begin{question} (Q17, Paper~IV, 1997) Use
Fourier transforms to find $g$ in the equation
\[5e^{-|t|}-8e^{-2|t|}+3e^{-3|t|}
=\int_{-\infty}^{\infty}g(\tau)e^{-|t-\tau|}\,d\tau.\]
\end{question}
\begin{question} (Q17, Paper~IV, 2000)
Assuming suitable decay of the function $w(x)$
as $x\rightarrow\pm\infty$, express the Fourier transforms
of $w'(x)$ and $xw(x)$ in terms of the Fourier transform of
$w(x)$.
Find the form of the Fourier transform of $w(x)$, if
\begin{equation*}
\tag*{($*$)}
x(w''(x)-w(x))+w'(x)=0.
\end{equation*}
By using the inversion formula and a suitable
change of variables, or otherwise, deduce that
\[w(x)=\frac{1}{\pi}\int_{0}^{\infty}
\cos(x\sinh u)\,du\]
is a solution of $(*)$.
\end{question}
\begin{center}
\bf{Part C}
\end{center}
\begin{question}(Q16, Paper~II, 2000) [This is not
really very hard but is a bit out of the ordinary
and a bit beyond the syllabus.]
The complex plane is cut along the real axis from $z=-1$
to $z=1$, and the branch of $f(z)=(z^{2}-1)^{1/2}$ is chosen
so that $f(z)$ is real and positive when $z$ is real
and $z>1$. Obtain expressions for $f(z)$ just above and
just below the cut and also when $|z|\gg 1$.
Show that
\[\int_{-1}^{1}\frac{\sqrt{1-x^{2}}}{1+x^{2}}\,dx
=\pi(\sqrt{2}-1),\]
where the square root gives positive values for $-10$ and that $f(t)=\theta(t)e^{-\gamma t}(1-e^{-t})$.
Find $K(t)$ for $t>0$ assuming that $K(t)=0$
for $t<0$. Explain why you can drop the assumption
$K(t)=0$ for $t<0$. Is it possible to find $K(t)$
for $t<0$ from the information given? Why?
When $g(t)=\delta'(t)$, use the expression previously found
for $K(t)$ to calculate $f(t)$ both
(i) by direct evaluation of the integral and
(ii) by calculating the Fourier transform of $g$
in this case and hence finding the Fourier transform
of $f$.
\noindent[Most students obtain different answers for~(i)
and~(ii). If this happens to you, the object of the game
is to discover what has gone wrong.]
\end{question}
\end{document}