How to show that $|\exp(z)-1|\le2|z|$ for $|z|\le 1$

Question

How to show that $|\exp(z)-1|\le2|z|$ for $|z|\le 1$

$\displaystyle|\exp(z)-1|=\big|\sum\limits_{k=1}^\infty\frac{z^k}{k!}\big|\le\sum\limits_{k=1}^\infty\frac{|z|^k}{k!}=|z|+\sum\limits_{k=2}^\infty\frac{|z|^k}{k!}$

now it remains to prove that the sum, the most right is $\le|z|$

Since $\sum\limits_{k=2}^\infty\frac{|z|^k}{k!}\le\sum\limits_{k=0}^\infty\left(\frac{|z|}{2}\right)^k-\frac{|z|}{2}-1+\frac{|z|^2}{4}$ (geometric series)

the RHS is $\displaystyle \frac{1}{1-\frac{|z|}{2}}-\frac{|z|}{2}-1+\frac{|z|^2}{4}=\frac{\frac{|z|}{2}}{\frac{2}{|z|}-1}+\frac{|z|^2}{4}\le|z|$

am I right ?

Answer 1

Yes, your proof is correct.

Here is an alternative one: $$e^z-1 = \int_0^z e^{\xi} \, d\xi = \int_0^1 e^{tz} \, dt$$ implies, by the triangle inequality, $$|e^z-1| \leq |z| \int_0^1 \underbrace{|e^{tz}|}_{e^{t Re z} \leq e^{t}} \, dt \leq |z| \cdot |e-1| \leq 2|z|$$