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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 $0\leq r <1$,

			\end{displaymath} (C.8.17)

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

			...\cos(x)+r^2}\hspace{20mm}0\leq r <\rho,
			\hspace{10mm}x\in \Re
			\end{displaymath} (C.8.18)


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

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


			\end{displaymath} (C.8.20)

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

			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

			...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

			\end{displaymath} (C.8.24)


			\end{displaymath} (C.8.25)

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

			\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.