Reciprocal of Complex Exponential

Theorem

Let $z \in \C$.

Let $\exp$ denote the complex exponential function.

Then:

$\dfrac 1 {\map \exp z} = \map \exp {-z}$


Proof

\(\ds \map \exp {-z}\) \(=\) \(\ds \dfrac {\map \exp {-z} } {\map \exp 0}\) as $\map \exp 0 = 1$ by Exponential of Zero
\(\ds \) \(=\) \(\ds \dfrac {\map \exp {-z} } {\map \exp {z - z} }\)
\(\ds \) \(=\) \(\ds \dfrac {\map \exp {-z} } {\map \exp z \, \map \exp {-z} }\) Exponential of Sum: Complex Numbers
\(\ds \) \(=\) \(\ds \dfrac 1 {\map \exp z}\)

$\blacksquare$


Sources

  • 2001: Christian Berg: Kompleks funktionsteori: $\S 1.5$
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