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

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SUMMARY

The equation j^{-p} = e^{-j\frac{p\pi}{2}} is confirmed as a valid exponential function. This conclusion is derived from the relationship e^{j\frac{\pi}{2}} = j, which establishes that raising j to the power of -p results in the equivalent expression e^{-j\frac{p\pi}{2}}. The mathematical manipulation is accurate and adheres to the properties of complex exponentiation.

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Homework Statement


I want to verify
[tex]j^{-p}=e^{-j\frac{p\pi}{2}}[/tex]


Homework Equations



[tex]e^{j\frac{\pi}{2}}=\cos(\frac{\pi}{2})+j\sin(\frac{\pi}{2})=j[/tex]

The Attempt at a Solution



[tex]j^{-p}=(e^{j\frac{\pi}{2}})^{-p}=e^{-j\frac{p\pi}{2}}[/tex]

Am I correct?
Thanks
 
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