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Math Help - Complex Number Question

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    Complex Number Question

    Hello there!

    It's my first time here, so I apologize in not being able to find the 'Complex Numbers' part of these forums.

    I must have forgotten how to solve these ones (ofcourse ), so here it is.

    (x+ai)^2=3-4i - Solve for {a,x} given that they are both Real numbers.

    Thanks, and if anyone knows the proper Complex Number forums page, please let me know
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    Quote Originally Posted by Tan90 View Post
    Hello there!

    It's my first time here, so I apologize in not being able to find the 'Complex Numbers' part of these forums.

    I must have forgotten how to solve these ones (ofcourse ), so here it is.

    (x+ai)^2=3-4i - Solve for {a,x} given that they are both Real numbers.

    Thanks, and if anyone knows the proper Complex Number forums page, please let me know
    Either the Algebra or Pre-Calculus sub-forums are fine.


    (x + ai)^2 = 3 - 4i

    x^2 + 2axi + a^2i^2 = 3 - 4i

    x^2 + 2axi - a^2 = 3 - 4i

    x^2 - a^2 + 2axi = 3-4i.


    Equating real and imaginary parts gives:

    x^2 - a^2 = 3 and 2ax = -4.


    From the second equation we find a = -\frac{2}{x}.

    Substituting into the first gives

    x^2 - \left(-\frac{2}{x}\right)^2 = 3

    x^2 - \frac{4}{x^2} = 3

    x^4 - 4 = 3x^2

    x^4 - 3x^2 - 4 = 0

    (x^2 - 4)(x^2 + 1) = 0

    (x - 2)(x + 2)(x^2 + 1) = 0.


    Since we know x is real, that means we can only accept the cases

    x - 2 = 0 and x + 2 = 0.

    Therefore x = 2 or x = -2.


    Since we know a = -\frac{2}{x}

    that means a = -1 or a = 1.


    So the solutions are:

    (x, a) = (2, -1) and (x, a) = (-2, 1).
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