# complex numbers - Equilateral triangle

• Mar 27th 2009, 04:13 PM
stud_02
complex numbers - Equilateral triangle
a,b,c are complex numbers.
if abc is an Equilateral triangle,show

$a^2 + b^2 + c^2 = ab + bc + ca$

• Mar 28th 2009, 01:46 AM
running-gag
Quote:

Originally Posted by stud_02
a,b,c are complex numbers.
if abc is an Equilateral triangle,show

$a^2 + b^2 + c^2 = ab + bc + ca$

Hi

Have a look to this link

http://www.mathhelpforum.com/math-he...x-numbers.html
• Mar 28th 2009, 02:33 AM
stud_02
thanks.is there any other way to prove it without vectors
• Mar 28th 2009, 07:55 AM
pankaj
Let $|a-b|=|b-c|=|c-a|=k$ (since a,b and c are vertices of an equilateral triangle)

$|a-b|^2=|b-c|^2=|c-a|^2=k^2$

$
(a-b)(\bar{a}-\bar{b})=(b-c)(\bar{b}-\bar{c})=(c-a)(\bar{c}-\bar{a})=k^2
$

(1) $\bar{a}-\bar{b}=\frac{k^2}{a-b}$

(2) $\bar{b}-\bar{c}=\frac{k^2}{b-c}$

(3) $
\bar{c}-\bar{a}=\frac{k^2}{c-a}
$

$
0=\frac{k^2}{a-b}+\frac{k^2}{b-c}+\frac{k^2}{c-a}
$

$
\frac{1}{a-b}+\frac{1}{b-c}+\frac{1}{c-a}=0
$

Taking LCM

$
a^2+b^2+c^2=ab+bc+ca
$