How to prove the inequality $|e^z-1| \geq |z|/2$ for sufficiently small $|z|$?

Question

How to prove the inequality $|e^z-1| \geq |z|/2$ for sufficiently small $|z|$?
I was thinking about the Taylor series for $e^z$, but I have no idea.
Any answers will be greatly appreciated.

Answer 1

By L'Hopital's Rule $\frac {e^{z}-1} z \to 1$ as $z \to 0$. Hence $|\frac {e^{z}-1} z|\geq \frac 1 2$ for $|z|$ sufficiently small.

Proof using power series: $e^{z}-1=z+\frac {z^{2}} {2!}+...$ so $|e^{z}-1| \geq |z|-|z|(\frac {|z|} {2!}+\frac {|z|^{2}} {3!}+...)$. Use the fact that $|\frac {|z|} {2!}+\frac {|z|^{2}} {3!}+...|\leq |z| e$ for $|z| <1$ Can you complete the proof now?