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

$\begin{array}{l}z=r,e^{i\theta },i=e^{i\pi /2}\\ \therefore iz=r{e}^{i(\pi /2+\theta )}=r(-\sin \theta +i\cos \theta )\\ =r(a+ib)\text{ where }a=-\sin \theta ,b=\cos \theta \\ \therefore e^{iz}=e^{ra}\cdot e^{ibr}\\ \left|e^{iz}\right|=e^{ra}\cdot 1=e^{-r\sin \theta }\end{array}$

$\therefore$ exponential function is always $+ ive$ and $\left|e^{i\theta }\right|=1$ .