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If $z=r e^{i \theta}$, then $\left|e^{i z}\right|=$
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Verified Answer
The correct answer is:
$e^{-r \sin \theta}$
We have,
$\begin{aligned}
z &=r e^{i \theta}=r(\cos \theta+i \sin \theta) \\
\therefore i z &=i r(\cos \theta+i \sin \theta) \\
&=i r \cos \theta+i^{2} r \sin \theta=-r \sin \theta+i r \cos \theta \\
\left|e^{i z}\right| &=\left|e^{(-r \sin \theta+i r \cos \theta)}\right| \\
&=\left|e^{-r \sin \theta}\right| \cdot|i r \cos \theta| \\
&=e^{-r \sin \theta}\left[\sqrt{\cos ^{2}(r \cos \theta)+\sin ^{2}(r \cos \theta)}\right] \\
&=e^{-r \sin \theta} \times 1=e^{-r \sin \theta}
\end{aligned}$
$\begin{aligned}
z &=r e^{i \theta}=r(\cos \theta+i \sin \theta) \\
\therefore i z &=i r(\cos \theta+i \sin \theta) \\
&=i r \cos \theta+i^{2} r \sin \theta=-r \sin \theta+i r \cos \theta \\
\left|e^{i z}\right| &=\left|e^{(-r \sin \theta+i r \cos \theta)}\right| \\
&=\left|e^{-r \sin \theta}\right| \cdot|i r \cos \theta| \\
&=e^{-r \sin \theta}\left[\sqrt{\cos ^{2}(r \cos \theta)+\sin ^{2}(r \cos \theta)}\right] \\
&=e^{-r \sin \theta} \times 1=e^{-r \sin \theta}
\end{aligned}$
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