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I want to prove that $$\int_0^{2\pi}\log|1-ae^{i\theta}|\, d\theta=0$$ for all $|a|\leq 1$. I can prove it easily for $|a|<1$ via power series expansion for $\log|1+(-a)e^{i\theta}|$ and then integrating term-by-term as the series is absolutely convergent. Whenever $|a|=1$, though, I'm having trouble proving much of anything. It seemed like the above technique would work for $a\neq1$ since it can be proven that $$\sum_{n=1}^\infty\frac{z^n}{n}$$is convergent on $S^1$ for $z\neq1$ (previous exercise), however, I wasn't sure I could interchange the summation and integral.

Will anyone provide a hint toward the solution (preferably without giving the entire solution away)? Thanks! Also, if there is a more elegant solution to prove the first portion, please feel free to open my mind a little bit. For those curious, this is Exercise $11$ of Chapter $3$ from Stein and Shakarchi's book.

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For $|a|<1$ you could also say it is true since $$\int_0^{2\pi} \log|1-ae^{i\theta}|\,d\theta = \mathrm{Re } \int_{|z|=1} \frac{\log(1-az)}{iz}\,dz = 0$$ by the residue theorem. –  Antonio Vargas Mar 2 '13 at 5:48
    
I feel like if I was in practice, I would know how to integrate a contour that passes through the branch cut or the branch point as well. :( –  Hurkyl Mar 2 '13 at 5:56
    
@AntonioVargas: Nice! I hadn't thought about that; I was trying to use the Residue theorem, but I didn't see how to apply it. Thanks for the insight :) –  Clayton Mar 2 '13 at 5:59
    
For $a=1$ simply indent the contour at $z=1$. In the limit it's either going to contribute nothing or cause the contour integral to blow up. –  Random Variable Mar 2 '13 at 6:54
    
@RandomVariable, Could you expand on that? It's not too hard to show the integral converges, so your observation would seal the deal. –  Antonio Vargas Mar 2 '13 at 7:02

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For $|a|<1$, the branch cut for $\log{(1-a z)}$ is on the real axis, for $z > 1/|a|$. Thus, we can write

$$\oint_{|z|=1} dz \frac{\log{(1-a z)}}{z} = 0$$

because $\frac{\log{(1-a z)}}{z}$ has no poles within $|z| \le 1$. Substitute $z=e^{i \theta}$ to get

$$i \int_0^{2 \pi} d\theta \log{(1-a e^{i \theta})} = i \int_0^{2 \pi} d\theta \log{|1-a e^{i \theta}|} - \int_0^{2 \pi} d\theta \: \arg{(1-a e^{i \theta})} = 0$$

Equating real and imaginary parts, we get the desired result for $|a| < 1$. For $|a|=1$, the circle $|z|=1$ passes through the branch cut so that Cauchy's theorem does not apply. In this case, we indent the circle $|z|=1$ near $z=1$ by an arc $\gamma$ on the circle $|z-1|=r$ to produce a contour $\Gamma$ and consider the limit $r \rightarrow 0$. On $\gamma$, we may show that

$$\left | \frac{\log{(1-z)}}{z} \right | \le \frac{\log{\frac{1}{r}}}{1-r}$$

because, for very small $r$, $\arg{(1-z)} \approx 0$ and $|z| > 1-r$ on $\gamma$. The magnitude of the integral over $\gamma$ is thus bounded by

$$\left | \int_{\gamma} dz \frac{\log{(1-z)}}{z} \right | \le 2 r \frac{\log{\frac{1}{r}}}{1-r} \arccos{\frac{r}{2}} $$

which goes to zero in the limit of $r \rightarrow 0$.

We may then show that, on the indented contour $\Gamma$:

$$\oint_{\Gamma} dz \frac{\log{(1-z)}}{z} = 0$$

By the same reasoning as above (equating real and imaginary parts), we see that

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

and the desired result follows.

Interestingly, though, we may show this explicitly. Let $z=e^{i \theta}$, $dz/z = i d\theta$, then

$$i \int_0^{2 \pi} d\theta \log{(1-e^{i \theta})} = 0$$

Now

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

It turns out that

$$\arg{(1-e^{i \theta})} = \arctan{\frac{\sin{\theta}}{1-\cos{\theta}}} = \arctan{(\cot{\frac{\theta}{2}})} = \frac{\pi}{2} - \frac{\theta}{2}$$

$$\int_0^{2 \pi} d\theta \: \arg{(1-e^{i \theta})}= \int_0^{2 \pi} d\theta \left ( \frac{\pi}{2} - \frac{\theta}{2} \right ) = \frac{\pi}{2} 2 \pi - \frac{1}{4} (2 \pi)^2 = 0 $$

$$\therefore \int_0^{2 \pi} d\theta \log{|1-e^{i \theta}|} = 0$$

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For $|a|=1$ but $a\ne 1$, use a change of variables. –  Braindead Jan 3 at 22:42

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