For which values of $z$ does the sequence $(e^{nz})$ converges? Where $z$ is a complex number. 
For which values of $z$ does the sequence $(e^{nz})$ converges? Where $z$ is a complex number.

I was able to figure it out for numbers with null imaginary part, but I get troubled when considering the imaginary part. Being $z = a + bi$.
With null imaginary part what we have is:
$$ e^{nz} = e^{n\cdot(a+bi)}  $$
As $b=0$, $$ e^{n\cdot(a+bi)} = e^{na}.$$
It converges if  $a\leq 0$, and diverges if $a <0$.
I could not develop the same argument when $b\neq 0$.
What we have is:
$$ e^{n\cdot(a+bi)} = e^{na}\cdot e^{nbi}$$
Which can be rewritten as:
$$ e^{na} \cdot [\cos (nb) + i\cdot \sin(nb)]$$
At this part I could not go any further. I tried thinking about the second term as being bounded but couldn't quite figure it out if it realy is.
Is this the right track? How can I show that $[\cos (nb) + i\cdot \sin(nb)]$ is bounded?

Put $e^z=:w$. Then $$e^{nz}=w^n$$ for all $n\geq0$. The complex sequence $\bigl(w^n\bigr)_{n\geq0}$ converges iff either $|w|<1$ or $w=1$. The first is the case if ${\rm Re}(z)<0$ and the second if   $z=2k\pi i$ for some $k\in{\mathbb Z}$.
