MHB Can You Separate Real and Imaginary Parts of $\sin^{-1}(e^{i\theta})$?

[Amer]

Is it possible to separate imaginary part from the real part in this question

$\sin ^{-1} ( e^{i\theta}) $

I tired to find u such that

$\sin u = e^{i\theta} $

$ \sin u = \cos \theta + i sin \theta $

$ \sin (x + iy) = \cos \theta + i \sin \theta $

$ \sin x \cos iy + \sin iy \cos x = \cos \theta + i \sin \theta $

$\sin x \cosh y + i \sinh y \cos x = \cos \theta + i \sin \theta $
but this is not easy

Thanks

 

[Dustinsfl]

Is it possible to separate imaginary part from the real part in this question

$\sin ^{-1} ( e^{i\theta}) $

I tired to find u such that

$\sin u = e^{i\theta} $

$ \sin u = \cos \theta + i sin \theta $

$ \sin (x + iy) = \cos \theta + i \sin \theta $

$ \sin x \cos iy + \sin iy \cos x = \cos \theta + i \sin \theta $

$\sin x \cosh y + i \sinh y \cos x = \cos \theta + i \sin \theta $
but this is not easy

Thanks

You just have. The real part is when $\sin x \cosh y = \cos\theta$.

 

[Sudharaka]

Is it possible to separate imaginary part from the real part in this question

$\sin ^{-1} ( e^{i\theta}) $

I tired to find u such that

$\sin u = e^{i\theta} $

$ \sin u = \cos \theta + i sin \theta $

$ \sin (x + iy) = \cos \theta + i \sin \theta $

$ \sin x \cos iy + \sin iy \cos x = \cos \theta + i \sin \theta $

$\sin x \cosh y + i \sinh y \cos x = \cos \theta + i \sin \theta $
but this is not easy

Thanks

Hi Amer, :)

Finding the real and imaginary parts seem not to be easy and this is what I got using Maxima. Hope this helps. :)

\[\mbox{Re}\left[\sin ^{-1} ( e^{i\theta})\right]=\mathrm{atan2}\left( \mathrm{sin}\left( \frac{\mathrm{atan2}\left( 0,1-{e}^{2\,i\,x}\right) }{2}\right) \,\sqrt{\left| {e}^{2\,i\,x}-1\right| }+{e}^{i\,x},\mathrm{cos}\left( \frac{\mathrm{atan2}\left( 0,1-{e}^{2\,i\,x}\right) }{2}\right) \,\sqrt{\left| {e}^{2\,i\,x}-1\right| }\right)\]

and

\[\mbox{Im}\left[\sin ^{-1} ( e^{i\theta})\right]=-\frac{\mathrm{log}\left( {\mathrm{cos}\left( \frac{\mathrm{atan2}\left( 0,1-{e}^{2\,i\,x}\right) }{2}\right) }^{2}\,\left| {e}^{2\,i\,x}-1\right| +{\left( \mathrm{sin}\left( \frac{\mathrm{atan2}\left( 0,1-{e}^{2\,i\,x}\right) }{2}\right) \,\sqrt{\left| {e}^{2\,i\,x}-1\right| }+{e}^{i\,x}\right) }^{2}\right) }{2}\]

where, \(\mbox{atan}2(y,x)\) is the value of \(\mbox{atan}\left(\frac{y}{x}\right)\) in the interval \([-\pi,\pi]\).

Kind Regards,
Sudharaka.