Bounds for Complex Exponential

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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:

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

$\blacksquare$


Also see