About complex power series I have a really big doubt. I'm trying to find all the values of $z$ for which the next power series converges: $$\sum_{n=0}^{\infty} \frac{z^{3n}}{8^{n}(1-in)} $$
Using the root test I have that $$\lim_{n\to \infty}\sqrt[n]{\left|\frac{z^{3n}}{8^{n}(1-in)}\right|} = \lim_{n\to\infty}\left|\frac{z^3}{8(1-in)^{1/n}} \right| = \left|\frac{z^3}{8}\right|<1 \Leftrightarrow |z|<2$$
So, here I haven't any problem. My problem is when I try to study the edge $|z|=2$. I been trying to use the Dirichlet's test, but the convergence of the serie depends of $\arg(z)$. If anyone can help me, really I appreciate it.

Denote $\varphi=\arg{z}.$ Then
$$\dfrac{\left(2e^{i\varphi}\right)^{3n}}{8^n (1-in)}=\dfrac{e^{3in\varphi}}{1-in}=\dfrac{e^{3in\varphi}(1+in)}{1+n^2}=e^{3in\varphi}\cdot\dfrac{1}{1+n^2}+ie^{3in\varphi}\cdot\dfrac{n}{1+n^2}.$$
The series $\sum\limits_{n=0}^{\infty}{e^{3in\varphi}\cdot\dfrac{1}{1+n^2}}$ is absolutely convergent, and  $\sum\limits_{n=0}^{\infty}{e^{3in\varphi}\cdot\dfrac{n}{1+n^2}}$ converges by the Dirichlet's test for all $\varphi \ne {\dfrac{2\pi k}{3}},\;\;k \in \mathbb{Z}$.
