# Thread: complex numbers - Equilateral triangle

1. ## complex numbers - Equilateral triangle

a,b,c are complex numbers.
if abc is an Equilateral triangle,show

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

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

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

Hi

Have a look to this link

http://www.mathhelpforum.com/math-he...x-numbers.html

3. thanks.is there any other way to prove it without vectors

4. Let $\displaystyle |a-b|=|b-c|=|c-a|=k$ (since a,b and c are vertices of an equilateral triangle)

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

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

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

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

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

Adding (1),(2) and (3)

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

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

Taking LCM

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