# Equilateral triangle and complex numbers

• March 12th 2011, 08:02 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, 08:39 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, 10:05 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, 03:11 PM
Glitch
Thanks guys.
• March 12th 2011, 04:20 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, 04:52 AM
HallsofIvy
Neither do I! It should be, of course,
$|w_1- w_2|= \sqrt{w_1- w_2)(\overline{w_1}- \overline{w_2})}= \sqrt{|w_1|^2- w_1\overline{w_2}- w_2\overline{w_1}+ |w_2|^2|}$