Question

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

• A. $e^{-r\sin \theta }$
• B. $e^{-i\sin \theta }$
• C. $e^{-r\cos \theta }$
• D. $e^{-\theta }$

Answer 1

The correct option is A $e^{-r\sin \theta }$

$z=r{e}^{i\theta }=r(\cos \theta +i\sin \theta )$
$\Rightarrow iz=ir(\cos \theta +i\sin \theta )$
$\Rightarrow iz=-r\sin \theta +ir\cos \theta$
$\Rightarrow e^{iz}=e^{-r\sin \theta +ir\cos \theta }$
$=e^{-r\sin \theta }{e}^{ri\cos \theta }$
$\Rightarrow |e^{iz}|=|e^{-r\sin \theta }||e^{ri\cos \theta }|$
$=e^{-r\sin \theta }|e^{i\alpha }|$, where $\alpha =r\cos \theta$
$=e^{-r\sin \theta }[{\cos }^2\alpha +{\sin }^2\alpha {]}^{\frac{1}{2}}$
$=e^{-r\sin \theta }$