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

Without loss of generality, assume that $r = 0$.



Write:

$\map f z = \map {B_{\rho_1} } z \dotsm \map {B_{\rho_n} } z \map g z$

where $\map {B_\rho} z := \dfrac {z - \rho} {1 - z \overline \rho}$ is the Blaschke factor at $\rho$.

So:

$\map g z$ is holomorphic and nonzero in $D_r$

and:

each $\map {B_{\rho_i} } z$ 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:

$\map h z = \map {B_{\rho_k} } z = \dfrac {z - \rho_k} {1 - z \overline {\rho_k} }$

for some $k \in \set {1, \ldots, n}$.

Note that:

$\size {\map h z} \equiv 1$ for $\size z = 1$

so the integral is $0$.

Also:

$\map {B_{\rho_k} } 0 = \rho_k$

so the equality holds in this case.




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

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



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.