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Thread: 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 $\displaystyle z=x+yi$ and $\displaystyle z^2=a+ib$ and a, b, x, and y are real, prove that
    $\displaystyle 2x^2=\sqrt{a^2+b^2}+a$

    How do i go about doing this? Other than that
    $\displaystyle a=x^2-y^2$ and $\displaystyle b=2xy$
    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 $\displaystyle z=x+yi$ and $\displaystyle z^2=a+ib$ and a, b, x, and y are real, prove that
    $\displaystyle 2x^2=\sqrt{a^2+b^2}+a$

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

    $\displaystyle y=\frac{b}{2x}$

    substitute this into the other equation to get:

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

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

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

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

    so:

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

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

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

    CB
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