Solve the following equation:

$\displaystyle \cos z = 3 - 2i$.

My book says that

$\displaystyle \cos^{-1}z= -i \log[z+i(1-z^2)^{\frac{1}{2}}]$. It also says that $\displaystyle (1-z^2)^{\frac{1}{2}}$ is a double valued fuction. How do I use this formula or is there a better way to do this? I don't see how to get the double value it is talking about.