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The complex numbers $z_1,z_2,z_3$ are vertices of an equilateral triangle with circumcentre $z_0$.The value of $z_1^2+z_2^2+z_3^2$ equals

$\begin{array}{1 1}(A)\;z_0^2&(B)\;3z_0^2\\(C)\;\large\frac{9}{2}\normalsize z_0^2&(D)\;\text{None of these}\end{array} $

1 Answer

Circumcentre and centroid of equilateral triangle coincides.
Thus,$z_0=\large\frac{1}{3}$$(z_1+z_2+z_3)$
$\Rightarrow (3z_0)^2=(z_1^2+z_2^2+z_3^2+2(z_1z_2+z_2z_3+z_1z_3)$
As the triangle is equilateral,
$z_1z_2+z_2z_3+z_1z_3=z_1^2+z_2^2+z_3^2$
$\Rightarrow gz_0^2=3(z_1^2+z_2^2+z_3^2)^2$
$\Rightarrow (z_1^2+z_2^2+z_3^2)^2=3z_0^2$
Hence (B) is the correct answer.
answered Apr 10, 2014 by sreemathi.v
 

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