Finding Omega: Evaluating sin^(-1)(3) on the Complex Plane

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SUMMARY

The discussion focuses on evaluating the expression sin^(-1)(3) on the complex plane by finding an expression for omega as a function of z, where z = sin(omega). The key identity used is sin(w) = (1/2) * [exp(i*w) - exp(-i*w)], which allows for the transformation of the sine function into exponential form. Participants clarify that the correct formulation for sine in terms of complex exponentials is sin(x) = (1/2i) * (e^(ix) - e^(-ix)). This leads to the conclusion that evaluating sin^(-1)(3) requires understanding complex analysis and the properties of the sine function.

PREREQUISITES
  • Complex analysis fundamentals
  • Understanding of exponential functions
  • Knowledge of trigonometric identities
  • Familiarity with the complex plane
NEXT STEPS
  • Study the properties of complex functions and their inverses
  • Learn about the complex logarithm and its applications
  • Explore the graphical representation of complex functions on the complex plane
  • Investigate the implications of evaluating inverse trigonometric functions in the complex domain
USEFUL FOR

Mathematicians, physics students, and anyone interested in complex analysis and the evaluation of trigonometric functions in the complex plane.

blueyellow

Homework Statement



if z= sin (omega) find an expression for omega as a function of z that can be used to evaluate all possible values of sin^(-1) (3). Plot these values on the complex plane

The Attempt at a Solution



z= sin (omega)
3= sin (omega)

I don't know how to proceed from here. Please help
 
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blueyellow said:

Homework Statement



if z= sin (omega) find an expression for omega as a function of z that can be used to evaluate all possible values of sin^(-1) (3). Plot these values on the complex plane

The Attempt at a Solution



z= sin (omega)
3= sin (omega)

I don't know how to proceed from here. Please help

For any w (real or complex) we have sin(w) = (1/2)*[exp(i*w) - exp(-i*w)], where i = sqrt(-1).

RGV
 


Actually, I believe the identity is \sin x = \frac{-i(e^{ix} - e^{-ix})}{2}.
 


Or, equivalentlty,
\frac{e^{ix}- e^{-ix}}{2i}
 

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