Argument of $\pi e^{-\frac{3i\pi}{2}}$ 
Find the argument of $\displaystyle \pi e^{-\frac{3i\pi}{2}}$

I thought the formula for the argument was $\arg{z} = i\log\frac{|z|}{z}$
In this case $|z|= \pi$, so it turns out that $-i \log(e^{-3i\pi/2}) = \frac{\pi}{2}.$
However, I don't understand why $-i \log(e^{-3i\pi/2}) = \frac{\pi}{2}$. 
Could someone please explain why. 
Also, is there a quicker method?

The argument is the angle that the vector makes with respect to the positive real axis. 
Using that $e^{i \pi} = \cos \pi + i \sin \pi$ and that $-\frac{3 \pi } {2} = \frac{\pi}{2}$, we see then that
$$ \textrm{exp} \left(\frac{3\pi i } {2}\right) = \textrm{exp} \left(\frac{\pi i}{2}\right) = \cos \frac{\pi}{2} + i \sin \frac{\pi}{2}$$
Thus the argument (angle) is $\frac{\pi}{2}$. 
