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Math Help - Equilateral triangle and complex numbers

  1. #1
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    Equilateral triangle and complex numbers

    The question:
    Show that the triangle in the complex plane whose verticies are the origin and the points w_1 and w_2 is equalateral if and only if w_1^2 + w_2^2 = w_1w_2

    I have a feeling that the reasoning behind this is quite simple. No matter what I've tried, I can't get the same result they do. Do I have to equate the distances between the vertices to be equal? Or is there some other property I should be exploiting? Thanks.
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  2. #2
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    Fixed math tags.

    The distance from the origin to w_1 is |w_1|= \sqrt{w_1\overline{w_1}}, the distance from the origin to w_2 is |w_2|= \sqrt{w_2\overline{w_2}}, and the distance from w_1 to w_2 is |w_1- w_2|= \sqrt{(w_1- w_2)(\overline{w_1}-\overline{w_2})}}= \sqrt{|w_1|^2+ w_1\overline{w_1}+ |w_2|^2}. For an equilateral triangle, those must all be equal.
    Last edited by mr fantastic; March 13th 2011 at 01:09 PM.
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  3. #3
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    I remember seeing other solutions, and after searching I was surprised to find my own advice, which works in one direction. This problem was also discussed here and here.
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  4. #4
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    Thanks guys.
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  5. #5
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    Quote Originally Posted by HallsofIvy View Post
    |w_1- w_2|= \sqrt{(w_1- w_2)(\overline{w_1}-\overline{w_2})}}= \sqrt{|w_1|^2+ w_1\overline{w_1}+ |w_2|^2}.
    I was just working through this, and I don't know how you got \sqrt{|w_1|^2+ w_1\overline{w_1}+ |w_2|^2}
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  6. #6
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    Neither do I! It should be, of course,
    |w_1- w_2|= \sqrt{w_1- w_2)(\overline{w_1}- \overline{w_2})}= \sqrt{|w_1|^2- w_1\overline{w_2}- w_2\overline{w_1}+ |w_2|^2|}
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