Proving sin⁻¹(ix): 2nπ ± i log (√1+x²+x)

[Suvadip]

I need to prove
$$sin^{-1}(ix)=2n\pi\pm i log (\sqrt{1+x^2}+x)$$

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

How to prove the other part. Please help

 

[chisigma]

I need to prove
$$sin^{-1}(ix)=2n\pi\pm i log (\sqrt{1+x^2}+x)$$

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

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

$\chi$ $\sigma$