Arccotangent Logarithmic Formulation

Theorem

$\arccot x = \dfrac 1 2 i \, \map \ln {\dfrac {1 + i x} {1 - i x} }$

where $\arccot x$ is the arccotangent and $i^2 = -1$.

Proof

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Assume $y \in \R$, $ -\dfrac \pi 2 \le y \le \dfrac \pi 2 $.

$\ds y$

$=$

$\ds \arccot x$

$\ds \leadstoandfrom \ \ $

$\ds x$

$=$

$\ds \cot y$

$\ds \leadstoandfrom \ \ $

$\ds x$

$=$

$\ds i \frac {1 + e^{2 i y} } {1 - e^{2 i y} }$

Euler's Cotangent Identity

$\ds \leadstoandfrom \ \ $

$\ds i x$

$=$

$\ds \frac {e^{2 i y} + 1} {e^{2 i y} - 1}$

$ i^2 = -1 $

$\ds \leadstoandfrom \ \ $

$\ds i x \paren {e^{2 i y} - 1}$

$=$

$\ds e^{2 i y} + 1$

$\ds \leadstoandfrom \ \ $

$\ds i x e^{2 i y} - i x$

$=$

$\ds e^{2 i y} + 1$

$\ds \leadstoandfrom \ \ $

$\ds e^{2 i y} - i x e^{2 i y}$

$=$

$\ds 1 + i x$

$\ds \leadstoandfrom \ \ $

$\ds e^{2 i y}$

$=$

$\ds \frac {1 + i x} {1 - i x}$

$\ds \leadstoandfrom \ \ $

$\ds y$

$=$

$\ds \frac 1 2 i \map \ln {\frac {1 + i x} {1 - i x} }$

$\blacksquare$

Also see

Arccotangent Logarithmic Formulation

Theorem

Proof

Also see

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