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Number of solutions to complex eq.

  1. Sep 1, 2007 #1
    1. The problem statement, all variables and given/known data

    How many solutions does

    z^2-iz(conjugate)=1/4

    have?

    2. The attempt at a solution

    z^2=(1/4)+y+ix

    Since the RHS is a complex number, the eq. has two solutions.

    Correct?
     
  2. jcsd
  3. Sep 1, 2007 #2

    dextercioby

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    I found 3 solutions. Why don't you solve the equation ?
     
  4. Sep 1, 2007 #3
    I learnt that the number of solutions equals the power of the eq., that is 2...
     
  5. Sep 1, 2007 #4

    dextercioby

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    Under what conditions ?
     
  6. Sep 1, 2007 #5
    w^n=z
     
  7. Sep 1, 2007 #6

    dextercioby

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    Okay, but did you solve the equation ?
     
  8. Sep 1, 2007 #7
    Can't make it further than this:

    (x^2-y^2-y-(1/4)+i(2xy-x)=0
     
    Last edited: Sep 1, 2007
  9. Sep 1, 2007 #8

    dextercioby

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    Okay then. What follows next ?
     
  10. Sep 1, 2007 #9
    Ah, y=2x since 2xy-x has got to be 0?

    But then...no idea.
     
  11. Sep 1, 2007 #10

    dextercioby

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    Well, if 2xy=x, then i see 2 solutions, either x=0, or y=1/2.
     
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