Limits in complex plane. Let $z$ and $z_n$ be complex numbers and assume $z_n \rightarrow z$.
It it true that 
$$\lim\limits_{n\to \infty}\left(1+\frac{z_n}{n}\right)^n=
\lim\limits_{n\to \infty}\left(1+\frac{z}{n}\right)^n  ?$$
Note that $\lim\limits_{n\to \infty}\left(1+\frac{z}{n}\right)^n$ exists and I take it as the definition of $e^z$. Further this relation is true (and easy) for real variables.
Note further that if the approach $z_n \rightarrow z$ is non tangential 
(in otherwords within a Stolz angle) then I have a proof of the above limit equality.
I would like to see either a proof of the general case or a counterexample.
Further what happens if $z_n \rightarrow z$ from outside the radius of $|z|$ ?
 A: Note that we have for $0\leq k\leq n$
 $$\frac{\binom{n}{k}}{n^k}=\frac{n(n-1)\cdots(n-k+1)}{k! n^k}\leq \frac{1}{k!}$$
Hence if we put
 $$\exp(z)-(1+\frac{z}{n})^n=\sum_{k\geq 0} a_{n,k}z^k$$
 all the $a_{n,k}$ are $\geq 0$. Let $R>0$, and $z\in \mathbb{C}$, $|z|\leq R$. We have:
$$|\exp(z)-(1+\frac{z}{n})^n|\leq \sum_{k\geq 0} |a_{n,k}||z|^k\leq \sum a_{n,k}R^k=\exp(R)-(1+\frac{R}{n})^n$$ 
If $z_n\to z$, there exists a $R$ such that $|z|\leq R$ and $|z_n|\leq R$ for all $n$, and hence we get $\displaystyle \exp(z_n)-(1+\frac{z_n}{n})^n\to 0$, and we easily conclude. 
A: Using the methods from this answer, we show the following:
$$
\begin{align}
\left|\frac{\left(1+\frac{z}n\right)^n}{\left(1+\frac{z+\delta}n\right)^n}\right|
&=\left|\,1-\frac\delta{n+z+\delta}\,\right|^n\tag{1a}\\
&\ge\left(1-\frac{|\delta|}{n-|z+\delta|}\right)^n\tag{1b}\\[3pt]
&\ge1-\frac{n\,|\delta|}{n-|z+\delta|}\tag{1c}
\end{align}
$$
Explanation:
$\text{(1a)}$: distribute the exponent over the quotient
$\phantom{\text{(1a): }}$simplify the quotient
$\text{(1b)}$: triangle inequality (twice)
$\text{(1c)}$: Bernoulli's Inequality
$$
\begin{align}
\left|\frac{\left(1+\frac{z+\delta}n\right)^n}{\left(1+\frac{z}n\right)^n}\right|
&=\left|\,1+\frac\delta{n+z}\,\right|^n\tag{2a}\\
&\ge\left(1-\frac{|\delta|}{n-|z|}\right)^n\tag{2b}\\[3pt]
&\ge1-\frac{n\,|\delta|}{n-|z|}\tag{2c}
\end{align}
$$
Explanation:
$\text{(2a)}$: distribute the exponent over the quotient
$\phantom{\text{(2a): }}$simplify the quotient
$\text{(2b)}$: triangle inequality (twice)
$\text{(2c)}$: Bernoulli's Inequality
Therefore,
$$
1-\frac{n\,|\delta|}{n-|z|}\le\left|\frac{\left(1+\frac{z+\delta}n\right)^n}{\left(1+\frac{z}n\right)^n}\right|\le\frac1{1-\frac{n\,|\delta|}{n-|z+\delta|}}\tag3
$$
Furthermore,
$$
\begin{align}
\left|\,\arg\left(\frac{\left(1+\frac{z+\delta}n\right)^n}{\left(1+\frac{z}n\right)^n}\right)\,\right|
&=n\left|\,\arg\left(\frac{1+\frac{z+\delta}n}{1+\frac{z}n}\right)\,\right|\tag{4a}\\
&=n\left|\,\arg\left(1+\frac{\delta}{n+z}\right)\,\right|\tag{4b}\\[3pt]
&\le n\sin^{-1}\left(\frac{|\delta|}{|n+z|}\right)\tag{4c}\\[3pt]
&\le\frac\pi2\frac{n\,|\delta|}{n-|z|}\tag{4d}
\end{align}
$$
Explanation:
$\text{(4a)}$: distribute the exponent
$\phantom{\text{(4a): }}$$\arg\left(z^n\right)=n\arg(z)$
$\text{(4b)}$: simplify the quotient
$\text{(4c)}$: $|\arg(1+z)|\le\sin^{-1}(|z|)$
$\text{(4d)}$: $\sin^{-1}(x)\le\frac\pi2x$ for $0\le x\le1$
$\phantom{\text{(4d):}}$ triangle inequality
For $|\delta|\lt\frac12$ and $n\ge2|z|+1$, $(3)$ gives the thin annulus
$$
1-2|\delta|\le\left|\frac{\left(1+\frac{z+\delta}n\right)^n}{\left(1+\frac{z}n\right)^n}\right|\le\frac1{1-2|\delta|}\tag5
$$
and $(4)$ gives the thin sector
$$
\left|\,\arg\left(\frac{\left(1+\frac{z+\delta}n\right)^n}{\left(1+\frac{z}n\right)^n}\right)\,\right|\le\pi|\delta|\tag6
$$
As $\delta\to0$, the intersection of the annulus and the sector tend to $1$:


Concerning $\bf{(4c)}$
In the justification of $\text{(4c)}$, it is claimed that
$$
|\arg(1+z)|\le\sin^{-1}(|z|)\tag7
$$
As illustrated below, the maximum of $\arg(1+z)$ is attained on the tangent to the circle of radius $|z|$ centered at $1$. In that arrangement, $\arg(1+z)=\sin^{-1}(|z|)$. Symmetry gives $(7)$.

A: I'm too lazy for inequalities today, so here's an outline. :) If you manipulate the limits, you'll see that you can set $z=0$ without loss of generality. Then it suffices to prove
$$n\log\left(1+\frac{z_n}{n}\right)\to0,$$
which follows from $\log(1+x)=x+o(x)$.
A: A more mundane approach:
Let $f_n(z) = ( 1+{ z \over n})^n$ and note that $f_n(z) \to e^z$ pointwise.
We see that $f'_n(z) \to e^z$ and $f''_n(z) = {n-2 \over n} ( 1+{ z \over n})^{n-2}$ and so $|f''_n(z)| \le  ( 1+{ |z| \over n})^{n-2} \le e^{|z|}$. In
particular, $f''_n(z)$ is bounded on bounded sets (the important part is uniformly in $n$).
Suppose $|w-z| \le 1$ and let $M$ be a bound on $f''_n(w)$,
then Taylor's theorem gives us the bound
$|f_n(w)-f_n(z)-f'_n(z) (w-z) | \le {1 \over 2} M |w-z|^2$, or
$|f_n(w)-f_n(z) | \le {1 \over 2} M |w-z|^2 + |f'_n(z) (w-z)|$.
Substituting $w=z_n$ and taking limits we get $f_n(z_n) \to e^z$.
