# complex inverse trig

• October 29th 2008, 06:21 PM
lllll
complex inverse trig
let $\sin(z) = 3$ what is $z$ equally to?

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

substituting z for 3 I get

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

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

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

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

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

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

for some reason this equation don't look right even, considering the fact that I have 2 possible values.
• October 29th 2008, 09:47 PM
mr fantastic
Quote:

Originally Posted by lllll
let $\sin(z) = 3$ what is $z$ equally to?

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

substituting z for 3 I get

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

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

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

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

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

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

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

$\ln (3 i + \sqrt{-8}) = \ln(3 i + i 2 \sqrt{2}) = \ln(i[3 + 2 \sqrt{2}]) = \ln (3 + 2 \sqrt{2}) + \frac{\pi}{2} i + 2 n \pi i$ where n is an integer.