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Thread: 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.


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

    $\displaystyle x^2 + 2axi + a^2i^2 = 3 - 4i$

    $\displaystyle x^2 + 2axi - a^2 = 3 - 4i$

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


    Equating real and imaginary parts gives:

    $\displaystyle x^2 - a^2 = 3$ and $\displaystyle 2ax = -4$.


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

    Substituting into the first gives

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

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

    $\displaystyle x^4 - 4 = 3x^2$

    $\displaystyle x^4 - 3x^2 - 4 = 0$

    $\displaystyle (x^2 - 4)(x^2 + 1) = 0$

    $\displaystyle (x - 2)(x + 2)(x^2 + 1) = 0$.


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

    $\displaystyle x - 2 = 0$ and $\displaystyle x + 2 = 0$.

    Therefore $\displaystyle x = 2$ or $\displaystyle x = -2$.


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

    that means $\displaystyle a = -1$ or $\displaystyle a = 1$.


    So the solutions are:

    $\displaystyle (x, a) = (2, -1)$ and $\displaystyle (x, a) = (-2, 1)$.
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