Here Log is the principal complex logarithm, Log$z = \log|z| + i$Arg$z$.
I have tried multiple approaches none of which look even remotely promising. I know that Log is discontinuous along the negative real axis and this was a cause of some minor problems.
As a follow up I'm supposed to show that $\lim_{n\to\infty}(1+z/n)^n = e^z$ which I have done assuming the result in the title.
Can someone give me hints or an outline of what I can try?
You have that
$$\text{Log}\left(1+\frac{z}n\right) = \log\left(\left|1+\frac{z}n\right|\right) + i\,\text{Arg}\left(1+\frac{z}n\right)$$
Now, let $z=a+bi$. Then
$$\text{Arg}\left(1+\frac{z}n\right) = \arctan\left(\frac{\frac{b}n}{1+\frac{a}n}\right)=\arctan\left(\frac{b}{n+a}\right)$$
From real analysis we know that
$$\arctan(x) = \sum_{k \geq 0}\frac{(-1)^k}{2k+1}\,x^{2k+1}=x-\frac{x^3}3+\frac{x^5}5-\dots$$
Hence,
\begin{align} n\cdot\arctan\left(\frac{b}{n+a}\right) &= n\,\frac{b}{n+a} - \frac{n}3\,\left(\frac{b}{n+a}\right)^3 + \frac{n}5\,\left(\frac{b}{n+a}\right)^5 - \dots\\ &\stackrel{n\to\infty}{\longrightarrow}\,\,b, \end{align}
which shows that $\Im(z)=\Im\left(\lim_{n\to\infty}n\,\text{Log}\left(1+\frac{z}n\right)\right)$.
For the real part, observe that
\begin{align} \left|1+\frac{z}n\right| &= \sqrt{{\left(1+\frac{a}n\right)}^2+{\left(\frac{b}n\right)}^2}\\ &= \sqrt{1+\left(\frac{2a}{n}+\frac{a^2+b^2}{n^2}\right)} \end{align}
Hence, $\log\left(\left|1+\frac{z}n\right|\right) = \frac12\,\log\left(1+\left(\frac{2a}{n}+\frac{a^2+b^2}{n^2}\right)\right)$. We know from real analysis that for $|x|<1$
$$\log(1+x) = \sum_{k\geq 1}\frac{(-1)^{k+1}}{k}\,x^k=x-\frac{x^2}2+\frac{x^3}3-\dots$$
Since $\frac{2a}{n}+\frac{a^2+b^2}{n^2} \to 0$ as $n\to\infty$, we have that for $n$ large enough:
\begin{align}n\cdot \log\left(\left|1+\frac{z}n\right|\right) &=\frac{1}2\,\left[n\,{\left(\frac{2a}{n}+\frac{a^2+b^2}{n^2}\right)} -\frac{n}2\,{\left(\frac{2a}{n}+\frac{a^2+b^2}{n^2}\right)}^2 +\frac{n}3\,\,{\left(\frac{2a}{n}+\frac{a^2+b^2}{n^2}\right)}^3 -\dots\right]\\ &\stackrel{n\to\infty}{\longrightarrow}\,\,\frac12\,[2a]=a. \end{align}
This shows that $\Re(z)=\Re\left(\lim_{n\to\infty}n\,\text{Log}\left(1+\frac{z}n\right)\right)$ and completes the proof. $\square$