Is the limit $\lim\limits_{x\to\infty} {i}^{-x}$ equal to $0$, or doesn't exist? Can someone show me if this limit exists:
$$\lim_{x\to \infty} {i}^{-x}=0$$
or it doesn't exist?
Here, $i$ is the unit imaginary part.
Thank you for any help.
 A: After choosing a branch $\log z$ of the complex log, you will get
$$i^{-x}=e^{-x\log i}.$$
Then $\log i= \ln|i|+i\arg i $, so
$$i^{-x} = e^{-x(\ln|i|+i\arg i)}=e^{-ix\arg i}$$
(since $\ln|i|=\ln 1=0$) .  We can assume $i$ has a well-defined argument in the given choice of branch, so that its argument is a constant value in whatever branch you select, say $\arg i=\theta$. Then, using Euler's identity, you get:
$$i^{-x}=e^{-x \ln i}=e^{-x(0+i\theta)}=e^{-ix\theta}=(\cos(x\theta)+i\sin(x\theta)).$$
Since $e^{i\cdot 0}=1 $ is fixed, the radius is fixed and you have a point in the unit circle.  
As others pointed out, when $x$ grows, for a fixed value of $\theta$, the expression: $\cos(x \theta)$ and $\sin(x\theta)$ , the positions (angles) within the unit circle will oscillate indefinitely (e.g., let $x=p/q\pi\theta$ for the "right" values of $p,q$ ; can do something a bit different if $\theta =0$ ), so the argument (complex) part of the expression $i^{-x}=U+iV$ will not converge, so the limit does not exist. 
A: $i^{-x} = \frac{1}{i^x}=(\frac{1}{i})^{x} = (-i)^x$. Now observe that $i = e^{\pi i/2} \implies (-i)^x = e^{3\pi i x /2}$. So informally, as $x$ gets larger and larger, you just keep spinning around on the unit circle.
