Suppose $z=x+iy$, prove that $\lim\limits_{n\to\infty}(1+\frac{z}{n})^n=e^x(\cos{y}+i\sin{y})$? Suppose $z=x+iy$, how to prove that $\lim\limits_{n\to\infty}(1+\frac{z}{n})^n=e^x(\cos{y}+i\sin{y})$? 
 A: Let consider separately 


*

*$Arg(1+z/n)$

*$|1+z/n|$
and use that


*

*$Arg(w^n)=nArg(w)$

*$|w^n|=|w|^n$


Notably


*

*$Arg\left(1+\frac z n\right)=Arg\left(1+\frac x n+i\frac y n\right)=\arctan\left(\frac{y}{n+x}\right)$

*$\left|1+\frac z n\right|=\left|1+\frac x n+i\frac y n\right|=\sqrt{\left(1+\frac x n\right)^2+\left(\frac y n\right)^2}$


and then


*

*$Arg\left(1+\frac z n\right)^n=n\arctan\left(\frac{y}{n+x}\right)\sim
   n\cdot \frac y n\to y$

*$\left|\left(1+\frac z n\right)^n\right|=\left|1+\frac z n\right|^n=\left(\left(1+\frac x n\right)^2+\left(\frac y n\right)^2\right)^n\to e^x$

A: I am avoid Complex Analysis, and trying to only use Real Analysis and ODEs.
It suffices to show that
$$
\frac{\left(1+\frac{x+iy}{n}\right)^n}{\left(1+\frac{x}{n}\right)^n}=\left(1+\frac{iy}{n+x}\right)^n\to \cos y+i\sin y.
$$
Let
$$
f_n(y)+ig_n(y)=\left(1+\frac{iy}{n+x}\right)^n.
$$
Then
$$
f_n'(y)+ig_n'(y)=\frac{n}{n+x}\big(g_{n-1}(y)-if_{n-1}(y\big).
$$
After showing that $f_n$ and $g_n$ converge locally uniformly in $\mathbb R$, to $f$ and $g$, respectively, we obtain that $f$ and $g$ satisfy the initial value problem (of a $2\times 2$ system of ODEs)
$$
f'=g,\,\,g'=-f,\,\,f(0)=1,\,\,g(0)=0,
$$
with unique solution
$$
f(y)=\cos y,\,\, g(y)=\sin y.
$$
