Finding Omega: Evaluating sin^(-1)(3) on 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
 
Physics news on Phys.org


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}
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top