Special Case Jensen's Formula

Question

This is the Lemma given in the book of Serge Lang Complex Analysis.

Lemma: Let $\alpha =ae^{i\varphi}$ with $|z|<a<R$. Prove that $$\int_o^{2\pi} \text{log}\left|e^{i\theta}-\frac{a}{R}e^{i\varphi} \right|d\theta=0.$$

In the proof of this lemma it says that, for $a<R$, the function $$\frac{\log(1-\frac{a}{R}z)}{z}$$ is analytic on $|z|\leq 1$, and the desired integral is $0$.

Question: How does $$\text{log}\left|e^{i\theta}-\frac{a}{R}e^{i\varphi} \right|$$ became $$\frac{\log(1-\frac{a}{R}z)}{z}?$$ What I know is that we are going to change $\theta$ into $z$ but I really cant figure it out.

Answer 1

Factor out an $e^{i\theta}$

$$\log|e^{i\theta} - ce^{i\psi}| = \log|e^{i\theta}(1 - ce^{i(\psi-\theta)})|$$

$$ = \log|e^{i\theta}| + \log|1 - ce^{i(\psi-\theta)}|$$

Now let $z=e^{i(\psi-\theta)}$. Then $dz =e^{i(\psi-\theta)}(-i)d\theta$. Or... $i\dfrac{dz}{z} = d\theta$.

So your integral becomes

$$i\int_{\exp(i\psi)}^{\exp(i(\psi-2\pi))}\dfrac{\log|e^{i\psi}z^{-1}| + \log|1-cz|}{z} dz$$

Answer 2

First note the following: \begin{align} \int^{2\pi}_0 \log\bigg|e^{i\theta}-\frac{\alpha}{R} \bigg| \,d\theta &= \int^{2\pi}_0 \log\bigg|e^{i\theta}\left(1-\frac{\alpha}{R}e^{-i\theta}\right) \bigg| \,d\theta\\ &=\int^{2\pi}_0 \log|e^{i\theta}|+\log\bigg|1-\frac{\alpha}{R}e^{-i\theta} \bigg| \,d\theta\\ &=\int^{2\pi}_0 \log\bigg|1-\frac{\alpha}{R}e^{-i\theta} \bigg| \,d\theta\\ \end{align} We also have: \begin{align} \int^{2\pi}_0 \log\bigg|1-\frac{\alpha}{R}e^{-i\theta} \bigg| \,d\theta = \text{Re}\left(\int^{2\pi}_0 \log\bigg(1-\frac{\alpha}{R}e^{-i\theta} \bigg) \,d\theta \right) \end{align}

Substitute $z=e^{-i\theta}$ so that $dz=-ie^{i\theta}d\theta$ and that means $i\frac{dz}{z}=d\theta$ hence: \begin{align} \int^{2\pi}_0 \log\bigg|1-\frac{\alpha}{R}e^{-i\theta} \bigg| \,d\theta &=\text{Re}\left(\int_{|z|=1} -i\frac{1}{z}\log\bigg(1-\frac{\alpha}{R}z \bigg) \,dz\right) \end{align} And that is what Lang did in the background.