# Cotangent Exponential Formulation

## Theorem

Let $z$ be a complex number such that:

$\forall k\in \Z$, $z \neq k \pi$

Let $\cot z$ denote the cotangent function and $i$ denote the imaginary unit, such that:

$i^2 = -1$

Then:

$\cot z = i \dfrac {e^{i z} + e^{-i z} } {e^{i z} - e^{-i z} }$

## Proof 1

We have, by hypothesis, that $z$ is a complex number such that:

$\forall k \in \Z: z \ne k \pi$

Therefore:

$\sin z \ne 0$

It follows from the definition of the complex cotangent function that:

$\cot z$

is well-defined.

Hence:

 $\ds \cot z$ $=$ $\ds \frac {\cos z} {\sin z}$ Definition of Complex Cotangent Function $\ds$ $=$ $\ds \frac {e^{i z} + e^{-i z} } 2 / \frac {e^{i z} - e^{-i z} } {2 i}$ Sine Exponential Formulation and Cosine Exponential Formulation $\ds$ $=$ $\ds i \frac {e^{i z} + e^{-i z} } {e^{i z} - e^{-i z} }$ multiplying numerator and denominator by $2 i$

$\blacksquare$

## Proof 2

We have, by hypothesis, that $z$ is a complex number such that:

$\forall k \in \Z: z \ne k \pi$

Therefore:

$\sin z \ne 0$

It follows from the definition of the complex cotangent function that:

$\cot z$

is well-defined.

Hence:

 $\ds \cot z$ $=$ $\ds \frac 1 {\tan z}$ Definition of Complex Cotangent Function $\ds$ $=$ $\ds 1 / \dfrac {e^{i z} - e^{-i z} } {i \paren {e^{i z} + e^{-i z} } }$ Tangent Exponential Formulation/Formulation 2 $\ds$ $=$ $\ds i \frac {e^{i z} + e^{-i z} } {e^{i z} - e^{-i z} }$ Definition of Reciprocal

$\blacksquare$

## Also defined as

This result is sometimes also presented as:

$\cot z = \dfrac {i \paren {e^{i z} + e^{-i z} } } {e^{i z} - e^{-i z} }$