TITLE: $i^{1/n}$ when $n \to \infty$
QUESTION [6 upvotes]: What would happen when $n \to \infty$ for $i^{1/n}$. Would it still be an imaginary number or would it simply be $i^0=1$?

REPLY [3 votes]: I want to respond to this provocative question as an algebraist, not an analyst. As many have noted, there are $n$ distinct roots of the equation $X^n=i$, equally spaced around the unit circle. And algebra refuses to distinguish among them: any $n$-th root of $i$ will be as good as any other.
If you’re willing to go along with me in this refusal, then you must agree that every complex number $a+bi$ on the unit circle, i.e. satisfying $a^2+b^2=1$, is a limit, as $n\to\infty$, of $n$-th roots of $i$, as long as those roots are  properly chosen.
For, let $a+bi$ be such a point on the unit circle. Among the $n$ numbers $\zeta$ satisfying $\zeta^n=i$, there is one, call it $\zeta_n$, that is at a distance of at most $\pi/n$ from $a+bi$. So if we ask for $\lim_n\zeta_n$, the value in answer is our $a+bi$.
( But at least there are limits. In my preferred environment, the $p$-adic, any sequence of $n$-th roots of $i$ will have no limit at all. )