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:
- $\exp z = \ds \sum_{n \mathop = 1}^\infty \frac {z^n} {n!}$
Thus
\(\ds \cmod {\exp z - 1 - z}\) | \(=\) | \(\ds \cmod {\sum_{n \mathop = 2}^\infty \frac {z^n} {n!} }\) | Linear Combination of Convergent Series | |||||||||||
\(\ds \) | \(\le\) | \(\ds \sum_{n \mathop = 2}^\infty \cmod {\frac {z^n} {n!} }\) | Triangle Inequality for Series | |||||||||||
\(\ds \) | \(\le\) | \(\ds \sum_{n \mathop = 2}^\infty \frac {\cmod z^n} 2\) | as $n \ge 2$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\cmod z^2 / 2} {1 - \cmod z}\) | Sum of Infinite Geometric Sequence | |||||||||||
\(\ds \) | \(\le\) | \(\ds \frac 1 2 \cmod z\) | as $\cmod z \le \dfrac 1 2$ |
By the Triangle Inequality:
- $\dfrac 1 2 \cmod z \le \cmod {\exp z - 1} \le \dfrac 3 2 \cmod z$
$\blacksquare$