Originally Posted by

**lllll** let $\displaystyle \sin(z) = 3$ what is $\displaystyle z$ equally to?

$\displaystyle \arcsin(z) = -i\log[i z+\sqrt{1-z^2}]$

substituting z for 3 I get

$\displaystyle \arcsin(3) = -i\log[i 3+\sqrt{-8}]$

I'm trying to get it into the form $\displaystyle \ln|x| +(x+2n)\pi i$

but what I'm confused about is the $\displaystyle \sqrt{-8}$ which gives $\displaystyle \sqrt{8} + \frac{1}{2}(\pi+2\pi n)i= \sqrt{8} + \left(\frac{\pi}{2} +\pi n \right)i$ where $\displaystyle n= 0,1$

now I'm not sure if I should take the above value and substitute it so that I would get:

$\displaystyle \arcsin(3) = -i\log\left[i 3+\sqrt{8} + \left(\frac{\pi}{2}\right)i \right]$ and

$\displaystyle \arcsin(3) = -i\log\left[i 3+\sqrt{8} + \left(\frac{3\pi}{2}\right)i \right]$

for some reason this equation don't look right even, considering the fact that I have 2 possible values.