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Math Help - Find a and b for complex number

  1. #1
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    Find a and b for complex number

    (a+bi)^2=4+4sqrt3i

    I got

    a^2-b^2+2abi=4+4sqrt3i

    a^2=b^2+4
    a=sqrt(b^2+4)

    2ab=4sqrt3
    a=4sqrt3/2b

    I can't seem to get the answer when I compare the values

    The answer given for a is +/- sqrt 6 and b is +/- sqrt2
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  2. #2
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    Re: Find a and b for complex number

    Quote Originally Posted by fuzzy View Post
    (a+bi)^2=4+4sqrt3i

    I got

    a^2-b^2+2abi=4+4sqrt3i

    a^2=b^2+4
    a=sqrt(b^2+4)

    2ab=4sqrt3
    a=4sqrt3/2b
    a= 2\sqrt{3}/b
    I would avoid square roots and start from this. Now a^2= 12/b^2 so that a^2+ b^2= 4 becomes \frac{12}{b^2}+ b^2= 4. Multiply both sides by b^2 to get 12+ b^4= 4b^2 of (b^2)^2- 4b^2+ 12= 0.

    That is a quadratic equation of b^2- and easily factorable. You will find two values for b^2 so 4 different values for b.

    I can't seem to get the answer when I compare the values

    The answer given for a is +/- sqrt 6 and b is +/- sqrt2
    Last edited by HallsofIvy; August 11th 2011 at 09:55 AM.
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  3. #3
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    Re: Find a and b for complex number

    This might be easiest to solve using polars.

    \displaystyle (a + ib)^2 = 4 + 4\sqrt{3}i = 8e^{i\frac{\pi}{3}}, so

    \displaystyle \begin{align*}a + ib &= \pm\left(8e^{i\frac{\pi}{3}}\right)^{\frac{1}{2}} \\ &= \pm 2\sqrt{2}e^{i\frac{\pi}{6}}\\ &= \pm 2\sqrt{2}\left(\cos{\frac{\pi}{6}} + i\sin{\frac{\pi}{6}}\right) \\ &= \pm 2\sqrt{2}\left(\frac{\sqrt{3}}{2} + \frac{1}{2}i\right) \\ &= \pm\left(\sqrt{6} + i\sqrt{2}\right) \end{align*}
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