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Question #e43d3

1 Answer

Apr 10, 2017

$i^{i} = e^{- \frac{π}{2}}$ $i^i = e^( - pi/2)$

Explanation:

Euler's Formula:

$e^{i x} = cos x + i sin x$ $e^(i x) = cos x + i sin x$

$\Rightarrow i = cos (\frac{π}{2}) + i sin (\frac{π}{2}) = e^{i \frac{π}{2}}$ $implies i = cos (pi/2) + i sin (pi/2) = e^( i pi/2)$

$i^{i} = {(e^{i \frac{π}{2}})}^{i} = e^{i \cdot i \frac{π}{2}} = e^{- \frac{π}{2}}$ $i^i = (e^( i pi/2))^i = e^( i cdot i pi/2) = e^( - pi/2)$

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