Proof of $\lim_{n\to \infty }(1+\frac{z}{n})^n=e^x(\cos y+i\sin y)$ Does anyone have proof that $\lim_{n\to \infty }\left(1+\frac{z}{n}\right)^n=e^x(\cos y+i\sin y)$ because the proof I have is unclear and/or incorrect.
$$z=x+iy \in \mathbb C$$
For a large enough $n$: $\left|\arg\left(1+\frac{1}{n}\right)\right|<\frac{\pi}{2}$ (Why/how?)
Lets look at the real sequences $z_n=\left(1+\frac{z}{n}\right)^n,|z_n| \text{ and } \arg z_n.$
It is proven that $\lim_{n\to \infty }|1+\frac{z}{n}|^n=e^x$
then it says, which makes no sense to me from here on(I'll highlight what lines are unclear): 
$$\arg (z_n)=\arg\left(\left(1+\frac{z}{n}\right)^n\right)=$$ 

$$n\arctan\left(\frac{y}{x+n}\right) \text{ then it's just some blank space and }
 \mod2\pi \to y \text{ and again some blank space} \mod 2\pi \text{
 when } n\to \infty, \text{therefore }\lim_{n\to \infty
 }\left(1+\frac{z}{n}\right)^n=e^x(\cos y+i\sin y) $$

 A: We know that principal value of the argument function of a complex number $z$ is defined as 
$$\text{Arg} (z)=\begin{cases}
\arctan\left(\frac{\text{Im}(z)}{\text{Re}(z)}\right)&,\text{Re}(z)>0\\\\
\arctan\left(\frac{\text{Im}(z)}{\text{Re}(z)}\right)+\pi&,\text{Re}(z)<0,\,\text{Im}(z)\ge 0\\\\
\arctan\left(\frac{\text{Im}(z)}{\text{Re}(z)}\right)-\pi&,\text{Re}(z)<0,\,\text{Im}(z)< 0\\\\
\pi/2&,\text{Re}(z)=0,\,\text{Im}(z)> 0\\\\
-\pi/2&,\text{Re}(z)=0,\,\text{Im}(z)< 0\\\\
\end{cases}$$
The argument of $z$ is 
$$\arg (z)=\text{Arg}(z)+2\ell \pi$$
for integer values of $\ell$. 
Now, for $z=x+iy$ with fixed values of $x$ and $y$ and $n>-x$, we have
$$\begin{align}
\arg\left(\left(1+\frac zn\right)^n\right)&=\arg\left(\left(1+\frac xn+i\frac yn\right)^n\right) \tag 1\\\\
&=n\arctan\left(\frac{y/n}{1+x/n}\right)+2\ell \pi \tag 2\\\\
&\to y+2\ell \pi \,\,\text{as}\,\,n\to \infty \tag 3
\end{align}$$
Therefore, since $\cos (y+2\ell \pi)=\cos y$ and $\sin (y+2\ell \pi)=\sin y$, we have
$$\lim_{n\to \infty}\left(1+\frac zn\right)^n=e^x\left(\cos y+i\sin y\right)$$
as was to be shown!

NOTE $1$:
In going from $(1)$ to $(2)$, we made use inductively of the relationship
$$\arg(z^2)=2\arg(z)+2\ell \pi \tag 4$$
We can verify $(4)$ as follow.  First, let $z=x+iy$.  Then, for $x^2-y^2>0$, and $x>0$, we have 
$$\begin{align}
\arg(z^2)&=\text{Arg}(z^2)+2\ell \pi\\\\
&=\arctan\left(\frac{2xy}{x^2-y^2}\right)+2\pi \ell\\\\
&=\arctan\left(\frac{2(y/x)}{1-(y/x)^2}\right)+2\pi \ell\\\\
&=2\arctan(y/x)+2\pi \ell\,\,\dots \,\text{where we used the double angle formula for the tangent}\\\\
&=2\arg(z)+2\ell \pi \tag 5
\end{align}$$
Now, let $z=\left(1+\frac xn\right)+i\left(\frac yn\right)$ in $(5)$.  Then, for $n$ sufficiently large, $\left(1+\frac xn\right)^2-\left(\frac yn\right)^2>0$ and $1+\frac xn >0$, we find  
$$\arg\left(z^2\right)=2\arg(z)+2\pi \ell$$
Proceeding inductively reveals that 
$$\arg(z^n)=n\arg (z)+2\ell \pi$$
as was to be shown.

NOTE $2$:
In going from $(2)$ to $(3)$, we made use of THIS ANSWER, in which I established the inequalities for the arctangent function
$$\frac{|x|}{\sqrt{1+x^2}} \le|\arctan(x)|\le |x|$$
Then, we have
$$\left|\frac{\frac{y}{1+x/n}}{\sqrt{1+\left(\frac{y/n}{1+x/n}\right)^2}} \right|\le \left| n\arctan\left(\frac{y/n}{1+x/n}\right)\right| \le \left|\frac{y}{1+x/n}\right|$$
whereupon applying the squeeze theorem reveals the limit
$$\lim_{n\to \infty}n\arctan\left(\frac{y/n}{1+x/n}\right)=y$$
as was to be shown!
A: I've always thought the binomial theorem and dominated convergence theorem are the way to go to prove this. For $n=1,2,\dots $ define
$$c_n(0) = 1, c_n(k) = \frac{n(n-1)\cdots (n-k+1)}{n^k},\, k = 1,2,\dots n, \,c_n(k) = 0\, \text {for}\,\, k> n.$$
Then the binomial theorem gives
$$\tag 1(1+z/n)^n = \sum_{k=0}^{\infty}c_n(k)\frac{z^k}{k!}.$$
For each fixed $k, |c_n(k)|\le 1$ for all $n,$ and $\lim_{n\to \infty}c_n(k) =1.$ Because $\sum |z|^k/k!<\infty$ for each $z,$ the limit of $(1)$ as $n\to \infty$ equals $\sum_{k=0}^{\infty}1\cdot\frac{z^k}{k!} = e^z$ by the DCT.
A: In my opinion this could be answered as follows:

Lemma: Let $z_n$ be a sequence of non-zero complex numbers and suppose
  that for each $n \in \mathbb{N}$ we have $\phi_n \in Arg(z_n)$ where
  $Arg(z_n)$ is the set of arguments (I usually denote by arg(z) the
  principal value). Assume that $\phi_n \to \phi$ and that $|z_n| \to \rho$ then we have $z_n \to \rho(cos \phi+isen\phi)$.

So once you've proved this lemma you can work as follows:
$lim |1+\frac{z}{n}|^n=
lim \sqrt{\left( (1+\frac{x}{n})^2 + (\frac{y}{n})^2\right)^n}=
lim \sqrt{\left(1+\frac{x^2/n+y^2/n+2x}{n}\right)^n}=(e^{2x})^{1/2}=e^x$
where I've used the real case of the limit and I've written $z=x+iy$.
Now by De Moivre's formula $n \ arg(1+\frac{z}{n}) \in Arg((1+\frac{z}{n})^n)$ and because of the previous lemma the only thing I need to proof is that $n \ arg(1+\frac{z}{n}) \to y$. 
You should remember at this point the value of the argument expressed in a general way which you can find here.
Using that form we have:
$n \ arg(1+\frac{z}{n}) = 2n \ arctg\left(\frac{y/n}{\sqrt{(1+x/n)^2+(y/n)^2}+(1+x/n)}\right)$ (1)
Note that this form could be wrong if $1+\frac{z}{n}$ was a negative real number  but clearly there exists $n_0$ such that for all $n \ge n_0$ $1+\frac{z}{n}$ is not a negative real number. 
Finally, if you apply that $arctg x$ and $x$ are equivalent infinitesimals when $x \to 0$ then in (1) you can put:
$2n \left(\frac{y/n}{\sqrt{(1+x/n)^2+(y/n)^2}+(1+x/n)}\right)$
which clearly converges to y.
If you apply the first lemma you are done.
