Help solving (complex) simultaneous equations

  • Thread starter KFC
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  • #1
KFC
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Homework Statement


I am trying to solving the following complex equation for [tex]x[/tex] and [tex]\theta[/tex]

[tex]a\sinh(2x) e^{-i\theta} + y\sinh^2x e^{-i2\theta} + y^*\cosh^2(x) = 0[/tex]

where [tex]a[/tex] is real constant, [tex]x[/tex] and [tex]\theta[/tex] are also real parameter. [tex]y[/tex] is complex number, [tex]y^*[/tex] is the complex conjugate.

Solve for [tex]x[/tex] and [tex]\theta[/tex] (in terms of y and a)

2. The attempt at a solution
Let
[tex]y = |y| e^{i\varphi}[/tex]

and multiply the equation with [tex]y[/tex]

[tex]ay\sinh(2x) e^{-i\theta} + y^2\sinh^2x e^{-i2\theta} + |y|^2\cosh^2(x) = 0[/tex]

Now let the real part and imaginary part equals ZERO.

[tex]
\begin{cases}
a\sinh(2x) |y|\cos(\theta-\varphi) + |y|^2\sinh^2(x)\cos(2\theta-2\varphi) + |y|^2\alpha^2 = 0, \\[3.8mm]
a\sinh(2x) |y|\sin(\theta-\varphi) + |y|^2\sinh^2(x)\sin(2\theta-2\varphi) = 0
\end{cases}
[/tex]

I tryied to solve that two days ago, I tried many way to simpliy that but still find no way to get the soluton. Could anyone give me some hints?

Thanks
 
Last edited:

Answers and Replies

  • #2
36,656
8,655
This doesn't make sense to me. How can you solve one equation for two variables? This seems to me like asking someone to solve y = 2x for x and y. You can solve the equation for y in terms of x, or you can solve for x in terms of y, but you can't solve it simultaneously for both variables.
 
  • #3
KFC
488
4
This doesn't make sense to me. How can you solve one equation for two variables? This seems to me like asking someone to solve y = 2x for x and y. You can solve the equation for y in terms of x, or you can solve for x in terms of y, but you can't solve it simultaneously for both variables.

How come. This is a equation for complex variable, the real part and imaginary part gives two equations.
 
  • #4
36,656
8,655
OK, I see.
[tex] a\sinh(2x) |y|\sin(\theta-\varphi) + |y|^2\sinh^2(x)\sin(2\theta-2\varphi) = 0[/tex]
In your 2nd equation, I think the sine arguments should be the other way around. Also, you can expand [itex]sin(2(\phi - \theta))[/itex] as [itex]2sin(\phi - \theta) cos(\phi - \theta)[/itex]. Then you'll have one factor the same in both terms, which hopefully leaves you with the other factor that you can do something with.
 

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