Find-arg-z-z-i-i-i-

Question Number 18961 by Tinkutara last updated on 02/Aug/17

Find \arg (z), z = i^{i^{i}} .

Answered by sma3l2996 last updated on 02/Aug/17

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

a r g (z) = \frac{π e^{- π / 2}}{2}

Commented by Tinkutara last updated on 02/Aug/17

Thank you very much Sir!