What is $0^{i}$? $$\lim_{n\to 0} n^{i} = \lim_{n\to 0} e^{i\log(n)} $$
I know that $0^{0}$ is generally undefined, but can equal one in the context of the empty set mapping to itself only one time. I realize that in terms of the equation above, the limit does not exist, but can $0^{i}$ be interpreted in a way to assign it a value? For the curious, I ran in to this when trying to calculate the imaginary-derivative of $\sin(x)$. 
 A: This was a comment, but @hjhjhj57 suggested that it might serve as an answer.
If you write the right-hand side of your equation as $\lim_{t\to−∞}e^{it}=\lim_{t\to−∞}(e^i)^t$, it’s completely clear that the limit doesn’t exist: you’re taking the number $e^i$, which is on the unit circle, and raising it to a large (but negative) power. You have a point that runs around the unit circle infinitely many times as $t\to−∞$, no limit at all. 
A: It is possible to interpret such expressions in many ways that can make sense.  The question is, what properties do we want such an interpretation to have?
$0^i = 0$ is a good choice, and maybe the only choice that makes concrete sense, since it follows the convention $0^x = 0$.  On the other hand, $0^{-1} = 0$ is clearly false (well, almost—see the discussion on goblin's answer), and $0^0=0$ is questionable, so this convention could be unwise when $x$ is not a positive real.

Digging deeper: One generally defines complex exponentiation as a multi-valued function: if $e^c = a$, then we can define $a^x = e^{cx}$.  This is not unique, since it depends on the choice of $c$, but it's a good way to think about quantities like $i^i$ (this is sometimes claimed to be $e^{-\pi/2}$, but it can be interpreted as $e^{-\pi/2 + 2\pi n}$ for any $n\in\mathbb{Z}$)
This approach breaks down for $0^i$, because $0$ has no natural logarithm in the complex numbers.  However, if we're comfortable calling $-\infty$ (or $-\infty + 2\pi i n$) a natural log of $0$, then we can say that $0^x = e^{-\infty \cdot x} = 0$ when $x$ has positive real part.
When $x$ has negative real part, this leads us to regard $0^x$ as a quantity with infinite magnitude and undefined argument.  When $x$ is imaginary, the argument is still undefined, but the magnitude is multi-valued rather than infinite.

My conclusion is that we should avoid assigning meaning to $0^i$.
Writing $|0^i| = 1$ may be sensible, however, under some circumstances.
In a general setting, I would be comfortable saying that $|0^i| = e^{2\pi n}$, for any $n\in\mathbb{Z}$.
A: We have:
$$\lim_{x\to0} e^{i \log x}=\lim_{x\to\infty} e^{-ix}=\lim_{x\to\infty} (-1)^{-x}=\lim_{x\to\infty} (-1)^{x}$$
The limit as you noticed, does not exist... But if you want to assign a value nevertheless... well, the mean value of $(-1)^n$ will be zero:
$$\lim _{x\to\infty} \frac 1{x}\int_0^x (-1)^x =0$$
A: Wait.
The first reasonable way to understand if a good definition of $0^i$ can be given, is to define
$$
0^i:=\lim_{z\to0}z^i=\lim_{r\to0}(re^{i\theta})^i=e^{-\theta}\lim_{r\to0}r^i
$$
But now $r^i=e^{i\log r}=\cos(\log r)+i\sin(\log r)$, and being $\log r\stackrel{r\to0}{\longrightarrow}-\infty$, we conclude the limit $\lim_{z\to0}z^i$ can't exist, thus $0^i$ can't be defined.
