Arctangent Logarithmic Formulation
Jump to navigation
Jump to search
Theorem
For any real number $x$:
- $\arctan x = \dfrac 1 2 i \map \ln {\dfrac {1 - i x} {1 + i x} }$
where $\arctan x$ is the arctangent and $i^2 = -1$.
Proof
Assume $y \in \R$, $ -\dfrac \pi 2 \le y \le \dfrac \pi 2 $.
| \(\ds y\) | \(=\) | \(\ds \arctan x\) | ||||||||||||
| \(\ds \leadstoandfrom \ \ \) | \(\ds x\) | \(=\) | \(\ds \tan y\) | |||||||||||
| \(\ds \leadstoandfrom \ \ \) | \(\ds x\) | \(=\) | \(\ds i \frac {1 - e^{2 i y} } {1 + e^{2 i y} }\) | Euler's Tangent Identity | ||||||||||
| \(\ds \leadstoandfrom \ \ \) | \(\ds i x\) | \(=\) | \(\ds \frac {e^{2 i y} - 1} {e^{2 i y} + 1}\) | $ i^2 = -1 $ | ||||||||||
| \(\ds \leadstoandfrom \ \ \) | \(\ds i x \paren {e^{2 i y} + 1}\) | \(=\) | \(\ds e^{2 i y} - 1\) | |||||||||||
| \(\ds \leadstoandfrom \ \ \) | \(\ds i x e^{2 i y} + i x\) | \(=\) | \(\ds e^{2 i y} - 1\) | |||||||||||
| \(\ds \leadstoandfrom \ \ \) | \(\ds e^{2 i y} - i x e^{2 i y}\) | \(=\) | \(\ds 1 + i x\) | |||||||||||
| \(\ds \leadstoandfrom \ \ \) | \(\ds e^{2 i y}\) | \(=\) | \(\ds \frac {1 + i x} {1 - i x}\) | |||||||||||
| \(\ds \leadstoandfrom \ \ \) | \(\ds e^{-2 i y}\) | \(=\) | \(\ds \frac {1 - i x} {1 + i x}\) | |||||||||||
| \(\ds \leadstoandfrom \ \ \) | \(\ds -2 i y\) | \(=\) | \(\ds \map \ln {\frac {1 - i x} {1 + i x} }\) | |||||||||||
| \(\ds \leadstoandfrom \ \ \) | \(\ds y\) | \(=\) | \(\ds \frac 1 2 i \map \ln {\frac {1 - i x} {1 + i x} }\) |
$\blacksquare$