If $z=r{e}^{i\theta }=r\left(\cos \theta +i\sin \theta \right)$

$\Rightarrow iz=ir\left(\cos \theta +i\sin \theta \right)$

$=-r\sin \theta +ir \cos \theta$

$\Rightarrow e^{iz}={\mathrm{e}}^{\left(-\mathrm{rsin}\theta +ir\cos \theta \right)}$

$={\mathrm{e}}^{-r\sin \theta }{\mathrm{e}}^{ir\cos \theta }$

$\Rightarrow \left|e^{iz}\right|=\left|e^{-r\sin \theta }\right|\left|e^{ri\cos \theta }\right|$

$\Rightarrow \left|e^{iz}\right|=\left|e^{-r\sin \theta }\right|\left[\cos \left(r\cos \theta \right)+i\sin \left(r\cos \theta \right)\right]$

$\Rightarrow \left|{\mathrm{e}}^{\mathrm{iz}}\right|={\mathrm{e}}^{-r\sin \theta }{\left[{\cos }^2\left(\mathrm{rcos\theta }\right)+{\sin }^2\left(\mathrm{rcos}\theta \right)\right]}^{\frac{1}{2}}$

$={\mathrm{e}}^{-r\sin \theta }$.