# Bounds for Complex Exponential

## Theorem

Let $\exp$ denote the complex exponential.

Let $z\in\C$ with $|z|\leq\frac12$.

Then $\frac 1 2 |z| \leq \left| \exp(z) -1 \right| \leq \frac32|z|$.

## Proof

By definition of complex exponential:

$\exp z = \displaystyle \sum_{n\mathop=1}^\infty\frac{z^n}{n!}$

Thus

 $\displaystyle \left\vert \exp z -1 - z \right\vert$ $=$ $\displaystyle \left\vert \sum_{n\mathop=2}^\infty \frac{z^n}{n!} \right\vert$ Linear Combination of Convergent Series $\displaystyle$ $\leq$ $\displaystyle \sum_{n\mathop=2}^\infty\left\vert \frac{z^n}{n!} \right\vert$ Triangle Inequality for Series $\displaystyle$ $\leq$ $\displaystyle \sum_{n\mathop=2}^\infty \frac{\vert z\vert^n }2$ $n\geq2$ $\displaystyle$ $=$ $\displaystyle \frac{\vert z\vert^2 /2}{1- \vert z\vert}$ Sum of Geometric Progression $\displaystyle$ $\leq$ $\displaystyle \frac12\vert z\vert$ $\vert z\vert \leq \frac12$

By the Triangle Inequality:

$\frac12 \vert z\vert\leq \left\vert \exp z - 1 \right\vert \leq \frac32\vert z\vert$

$\blacksquare$