Jensen's Formula/Proof 3

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Theorem

Let $S$ be an open subset of the complex plane containing the closed disk:

$D_r = \set {z \in \C : \cmod z \le r}$

of radius $r$ about $0$.

Let $f: S \to \C$ be holomorphic on $S$.

Let $f$ have no zeroes on the circle $\cmod z = r$.

Let $\map f 0 \ne 0$.

Let $\rho_1, \ldots, \rho_n$ be the zeroes of $f$ in $D_r$, counted with multiplicity.

Then:

$(1): \quad \ds \frac 1 {2 \pi} \int_0^{2 \pi} \ln \cmod {\map f {r e^{i \theta} } } \rd \theta = \ln \cmod {\map f 0} + \sum_{k \mathop = 1}^n \paren {\ln r - \ln \size {\rho_k} }$

Proof

This article, or a section of it, needs explaining. In particular: I don't see why $r = 0$ is WLOG? You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by explaining it. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Explain}} from the code.

Assume without loss of generality that $r = 0$.

Write $\map f z = B_{\rho_1}(z) \dotsm B_{\rho_n}(z) \map g z$,

where $B_\rho(z) := \frac{z - \rho}{1 - z \overline{\rho} }$ is the Blaschke factor at $\rho$, so $\map g z$ is holomorphic and nonzero in $D_r$, and each $B_{\rho_i}(z)$ is holomorphic and has a simple zero $\rho_i$ inside $D_r$.

It is sufficient to check the equality for each factor of $f$ in this expansion.

First let $\map h z = B_{\rho_k}(z) = \frac {z - \rho_k} {1 - z \overline{\rho_k} }$ for some $k \in \set {1, \ldots, n}$.

Note that $|\map h z| \equiv 1$ for $|z| = 1$, so the integral is 0. Also $B_{\rho_k}(0) = \rho_k$, so the equality holds in this case.

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To show equality for $\map g z$, first observe that by the Residue Theorem:

$\ds \int_{\size z = r} \frac {\ln \map g z} z \rd z = 2 \pi i \ln \map g 0$

This article, or a section of it, needs explaining. In particular: Is $\ln \map g z$ well-defined? You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by explaining it. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Explain}} from the code.

Therefore substituting $z = r e^{i \theta}$ we have

$\ds 2 \pi i \ln \map g 0 = i \int_0^{2 \pi} \ln \map g {r e^{i \theta} } \rd \theta$

Comparing the imaginary parts of this equality we see that:

$\ds \frac 1 {2 \pi} \int_0^{2 \pi} \ln \size {\map g {r e^{i \theta} } } \rd \theta = \ln \size {\map g 0}$

as required.

$\blacksquare$

Source of Name

This entry was named for Johan Jensen.