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

  • Context: MHB 
  • Thread starter Thread starter Amer
  • Start date Start date
  • Tags Tags
    Imaginary
Click For Summary
SUMMARY

The discussion focuses on separating the real and imaginary parts of the expression $\sin^{-1}(e^{i\theta})$. Participants explore the relationship $\sin u = e^{i\theta}$, leading to the equations $\sin x \cosh y = \cos \theta$ and $\sinh y \cos x = \sin \theta$. The final results for the real and imaginary components are derived using Maxima, yielding specific expressions involving $\mathrm{atan2}$ and logarithmic functions. This analysis confirms the complexity of isolating these components in the context of complex analysis.

PREREQUISITES
  • Understanding of complex functions, specifically $\sin$ and $\sin^{-1}$.
  • Familiarity with hyperbolic functions such as $\sinh$ and $\cosh$.
  • Knowledge of the $\mathrm{atan2}$ function and its application in determining angles in the complex plane.
  • Experience with symbolic computation tools like Maxima for mathematical analysis.
NEXT STEPS
  • Study the properties of inverse trigonometric functions in complex analysis.
  • Learn how to utilize Maxima for symbolic computation of complex expressions.
  • Research the applications of $\mathrm{atan2}$ in polar coordinates and complex number analysis.
  • Explore the relationship between hyperbolic and trigonometric functions in complex contexts.
USEFUL FOR

Mathematicians, students of complex analysis, and anyone interested in the properties of inverse trigonometric functions and their applications in engineering and physics.

Amer
Messages
259
Reaction score
0
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
 
Physics news on Phys.org
Amer said:
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 $\sin x \cosh y = \cos\theta$.
 
Amer said:
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.
 

Similar threads

  • · Replies 13 ·
Replies
13
Views
6K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
Replies
7
Views
3K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
7K
  • · Replies 5 ·
Replies
5
Views
2K