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C. Results from Analytic Function Theory

C.8.3 Poisson's Integral for the Unit Disk

Theorem C.10  Let $f(z)$ be analytic inside the unit disk. Then, if $z_0=re^{j\theta}$, with $0\leq r <1$,


\begin{displaymath} f(z_0)=\frac{1}{2\pi}\int_0^{2\pi}P_{1,r}(\theta-\omega)f(e^{j\omega})d\omega \end{displaymath} (C.8.17)

where $P_{1,r}(x)$ is the Poisson kernel defined by


\begin{displaymath} P_{\rho,r}(x)\stackrel{\rm\triangle}{=}\frac{\rho^2-r^2}{\rh... ...\cos(x)+r^2}\hspace{20mm}0\leq r <\rho, \hspace{10mm}x\in \Re \end{displaymath} (C.8.18)

Proof

Consider the unit circle $C$. Then, using Theorem C.8, we have that


\begin{displaymath} f(z_0)=\frac{1}{2\pi j}\oint_{C}\frac{f(z)}{z-z_0}dz \end{displaymath} (C.8.19)

Define


\begin{displaymath}z_1\stackrel{\rm\triangle}{=}\frac{1}{r}e^{j\theta} \end{displaymath} (C.8.20)

Because $z_1$ is outside the region encircled by $C$, the application of Theorem C.8 yields


\begin{displaymath} 0=\frac{1}{2\pi j} \oint_{C}\frac{f(z)}{z-z_1}dz \end{displaymath} (C.8.21)

Subtracting (C.8.21) from (C.8.19) and changing the variable of integration, we obtain


\begin{displaymath}f(z_0)=\frac{1}{2\pi}\int_0^{2\pi}f(e^{j\omega})e^{j\omega} \... ...j\theta}} - \frac{r}{re^{j\omega}-e^{j\theta}} \right ]d\omega \end{displaymath} (C.8.22)

from which the result follows.

$\Box \Box \Box $

Consider now a function $g(z)$ which is analytic outside the unit disk. We can then define a function $f(z)$ such that


\begin{displaymath}f(z)\stackrel{\rm\triangle}{=}g\left(\frac{1}{z} \right) \end{displaymath} (C.8.23)

Assume that one is interested in obtaining an expression for $g(\zeta_0)$, where $\zeta_0=re^{j\theta}$, $r>1$. The problem is then to obtain an expression for $f\left(\frac{1}{\zeta_0} \right)$. Thus, if we define $z_0\stackrel{\rm\triangle}{=}\frac{1}{\zeta_0}=\frac{1}{r}e^{-j\theta}$, we have, on applying Theorem C.10, that


\begin{displaymath}g(\zeta_0)=\frac{1}{2\pi}\int_0^{2\pi}P_{1,\frac{1}{r}}(-\theta-\omega)g(e^{-j\omega})d\omega \end{displaymath} (C.8.24)

where


\begin{displaymath}P_{1,\frac{1}{r}}(-\theta-\omega)=\frac{r^2-1}{r^2-2rcos(\theta+\omega)+1} \end{displaymath} (C.8.25)

If, finally, we make the change in the integration variable $\omega=-\nu$, the following result is obtained.


\begin{displaymath}g(re^{j\theta})=\frac{1}{2\pi}\int_0^{2\pi}\frac{r^2-1}{r^2-2rcos(\theta-\nu)+1} g(e^{j\nu})d\nu \end{displaymath} (C.8.26)

Thus, Poisson's integral for the unit disk can also be applied to functions of a complex variable which are analytic outside the unit circle.