$\sum\limits_{n=0}^{\infty} {e^{-\mathrm{i}nz}}$ is absolutely convergent in $D=\{\mathrm{Im}(z)<0\}$, does it diverges in $\{\mathrm{Im(z)}\geq0\}$?

Question

Consider $\sum\limits_{n=0}^{\infty} {e^{-\mathrm{i}nz}}$

1. Prove that the series is absolutely convergent in $D=\{\mathrm{Im}(z)<0\}$, and divergent in $\{\mathrm{Im}(z)\geq0\}$.

I evaluated the expression and got this, and I am not sure what is the explanation for getting the convergence but I think that it has something with the geometric series: I wrote this , but I am not sure that is satisfies the question, and for diverging I am also not sure how to explain...

2. Prove that the series is absolutely and uniformly converges in closed circles that are included in D.

Answer 1

Computing the absolute value, we see that $$|e^{-i n z}|=|e^{n\Im(z)}e^{-in\Re(z)}|=e^{n\Im(z)}=\left(e^{\Im(z)}\right)^n$$

Thus if $\Im(z)>0$ we see that $\sum |e^{-inz}|$ converges (since its a geometric series) , while if $\Im(z)\ge 0$ the absolute value of $e^{-inz}$ does not go to zero, which means that the series $\sum e^{-inz}$ diverges.