Cotangent Exponential Formulation

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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:

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


Also see


Sources

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