Finding $z$ for Complex Convergence I am having an issue understanding how to go about solving a problem regarding complex sequences.  The problem is as follows:

Find a $z$ for which the following sequence converges: $f_{n} (z) =e^{nz}$ 

My attempt thus far is:
$$f_{n} (z) = e^{nz} = e^{nx} \cdot (\cos(ny) + i\sin(ny))$$
Then I was planning to use the fact that both the real part and the complex part must converge if the sequence converges, so:
$$\text{Re}(f_{n}(z)) = e^{nx}\cos(ny)$$
$$\text{Im}(f_{n}(z)) = e^{nx}i\sin(ny)$$
But where do I go from here?  

Let $z=x+iy$.  Then, note that $f_n(z)=e^{nx}e^{iny}$.  
The real and imaginary parts of the sequence $f_n(z)$ are given respectively by 
$$\text{Re}(f_n(z))=e^{nx}\cos(ny)$$
and
$$\text{Im}(f_n(z))=e^{nx}\sin(ny)$$
Note that if $x< 0$, the exponential term $e^{nx}$ approaches zero as $n\to \infty$.  Since both the sine and cosine functions are bounded in absolute value by $1$, both real and imaginary parts of the sequence converge to zero for $x<0$.
If $x=0$ and $y=2\pi \ell$ for any integer $\ell$, then the sequence converges to $1$.
