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
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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 $ | ||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds i x \paren {e^{2 i y} - 1}\) | \(=\) | \(\ds e^{2 i y} + 1\) | |||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds i x e^{2 i y} - i x\) | \(=\) | \(\ds e^{2 i y} + 1\) | |||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds e^{2 i y} - i x e^{2 i y}\) | \(=\) | \(\ds 1 + i x\) | |||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds e^{2 i y}\) | \(=\) | \(\ds \frac {1 + i x} {1 - i x}\) | |||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds y\) | \(=\) | \(\ds \frac 1 2 i \map \ln {\frac {1 + i x} {1 - i x} }\) |
$\blacksquare$