Proof for $|e^z-1| \leq e^{|z|}-1 \leq |z|e^{|z|}$ and $|e^z| = e^{Re(z)}$

Question

Prove that

$$|e^z-1| \leq e^{|z|}-1 \leq |z|e^{|z|}$$ and $$|e^z| = e^{Re(z)}$$

For the inequation $|e^z-1| \leq e^{|z|}-1$, I was thinking that if $z<0$ the left part would be smaller than the right part and if $z>0$ it would be equal.

Answer 1

Hint

$$|e^z-1|=\left|\sum^\infty_{n=1}\frac{z^n}{n!}\right|\leq \sum^\infty_{n=1}\frac{|z|^n}{n!} $$

If $z=x+iy$ $e^{z}=e^{x}e^{iy}$ and $|e^{iy}|=1$

For the las inequality, it suffices to assume $z$ is a nonnegative number, that is show $e^t-1\leq te^t$ for all $t\geq0$. The mean value theorem may be useful.