If z=r e^ i 0 , then |e^ i z |=

MCQ

If $z=r e^{i 0}$, then $\left|e^{i z}\right|=$

A
$e^{r \sin \theta}$
✓
$e^{-r \sin 0}$
C
$e ^{-r \cos \theta}$
D
$e^{t \cos \theta}$

✓

Answer

Correct option: B.

$e^{-r \sin 0}$

(B)
$z=r e^{i \theta}=r(\cos \theta+i \sin \theta)$
$\Rightarrow iz = ir (\cos \theta+ i \sin \theta)=- r \sin \theta+ ir \cos \theta$
$\Rightarrow e ^{1 z}= e ^{(- r \sin \theta+ ir \cos \theta)}= e ^{- r \sin \theta} e ^{ ri \cos \theta}$
$\Rightarrow\left| e ^{ iz }\right|=\left| e ^{- r \sin \theta} \| e ^{ ri \cos \theta}\right|$
$=\left| e ^{- r \sin \theta}\right||\cos ( r \cos \theta)+ i \sin ( r \cos \theta)|$
$= e ^{- r \sin \theta}\left[\left\{\cos ^2( r \cos \theta)+\sin ^2( r \cos \theta)\right\}\right]^{\frac{1}{2}}$
$= e ^{- r \sin \theta} \quad \ldots\left[\because \cos ^2 \theta+\sin ^2 \theta=1\right]$

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