Jensen's Formula/Proof 2

Proof
Write
 * $\displaystyle f \left({z}\right) = \frac{ r^2 - z \overline{\rho_1} } {r \left({z - \rho_1}\right)} \cdots \frac{r^2 - z \overline {\rho_n} } {r \left({z - \rho_n}\right)} g \left({z}\right)$

so $g \left({z}\right) \ne 0$ for $z \in D_r$.

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

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

and:
 * $\left\vert{\dfrac z r}\right\vert = 1$

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

So:

so the LHS is $0$.

Moreover:
 * $\displaystyle \ln \left\vert{\frac {r^2 - 0 \overline{\rho_k} } {r \left({0 - \rho_k}\right)} }\right\vert = \ln \left\vert{\rho_k}\right\vert - \ln r$

so the RHS is $0$.

Therefore, the formula holds for $\displaystyle\frac{r^2-z \overline{\rho_k}}{r(z - \rho_k)}$.

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

it follows that $g \left({z}\right)$ is holomorphic without zeroes on $D_r$.

So the RHS is $\ln \left\vert{g \left({0}\right)}\right\vert$.

On the other hand, by the mean value property:
 * $\displaystyle \frac 1 {2 \pi} \int_0^{2 \pi} \ln \left\vert{g \left({r e^{i \theta} }\right)}\right\vert \, \mathrm d \theta = \ln g \left({0}\right)$

as required.