Q:

Complex analysis : If $z =re^{i\theta}$, then prove that $|e^{iz}| =e^{-r\sin\theta}$

Problem : 
If $z =re^{i\theta}$, then prove that $|e^{iz} | =e^{-r\sin\theta}$ 
My working : 
$z = re^{i\theta} = r(\cos\theta + i\sin\theta)$ 
$\Rightarrow iz = ir(\cos\theta +\sin\theta) $ = $-r\sin\theta +ir\cos\theta $
$\Rightarrow e^{iz} =e^{(-r\sin\theta +ir\cos\theta)} = e^{-\sin\theta} e^{ir\cos\theta} $ 
$\Rightarrow |e^{iz}| = |e^{-r\sin\theta}||e^{ri\cos\theta}|$ 
Now please guide how to proceed further to get the result... thanks.

A:

You're almost there, Now just use the fact that $|e^{ix}|=1$ for all real $x$, and that $e^a>0$ for all $a\in\mathbb{R}$.

