Question

If $z=r{e}^{i\theta }$, then $|e^{iz}|=$

• A. $e^{rsin\theta }$
• B. $e^{-rsin\theta }$
• C. $e^{-rcos\theta }$
• D. $e^{rcos\theta }$

Answer 1

Answer: (B)

$e^{-rsin\theta }$

If $z=r{e}^{i\theta }=r(cos\theta +isin\theta )$
$\Rightarrow iz=ir(cos\theta +isin\theta )=-rsin\theta +ircos\theta$
or $e^{iz}=e^{(-rsin\theta +ircos\theta )}=e^{-rsin\theta }{e}^{ricos\theta }$
or $|e^{iz}|=|e^{-rsin\theta }||e^{ricos\theta }|$
$=e^{-rsin\theta }\left[co{s}^2\right(rcos\theta )+si{n}^2(rcos\theta ){]}^{1/2}=e^{-rsin\theta }$