Arccotangent Logarithmic Formulation

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Theorem

For any real number $x$:

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

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


Proof

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

\(\displaystyle y\) \(=\) \(\displaystyle \arccot x\)
\(\displaystyle \leadstoandfrom \ \ \) \(\displaystyle x\) \(=\) \(\displaystyle \cot y\)
\(\displaystyle \leadstoandfrom \ \ \) \(\displaystyle x\) \(=\) \(\displaystyle i \frac {1 + e^{2 i y} } {1 - e^{2 i y} }\) Cotangent Exponential Formulation
\(\displaystyle \leadstoandfrom \ \ \) \(\displaystyle i x\) \(=\) \(\displaystyle \frac {e^{2 i y} + 1} {e^{2 i y} - 1}\) $ i^2 = -1 $
\(\displaystyle \leadstoandfrom \ \ \) \(\displaystyle i x \paren {e^{2 i y} - 1}\) \(=\) \(\displaystyle e^{2 i y} + 1\)
\(\displaystyle \leadstoandfrom \ \ \) \(\displaystyle i x e^{2 i y} - i x\) \(=\) \(\displaystyle e^{2 i y} + 1\)
\(\displaystyle \leadstoandfrom \ \ \) \(\displaystyle e^{2 i y} - i x e^{2 i y}\) \(=\) \(\displaystyle 1 + i x\)
\(\displaystyle \leadstoandfrom \ \ \) \(\displaystyle e^{2 i y}\) \(=\) \(\displaystyle \frac {1 + i x} {1 - i x}\)
\(\displaystyle \leadstoandfrom \ \ \) \(\displaystyle y\) \(=\) \(\displaystyle \frac 1 2 i \map \ln {\frac {1 + i x} {1 - i x} }\)

$\blacksquare$


Also see