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Math Help - Help - Square rooting a complex number

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    Help - Square rooting a complex number

    z^2 = 2√3 + 2j

    I just need pushing in the right direction?? I am sure its simpler than I think it is!!

    Thank you
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Charlieengineer84 View Post
    z^2 = 2√3 + 2j

    I just need pushing in the right direction?? I am sure its simpler than I think it is!!

    Thank you
    One way: let z = x + iy where x,y \in \mathbb{R}, then we have

    (x + iy)^2 = 2 \sqrt 3 + 2i

    this implies

    x^2 - y^2 = 2 \sqrt 3 and,

    2xy = 2

    Solve this system simultaneously for x and y and you will find z.


    Another way: use polar coordinates (or the cis form if you prefer)

    Let z = re^{i \theta} \implies z^2 = r^2 e^{2 i \theta}.

    Thus, z^2 = 2 \sqrt 3 + 2 i

    \Rightarrow r^2 e^{2i \theta} = 4 \left( \frac {\sqrt 3}2 + \frac 12i \right) = 4e^{i(\pi / 6 + 2k \pi)}

    Now, solve for r and \theta ( k = 0, 1)
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