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6. Complex function theory

6.1 Functions of complex variables

Complex function theory deals with complex functions of a complex variable. Some definitions:

$f$ is analytical on $\cal G$ if $f$ is continuous and differentiable on $\cal G$ .

A Jordan curve is a curve that is closed and singular.

If K is a curve in $\mathbb{C}$ with parameter equation $z=\phi(t)=x(t)+iy(t)$ , $a\leq t\leq b$ , than the length $L$ of K is given by: $L=\int\limits_a^b \sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}dt= \int\limits_a^b\left|\frac{dz}{dt}\right|dt=\int\limits_a^b|\phi'(t)|dt$ The derivative of $f$ in point $z=a$ is: $f'(a)=\lim_{z\rightarrow a}\frac{f(z)-f(a)}{z-a}$ If $f(z)=u(x,y)+iv(x,y)$ the derivative is: $f'(z)=\frac{\partial u}{\partial x}+i\frac{\partial v}{\partial x}=-i\frac{\partial u}{\partial y}+\frac{\partial v}{\partial y}$ Setting both results equal yields the equations of Cauchy-Riemann: $\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}~~~,~~~\frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}$ These equations imply that $\nabla^2u=\nabla^2v=0$ . $f$ is analytical if $u$ and $v$ satisfy these equations.

6.2 Complex integration

6.2.1 Cauchy's integral formula

Let

$K$ be a curve described by

$z=\phi(t)$ on

$a\leq t\leq b$ and

$f(z)$ is continuous on

$K$ . Than the integral of

$f$ over

$K$ is:

$\int\limits_Kf(z)dz=\int\limits_a^bf(\phi(t))\dot{\phi}(t)dt \stackrel{f\;\mbox{continuous}}{=}F(b)-F(a)$ Lemma: let

$K$ be the circle with center

$a$ and radius

$r$ taken in a positive direction. Than holds for integer

$m$ :

$\frac{1}{2\pi i}\oint\limits_K \frac{dz}{(z-a)^m}=\left\{ \begin{array}{l} 0~~\mbox{if}~~m\neq1\\ 1~~\mbox{if}~~m=1 \end{array}\right.$ Theorem: if

$L$ is the length of curve

$K$ and if

$|f(z)|\leq M$ for

$z\in K$ , than, if the integral exists, holds:

$\left|\int\limits_K f(z)dz\right|\leq ML$ Theorem: let

$f$ be continuous on an area

$G$ and let

$p$ be a fixed point of

$G$ . Let

$F(z)=\int_p^zf(\xi)d\xi$ for all

$z\in G$ only depend on

$z$ and not on the integration path. Than

$F(z)$ is analytical on

$G$ with

$F'(z)=f(z)$ .

This leads to two equivalent formulations of the main theorem of complex integration: let the function $f$ be analytical on an area $G$ . Let $K$ and $K'$ be two curves with the same starting - and end points, which can be transformed into each other by continous deformation within $G$ . Let $B$ be a Jordan curve. Than holds $\int\limits_Kf(z)dz=\int\limits_{K'}f(z)dz\Leftrightarrow\oint\limits_Bf(z)dz=0$ By applying the main theorem on ${\rm e}^{iz}/z$ one can derive that $\int\limits_0^\infty\frac{\sin(x)}{x}dx=\frac{\pi}{2}$

6.2.2 Residue

A point

$a\in\mathbb{C}$ is a regular point of a function

$f(z)$ if

$f$ is analytical in

$a$ . Otherwise

$a$ is a singular point or pole of

$f(z)$ . The residue of

$f$ in

$a$ is defined by

$\mathop{\rm Res}\limits_{z=a}f(z)=\frac{1}{2\pi i}\oint\limits_Kf(z)dz$ where

$K$ is a Jordan curve which encloses

$a$ in positive direction. The residue is 0 in regular points, in singular points it can be both 0 and

$\neq0$ . Cauchy's residue proposition is: let

$f$ be analytical within and on a Jordan curve

$K$ except in a finite number of singular points

$a_i$ within

$K$ . Than, if

$K$ is taken in a positive direction, holds:

$\frac{1}{2\pi i}\oint\limits_Kf(z)dz=\sum_{k=1}^n\mathop{\rm Res}\limits_{z=a_k}f(z)$ Lemma: let the function

$f$ be analytical in

$a$ , than holds:

$\mathop{\rm Res}\limits_{z=a}\frac{f(z)}{z-a}=f(a)$ This leads to Cauchy's integral theorem: if

$F$ is analytical on the Jordan curve

$K$ , which is taken in a positive direction, holds:

$\frac{1}{2\pi i}\oint\limits_K\frac{f(z)}{z-a}dz=\left\{\begin{array}{l} f(a)~~\mbox{if}~~a~~\mbox{inside}~~K\\ 0~~\mbox{if}~~a~~\mbox{outside}~~K \end{array}\right.$ Theorem: let

$K$ be a curve (

$K$ need not be closed) and let

$\phi(\xi)$ be continuous on

$K$ . Than the function

$f(z)=\int\limits_K\frac{\phi(\xi)d\xi}{\xi-z}$ is analytical with

$n$ -th derivative

$f^{(n)}(z)=n!\int\limits_K\frac{\phi(\xi)d\xi}{(\xi-z)^{n+1}}$ Theorem: let

$K$ be a curve and

$G$ an area. Let

$\phi(\xi,z)$ be defined for

$\xi\in K$ ,

$z\in G$ , with the following properties:

$\phi(\xi,z)$ is limited, this means $|\phi(\xi,z)|\leq M$ for $\xi\in K$ , $z\in G$ ,
For fixed $\xi\in K$ , $\phi(\xi,z)$ is an analytical function of $z$ on $G$ ,
For fixed $z\in G$ the functions $\phi(\xi,z)$ and $\partial\phi(\xi,z)/\partial z$ are continuous functions of $\xi$ on $K$ .

Than the function $f(z)=\int\limits_K\phi(\xi,z)d\xi$ is analytical with derivative $f'(z)=\int\limits_K\frac{\partial\phi(\xi,z)}{\partial z}d\xi$ Cauchy's inequality: let $f(z)$ be an analytical function within and on the circle $C:|z-a|=R$ and let $|f(z)|\leq M$ for $z\in C$ . Than holds $\left|f^{(n)}(a)\right|\leq\frac{Mn!}{R^n}$

6.3 Analytical functions definied by series

The series

$\sum f_n(z)$ is called pointwise convergent on an area

$G$ with sum

$F(z)$ if

$\forall_{\varepsilon>0}\forall_{z\in G}\exists_{N_0\in I\hspace{-1mm}R}\forall_{n>n_0} \left[~\left|f(z)-\sum_{n=1}^Nf_n(z)\right|<\varepsilon\right]$ The series is called uniform convergent if

$\forall_{\varepsilon>0}\exists_{N_0\in I\hspace{-1mm}R}\forall_{n>n_0}\exists_{z\in G} \left[~\left|f(z)-\sum_{n=1}^Nf_n(z)\right|<\varepsilon\right]$ Uniform convergence implies pointwise convergence, the opposite is not necessary.

Theorem: let the power series $\sum\limits_{n=0}^\infty a_nz^n$ have a radius of convergence $R$ . $R$ is the distance to the first non-essential singularity.

If $\displaystyle\lim_{n\rightarrow\infty}\sqrt[n]{|a_n|}=L$ exists, than $R=1/L$ .
If $\displaystyle\lim_{n\rightarrow\infty}|a_{n+1}|/|a_n|=L$ exists, than $R=1/L$ .

If these limits both don't exist one can find $R$ with the formula of Cauchy-Hadamard: $\frac{1}{R}=\lim_{n\rightarrow\infty}{\rm sup}\sqrt[n]{|a_n|}$

6.4 Laurent series

Taylor's theorem: let

$f$ be analytical in an area

$G$ and let point

$a\in G$ has distance

$r$ to the boundary of

$G$ . Than

$f(z)$ can be expanded into the Taylor series near

$a$ :

$f(z)=\sum_{n=0}^\infty c_n(z-a)^n~~~\mbox{with}~~~c_n=\frac{f^{(n)}(a)}{n!}$ valid for

$|z-a| < r$ . The radius of convergence of the Taylor series is

$\geq r$ . If

$f$ has a pole of order

$k$ in

$a$ than

$c_1,...,c_{k-1}=0$ ,

$c_k\neq0$ .

Theorem of Laurent: let $f$ be analytical in the circular area $G:r < |z-a| < R$ . Than $f(z)$ can be expanded into a Laurent series with center $a$ : $f(z)=\sum_{n=-\infty}^\infty c_n(z-a)^n~~~\mbox{with}~~~ c_n=\frac{1}{2\pi i}\oint\limits_K\frac{f(w)dw}{(w-a)^{n+1}}~~,~~n\in Z\hspace{-1ex}Z$ valid for $r<|z-a| The principal part of a Laurent series is: $\sum\limits_{n=1}^\infty c_{-n}(z-a)^{-n}$ . One can classify singular points with this. There are 3 cases:

There is no principal part. Than $a$ is a non-essential singularity. Define $f(a)=c_0$ and the series is also valid for $|z-a| < R$ and $f$ is analytical in $a$ .
The principal part contains a finite number of terms. Than there exists a $k\in I\hspace{-1mm}N$ so that $\lim\limits_{z\rightarrow a}(z-a)^kf(z)=c_{-k}\neq0$ . Than the function $g(z)=(z-a)^kf(z)$ has a non-essential singularity in $a$ . One speaks of a pole of order $k$ in $z=a$ .
The principal part contains an infinite number of terms. Then, $a$ is an essential singular point of $f$ , such as $\exp(1/z)$ for $z=0$ .

If $f$ and $g$ are analytical, $f(a)\neq0$ , $g(a)=0$ , $g'(a)\neq0$ than $f(z)/g(z)$ has a simple pole (i.e. a pole of order 1) in $z=a$ with $\mathop{\rm Res}\limits_{z=a}\frac{f(z)}{g(z)}=\frac{f(a)}{g'(a)}$

6.5 Jordan's theorem

Residues are often used when solving definite integrals. We define the notations

$C_\rho^+=\{z||z|=\rho,\Im(z)\geq0\}$ and

$C_\rho^-=\{z||z|=\rho,\Im(z)\leq0\}$ and

$M^+(\rho,f)=\mathop{\rm max}\limits_{z\in C_\rho^+}|f(z)|$ ,

$M^-(\rho,f)=\mathop{\rm max}\limits_{z\in C_\rho^-}|f(z)|$ . We assume that

$f(z)$ is analytical for

$\Im(z)>0$ with a possible exception of a finite number of singular points which do not lie on the real axis,

$\lim\limits_{\rho\rightarrow\infty}\rho M^+(\rho,f)=0$ and that the integral exists, than

$\int\limits_{-\infty}^\infty f(x)dx=2\pi i\sum{\rm Res}f(z)~~~\mbox{in}~~~\Im(z)>0$ Replace

$M^+$ by

$M^-$ in the conditions above and it follows that:

$\int\limits_{-\infty}^\infty f(x)dx=-2\pi i\sum{\rm Res}f(z)~~~\mbox{in}~~~\Im(z)<0$ Jordan's lemma: let

$f$ be continuous for

$|z|\geq R$ ,

$\Im(z)\geq0$ and

$\lim\limits_{\rho\rightarrow\infty}M^+(\rho,f)=0$ . Than holds for

$\alpha>0$

$\lim_{\rho\rightarrow\infty}\int\limits_{C_\rho^+}f(z){\rm e}^{i\alpha z}dz=0$ Let

$f$ be continuous for

$|z|\geq R$ ,

$\Im(z)\leq0$ and

$\lim\limits_{\rho\rightarrow\infty}M^-(\rho,f)=0$ . Than holds for

$\alpha<0$

$\lim_{\rho\rightarrow\infty}\int\limits_{C_\rho^-}f(z){\rm e}^{i\alpha z}dz=0$ Let

$z=a$ be a simple pole of

$f(z)$ and let

$C_\delta$ be the half circle

$|z-a|=\delta,0\leq{\rm arg}(z-a)\leq\pi$ , taken from

$a+\delta$ to

$a-\delta$ . Than is

$\lim_{\delta\downarrow0}\frac{1}{2\pi i}\int\limits_{C_\delta}f(z)dz=\frac{1}{2}\mathop{\rm Res}\limits_{z=a}f(z)$