Can someone explain why the last equality is possible?
\[\large{a^b=(re^{i\theta })^b}\]\[\large{=(e^{ln~r+i\theta})^b=e^{(ln~r+i\theta)b}} \]

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@2bornot2b ,
\[\huge r = e^{\ln(r)}\]
Using this the result is got

As far as I know \(\large(e^z)^w\ne e^{zw}\) rather \(\large(e^z)^w=e^{(z+2n\pi i)w} \)

@shivam_bhalla I am talking about the equality that appears next to the one you have talked about.

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