Why does e^-im(3pi/2) equal i^m?

[fluidistic]

Homework Statement

I'm trying to follow some solution to an exercise in physics and apparently $e^{-im \frac{3\pi}{2}}=i^m$ where $m \in \mathbb{Z}$.
I don't realize why this is true.

Homework Equations

Euler's formula.

The Attempt at a Solution

I applied Euler's formula but this is still a mistery.
$i^m=\cos \left ( \frac{3\pi m}{2} \right ) -i \sin \left ( \frac{3\pi m }{2} \right )$.
I've checked the formula for m=1 and 2, it works. I must be missing the obvious, but I'm very tired physically and mentally.
Thanks for any help.

Edit: I found it. I drew a mental sketch of $e^{-i \frac{3\pi }{2}}$, it's "i" in the complex plane. Then just elevate this to the power m and the job is done.

 

[clamtrox]

What you can do is to use
$$i^m = e^{\ln(i^m)} = e^{m \ln i}$$
and then calculate what is $\ln i$

 

[HallsofIvy]

$e^{im\frac{-3\pi}{2}}= \left(e^{i\frac{-3\pi}{2}}\right)^m$

And, of course, $e^{i\frac{-3\pi}{2}}= i$ so that expression is just $i^m$.

If you are not clear that $e^{i\frac{3\pi}{2}}= i$, recall that $e^{i\theta}$, for any real $\theta$, lies on the unit circle ($|e^{i\theta}|= 1$ at angle $\theta$ measured counter clockwise from the positive real axis.
$e^{i\frac{-3\pi}{2}}$ lies on the unit circle, an angle $3\pi/2$ measured clockwise from the positive real axis.

Another way to see this is to recall that $x^{-1}= 1/x$ and that $1/i= -i$.