Equilateral triangle and complex numbers

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March 12th 2011, 07:02 AM
Glitch

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.
March 12th 2011, 07:39 AM
HallsofIvy

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.
March 12th 2011, 09:05 AM
emakarov

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.
March 12th 2011, 02:11 PM
Glitch

Thanks guys.
March 12th 2011, 03:20 PM
Glitch

Quote:

Originally Posted by HallsofIvy

$|w_1- w_2|= \sqrt{(w_1- w_2)(\overline{w_1}-\overline{w_2})}}= \sqrt{|w_1|^2+ w_1\overline{w_1}+ |w_2|^2}$ .

I was just working through this, and I don't know how you got $\sqrt{|w_1|^2+ w_1\overline{w_1}+ |w_2|^2}$
March 13th 2011, 03:52 AM
HallsofIvy

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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