mth305.tex

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\begin{document}
\title{Complex Analysis}
\author{Ronald Mangang}
\maketitle
\begin{abstract}
\noindent Complex analysis investigates functions of complex variables.
Holomorphic functions (also called analytic functions) are the heart
of complex analysis. These functions behave very nicely --- for instance,
they are infinitely differentiable and are equal to their own Taylor
series. These notes are taken during a spring 2022 undergrad course
of MTH305: Complex Analysis course at IISER Mohali. 
\end{abstract}
\newpage

\section{Complex numbers}

\subsection{Algebra of complex numbers}

Using elementary algebra, one finds the square root of a complex number
$\alpha+i\beta$ as
\begin{equation}
\sqrt{\alpha+i\beta}=\pm\left(\sqrt{\frac{\alpha+\sqrt{\alpha^{2}+\beta^{2}}}{2}}+i\frac{\beta}{|\beta|}\sqrt{\frac{-\alpha+\sqrt{\alpha^{2}+\beta^{2}}}{2}}\right)
\end{equation}
provided that $\beta\ne0$. If $\beta=0$ the square roots are $\pm\sqrt{\alpha}$
if $\alpha\geq0$, $\pm i\sqrt{-\alpha}$ if $\alpha<0$.

A complex number $\alpha+i\beta$ may be represented in various ways.
It can be associated to the coordinate $(\alpha,\beta)$ in the two-dimensional
plane or the matrix
\[
\begin{pmatrix}\alpha & \beta\\
-\beta & \alpha
\end{pmatrix}.
\]
One can verify that these representations are all isomorphic to one
another. $\mathbb{C}$ can also be thought as the field $\mathbb{R}[x]/(x^{2}+1)$.

The absolute value (or modulus) has the following properties
\begin{align}
|z_{1}+z_{2}|^{2} & =|z_{1}|^{2}+|z_{2}|^{2}+2\text{ Re }z_{1}\bar{z_{2}},\\
|z_{1}-z_{2}|^{2} & =|z_{1}|^{2}+|z_{2}|^{2}-2\text{ Re }z_{1}\bar{z_{2}}.
\end{align}
The triangle inequality is particularly important and shows that $\mathbb{C}$
is a metric space.
\begin{equation}
|z_{1}+z_{2}|\leq|z_{1}|+|z_{2}|.
\end{equation}
Cauchy-Schwarz inequality states that
\begin{equation}
|\sum_{j=1}^{n}a_{j}b_{j}|^{2}\leq\sum_{j=1}^{n}|a_{j}|^{2}\sum_{j=1}^{n}|b_{j}|^{2}.
\end{equation}
for complex numbers $a_{j}$ and $b_{j}$.

\subsection{Geometry of complex numbers}

\textbf{de Moivre's formula. }The following formula gives an easy
way to express $\cos n\phi$ and $\sin n\phi$ in terms of $\cos\phi$
and $\sin\phi$.
\begin{equation}
(\cos\phi+i\sin\phi)^{n}=\cos n\phi+i\sin n\phi.
\end{equation}
It follows from the above formula that the $n$th roots of a complex
number $z=r(\cos\phi+i\sin\phi)$ are given by
\begin{equation}
\sqrt[n]{z}=\sqrt[n]{r}\left[\cos\left(\frac{\phi}{n}+k\frac{2\pi}{n}\right)+\sin\left(\frac{\phi}{n}+k\frac{2\pi}{n}\right)\right].
\end{equation}
where $k=0,1,\ldots,n-1$. These $n$th roots are the vertices of
a regular $n$-gon. For a particularly interesting case, set
\begin{equation}
\omega=\cos\frac{2\pi}{n}+i\sin\frac{2\pi}{n}.
\end{equation}
Then the $n$th roots of unity are $1,\omega,\omega^{2},\ldots,\omega^{n-1}$.

\textbf{Analytic geometry. }Often analytic geometry on the complex
plane gives an easier or more elegant way to solve problems than using
$x$ and $y$ in $\mathbb{R}^{2}.$

A directed line $z=a+bt$ determines a right half plane consisting
of all points $z$ with $\text{Im }(z-a)/b<0$ and a left half plane
with $\text{Im }(z-a)/b>0$.

\textbf{The Riemann sphere. }Consider the unit sphere $S$ in the
three-dimensional space given by $x_{1}^{2}+x_{2}^{2}+x_{3}^{3}=1$.
To each $(x_{1},x_{2},x_{3})$ on $S$ except $(0,0,1)$ we associate
a complex number 
\begin{equation}
z=\frac{x_{1}+ix_{2}}{1-x_{3}}.
\end{equation}
This correspondence is one-one. A little computation yields
\begin{equation}
x_{1}=\frac{z+\bar{z}}{1+|z|^{2}},\ x_{2}=\frac{z-\bar{z}}{i(1+|z|^{2})},\ x_{3}=\frac{|z|^{2}-1}{|z|^{2}+1}.
\end{equation}
The correspondence is completed by associating the point at $\infty$
to $(0,0,1)$. This way we regard the sphere as a representation of
the extended complex plane. The hemisphere $x_{3}<0$ corresponds
to the disk $|z|<1$ and the hemisphere $x_{3}>0$ to its outside
$|z|>1.$

\section{Holomorphic functions}

\subsection{Differentiation of complex functions}

Suppose $f$ is a complex-valued function of a single complex variable
$z$. The derivative of $f$ at a point $z_{0}$ in its domain is
defined by the limit:
\begin{equation}
f^{\prime}(z_{0})=\lim_{z\to z_{0}}\frac{f(z)-f(z_{0})}{z-z_{0}}.
\end{equation}
For the limit to exist, it must have the same value for any sequence
of complex values for $z$ that approach $z_{0}$ on the complex plane.
Just like real differentiability, complex differentiability is linear
and obeys the product rule, quotient rule and the chain rule.

The function $f$ is said to be holomorphic on an open set $U$ if
$f$ is complex differentiable at every point in $U$. If $f$ is
complex differentiable on some open neighbourhood of $z_{0}$, $f$
is said to be holomorphic at $z_{0}$.
\begin{example*}
\begin{flushleft}
The function $f(z)=|z|^{2}$ is complex differentiable at exactly
one point $(z_{0}=0)$. Thus it is not holomorphic at $0$.
\par\end{flushleft}
\end{example*}

\subsection{Cauchy-Riemann equations}

Suppose $f(x+iy)=u+iv$ is a holomorphic function. Then
\begin{equation}
\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}\ \text{ and }\ \frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}.
\end{equation}
These are called the Cauchy-Riemann equations.

A holomorphic function also satisfies:
\begin{equation}
\frac{\partial f}{\partial\bar{z}}=0.
\end{equation}
which implies that $f$ is functionally independent from $\bar{z}$.

If a function $f(z)=u+iv$ satisfies the Cauchy-Riemann equations
and if the partial derivates are also continuous then $f$ is holomorphic.

The derivative of $f$ is given by:
\begin{equation}
f^{\prime}(z)=\frac{\partial u}{\partial x}+i\frac{\partial v}{\partial x}.
\end{equation}
Using the Cauchy-Riemann equations, $f^{\prime}(z)$ can be written
in four different ways. Also,
\[
|f^{\prime}(z)|^{2}=\left(\frac{\partial u}{\partial x}\right)^{2}+\left(\frac{\partial u}{\partial y}\right)^{2}=\frac{\partial u}{\partial x}\frac{\partial v}{\partial y}-\frac{\partial u}{\partial y}\frac{\partial v}{\partial x}.
\]
That is, $|f^{\prime}(z)|^{2}$ is the Jacobian of $u$ and $v$ with
respect to $x$ and $y$.

\textbf{Laplace's equation.} If $f(z)=u+iv$ is holomorphic then $u(x,y)$
and $v(x,y)$ satisfy the Laplace's equation:
\begin{equation}
\frac{\partial^{2}u}{\partial x^{2}}+\frac{\partial^{2}v}{\partial y^{2}}=0\ \text{ and }\ \frac{\partial^{2}v}{\partial x^{2}}+\frac{\partial^{2}v}{\partial y^{2}}=0.
\end{equation}

Any function that satisfies the Laplace's equation is said to be harmonic.
If two harmonic functions $u$ and $v$ satisfy the Cauchy-Riemann
equations, then $v$ is said to be the conjugate harmonic function
of $u$ and $u$ is the conjugate harmonic function of $-v$.

The conjugate of a harmonic function can be found by integration.
It is determined only upto an additive constant.

There is another way to compute the holomorphic function $f(z)$ whose
real part is a given harmonic function $u(x,y)$ without use of integration.
We treat $\bar{f}(\bar{z})$, the conjugate of $f(z)$, as a function
of $\bar{z}$ only and write
\[
u(x,y)=\frac{1}{2}\left[f(x+iy)+\bar{f}(x-iy)\right].
\]
This is a formal identity and it holds even when $x$ and $y$ are
complex. If we substitute $x=z/2,y=z/2i$ we obtain
\[
u\left(\frac{z}{2},\frac{z}{2i}\right)=\frac{1}{2}\left[f(z)+\bar{f}(0)\right].
\]
Since $f(z)$ is only determined up to a purely imaginary constant,
we assume that $f(0)$ is real, which implies $\bar{f}(0)=u(0,0)$.
Then $f(z)$ can be computed by
\[
f(z)=2u\left(\frac{z}{2},\frac{z}{2i}\right)-u(0,0).
\]
A purely imaginary constant can be added at will.

\subsection{Polynomials and rational functions}

\textbf{Polynomials. }Consider the $n$th degree polynomial in complex
coefficients:
\begin{equation}
P(z)=a_{0}+a_{1}z+\cdots+a_{n}z^{n}.\label{eq:poly}
\end{equation}
where $a_{n}\neq0$. It is trivial to see that $P(z)$ is holormorphic.
We can write
\begin{equation}
P(z)=a_{n}(z-\alpha_{1})\cdots(z-\alpha_{n}).\label{eq:polyfacto}
\end{equation}
This factorization is unique except for the order of the factors.
If $\alpha_{i}$ repeats $k$ times $(k\leq n)$ then we say the order
of the zero $\alpha_{i}$ is $k$.

Suppose $\alpha$ is a zero of order $k$. Then $P(z)=(z-\alpha)^{k}P_{k}(z)$
where $P_{k}(\alpha)\neq0$. Also, $P(\alpha)=P^{\prime}(\alpha)=\cdots=P^{(k-1)}(\alpha)=0$
and $P^{(k)}(\alpha)\neq0$ (by successive differentiation). A simple
zero is a zero of order $1$ and it satisfies $P(\alpha)=0,P^{\prime}(\alpha)\neq0$.
\begin{thm}[Lucas's theorem]
If all zeroes of a polynomial $P(z)$ lie in a half plane, then all
zeroes of the derivative $P^{\prime}(z)$ lie in the same half plane.
\end{thm}

\begin{proof}
From \ref{eq:polyfacto} we obtain
\begin{equation}
\frac{P^{\prime}(z)}{P(z)}=\frac{1}{z-\alpha_{1}}+\cdots+\frac{1}{z-\alpha_{n}}.
\end{equation}

{[}ADD MORE CONTENT{]}
\end{proof}
\textbf{Rational functions. }Consider the rational function
\[
R(z)=\frac{P(z)}{Q(z)}.
\]
where $P(z)$ and $Q(z)$ has no common factors. We put $R(z)=\infty$
when $Q(z)=0$ such that $R(z)$ is continuous in the extended complex
plane. The zeroes of $Q(z)$ are poles of $R(z)$ and the order of
a pole is equal to the order of the corresponding zero of $Q(z)$.

The derivative
\begin{equation}
R^{\prime}(z)=\frac{P^{\prime}(z)Q(z)-Q^{\prime}(z)P(z)}{Q(z)^{2}}.
\end{equation}
exists when $Q(z)\ne0$. As a rational function, $R^{\prime}(z)$
has the same poles as that of $Q(z)$.

To define $R(\infty)$ we put $R(1/z)=R_{1}(z)$ and equate $R(\infty)=R_{1}(0).$
Suppose
\[
R(z)=\frac{a_{0}+a_{1}z+\cdots+a_{n}z^{n}}{b_{0}+b_{1}z+\cdots+b_{m}z^{m}}.
\]
Then 
\[
R_{1}(z)=z^{m-n}\frac{a_{0}z^{n}+\cdots+a_{n}}{b_{0}z^{m}+\cdots+b_{m}}.
\]
If $m>n$ $R(z)$ has a zero of order $m-n$ at $\infty$, if $m<n$
the point at $\infty$ is a pole of order $n-m$ and if $m=n$ $R(\infty)=a_{n}/b_{n}$.

The number of zeroes, including those at $\infty$, is equal to $\max(m,n)$.
The number of poles is the same. This common number is called the
order of $R(z)$.

$R(z)-a$ and $R(z)$ have the same order for a constant $a$. Thus
a rational funcrtion $R(z)$ of order $p$ has $p$ zeroes and $p$
poles, and every equation $R(z)=a$ has exactly $p$ roots.

Rational functions of order $1$ are called linear transformations.
The linear transformation $z+a$ is called a parallel translation
and $1/z$ is an inversion.

\textbf{Partial fractions. }Suppose $Q(z)$ is a polynomial with distinct
roots $\alpha_{1},\ldots,\alpha_{n}$ and if $P(z)$ is a polynomial
of degree $<n$ then
\begin{equation}
\frac{P(z)}{Q(z)}=\sum_{k=1}^{n}\frac{P(\alpha_{k})}{Q^{\prime}(\alpha_{k})(z-\alpha_{k})}.
\end{equation}
This is also called Heavyside's cover-up method of decomposing a rational
function into partial fractions. We can also write
\[
\frac{P(z)}{Q(z)}=\sum_{k=1}^{n}\frac{A_{k}}{(z-\alpha_{k})}
\]
where $A_{k}$ are constants and use the method of undetermined coefficients
to compute $A_{k}$. If $\alpha_{s}$ is a zero of order $p$ and
every other zero is of order $1$ then we put
\[
\frac{P(z)}{Q(z)}=\sum_{k=1,k\ne s}^{n}\frac{A_{k}}{(z-\alpha_{k})}+\frac{A_{s1}}{(z-\alpha_{s})}+\frac{A_{s2}}{(z-\alpha_{s})^{2}}+\cdots+\frac{A_{sp}}{(z-\alpha_{s})^{p}}.
\]
We do the same for every zero of order $>1$.

\subsection{Power series}

\textbf{Sequences. }Often it becomes difficult to prove convergence
of sequence through explicitly determining its limit. An easier way
is to prove that it is a Cauchy sequence.

Suppose $\{b_{n}\}$ and $\{a_{n}\}$ are two sequences such that
$|b_{m}-b_{n}|\leq|a_{m}-a_{n}|$ for all pairs of subscripts then
under Cauchy's condition, if $\{a_{n}\}$ is a Cauchy sequence, so
is $\{b_{n}\}$.

\textbf{Series. }To an infinite series
\begin{equation}
a_{1}+a_{2}+\cdots+a_{n}+\cdots\label{eq:series}
\end{equation}
we associate a sequence of partial sums
\begin{equation}
s_{n}=a_{1}+a_{2}+\cdots+a_{n}.
\end{equation}
The series converges iff the sequence of its partial sums converges
and the limit of the sequence is the sum of the series.

The series \ref{eq:series} can be associated with another series
\begin{equation}
|a_{1}|+|a_{2}|+\cdots+|a_{n}|+\cdots.\label{eq:abseries}
\end{equation}
Since $|a_{n}+a_{n+1}+\cdots+a_{n+p}|\leq|a_{n}|+|a_{n+1}|+\cdots+|a_{n+p}|$,
the convergence of \ref{eq:abseries} implies that the original series
\ref{eq:series} is convergent. A series with the property that the
series formed by the absolute values of the terms converges is said
to be absolutely convergent.

\textbf{Uniform convergence vs. pointwise convergence. }Consider the
limit
\[
\lim_{n\to\infty}\left(1+\frac{1}{n}\right)x=x.
\]
This is true for all $x$. However, to have $|(1+1/n)x-x|=|x|/n<\epsilon$
for $n\geq n_{0}$ it is necessary that $n_{0}>|x|/\epsilon$. Such
an $n_{0}$ exists for every fixed $x$ but the requirement cannot
be met simultaneously for all $x$. This is the case of pointwise
convergence.

A sequence $\{f_{n}(x)\}$ converges uniformly to $f(x)$ on the set
$E$ if to every $\epsilon>0$ there exists an $n_{0}$ such that
$|f_{n}(x)-f(x)|<\epsilon$ for all $n\geq n_{0}$ and all $x$ in
$E$. Unlike pointwise convergence, $n_{0}$ does not depend of $x$.

Cauchy's condition for uniform convergence would then be: the sequence
$\{f_{n}(x)\}$ converges uniformly on $E$ iff to every $\epsilon>0$
there exists an $n_{0}$ such that $|f_{m}(x)-f_{n}(x)|<\epsilon$
for all $m,n\geq n_{0}$ and all $x$ in $E$.

Also if $|f_{m}(x)-f_{n}(x)|\leq|a_{m}-a_{n}|$ on $E$ and $\{a_{n}\}$
is convergent then $\{f_{n}(x)\}$ converges uniformly on $E$.

A series with variable terms 
\[
f_{1}(x)+f_{2}(x)+\cdots+f_{n}(x)+\cdots
\]
has the series with positive terms 
\[
a_{1}+a_{2}+\cdots+a_{n}+\cdots
\]
for a majorant if $|f_{n}(x)|\leq Ma_{n}$ for some constant $M$
and for all sufficiently large $n$. The first series is a minorant
of the second. Then
\[
|f_{n}(x)+f_{n+1}(x)+\cdots+f_{n+p}(x)|\leq M(a_{n}+a_{n+1}+\cdots+a_{n+p}).
\]
If the majorant converges, the minorant converges uniformly. This
is called the Weierstrass M test.

\textbf{Power series. }For every power series
\begin{equation}
a_{0}+a_{1}z+\cdots+a_{n}z^{n}+\cdots,
\end{equation}
where $a_{i},z\in\mathbb{C}$, there exists $R$, $0\leq R\leq\infty$,
called the radius of covergence such that
\begin{enumerate}
\item The series converges absolutely for every $z$ with $|z|<R$. If $0\leq\rho<R$
the convergence is uniform for $|z|\leq\rho$.
\item If $|z|>R$ the series diverges.
\item In $|z|<R$ the sum of the series is analytic whose derivative can
be obtained by termwise differentiation and the derived series has
the same radius of convergence.
\end{enumerate}
Hadamard's formula gives
\begin{equation}
\frac{1}{R}=\lim_{n\to\infty}\sup\sqrt[n]{|a_{n}|}.
\end{equation}
It follows that a power series with positive $R$ has derivatives
of all orders.
\begin{thm}[Abel's limit theorem]
 If $\sum_{n=0}^{\infty}a_{n}$ converges, then $f(z)=\sum_{n=0}^{\infty}a_{n}z^{n}\to f(1)$
as $z\to1$ in such a way that $|1-z|/(1-|z|)$ remains bounded. 
\end{thm}


\subsection{Euler's formula}

\textbf{The exponential. }The solution to the differential equation
$f^{\prime}(z)=f(z)$ with the initial value $f(0)=1$ can be found
by representing $f(z)$ as a power series and differentiating it.
We call this solution the exponential $e^{z}$. We find that
\begin{equation}
e^{z}=1+z+\frac{z^{2}}{2!}+\cdots+\frac{z^{n}}{n!}+\cdots
\end{equation}
Since $\sqrt[n]{n!}\to\infty$ the exponential $e^{z}$ converges
in the whole complex plane.

A consequence of the differential equation is
\begin{equation}
e^{a+b}=e^{a}\cdot e^{b}.
\end{equation}
Since $e^{z}\cdot e^{-z}=1$ it is clear that $e^{z}$ is never $0$.
Since the series has real coefficients $e^{\bar{z}}$ is the complex
conjugate of $e^{z}$. Hence $|e^{iy}|^{2}=e^{iy}\cdot e^{-iy}=1$
and $|e^{x+iy}|=e^{x}$.

\textbf{The trigonometric functions. }The trigonometric functions
are defined by
\begin{equation}
\cos z=\frac{e^{iz}+e^{-iz}}{2},\ \sin z=\frac{e^{iz}-e^{-iz}}{2i}.
\end{equation}
Using the expansion of $e^{z}$ we obtain
\begin{align*}
\cos z & =1-\frac{z^{2}}{2!}+\frac{z^{4}}{4!}-\cdots\\
\sin z & =z-\frac{z^{3}}{3!}+\frac{z^{5}}{5!}-\cdots
\end{align*}

Also we obtain Euler's formula
\begin{equation}
e^{iz}=\cos z+i\sin z.
\end{equation}
All the trigonometric identities including the definitions of other
trigonometric functions can be derived. All trigonometric functions
are rational functions of $e^{iz}.$

\textbf{The logarithm. }By definition, $z=\log w$ is a root of the
equation $e^{z}=w$. For $w\ne0$ the equation $e^{x+iy}=w$ is equivalent
to 
\begin{equation}
e^{x}=|w|,\ \ \ \ e^{iy}=\frac{w}{|w|}.
\end{equation}
Then $x=\log|w|$. Also there is only one $y$ satisfying the above
equation in the interval $0\leq y<2\pi.$ Of course, it has period
$2\pi$. Thus
\[
\log w=\log|w|+i\arg w.
\]
Every complex number other than $0$ has infinitely many logarithms
which differ from each other by multiples of $2\pi i$. Also if $a\ne0$
we write
\[
a^{b}=e^{b\log a}.
\]
See that if \textbf{$a$ }is complex and \textbf{$b$ }is a rational
number in the reduced form $p/q$ then $a^{b}$ will have exactly
$q$ values.

The addition theorem of the exponential function yields
\begin{align*}
\log(z_{1}z_{2}) & =\log z_{1}+\log z_{2}\\
\arg(z_{1}z_{2}) & =\arg z_{1}+\arg z_{2},
\end{align*}
in the sense that both sides represent the same infinite set of complex
numbers. If we want to compare a value on the left with a value on
the right, then we can merely assert that they differ by a multiple
of $2\pi i$ (or $2\pi)$.

\textbf{The inverse trigonometric functions. }The inverse cosine can
be obtained by solving the equation
\[
\cos z=\frac{1}{2}(e^{iz}+e^{-iz})=w
\]
which is a quadratic equation in $e^{iz}$ with the roots $e^{iz}=w\pm\sqrt{w^{2}-1}$
and hence
\[
z=\arccos w=-i\log(w\pm\sqrt{w^{2}-1}).
\]
Since $w+\sqrt{w^{2}-1}$ and $w-\sqrt{w^{2}-1}$ are reciprocal numbers
we may write
\[
\arccos w=\pm i\log(w+\sqrt{w^{2}-1}).
\]
The inverse sine can then be defined by 
\[
\arcsin w=\frac{\pi}{2}-\arccos w.
\]


\section{Mappings}
\end{document}