6
$\begingroup$

This is a homework problem for a second course in complex analysis. I've done a good bit of head-bashing and I'm still not sure how to solve it-- so I might just be missing something here. The task is to show that given $|a|<1$, $\int_0^{2\pi} \log|1-ae^{i\theta}|d\theta=0.$

So right off the bat we can let $z=e^{i\theta}$ so that $\int_0^{2\pi} \log|1-ae^{i\theta}|d\theta=\int_{|z|=1} \log|1-az|\frac{dz}{iz}=-i\int_{|z|=1} \log|1-az|{dz}.$

After that I'm not sure if using the residue theorem is the way to go?

  • 1
    Thanks for the help everyone! Just one more question: apparently equality holds if $a=1$, but I don't really see how this works. Could anyone care to elaborate$a$bit?2012-03-01

2 Answers 2

4

Hint:

Note that $\log|1-ae^{i\theta}|$ is the real part of $\log(1-ae^{i\theta})$. Then try differentiating with respect to $a$. Then notice that integrating around the unit circle $ \frac1i\oint\frac{\mathrm{d}z}{1-az}=0 $ when $|a|<1$.

  • 0
    nice............+1@robjohn2016-01-03
4

Consider the function $\log(1-a\,z)$ and think mean value.