4-1.Complex numbers
normal

Let $z$ =${i^{2i}}$ , then $|z|$ is (where $i$ =$\sqrt { - 1}$ )

A

$1$

B

${e^\pi }$

C

${e^{ - \pi }}$

D

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

Solution

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

Standard 11
Mathematics

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