Jensen's Formula/Proof 3

Proof
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.

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


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

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.