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

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?

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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 $|a|<1$.

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