complex numbers - Equilateral triangle

**stud_02** · March 27th 2009, 04:13 PM

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

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

i cant thin of a equation about an Equilateral triangle. please help

**running-gag** · March 28th 2009, 01:46 AM

Originally Posted by stud_02

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

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

i cant thin of a equation about an Equilateral triangle. please help

Hi

Have a look to this link

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

**stud_02** · March 28th 2009, 02:33 AM

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

**pankaj** · March 28th 2009, 07:55 AM

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} $

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

$ 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 $

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