If z=r $e^{i θ}$ ,then | $e^{i z}$ |=

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Solution

The correct option is B

if z=r $e^{i θ}$ =r(cos $θ$ + isin $θ$ )

$\Rightarrow$ iz=ir(cos $θ$ + isin $θ$ )=-rcos $θ$ + irsin $θ$

or $e^{i z}$ = $e^{- r c o s θ + i r s i n θ}$ = $e^{- s i n θ} e^{r i c o s θ}$

or | $e^{i z}$ |= $| e^{- r s i n θ} | | e^{i r c o s θ} |$

= $e^{- r s i n θ}$ $[c o s^{2} (r c o s θ) + s i n^{2} (r c o s θ)]^{2}$ = $e^{- r s i n θ}$

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