Solve for z = a + bi, where a and b are real numbers. (z minus conjugate.z) ^ 2 = 2 z + 6i?

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  • 1 decade ago
    Favorite Answer

    (z - z̅ )² = 2z + 6i

    Remember that the conjugate of the complex number

    a + bi

    Is

    a - bi

    So, if z = a + bi

    Then z̅ = a - bi

    And, z - z̅ = (a + bi) - (a - bi)

    = a + bi - a + bi

    = 2bi

    So, we have

    (2bi)² = (z - z̅ )² = 2z + 6i

    But, 2z + 6i

    = 2(a + bi) + 6i

    = 2a + 2bi + 6i

    And,

    (2bi)² = 4b²i² = -4b²

    So, we must have

    -4b² = 2a + 2bi + 6i = 2a + (2b + 6)i

    The left hand side is real, so the right hand side must be as well.

    In other words, the imaginary part on the right hand side,

    (2b + 6)i, must be 0

    So, we have

    2b + 6 = 0

    ⇒ 2b = -6

    ⇒ b = -3

    Substituting this into the equation

    -4b² = 2a + (2b + 6)i, we get

    -4(-3)² = 2a

    ⇒ -4(9) = 2a

    ⇒ -36 = 2a

    ⇒ a = -18

    So we have

    z = a + bi

    = -18 - 3i

  • Jake.
    Lv 5
    1 decade ago

    (a + bi - a + bi)^2 = 2z + 6i

    (2bi)^2 = 2z + 6i

    4b^2i^2 = 2z + 6i

    -4b^2 = 2z + 6i

    2z = -4b^2 - 6i

    z = -2b^2 - 3i

    b = -3

    therefore

    z = -18 - 3i

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