Inverse hyperbolic functions.

Theorem 4.8.1   For any complex number $ z$ :

$\displaystyle \sinh^{-1} z$ $\displaystyle = \log (z + (z^2+1)^{1/2})$    
$\displaystyle \cosh^{-1} z$ $\displaystyle =\log (z +(z^2-1)^{1/2})$    
$\displaystyle \tanh^{-1} z = \frac 12 \log \frac {1+z}{1-z}$    

Example 4.8.2  

$\displaystyle \cosh^{-1} (1)$ $\displaystyle =\log (1 +0^{1/2}) = \log 1 = 2k\pi i, \; k \in \mathbb{Z}.$    
$\displaystyle \tanh^{-1}(i)$ $\displaystyle = \left( \frac {\pi}{4}+ 2k \pi \right) \; i$    



Noah Dana-Picard 2007-12-24