Jensen's Formula/Proof 1

Proof
Write:
 * $\map f z = \paren {z - \rho_1} \dotsm \paren {z - \rho_n} \map g z$

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

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

First let:
 * $\map h z = z - \rho_k$

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

Making use of the substitution $u = r e^{i \theta} - \rho_k$ we find that:


 * $\ds \frac 1 {2 \pi} \int_0^{2 \pi} \ln \size {\map h {r e^{i \theta} } } \rd \theta = \frac 1 {2 \pi i} \int_\gamma \frac {\ln \size u} {u + \rho_k} \rd u$

where $\gamma$ is a circle of radius $r$ centred at $-\rho_k$, traversed anticlockwise.

On this circle, $\ln \size u = \ln r$ is constant, and we have that:


 * $\ds \int_\gamma \frac 1 {u + \rho_k} \rd u = \int_{\size z \mathop = r} \frac {\d u} u = 2 \pi i$

Therefore the of $(1)$ is $\ln r$ as required.

To show equality for $\map g z$, first observe that by Cauchy's 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.