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

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.

• $a+bi$ isn’t really approaching $0$, just the imaginary axis. You need both $a$ and $b$ to go to zero. Dec 26, 2020 at 19:28
• @Tavish Yeah, my bad. I wrote $n=a+bi$ instead of $n=ae^{ib}$. Fixed. Still getting weird results, though.
– Rain
Dec 26, 2020 at 19:38

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 $$n$$.
• Ah, thank you! It seems so easy; I was taking the wrong approach separating the phase from the amplitude and computing the limits separately. Also forgot that the limit of $e^{iy}$ as $y$ approaches zero is not infinity. There goes my maths cred. :)
• @Rain :) ${}{}$ Dec 26, 2020 at 19:43