Complex integration and Gauss mean value theorem

Question

I'm trying to show that $\frac{1}{2\pi} \int_0^{2\pi} \log | 1-ae^{i\theta}|d\theta=0$ for $|a|<1$ implies $\frac{1}{2\pi} \int_0^{2\pi} \log | a-e^{i\theta}|d\theta=0$

Answer 1

Here is a hint: $$|a - e^{i\theta}| = |a e^{i\theta}e^{-i \theta} - e^{i\theta}| = |a e^{-i\theta} -1|$$ and $$\int_0^{2\pi} \log| 1 - ae^{i \theta}| \, d\theta = \int_0^{2\pi} \log| 1 - ae^{-i \theta}| \, d\theta.$$

Answer 2

Another way to see this is to note that for $a$ real valued we have

$$|1-ae^{i\theta}|=\sqrt{1+a^2-2a\cos \theta}$$

and

$$|a-e^{i\theta}|=\sqrt{1+a^2-2a\cos \theta}$$

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Note that if $a$ is complex, then we have

$$|1-ae^{i\theta}|=\sqrt{1+|a|^2-2|a|\cos (\arg(a)+\theta)}$$

and

$$|a-e^{i\theta}|=\sqrt{1+|a|^2-2|a|\cos(\arg(a)- \theta)}$$

Exploiting the $2\pi$-periodicity and evenness of the cosine function, we find that

$$\begin{align} \int_0^{2\pi} |a-e^{i\theta}|\,d\theta&=\int_0^{2\pi} \sqrt{1+|a|^2-2|a|\cos(\arg(a)- \theta)}\,d\theta \tag 1\\\\ &=\int_0^{2\pi} \sqrt{1+|a|^2-2|a|\cos(\arg(a)+ \theta)}\,d\theta \tag 2\\\\ &=\int_0^{2\pi} |1-ae^{i\theta}|\,d\theta \end{align}$$

where in going from $(1)$ to $(2)$ we enforced the substitution $\theta \to 2\pi-\theta$.