Limit with number $e$ and complex number This is my first question here. I hope that I spend here a lot of fantastic time.
How to proof that fact?
$$\lim_{n\to\infty} \left(1+\frac{z}{n}\right)^{n}=e^{z}$$
where $z \in \mathbb{C}$ and $e^z$ is defined by its power series.
I have one hint: find the limit of abs. value and arguments,
but i don't know how to use it to solve that problem.
Thank you for help.
Before I try solve this problem, I proofed that $$e^z=e^{x}(\cos y + i \sin y)$$
where $z=x+yi$ ,maybe this help.

I like to prove it as follows: 
If $g(z)=\sum a_i z^i$ where $a_0=1$ and $a_1=0$, with $g$ entire (not sure you need entire,) then:
$$\lim_{n\to \infty}\left(g\left(\frac z n\right)\right)^n = 1$$
Then define $g(z)=(1+z)e^{-z}$.
You'd need to first show that $(e^w)^n=e^{wn}$ for that proof to work. This latter can be proved by induction if you first prove $e^{z_1}\cdot e^{z_2}=e^{z_1+z_2}$.
This approach does not use your hint, however.
