Euler's Cotangent Identity/Proof 1

Theorem

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

Proof

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

\(\ds \)

\(=\)

\(\ds \frac {e^{i z} + e^{-i z} } 2 / \frac {e^{i z} - e^{-i z} } {2 i}\)

Euler's Sine Identity and Euler's Cosine Identity

\(\ds \)

\(=\)

\(\ds i \frac {e^{i z} + e^{-i z} } {e^{i z} - e^{-i z} }\)

multiplying numerator and denominator by $2 i$

$\blacksquare$