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Math Help - Complex quadratic equation

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    Complex quadratic equation

    Find the complex solutions of z^2 + (1 + 3i)z − 1 + (3/4) i = 0

    Im not sure how to solve this, any help would be much appreciated

    Thanks
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    Quote Originally Posted by callumh167 View Post
    Find the complex solutions of z^2 + (1 + 3i)z − 1 + (3/4) i = 0

    Im not sure how to solve this, any help would be much appreciated

    Thanks
    One way:
    Let z = a + ib, then z^2 = (a^2 - b^2) + 2iab, so
    (a^2 - b^2) + 2iab + (1 + 3i)(a + ib) + \frac{3}{4}i = 0

    (a^2 - b^2) + 2iab + (a + ib + 3ia - 3b) + \frac{3}{4}i = 0

    Resolving this into real and complex components:
    a^2 - b^2 + a - 3b = 0
    and
    2ab + b + 3a + \frac{3}{4} = 0

    Now solve the system of equations.

    -Dan
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