Question

Find the modulus and argument of i^i.

Answer 1

Answer:

$\bold{e^{-\frac{\pi}{2}}}$

Step-by-step explanation

$e^{i\theta}=cos\theta+isin\theta\\\\i = cos\frac{\pi}{2}+isin\frac{\pi}{2}=e^{i\frac{\pi}{2}}\\\\Now\\\\i^i=(e^{i\frac{\pi}{2}})^i=e^{i^2\frac{\pi}{2}}=e^{-\frac{\pi}{2}}$

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