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Thread: square root of a complex number

  1. #1
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    square root of a complex number

    Hi folks,


    I am trying to find the square roots of the complex number $3 + 4i$


    Suppose the square root is a + ib where a and b $\in \mathcal{\ R}$


    $3 + 4i = (a + ib)^2$


    $3 + 4i = a^2 - b^2 + 2abi$


    so,


    $3 = a^2 - b^2$ ......... (1)


    and


    $4 = 2ab$ ................(2)


    sub (2) in (1)


    $3 = (\dfrac{2}{b})^2 - b^2$


    $3 = \dfrac{4}{b^2} - b^2$


    $3b^2 = 4 - b^4$


    $b^4 + 3b^2 - 4 = 0$


    let $y = b^2$


    $y^2 + 3y - 4 = 0$


    $y = 1$ or $y = -4$


    i.e.


    $b^2 = 1 $ or $b^2 = -4$


    $b = \pm 1$ or $b = \sqrt{4}\sqrt{-1} = \pm 2i$


    so I have $b = +1 + 2i$ or $b = -1 + 2i$ or $b = +1 - 2i$ or $b = -1 - 2i$


    from equation (2) for each of the 4 values of b, I get a corresponding value of a


    $a = \dfrac{2}{b}$


    $a = \dfrac{2}{1 + 2i}$


    $a = \dfrac{2}{1 + 2i} . \dfrac{1 - 2i}{1 - 2i}$


    $a = \dfrac{2 - 4i}{1 - 4i^2}$


    $a = \dfrac{2}{5} - \dfrac{4}{5}i$


    so


    $a + ib = (1 + 2i) + i(\dfrac{2}{5} - \dfrac{4}{5}i)$


    $a + ib = 1 + 2i + \dfrac{2}{5}i - \dfrac{4}{5}$


    $a + ib = \dfrac{1}{5} + \dfrac{12}{5}i $


    this is the first of 4 square roots


    unfortunately these do not match the answers which are $\pm (2 + i)$


    where am I going wrong?
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  2. #2
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    Re: square root of a complex number

    $b^2=-4$ has no solutions as $b$ is a real number
    Thanks from s_ingram
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  3. #3
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    Re: square root of a complex number

    Yes, I can see where I have gone wrong. If I could, I would delete the post.
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