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Algebra in the complex plane

  1. Jan 12, 2017 #1
    1. The problem statement, all variables and given/known data
    Solve each equation for z=a+ib
    [itex] z^{*2}=4z [/itex]
    where z* is the complex conjugate

    3. The attempt at a solution
    I wrote z and z* in terms of x and iy , and tried solving for x and y, but I get quartic terms for y, it doesnt look like it will boil down, It was like over 2 pages of algera, I dont think that is how it is supposed to be done, I tired some alternative forms of the modulus and it didnt go anywhere, Are my approaches the right way, or do I need some clever substitution, or can it be solved with Eulers formula ?
     
  2. jcsd
  3. Jan 12, 2017 #2

    Ray Vickson

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    Show us the equations in x and y that you actually get. What you are claiming sounds wrong to me.
     
  4. Jan 12, 2017 #3

    Mark44

    Staff: Mentor

    If z = a + bi, then ##\bar{z} = a - bi##
    If I substitute these into the given equation, I don't get a quartic, but I do get a solution fairly quickly. Please show what you did.
     
  5. Jan 12, 2017 #4

    PeroK

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    I would automatically use the polar form here.
     
  6. Jan 12, 2017 #5

    Mark44

    Staff: Mentor

    That's a possibility, but it isn't necessary here.
     
  7. Jan 12, 2017 #6
    oh ok I think I got it
    sub in x+iy and x-iy
    then equate real and imaginary
    Real = [itex] x^2+y^2=4x [/itex]
    Im = [itex] -2xy=4y [/itex]
    I solved for x then put it in the other equation, then I solved for y,
    and I got the answer in the back of the book, thanks for your posts
     
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