how to find $\int_{0}^{2\pi} \log|1-ae^{i\theta}|\,\mathrm{d}\theta=0$,for a≤1

Question

This a problem from Stein's Complex analysis. I just don't know how solve it. $$ \int_{0}^{2\pi}\log|1-ae^{i\theta}| \,\mathrm{d}\theta=0, $$ for $a\le 1$, especially the case $a=1$.

Answer 1

HINT: When $|a|<1$ the conclusion is trivial since the integrand is the real part of the function $$ z\mapsto\log z $$ for $z$ belonging to a circle centered in $z_0=1$ and radius $|a|<1$, thus we would move remaining inside the slit plane $\Bbb C\setminus]-\infty,0]$ thus we would integrate an holomorphic function and by Residue theorem we would have the result.

Thus we have proved that $$ \int_0^{2\pi}\log(1-ae^{i\theta})\,d\theta=0 $$ holds for all $|a|<1$. Let's look at the case $|a|=1$; I know the proof by Rudin, so here's a (sketch of a) different one.

Now we can consider $a$ to be real and positive, since it is multiplied by $e^{i\theta},\;\;\theta\in[0,2\pi]$.

Observing then that $$ a\mapsto \int_0^{2\pi}\log(1-ae^{i\theta})\,d\theta $$ is a continuous function for $a\in]0,1[$, which is identically $0$, then passing to the limit for $a\to1^-$, it will be $0$ too.

Answer 2

Thanks to Mr Joe's hint, i had thought out a solution,we choose a toy counter as the graph shown:the big semicircle is centered at 0,with radius 1, the small one with the radius $\epsilon$,consider the function $f(z)=\frac{Log(1-az)}{z}$ which has no poles in the counter ,hence $$\int_{C}f=0$$. $$\int_{-1}^{-\epsilon}frac{log(1-ax)}{x}=I_{1}$$ the case is same on the part of $[\epsilon,1]$,denote it by $I_{2}$,note that they are both real. $$\int _{-pi}^{0}iLog(1-ae^{i\theta})d\theta$$ tends to 0,as $\epsilon$ tends to 1, $$\int_{0}^{\pi}iLog(1-ae^{i\theta})d\theta=i\int_{0}^{\pi}log|1-ae^{i\theta}|-\int_{0}^{\pi}g(\theta)d\theta=iI_{4}+I_{3}$$. Note that,as $\epsilon$ tends to 0, $$\int_{C}fdz\rightarrow I_{1}+I_{2}+I_{3}+iI_{4}=0$$, since $I1~I4$ are all real,that implies I4=0,i.e,$$\int_{0}^{2\pi}log|1-ae^{i\theta}|d\theta=0$$ as was to be shown, For the case a=1,it suffices to take the big semicircle with radius $1-\epsilon$