An elegant proof of the identity $\int_0^{2\pi} \log \left| 1-ae^{i\theta} \right|d\theta=0$ when $|a|\leq 1, a \in \mathbb{C}$.

Asked Jan 07 '17 at 12:38

Active Jan 07 '17 at 12:38

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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.

asked Jan 07 '17 at 12:38

HyJu

3

$f(z)=1-az$ is analytic and nonzero in the disc, then $\log|f(z)|$ is harmonic. By Mean Value Theorem for harmonic functions the integral is $2\pi\log 1=0$. – A.Γ. Jan 07 '17 at 13:00
@A.G. That's cool. But I'm not sure whether we can use the argument in the case $|a|=1$. – HyJu Jan 07 '17 at 13:05
1

You are right, for $|a|=1$ it's a bit trickier. – A.Γ. Jan 07 '17 at 13:14
@A.G. Thanks a lot! – HyJu Jan 07 '17 at 15:56
More here and here – A.Γ. Jan 07 '17 at 23:20

An elegant proof of the identity $\int_0^{2\pi} \log \left| 1-ae^{i\theta} \right|d\theta=0$ when $|a|\leq 1, a \in \mathbb{C}$.

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