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.
Best Answer
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 $n$.