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C. Results from Analytic Function Theory

C.8.4 Poisson-Jensen Formula for the Unit Disk

Lemma C.2  Consider a function $g(z)$ having the following properties:

$g(z)$ is analytic on the closed unit disk;
$g(z)$ does not vanish on the unit circle;
$g(z)$ has zeros in the open unit disk, located at $\bar{\alpha}_1, \bar{\alpha}_2,
		    \ldots, \bar{\alpha}_{\bar{n}}$.

Consider also a point $z_0=re^{j\theta}$ such that $r<1$; then

			\ln \vert g(z_0)\vert=\sum_{i=1}^{\bar{n}}\ln \left \vert \...
			...pi}P_{1,r}(\theta-\omega)\ln \vert g(e^{j\omega})\vert d\omega
			\end{displaymath} (C.8.27)



\begin{displaymath}\tilde{g}(z)\stackrel{\rm\triangle}{=}g(z)\prod_{i=1}^n \frac{1-\bar{\alpha}_i^*z}{z-\bar{\alpha}_i}
			\end{displaymath} (C.8.28)

Then $\ln \tilde{g}(z)$ is analytic on the closed unit disk. If we now apply Theorem C.10 to $\ln \tilde{g}(z)$, we obtain

\begin{displaymath}\ln \tilde{g}(z_0)= \ln {g(z_0)}+
			\sum_{i=1}^n\ln \left ( \fr...
			...^{2\pi}P_{1,r}(\theta-\omega)\ln \tilde{g}(e^{j\omega})d\omega
			\end{displaymath} (C.8.29)

We also recall that, if $x$ is any complex number, then $\ln
		  x=ln\vert x\vert+j\angle x$. Thus the result follows upon equating real parts in the equation above and noting that

\begin{displaymath}\ln \left \vert\tilde{g}(e^{j\omega}) \right\vert=\ln \left \vert g(e^{j\omega}) \right\vert
			\end{displaymath} (C.8.30)

$\Box \Box \Box $

Theorem C.11 (Jensen's formula for the unit disk)  Let $f(z)$ and $g(z)$ be analytic functions on the unit disk. Assume that the zeros of $f(z)$ and $g(z)$ on the unit disk are $\bar{\alpha}_1, \bar{\alpha}_2,
		  \ldots, \bar{\alpha}_{\bar{n}}$ and $\bar{\beta}_1, \bar{\beta}_2, \ldots, \bar{\beta}_{\bar{m}}$respectively, where none of these zeros lie on the unit circle.


			h(z)\stackrel{\rm\triangle}{=}z^\lambda \frac{f(z)}{g(z)} \hspace{10mm}\lambda\in \Re
			\end{displaymath} (C.8.31)


			\frac{1}{2\pi}\int_0^{2\pi}\ln \vert h(e^{j\omega})\vert d\...{\alpha}_1\bar{\alpha}_2\ldots \bar{\alpha}_{\bar{n}} \vert}
			\end{displaymath} (C.8.32)


We first note that $\ln \vert h(z)\vert=\lambda \ln \vert z\vert+\ln \vert f(z)\vert-\ln
		\vert g(z)\vert$. We then apply the Poisson-Jensen formula to $f(z)$ and $g(z)$ at $z_0=0$ to obtain

\begin{displaymath}P_{1,r}(x)=P_{1,0}(x)=1; \hspace{10mm}
			\ln \left \vert \frac...
			...-\bar{\beta}_i^*z_0 } \right
			\end{displaymath} (C.8.33)

We thus have that

$\displaystyle \ln\vert f(0)\vert=\sum_{i=1}^n\ln\vert\bar{\alpha}_i\vert- \frac{1}{2\pi}\int_0^{2\pi}\ln
			\vert f(e^{j\omega})\vert d\omega$     (C.8.34)
$\displaystyle \ln\vert g(0)\vert=\sum_{i=1}^n\ln\vert\bar{\alpha}_i\vert- \frac{1}{2\pi}\int_0^{2\pi}\ln
			\vert g(e^{j\omega})\vert d\omega$     (C.8.35)

The result follows upon subtracting Equation (C.8.35) from (C.8.34), and noting that

\begin{displaymath}\frac{\lambda}{2\pi}\int_0^{2\pi}\ln\left\vert e^{j\omega}\right\vert d\omega=0
			\end{displaymath} (C.8.36)

$\Box \Box \Box $

Remark C.3  Further insights can be obtained from Equation (C.8.32) if we assume that, in (C.8.31), $f(z)$ and $g(z)$ are polynomials;

$\displaystyle f(z)$ $\textstyle =$ $\displaystyle K_f\prod_{i=1}^n (z-\alpha_i)$ (C.8.37)
$\displaystyle g(z)$ $\textstyle =$ $\displaystyle \prod_{i=1}^n (z-\beta_i)$ (C.8.38)


\begin{displaymath}\left \vert\frac{f(0)}{g(0)} \right \vert=\vert K_f\vert \lef...{\prod_{i=1}^n \alpha_i }
			{\prod_{i=1}^m \beta_i}\right\vert
			\end{displaymath} (C.8.39)

Thus, $\alpha_1, \alpha_2, \ldots \alpha_n$ and $\beta_1, \beta_2, \ldots \beta_m$ are all the zeros and all the poles of $h(z)$, respectively, that have nonzero magnitude.

This allows Equation (C.8.32) to be rewritten as

			\frac{1}{2\pi}\int_0^{2\pi}\ln \vert h(e^{j\omega})\vert d\...
			...a_{nu}' \vert}
			{\vert\beta_1'\beta_2'\ldots \beta_{mu}'\vert}
			\end{displaymath} (C.8.40)

where $\alpha_1', \alpha_2', \ldots \alpha_{nu}'$ and $\beta_1', \beta_2', \ldots \beta_{mu}'$ are the zeros and the poles of $h(z)$, respectively, that lie outside the unit circle .

$\Box \Box \Box $