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Thread: proof

  1. #1
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    proof

    w1 =   - 1/2 + √3/2 i and w2 =(w1)^2

    show that [/FONT][/COLOR]   x^2 +xy+y^2=(x-w1y)(x-w2y)

    I am able to find (w1)^2
    But that is all I can do
    Last edited by markosheehan; Jun 9th 2017 at 03:14 AM.
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  2. #2
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    Re: proof

    Ignore the orginal post
    w1 = - 1/2 + (√3/2)i and w2 =(w1)^2

    show that x^2 +xy+y^2=(x-w1y)(x-w2y)

    I am able to find (w1)^2
    But that is all I can do
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  3. #3
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    Re: proof

    So you know that w_1= -1/2+ \sqrt{3}{2}i and w_2= w_1^2= -1/2- \sqrt{3}/2 i. To show the final result, work from right to left. x- w_1y= x-(-1/2+ (\sqrt{3}/2)yi) and x- w_2y= x-(-1/2-(\sqrt{3}/2)yi).

    (x- w_1y)(x- w_2y)= [(x+ (1/2)y)+ (\sqrt{3}/2}yi][(x+ (1/2)y)- (\sqrt{3}/2)yi.

    That is of the form (a+ b)(a- b)= a^2- b^2 with a= x+ (1/2)y and b= (\sqrt{3}/2)yi so is equal to (x+ (1/2)y)^2- ((\sqrt{3}/2)yi)= x^2+ xy+ y^2/4- (3/4)(-y^2)= x^2+ xy+ y^2.
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  4. #4
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    Re: proof

    I get it now thanks. I was stuck on
    (a-b)^2 =(a-b)(a+b)=a^2-b^2
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  5. #5
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    Re: proof

    actually I am a bit confused.
    x^2 +xy+y^2 is not in the form of a^2-b^2 so how can you let the components equal a^2-b^2
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  6. #6
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    Re: proof

    Quote Originally Posted by markosheehan View Post
    I get it now thanks. I was stuck on
    (a-b)^2 =(a-b)(a+b)=a^2-b^2
    This is false. The second equality is true: $(a-b)(a+b) = a^2-b^2$, but the first is not. $(a-b)^2 = a^2-2ab+b^2 \neq a^2-b^2$.
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  7. #7
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    Re: proof

    Quote Originally Posted by markosheehan View Post
    w1 =   - 1/2 + √3/2 i and w2 =(w1)^2
    show that [/FONT][/COLOR]   x^2 +xy+y^2=(x-w1y)(x-w2y)
    I am able to find (w1)^2
    But that is all I can do
    Quote Originally Posted by markosheehan View Post
    I get it now thanks. I was stuck on
    (a-b)^2 =(a-b)(a+b)=a^2-b^2
    Quote Originally Posted by markosheehan View Post
    actually I am a bit confused.
    x^2 +xy+y^2 is not in the form of a^2-b^2 so how can you let the components equal a^2-b^2
    @markosheehan, if you are a bit confused, I am totally confused.

    Please, please tell us all exactly what it is that you trying to do, show, or prove.
    Thanks from markosheehan
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  8. #8
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    Re: proof

    w1 = - 1/2 + (√3/2)i and w2 =(w1)^2

    show that x^2 +xy+y^2=(x-w1y)(x-w2y)
    $(x-w_1y)(x-w_2y) = x^2 - \color{red}{(w_1+w_2)}xy + \color{blue}{w_1w_2}y^2$

    $\color{red}{w_1 + w_2} = w_1 + w_1^2 = \left(\dfrac{-1+i\sqrt{3}}{2}\right) + \left(\dfrac{-2-2i\sqrt{3}}{4}\right) = \dfrac{-2+2i\sqrt{3}-2-2i\sqrt{3}}{4} = \color{red}{-1}$

    $\color{blue}{w_1w_2} = w_1 w_1^2 = \left(\dfrac{-1+i\sqrt{3}}{2}\right) \cdot \left(\dfrac{-2-2i\sqrt{3}}{4}\right) = \dfrac{2 +2i\sqrt{3}-2i\sqrt{3}+6}{8} = \color{blue}{1}$
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  9. #9
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    Re: proof

    I get it now.I am Sorry for being confusing
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