# Bounds for Complex Exponential

## Theorem

Let $\exp$ denote the complex exponential.

Let $z \in \C$ with $\cmod z \le \dfrac 1 2$.

Then

$\dfrac 1 2 \cmod z \le \cmod {\exp z - 1} \le \dfrac 3 2 \cmod z$

## Proof

By definition of complex exponential:

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

Thus

 $\displaystyle \cmod {\exp z - 1 - z}$ $=$ $\displaystyle \cmod {\sum_{n \mathop = 2}^\infty \frac {z^n} {n!} }$ Linear Combination of Convergent Series $\displaystyle$ $\le$ $\displaystyle \sum_{n \mathop = 2}^\infty \cmod {\frac {z^n} {n!} }$ Triangle Inequality for Series $\displaystyle$ $\le$ $\displaystyle \sum_{n \mathop = 2}^\infty \frac {\cmod z^n} 2$ as $n \ge 2$ $\displaystyle$ $=$ $\displaystyle \frac {\cmod z^2 / 2} {1 - \cmod z}$ Sum of Geometric Sequence $\displaystyle$ $\le$ $\displaystyle \frac 1 2 \cmod z$ as $\cmod z \le \dfrac 1 2$

By the Triangle Inequality:

$\dfrac 1 2 \cmod z \le \cmod {\exp z - 1} \le \dfrac 3 2 \cmod z$

$\blacksquare$