Jensen's Formula/Proof 2

Theorem

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

$D_r = \set {z \in \C : \cmod z \le r}$

of radius $r$ about $0$.

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

Let $f$ have no zeroes on the circle $\cmod z = r$.

Let $\map f 0 \ne 0$.

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

Then:

$(1): \quad \ds \frac 1 {2 \pi} \int_0^{2 \pi} \ln \cmod {\map f {r e^{i \theta} } } \rd \theta = \ln \cmod {\map f 0} + \sum_{k \mathop = 1}^n \paren {\ln r - \ln \size {\rho_k} }$

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 $z \in D_r$.

This article, or a section of it, needs explaining. In particular: Why can you write like this? Looks wrong, because the zeros do not correspond. For example, insert $z=\rho_1$ in both sides. You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by explaining it. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Explain}} from the code.

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:

\(\ds \cmod {\frac {r^2 - z \overline {\rho_k} } {r \paren {z - \rho_k} } }\)

\(=\)

\(\ds \frac {\cmod {r^2 - \overline z \rho_k} } {\cmod {r \paren {z - \rho_k} } }\)

\(\ds \)

\(=\)

\(\ds \frac {\cmod {1 - \overline z \rho_k / r^2} } {\cmod {z / r - \rho_k / r} }\)

\(\ds \)

\(=\)

\(\ds \frac {\cmod {1 - \overline z \rho_k / r^2} } {\cmod {1 - \rho_k / z} }\)

\(\ds \)

\(=\)

\(\ds \frac {\cmod {1 - \overline z \rho_k / r^2} } {\cmod {1 - \overline z \rho_k / r^2} }\)

\(\ds \)

\(=\)

\(\ds 1\)

so the left hand side is $0$.

This article, or a section of it, needs explaining. In particular: What are you doing? The calculation above looks wrong and the conclusion that left hand side is zero because right hand side is one does not make sense. Please explain more. You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by explaining it. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Explain}} from the code.

Moreover:

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

so the right hand side 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 right hand side is $\ln \size {\map g 0}$.

On the other hand, by the mean value property:

$\ds \dfrac 1 {2 \pi} \int_0^{2 \pi} \ln \size {\map g {r e^{i \theta} } } \rd \theta = \ln \map g 0$

as required.

$\blacksquare$

Source of Name

This entry was named for Johan Jensen.