What is the right treatment for $0^i$? I need to calculate a limit of a complex expression (had it in a physics research) that contains a term $(r-b)^p$ for $r\rightarrow b+$ where $r,b$ are reals, and $p$ is complex, let's suppose for simplicity $p=i$.
At the beginning, I put it equals to $0$ as an obvious thing, but strangely enough, "Mathematica" gives me something strange: $e^{2i\operatorname{Interval}[0,\pi]}$, then I thought maybe I was wrong with "obviousity" of the result, because we can actually put (formally) $0=e^{-\infty+i\phi}$, then if we use $a^b=e^{b\log a}$ we have $0^i=e^{-i\infty-\phi}$ (where all $2\pi k$ included in the infinity or $\phi$).
This result seems to be very strange for me (it will introduce a new parameter $\phi$ into my theory), and I found no example for this, especially that I know one needs to be careful with branches when making such tricks, but I have no clue if that is correct or wrong, any help will be appreciated.
 A: The problem is that $x^i$ is only defined as $e^{i\log x}$. 
When $x$ gets close to zero, $\log x$ is of the form $a+bi$ where $a$ is a very large negative number and $b$ can be restricted to $(-\pi,\pi]$. Then $e^{i\log x}$ is $e^{-b} e^{ia}$.Note then that if we let $x$ approach zero along a line from one side at angle $\theta$, that means the modulous of $e^{i\log x}$ will be $e^{-\theta}$, which is constant, so it does not approach zero (and it does not even converge - $x^i$ basically whizzes around a circle when $x\to 0$ along a straight line.)
Exponentiation $x^u$ near $x=0$ just is not well-behaved.
And this doesn't even take into account that $\log x$ is more naturally a multi-valued function.
Even restricting $x$ to $\mathbb R^+$ with $\log x$ the normal real natural logarithm, $$x^i=e^{i\log x}$$
So $|x^i|=1$, and as $x\to 0+$, $x^i$ spins around the unit circle clockwise and certainly does not have a limit.
A: The complex exponential $(x+iy)^p$ is not defined uniquely. It is $(x+iy)^p = \exp(p\log(x+iy))$ where the logarithm is defined up to a multiple of $2\pi i$. Hence your formula is not uniquely determined and you cannot say that it tends to zero...
A: To make $\ln z$ unique, one needs to slit the complex plane and it is common to use only the main branch on $\mathbb C\setminus\mathbb R_{\le0}$.
As in your specific problem $r,b$ are always reals and $r>b$, we can write $r-b=e^t$ with $t\in \mathbb R$ and $t\to-\infty$.
Then $(r-b)^p=e^{pt}$. If $\Re p=0$ (but $\Im p\ne 0$) this indeed keeps rotating around on the unit circle so that $\lim_{r\to b^+}(r-b)^p$ does not exist.
If $\Re p>0$, however, then $|(r-b)^p|=|e^{pt}|=e^{t\Re p}\to 0$. And if $\Re p<0$, then $|(r-b)^p|\to\infty$. In both these latter cases we still have that rotation phenomenon if $\Im p\ne 0$.
