(C) We know that $i = e^{i\pi/2}$.Substituting this into the expression for $z$:$z = (e^{i\pi/2})^{2i} = e^{(i\pi/2) \times (2i)} = e^{i^2 \pi} = e^{-

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(C) We know that $i = e^{i\pi/2}$.Substituting this into the expression for $z$:$z = (e^{i\pi/2})^{2i} = e^{(i\pi/2) \times (2i)} = e^{i^2 \pi} = e^{-\pi}$.Since $e^{-\pi}$ is a real positive number,its modulus is $|z| = |e^{-\pi}| = e^{-\pi}$.

Class 11 Mathematics 4-1.Complex numbers Square root, Representation and Logarithm of complex numbers

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

A
$1$
B
$e^{\pi}$
C
$e^{-\pi}$
D
$e^{\pi/2}$

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4-1.Complex numbers

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Square root, Representation and Logarithm of complex numbers

$\sqrt{12-\sqrt{68+48 \sqrt{2}}}$ is equal to :

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$\sec h^{-1}\left(\frac{1}{2}\right)-\operatorname{cosec} h^{-1}\left(\frac{3}{4}\right)$ is equal to

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

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$\sqrt{12-\sqrt{68+48 \sqrt{2}}}$ is equal to :

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