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.