Does the limit definition of $e$ work if the quantity that approaches 0 (or $\infty$) is complex?

Question

Does the definition

$$e\equiv\displaystyle\lim_{n\to0}\left(1+n\right)^{1/n}$$

or the equivalent

$$e\equiv\displaystyle\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n$$

work if $n$ is complex? (I'm partial to the first one, as the second one would probably require something like $|n|\to\infty$ and I'm always wary around $\infty$ anyway, but either works.)

I've thought about writing $n$ as $|n|e^{i\arg(n)}$ (I'm not trying to define $e$, just trying to use one of the above definitions to equate something to $e$ in my calculations). For brevity, I'll write $a=|n|,b=\arg(n)$. The first definition of $e$ then becomes

\begin{eqnarray} \displaystyle\lim_{n\to0}\left(1+n\right)^{1/n} & = & \displaystyle\lim_{a\to0}\left(1+ae^{ib}\right)^{1/ae^{ib}}\\ & = & \displaystyle\lim_{a\to0}e^{ib/ae^{ib}}\left(\left(1+a\right)^{1/a}\right)^{1/e^{ib}}\\ & = & \displaystyle\lim_{a\to0}e^{ib/ae^{ib}}\displaystyle\lim_{a\to0}\left(\left(1+a\right)^{1/a}\right)^{1/e^{ib}}\\ & = & \infty e^{e^{-ib}}\\ & = & \infty, \end{eqnarray}

but I find it hard to believe that $n$ being complex will turn the limit from $e$ to $\infty$. (I've changed $n\to0$ to $a\to0$ because the only way for $n$ to approach $0$ is for its magnitude to approach $0$; we don't really care about what its phase is doing in this case, right?)

Thanks for any help.

Edit: My maths were wrong because I'm an idiot.

Answer 1

You need the whole complex number to go to the origin, for the definition to hold. One way of doing it is writing $n=re^{i\theta}$ with $r\to 0$.

$$\lim_{r\to 0} (1+re^{i\theta} )^{1/re^{i\theta}} \\ = \bigg(\lim_{r\to 0} (1+re^{i\theta})^{1 /r} \bigg)^{e^{-i\theta}} \\ =\big( e^{e^{i\theta}} \big)^{e^{-i\theta}} \\ = e$$ where I have used the standard result $\lim_{n\to 0} (1+an)^{1/n} = e^a$ for real $n$.