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Inverse of sine(ix)

suvadip

Member
Feb 21, 2013
69
I need to prove
\(\displaystyle sin^{-1}(ix)=2n\pi\pm i log (\sqrt{1+x^2}+x)\)

I can prove \(\displaystyle sin^{-1}(ix)=2n\pi+ i log (\sqrt{1+x^2}+x)\)

How to prove the other part. Please help
 

chisigma

Well-known member
Feb 13, 2012
1,704
I need to prove
\(\displaystyle sin^{-1}(ix)=2n\pi\pm i log (\sqrt{1+x^2}+x)\)

I can prove \(\displaystyle sin^{-1}(ix)=2n\pi+ i log (\sqrt{1+x^2}+x)\)

How to prove the other part. Please help
You have to find the z for which is...

$\displaystyle \sin z = \frac{e^{i\ z}-e^{- i\ z}}{2\ i} = i\ x$ (1)

Setting in (1) $\displaystyle e^{i\ z}=y$ You arrive to the equation...


$\displaystyle y^{2} + 2\ x\ y -1 =0 $ (2)


... which is solved for $\displaystyle y= - x \pm \sqrt{1+x^{2}}$ so that is...


$\displaystyle \sin^{-1} (i\ x) = 2\ \pi\ i\ n - i\ \ln (- x \pm \sqrt{1+x^{2}}) = 2\ \pi\ i\ n + i\ \ln (x \pm \sqrt{1+x^{2}})$ (3)

Kind regards

$\chi$ $\sigma$