Inverse trigonometric functions.

Theorem 4.7.1 For any complex number :

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

Example 4.7.2

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