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Math Help - if z= sin (omega) find an expression for omega as a function of z that can be used...

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    if z= sin (omega) find an expression for omega as a function of z that can be used...

    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

    so:

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

    but this doesn't seem to be leading anywhere. please help
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    MHF Contributor chisigma's Avatar
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    Re: if z= sin (omega) find an expression for omega as a function of z that can be use

    Quote Originally Posted by blueyellow View Post
    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

    so:

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

    but this doesn't seem to be leading anywhere. please help
    Setting \omega= i\ \gamma the equation becomes...

    \sin \omega = \sin (i\ \gamma)= i\ \sinh \gamma= 3 \implies e^{\gamma} - \frac{1}{e^{\gamma}}= -6\ i (1)

    Setting now s= e^{\gamma} the (1) becomes...

    s - \frac{1}{s}= - 6\ i \implies s^{2} + 6\ i\ s -1=0 (2)

    The (2) is an ordinary quadratic and if s_{0} is a solution of (2), then \omega_{0}= -i\ \ln s_{0} is solution of (1)...

    Kind regards

    \chi \sigma
    Last edited by chisigma; July 25th 2011 at 09:32 AM.
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  3. #3
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    Re: if z= sin (omega) find an expression for omega as a function of z that can be use

    Quote Originally Posted by blueyellow View Post
    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
    Here is the standard representation:
    \arcsin (z) =  - i\log \left[ {iz + \left( {1 - z^2 } \right)^{\frac{1}{2}} } \right]
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    MHF Contributor chisigma's Avatar
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    Re: if z= sin (omega) find an expression for omega as a function of z that can be use

    Quote Originally Posted by Plato View Post
    Here is the standard representation:
    \arcsin (z) = - i\log \left[ {iz + \left( {1 - z^2 } \right)^{\frac{1}{2}} } \right]
    Proceeding as in my post the equation \sin \omega= z conducts to the equation in s...

    s^{2} +2\ i\ z\ s -1=0 (1)

    ... the solution of which are...

    s= -i\ z \pm \sqrt{1-z^{2}} (2)

    ... so that [if no mistakes of me...] it would be...

    \omega= \sin^{-1} z = -i\ \ln (-i\ z \pm \sqrt{1-z^{2}}) (3)

    Kind regards

    \chi \sigma
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