Question

If $z=r{e}^{i\theta }$, then prove that $\left|e^{iz}\right|=e^{-r}sin\theta$.

Answer 1

$z=r{e}^{i\theta }$
$=r(cos\theta +isin\theta )$
$z^{\prime }=e^{iz}$
$=e^{ir(cos\theta +isin\theta )}$
$=e^{r(-sin\theta +icos\theta )}$
$=e^{-rsin\theta }.e^{icos\theta }$
$=R.e^{i\varphi }$ where $R=e^{-rsin\theta }$ and $\varphi =cos\theta$
Hence
$|z^{\prime }|$
$=|e^{iz}|$
$=R$
$=e^{-rsin\theta }$