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: $\S 1.5$