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Math Help - complex conjugates

  1. #1
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    complex conjugates

    Right, this comes in 2 parts, the 1st part is that this problem is part (b) of a question, in part (a) you are given z= 3+5i however i don't know if you can/need to use this fact for this part of the question (gotta love my teachers ). second part is the question itself:

    Solve the equation z^2 + 2z*= -3, where z* denotes the complex conjugate of z.

    thanks in advance for any help, really appreciate it.
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  2. #2
    Super Member flyingsquirrel's Avatar
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    Hello
    Quote Originally Posted by snowmonkey12 View Post
    Right, this comes in 2 parts, the 1st part is that this problem is part (b) of a question, in part (a) you are given z= 3+5i however i don't know if you can/need to use this fact for this part of the question (gotta love my teachers ). second part is the question itself:

    Solve the equation z^2 + 2z*= -3, where z* denotes the complex conjugate of z.

    thanks in advance for any help, really appreciate it.
    A possible approach is to work with z^2+2z*=-3 and with the conjugate of this equality :

    This gives us two equations :



    Adding and subtracting :



    From (4) you can get the value of z+z*=2Re(z) which can then be used to get Im(z) from (3). Good luck !
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  3. #3
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    Another way to do it:

    To solve
    z + 2z* = -3

    Let z = (x + iy); z* = (x - iy)

    (x + iy) + 2(x-iy) = -3
    x + 2xiy -y + 2x - 2iy = -3

    Equating Real and Imaginary parts:
    Real: x + 2x - y = -3
    Imaginary: 2xiy - 2iy = 0

    From imaginary part:
    2iy (x-1) = 0
    So x = 1 or y =0

    Substituting y = 0 into the real part:
    x + 2x + 3 = 0.
    This quadratic has no real roots. We know that x has to be an integer, so this will not give us a solution.

    Substituting x = 1 into the real part:
    y = 6
    y= √6

    Thus z = x + iy = 1 + √6i
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