Is j^{-p}=e^{-j\frac{p\pi}{2}} a valid exponential function?

[yungman]

Homework Statement

I want to verify
$$j^{-p}=e^{-j\frac{p\pi}{2}}$$

Homework Equations

$$e^{j\frac{\pi}{2}}=\cos(\frac{\pi}{2})+j\sin(\frac{\pi}{2})=j$$

The Attempt at a Solution

$$j^{-p}=(e^{j\frac{\pi}{2}})^{-p}=e^{-j\frac{p\pi}{2}}$$

Am I correct?
Thanks

 

[dirk_mec1]

Yes.