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

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

By the Triangle Inequality:

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

$\blacksquare$


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