# Inequality $|e^z -1| \le 2 |z|$ for complex $z$ with $|z|\le1$

I am trying to prove that for $z \in \mathbb C, |z|\le 1$:

$$|e^z -1| \le 2 |z|$$

But I'm stuck and I need help. I showed that for all $z$:

$|e^z -1| \le |z|e^{|z|}$

but it does not seem useful. Also, $|e^z -1| \le \sum_{n \ge 1}{|z|^n \over n!}$ does not seem useful.

Could someone show me how one would prove this inequality?

Let $|z| \le 1$ and $N \in \Bbb N$. Then

$$\left|\sum_{n = 1}^N \frac{z^n}{n!}\right| \le \sum_{n = 1}^N \frac{|z|^n}{n!} \le \sum_{n = 1}^N \frac{|z|^n}{2^{n-1}} \le |z| \sum_{n = 1}^\infty \left(\frac{1}{2}\right)^{n-1} = 2|z|.$$

Letting $N\to \infty$ results in

$$|e^z - 1| \le 2|z|.$$

• Thank you for your help. I understand it now. – Anna Apr 20 '15 at 8:06

$\frac {e^z - 1}{z}= 1 + z/2 + z^2/3! + \cdots .$ For $|z|\le 1,$ the absolute value of this is $\le 1 + 1/2 + 1/3! + \cdots = e - 1 <2.$

• But $2 |z| \le 2$... – Anna Apr 20 '15 at 8:05

We have $|z|\le 1$. $$\sum_{n\ge 1} \frac{|z|^n}{n!} \overset{\text{?}}\le 2|z|$$ $$\sum_{n\ge1} \frac{|z|^{n-1}}{n!} \overset{\text{?}}\le 2$$ $$\sum_{n\ge 2} \frac{|z|^{n-1}}{n!} \overset{\text{?}}\le 1$$ $$\frac{|z|} 2 + \underbrace{\frac{|z|^2} 6 + \frac{|z|^3} {24} + \frac{|z|^4}{120} + \cdots} \overset{\text{?}}\le 1$$ Can one show that the part over the $\underbrace{\text{underbrace}}$ is bounded above by a geometric series with first term $1/6$ and common ratio $1/4$? The sum of that series is $2/9$ and the first term above is $\le 1/2$.

$g(z)=\frac{e^z-1}{z}$ is a holomorphic map in the unit disk, hence by the Schwarz lemma: $$\forall z\in\mathbb{C}:|z|\leq 1,\qquad |g(z)-1|\leq |z|\tag{1}$$ that implies: $$\forall z\in\mathbb{C}:|z|\leq 1,\qquad |g(z)|\leq 1+|z|\tag{2}$$ and finally: $$\forall z\in\mathbb{C}:|z|\leq 1,\qquad |e^z-1|\leq |z|+|z|^2\tag{3}$$ that is stronger than needed.