# Separate imaginary from real

#### Amer

##### Active member
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

#### dwsmith

##### Well-known member
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

##### Well-known member
MHB Math Helper
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.