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Math Help - Complex numbers

  1. #1
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    Complex numbers

    I don't know where to post this, so I'll put it here.
    If z=x+yi and z^2=a+ib and a, b, x, and y are real, prove that
    2x^2=\sqrt{a^2+b^2}+a

    How do i go about doing this? Other than that
    a=x^2-y^2 and b=2xy
    Thanks
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by arze View Post
    I don't know where to post this, so I'll put it here.
    If z=x+yi and z^2=a+ib and a, b, x, and y are real, prove that
    2x^2=\sqrt{a^2+b^2}+a

    How do i go about doing this? Other than that
    a=x^2-y^2 and b=2xy
    Thanks
    you have:

     y=\frac{b}{2x}

    substitute this into the other equation to get:

    x^2-\frac{b^2}{4x^2}-a=0

    Multiply this through by x^2 to get a quadratic in x^2, and solve for x^2 using the quadratic formula and discard the negative solution.

    CB
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  3. #3
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    so we have 4x^2-ax^2-b^2=0
    x^2=\frac{a\pm\sqrt{a^2+16b^2}}{8}
    then 2x^2=\frac{a+\sqrt{a^2+16b^2}}{4}
    correct?
    how does this become 2x^2=\sqrt{a^2+b^2}+a?
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by arze View Post
    so we have 4x^2-ax^2-b^2=0
    x^2=\frac{a\pm\sqrt{a^2+16b^2}}{8}
    then 2x^2=\frac{a+\sqrt{a^2+16b^2}}{4}
    correct?
    how does this become 2x^2=\sqrt{a^2+b^2}+a?
    You get:

    (x^2)^2-a(x^2)-\frac{b^2}{4}=0

    so:

    x^2=\frac{a\pm\sqrt{a^2+\frac{4b^2}{4}}}{2}=\frac{  a\pm\sqrt{a^2+b^2}}{2}

    Now as x^2 is intrinsically positive we discard the negative root in the above to get:

    x^2=\frac{a+\sqrt{a^2+b^2}}{2}

    CB
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