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?

See $π$ is  the radius so it doesnt matter in so lets start our complex number is $$\frac{π}{e^{i3π/2}}=\frac{π}{-i}=0+(-π.)(i)$$  after rationalising . So now the theta becomes as it is third quadrant $\theta=π+tan^-{1}((-π)/0)=π-π/2=π/2$ thats all and hint $log(x+iy)=log(i\theta)+|z|$ where $\theta$ is its argument and z is its radius from origin.
