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Is there any fantastic way to proving this identity?

$$\int_0^{2\pi} \log \left| 1-ae^{i\theta}\right|d\theta =0$$

I used power series approach(we have $2\log |1-z|=\log (1-z)(1-\bar{z})$), which was unsuccessful to show the result when $|a|=1$ because the series $\log(1-z)=-\sum_{n=1}^{\infty} z^n/n$ does not converge uniformly for $|z|=1$.

Any solution is appreciated. Thanks.

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    $f(z)=1-az$ is analytic and nonzero in the disc, then $\log|f(z)|$ is harmonic. By Mean Value Theorem for harmonic functions the integral is $2\pi\log 1=0$. – 2017-01-07
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    @A.G. That's cool. But I'm not sure whether we can use the argument in the case $|a|=1$. – 2017-01-07
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    You are right, for $|a|=1$ it's [a bit trickier](http://math.stackexchange.com/a/1780372/253273). – 2017-01-07
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    @A.G. Thanks a lot! – 2017-01-07
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    More [here](http://math.stackexchange.com/questions/244059/int-02-pi-log1-aei-thetad-theta-0-when-a-1?rq=1) and [here](http://math.stackexchange.com/a/184874/253273) – 2017-01-07

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