i to the - i

Answers (1)

Here is the solution:

\mathrm{i}^{\mathrm{-i}}=\left(\mathrm{e}^{\mathrm{i} \ \frac{\pi}{2} }\right)^{\mathrm{-i}}=\mathrm{e}^{\mathrm{-i}^{2} \frac{\pi}{2} }=\mathrm{e}^{\frac{\pi}{2}}

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