Jensen's Formula/Proof 1

Theorem

Let $S$ be an open subset of the complex plane containing the closed disk:

$D_r = \left\{ {z \in \C : \left\vert{z}\right\vert \le r}\right\}$

of radius $r$ about $0$.

Let $f: S \to \C$ be holomorphic on $S$.

Let $f$ have no zeroes on the circle $\left\vert{z}\right\vert = r$.

Let $f \left({0}\right) \ne 0$.

Let $\rho_1, \ldots, \rho_n$ be the zeroes of $f$ in $D_r$, counted with multiplicity.

Then:

$(1): \quad \displaystyle \frac 1 {2 \pi} \int_0^{2 \pi} \ln \left\vert{f \left({r e^{i \theta} }\right)}\right\vert \, \mathrm d \theta = \ln \left\vert{f \left({0}\right)}\right\vert + \sum_{k \mathop = 1}^n \left({\ln r - \ln \left\vert{\rho_k}\right\vert}\right)$

Proof

Write $f \left({z}\right) = \left({z - \rho_1}\right) \cdots \left({z - \rho_n}\right) g \left({z}\right)$, so $g \left({z}\right) \ne 0$ for $z \in D_r$.

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

First let $h \left({z}\right) = z - \rho_k$ for some $k \in \left\{{1, \ldots, n}\right\}$.

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

$\displaystyle \frac 1 {2 \pi} \int_0^{2 \pi} \ln \left\vert{h \left({r e^{i \theta} }\right)}\right\vert \ \mathrm d \theta = \frac 1 {2 \pi i} \int_\gamma \frac {\ln \left\vert{u}\right\vert} {u + \rho_k} \, \mathrm d u$

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

On this circle, $\ln \left\vert{u}\right\vert = \ln r$ is constant, and we have that:

$\displaystyle \int_\gamma \frac 1 {u + \rho_k} \ \mathrm d u = \int_{\left\vert{z}\right\vert \mathop = r} \frac {\mathrm d u} u = 2 \pi i$

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

To show equality for $g \left({z}\right)$, first observe that by the Residue Theorem:

$\displaystyle \int_{\left\vert{z}\right\vert = r} \frac {\ln g \left({z}\right)} z \, \mathrm d z = 2 \pi i \ln g \left({0}\right)$

Therefore substituting $z = r e^{i \theta}$ we have

$\displaystyle 2 \pi i \ln g \left({0}\right) = i \int_0^{2 \pi} \ln g \left({r e^{i \theta} }\right) \, \mathrm d \theta$

Comparing the imaginary parts of this equality we see that:

$\displaystyle \frac 1 {2 \pi} \int_0^{2 \pi} \ln \left\vert{g \left({r e^{i \theta} }\right)}\right\vert \, \mathrm d \theta = \ln \left\vert{g \left({0}\right)}\right\vert$

as required.

$\blacksquare$

Source of Name

This entry was named for Johan Jensen.