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

$$\mathrm{Find}\arg \left(z\right),z=i^{i^i}.$$

Answered by sma3l2996 last updated on 02/Aug/17

$$z=i^{{\left(e^{i\pi /2}\right)}^i}=i^{e^{-\pi /2}}={\left(e^{i\pi /2}\right)}^{e^{-\pi /2}}=e^{i\frac{\pi e^{-\pi /2}}{2}}$$$$arg\left(z\right)=\frac{\pi e^{-\pi /2}}{2}$$

Commented by Tinkutara last updated on 02/Aug/17

$$\mathrm{Thank}\mathrm{you}\mathrm{very}\mathrm{much}\mathrm{Sir}!$$

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