# $\int_0^{2\pi} \log|1-ae^{i\theta}|d\theta=0$ when $|a| = 1$

I'm taking a course on complex analysis, and I've been working on this problem forever, but can't seem to figure out. The question asks you to show the following:

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

when $|a|<1$. I was able to do this by showing that $\int_0^{2\pi} \log(1-ae^{i\theta})\; \mathrm d\theta=0$ by substituting $z=e^{i\theta}$ and then calculating the residue of $\frac{\log(1-az)}{zi}$ at the origin to be zero.

However, the question further asks you to show the same equation holds when $|a|=1$. Can someone help me with this? Thank you in advance.

My attempt at solution ->

Define $F(a) = \int_0^{2\pi} \log(1-ae^{i\theta})\; \mathrm d\theta$

Then, it can be shown that $F(a)$ is holomorphic in $|a|<1$. Given that $F(a)=0$ on the entire $|a|<1$, we can say that limit of $F(a)$ is $0$ when $|a|$ approaches 1.

Therefore, since $F(a)$ is holomorphic in $|a|<1$ and converges uniformly to $0$ as $|a|$ approaches $1$, we can extend $F(a)$ to the boundary $|a|=1$ ((Is this true? Or in general, when can you "continuously extend" a holomorphic function defined on an open set to the boundary of the open set?))

Thus, we have a function $F(a)$ that takes on the value $0$ for $|a|<=1$. Since $\int_0^{2\pi} \log|1-ae^{i\theta}|\; \mathrm d\theta$ is the real part of $\int_0^{2\pi} \log(1-ae^{i\theta})\; \mathrm d\theta$, it follows that $\int_0^{2\pi} \log|1-ae^{i\theta}|\; \mathrm d\theta=0$ for $|a| = 1$ as well.

Could you let me know if the steps are correct? I'm very iffy about the "continuous extension" that I did.

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Read the lemma, and in particular its corollay, in the answer here math.stackexchange.com/questions/83828/… – DonAntonio Nov 25 '12 at 2:45
Now you just have to take the real part, as $\mbox{Re}\log w=\log|w|$. – 1015 Apr 6 '13 at 16:13

Your integral can be re-written as $$I = \int_{0}^{2\pi} \log |2 \sin (u/2)| du$$ Now we make a change of variables and then use the formula $\sin(u) = 2\sin(u/2)\cos(u/2)$ to get \begin{align*} I & = 2 \int_{0}^{\pi}\log |2\sin(u)| du \\ & = 2\int_{0}^{\pi} |4\sin(u/2)\cos(u/2)| du \\ & = 2 \int_{0}^{\pi} \log |2\sin(u/2)| du + 2\int_{0}^{\pi} \log |2\cos(u/2)| du = I + I \end{align*} so $I = 0$. The last equality is seen by drawing a picture of the values traversed by each of the functions $|2\sin(u/2)|$ and $|2\cos(u/2)|$ in the ranges $0 < u < \pi$ and $\pi < u < 2\pi$.
You might want to explain what happened to the complex number $a$. – Did Apr 6 '13 at 16:20