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Homework Help: Complex numbers and completing the square

  1. Nov 27, 2014 #1
    1. The problem statement, all variables and given/known data

    let z' = (a,b), find z in C such that z^2 = z'

    2. Relevant equations

    3. The attempt at a solution

    let z = (x,y) then z^2 = (x^2-y^2, 2xy)
    since z^2 = z', we have,

    (x^2-y^2, 2xy) = (a,b)

    comparing real and imaginary components we have;

    x^2-y^2 = a,
    2xy = b.

    Now, this is where i am stuck.
    i know i have to find z in terms of a and b.
    here is an attempt at what to do next;

    subtracting the second equation from the first,

    x^2 - 2xy - y^2 = a - b

    completing the square,

    (x - y)^2 - 2y^2 = a - b.


    Not asking for the answer or anything, just a push in the right direction.
  2. jcsd
  3. Nov 27, 2014 #2


    Staff: Mentor

    You have z' as a+bi and you know then that z is a-bi so shouldn't you just solve for a in terms of b or vice versa?
  4. Nov 28, 2014 #3

    Ray Vickson

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    Homework Helper

    Write ##z' = a + ib## in polar form. Then, finding the roots of ##z^2 = z'## is straightforward.
  5. Nov 28, 2014 #4
    It might be worth calculating |z| and |z2|. Also investigating what (kz)2 looks like and thinking about the implications for z.
  6. Nov 28, 2014 #5
    Thanks guys! My text book hadn't yet defined polar form so I was reluctant to use it. I solved in the way I was progressing via substitution.
  7. Nov 29, 2014 #6


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    Science Advisor

    I don't think it helps to convert to polar form. What you already have looks great to me except that you have written z as x+ yi and z' as a+ bi. So exactly what does z' mean here? You are treating it as just "some other" complex number but I would have interpreted it as the complex conjugate of z. If that is correct then the problem is much easier than what you are doing.
  8. Nov 30, 2014 #7
    no, some other complex number (not complex conjugate), i should have denoted it as ## z_0 ## or something similar!
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