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.
 A: Expand using the binomial formula: $\displaystyle \left(1+\frac{z}{n}\right)^n = \sum_{k=0}^n {n\choose k}\left( \frac{z}{n}\right)^k = \sum_{k=0}^\infty E_k^n$ where we define $\displaystyle E_k^n = {n\choose k}\left( \frac{z}{n}\right)^k$ for $k \le n$ and $= 0$ otherwise
We want $\displaystyle \sum_{k=0}^\infty E_k^n$ to converge to $\displaystyle \sum _{k=0}^\infty \frac{z^k}{k!}$ as $n \to \infty$ To do that we will show $\displaystyle E^n_k \to \frac{z^k}{k!}$ as $n \to \infty$
$\displaystyle E^n _k = \frac{n!}{k!(n-k)!}\left( \frac{z}{n}\right)^k = \frac{n!}{k!(n-k)!}  \frac{z^k}{n^k} = \frac{n!}{n^k (n-k)!}\frac{z^k}{k!} =\frac{n}{n} \frac{(n-1)}{n}\cdot \ldots \cdot \frac{(n-k+1)}{n}\frac{z^k}{k!} $
Therefore we just have to prove $\displaystyle \frac{n}{n} \frac{(n-1)}{n}\cdot \ldots \cdot \frac{(n-k+1)}{n} \to 1$. The number of terms to multiply is constant and equal to $k$. So there is no problem with invoking how each of them goes to $1$ seprately, and that limits commute with multiplication. 
A: 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.
A: Here is an approach.

$$\lim_{n\to \infty} \left(1+\frac{z}{n}\right)^{n} =\lim_{n\to \infty} \sum_{k=0 }^{n} {n\choose k} \frac{z^k}{n^k}. $$

Now try to prove that 

$$\lim_{n\to \infty} \frac{{n\choose k}}{n^k} = \frac{1}{k!}. $$

Note:

$$ {n\choose k} = \frac{n!}{(n-k)! k!}. $$

A: With your starting point my immediate approach would be something like


*

*Define the logarithm as the inverse function of $e^z$, on some domain that includes $1$, such that $\log 1 = 0$. (The amount of legwork this requires if you don't already have a suitable logarithm depends on whether you have the inverse function theorem, etc.)

*Define $a^b = e^{b\log a}$ when $a$ is in the domain of your logarithm. Prove that this satisfies enough of the usual power rules to agree with the usual raising to integer powers. This is also enough to evaluate $(1+z/n)^n$ when $n$ is large enough, no matter what $z$ is.

*Now $$\lim_{h\to 0} (1+hz)^{1/h} = \lim_{h\to 0}e^{\log(1+hz)/h} = e^{\lim_{h\to0}\log(1+hz)/h} = e^z $$
because $\lim_{h\to 0}\frac{\log(1+hz)}h$ is the definition of $\frac{d}{dy}\log(1+zy)$ at $y=0$, and this derivative can be evaluated symbolically using the chain rule and the rule for the derivative of an inverse function using what we already know.

*Subsitute $n=1/h$ in the above limit.
A: Another very simple answer using a powerful theorem is described by the following steps:


*

*First define :
$$ f(z)=\sum_{n=0}^{+\infty}\frac{z^n}{n!}\ \ \ \ \ \ \ \ \   g(z)=\lim_{n\to +\infty}\big (1+\frac{z}{n}\big)^n$$

*Second prove that $f$ and $g$ coincides over real numbers

*Conclude using the identity theorem of power series
