Complex Analysis

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1 Integral of a function around a closed loop

Let \(f\) be a meromorphic function, and \(\mathcal{C}\) a closed loop. Then

\begin{equation} \iint_{\mathcal{C}}f(z)dz = \sum \operatorname{Res}_f(a_i) \end{equation}

where the \(a_i\) are the poles of \(f\), and \(\operatorname{Res}_f(a_i)\) is the coefficient of \(z^{-1}\) in the power series expansion of \(f\) at \(a_i\).

2 Cauchy's integral formula

\begin{equation} \frac{1}{2\pi i}\iint \frac{f(\zeta)}{\zeta-z}d\zeta = f(z) \end{equation}

In the case \(z=0\), and if \(\zeta\) travels along a circle around the origin at constant radius, this is

\begin{equation} f(0)=\frac{1}{2\pi} \oint f(Re^{i\theta})d\zeta \end{equation}

Weighted limits: What is a weighted limit?

3 Jensen's formula

Let \(D_R\) denote the disk of radius \(R\) centered at the origin. Let \(g\) be a function on an open neighborhood of \(D_R\); suppose \(g\) vanishes nowhere. Then \(g\) admits a logarithm on \(D_R\), i.e. we have

\begin{equation} g(z) = e^{ih(z)} \end{equation}

where \(h\) is defined on an open neighborhood of \(D_R\). The imaginary part of \(h\) controls the magnitude of \(g\) and the real part of \(h\) controls the argument (modulus) of \(g\), i.e., \(\left\lvert g(z) \right\rvert = e^{- \operatorname{Im}(h(z))}\), \(\operatorname{Arg}(g(z))= \operatorname{Re}(h(z))\).

Or, to say this another way,

\begin{equation} -\ln \left\lvert g(z) \right\rvert = \operatorname{Im}(h(z)) \end{equation}

Now, by Cauchy's integral formula,

\begin{equation} h(0) = \frac{1}{2\pi}\int_{\theta=0}^{2\pi} h(Re^{i\theta})d\theta \end{equation}

and this splits into the two integrals

\begin{equation} \operatorname{Re}(h(0)) = \frac{1}{2\pi}\int_{\theta=0}^{2\pi} \operatorname{Re}(h(Re^{i\theta})d\theta \end{equation} \begin{equation} \operatorname{Im}(h(0)) = \frac{1}{2\pi}\int_{\theta=0}^{2\pi} \operatorname{Im}(h(Re^{i\theta})d\theta \end{equation}

the latter of which is equivalent to

\begin{equation}\label{jensen-nonvanishing} -\ln \lvert g(0)\rvert = \frac{1}{2\pi}\int_{\theta=0}^{2\pi} -\ln \lvert g(Re^{i\theta})\rvert d\theta \end{equation}

In the case where \(g\) does have zeroes, this equation can be modified to account for it. Consider the case \(g= z-z_0\). We have

\begin{equation} -\ln \lvert g(0)\rvert = -\ln \lvert z_0\rvert \end{equation}

On the other hand,

\begin{equation} \frac{1}{2\pi}\int_{\theta=0}^{2\pi} -\ln \lvert Re^{i\theta} - z_0\rvert d\theta = \frac{1}{2\pi}\int_{\theta=0}^{2\pi} - \ln\left \lvert e^{i\theta} -\frac{z_0}{R}\right \rvert d\theta - \ln R \end{equation}

Introduce the notation \(a= \frac{z_0}{R}\); then \(|a|<1\). Actually \(\int_{\theta=0}^{2\pi} \ln \lvert e^{i\theta} - a\rvert d\theta=0\). To see this, first introduce the substitution \(\theta\mapsto -\theta\), and rewrite \(\ln \lvert e^{-i\theta}- a \rvert =\ln \lvert e^{-i\theta}(1-ae^{i\theta}) \rvert = \ln \lvert 1 - ae^{i\theta}\rvert\). So we must show that

\begin{equation} \int -\ln \lvert 1 - ae^{i\theta} \rvert d\theta =0 \end{equation}

But now we can reduce to the case of \ref{jensen-nonvanishing}, by writing \(F(z) = 1-az\); \(F(z)\) vanishes nowhere on the closed unit disk, and so by \ref{jensen-nonvanishing} we should have

\begin{equation} \int_{\theta=0}^{2\pi} -\ln \lvert F(e^{i\theta}) \rvert d\theta = -\ln \lvert F(0)\rvert = -\ln 1 = 0 \end{equation}

Thus, in the case \(g(z)=z-z_0\) we have \(\int_{\theta=0}^{2\pi} -\ln \lvert g(Re^{i\theta})\rvert d\theta = -\ln R\), and so

\begin{equation} -\ln \lvert g(0)\rvert = -\ln \lvert z_0\rvert + 0 = -\ln \lvert z_0\rvert + \left( \frac{1}{2\pi}\int_{\theta=0}^{2\pi} -\ln \lvert g(Re^{i\theta})\rvert d\theta - \ln R \right) = - \ln \frac{\lvert z_0\rvert}{R} - \int_{\theta=0}^{2\pi}\ln \lvert g(Re^{i\theta})\rvert d\theta \end{equation}

Then for an arbitrary function \(f\), writing \(f = g\cdot (z-z_0)\dots (z-z_k)\) for \(g\) nowhere vanishing, it is not hard to see by combining the above equations that (dropping negative signs)

\begin{equation} \ln \lvert f(0)\rvert = \sum \ln \frac{\lvert z_k\rvert }{R} + \int_{\theta=0}^{2\pi} \ln \lvert f(Re^{i\theta})\rvert d\theta \end{equation}

Author: Patrick Nicodemus

Created: 2021-08-20 Fri 17:09

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