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

• Jul 25th 2011, 08:07 AM
blueyellow
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)

• Jul 25th 2011, 08:39 AM
chisigma
Re: if z= sin (omega) find an expression for omega as a function of z that can be use
Quote:

Originally Posted by blueyellow
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)

Setting $\displaystyle \omega= i\ \gamma$ the equation becomes...

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

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

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

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

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$
• Jul 25th 2011, 08:43 AM
Plato
Re: if z= sin (omega) find an expression for omega as a function of z that can be use
Quote:

Originally Posted by blueyellow
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:
$\displaystyle \arcsin (z) = - i\log \left[ {iz + \left( {1 - z^2 } \right)^{\frac{1}{2}} } \right]$
• Jul 25th 2011, 09:46 AM
chisigma
Re: if z= sin (omega) find an expression for omega as a function of z that can be use
Quote:

Originally Posted by Plato
Here is the standard representation:
$\displaystyle \arcsin (z) = - i\log \left[ {iz + \left( {1 - z^2 } \right)^{\frac{1}{2}} } \right]$

Proceeding as in my post the equation $\displaystyle \sin \omega= z$ conducts to the equation in s...

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

... the solution of which are...

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

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

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

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$