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Complex Numbers

  1. Aug 17, 2013 #1
    1. The problem statement, all variables and given/known data
    Hi there, you can see from my nickname that I am a noob in maths :D.
    So, here should is one problem that I cannot solve, even though I know some basics of complex numbers. Its the 2nd problem from the revision exercises, so please be gentle :)


    2. Relevant equations
    Find x and y :
    (jy/(jx-1)-((3y+4j)/(3x+y)))=0
    x = +-3/2 y = +-2
    which probably means that at the end I have to end up with x^2 and y^2;
    I am using j, since I am studying Electronics and in our math course we use j instead of i.


    3. The attempt at a solution
    I know that when we have j/i in the denominator, we have to multiply by its conjugate. I tried that and I don't seem to find the right answers.
    Tried to eliminate the denominator like normal equation but it gets nastier and I am not even close to the solution.
    Thanks
     
  2. jcsd
  3. Aug 17, 2013 #2

    haruspex

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    I assume you are given that x and y are real. That means getting rid of the complex term in the denominators (only one such in this case) will be useful, since you can then write the real and imaginary parts in separate equations. So please post your attempt at this.
     
  4. Aug 17, 2013 #3

    lurflurf

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    clear denominators
    $$\require{cancel}\frac{\jmath \, y}{\jmath \, x -1}-\frac{3y+4\jmath}{3x+y}=0\\
    \frac{\jmath \, y}{\jmath \, x -1}(\jmath \, x -1)(3x+y)-\frac{3y+4\jmath}{3x+y}(\jmath \, x -1)(3x+y)=0(\jmath \, x -1)(3x+y)\\
    \frac{\jmath \, y}{\cancel{\jmath \, x -1}}(\cancel{\jmath \, x -1})(3x+y)-\frac{3y+4\jmath}{\cancel{3x+y}}(\jmath \, x -1)\cancel{(3x+y)}=0(\jmath \, x -1)(3x+y)\\
    (\jmath \, y)(3x+y)-(3y+4\jmath)(\jmath \, x -1)=0$$
     
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