Inverse trigonometric functions.

Theorem 4.7.1   For any complex number $ z$ :

$\displaystyle \sin^{-1} z$ $\displaystyle = -i \; \log (i \; z + (1-z^2)^{1/2})$    
$\displaystyle \cos^{-1} z$ $\displaystyle =-i \; \log (z +i \;(1-z^2)^{1/2})$    
$\displaystyle \tan^{-1} z = \frac i2 \log \frac {i+z}{i-z}$    

Example 4.7.2       

$\displaystyle \sin^{-1}(0)$ $\displaystyle = -i \log (1) = -i (\ln 1 + i \; 2k \pi) = 2k \pi, \; k\in \mathbb{Z};$    
$\displaystyle \sin^{-1}(1/2)$ $\displaystyle = -i \; \log \left[ i \; \frac 12 + \left( 1-\frac 14 \right)^{1/2} \right]$    
$\displaystyle \quad$ $\displaystyle = -i \log \left(\frac{\sqrt{3}}{2} + \frac 12 i \right)$    
$\displaystyle \quad$ $\displaystyle = -i \left[ \ln 1 + i \left(\frac {\pi}{3}+ 2k \pi \right) \right]$    
$\displaystyle \quad$ $\displaystyle = - \frac {\pi}{3}+ 2k \pi$    
$\displaystyle \sin^{-1}(i)$ $\displaystyle = \left( \frac {\pi}{2} + 2k \pi \right) \; i.$    

Noah Dana-Picard 2007-12-24