Prove $e^{|z|}-1\geq|e^z-1|$

Question

Hi Does anyone know how to prove $e^{|z|}-1\geq|e^z-1|$ for any complex number $z$? I can see from the graph it must be right but have no idea to prove it. Thanks.

Answer 1

$|e^z-1|= |\sum_{n=1}^{\infty}\frac{z^n}{n!}| \le \sum_{n=1}^{\infty}\frac{|z|^n}{n!}=e^{|z|}-1$.