Convergence of $\sum_{n=1}^{\infty}\frac{1}{n}\left(\frac{z-\alpha}{1-\overline{\alpha}z}\right)^n$ 
Find all the values of $z \in \Bbb{C}$ for which $$\sum_{n=1}^{\infty}\frac{1}{n}\left(\frac{z-\alpha}{1-\overline{\alpha}z}\right)^n$$ with $|\alpha|<1$
Converges.

So I defined $w=\frac{z-\alpha}{1-\overline{\alpha}z}$ and my series $\sum\frac{1}{n}w^n$ converges for every $w\in \Bbb{C}$ with $|w|\leq 1$, $w\neq 1$.
However I have to find now all the $z\in \Bbb{C}$ such that $w=\frac{z-\alpha}{1-\overline{\alpha}z}$ with $|w|\leq 1$, $w\neq 1$, which is kind of a Mobius transformation problem, but Im quite lost, how can I proceed? What should I compute?

You can check that 
$$
\left| \frac{z-\alpha}{1-\bar \alpha z} \right| < 1
$$
precisely when $|z|<1$. (See for example this, but there are many many others on this site as well.)
Hence the series converges for $|z| < 1$ (and diverges for $|z| > 1$ where $\left| \frac{z-\alpha}{1-\bar \alpha z} \right| > 1$).
Finally, by Dirichlet's test, $\sum_{k=1}^\infty z^n/n$ converges for $|z| = 1$ unless $z=1$. All that remains is to find the values of $z$ on the circle for which $\frac{z-\alpha}{1-\bar \alpha z} = 1$.
