# Bounds for Complex Exponential

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## 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:

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

$\blacksquare$