Cotangent Exponential Formulation

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Theorem

Let $z$ be a complex number.

Let $\cot z$ denote the cotangent function and $i$ denote the imaginary unit: $i^2 = -1$.

Then:

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


Proof 1

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

$\blacksquare$


Proof 2

\(\displaystyle \cot z\) \(=\) \(\displaystyle \frac 1 {\tan z}\) Definition of Complex Cotangent Function
\(\displaystyle \) \(=\) \(\displaystyle 1 / \dfrac {e^{i z} - e^{-i z} } {i \left({e^{i z} + e^{-i z} }\right)}\) Tangent Exponential Formulation/Formulation 2
\(\displaystyle \) \(=\) \(\displaystyle i \frac {e^{i z} + e^{-i z} } {e^{i z} - e^{-i z} }\)

$\blacksquare$


Also defined as

This result is sometimes also presented as:

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


Also see


Sources