Jensen's Formula/Proof 2

Proof
Write
 * $\map f z = \dfrac {r^2 - z \overline {\rho_1} } {r \paren {z - \rho_1} } \cdots \dfrac {r^2 - z \overline {\rho_n} } {r \paren {z - \rho_n} } \map g z$

so $\map g z \ne 0$ for $z \in D_r$.

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

When $\cmod z = r$, we have:
 * $\dfrac 1 z = \dfrac {\overline z} {r^2}$

and:
 * $\cmod {\dfrac z r} = 1$

where $\overline z$ denotes the complex conjugate of $z$.

So:

so the is $0$.

Moreover:
 * $\\ln \cmod {\dfrac {r^2 - 0 \overline {\rho_k} } {r \paren {0 - \rho_k} } } = \ln \cmod {\rho_k} - \ln r$

so the is $0$.

Therefore, the formula holds for $\dfrac {r^2 - z \overline {\rho_k} } {r \paren {z - \rho_k} }$.

Since:
 * $r^2 - z \overline {\rho_i} = 0 \implies \cmod z = r^2 / \cmod {\overline {\rho_i} } > r$

it follows that $\map g z$ is holomorphic without zeroes on $D_r$.

So the is $\ln \size {\map g 0}$.

On the other hand, by the mean value property:
 * $\displaystyle \dfrac 1 {2 \pi} \int_0^{2 \pi} \ln \size {\map g {r e^{i \theta} } } \rd \theta = \ln \map g 0$

as required.