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If complex numbers $z_1,z_2,z_3$ represents vertices of an equilateral triangle and $\mid z_1\mid=\mid z_2\mid=\mid z_3\mid$ then

$\begin{array}{1 1} (A) z_1+z_2+z_3=0 \\(B) z_1+z_2+z_3 \neq 0 \\ (C) z_1+z_2+z_3=\large\frac{\sqrt 3}{4}\normalsize z_1 \\ (D) None\; of\; these \end{array}$

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Let $\mid z_1\mid=\mid z_2\mid=\mid z_3\mid=k$
$\Rightarrow z_1,z_2,z_3$ lies on a circle with centre as origin and radius k.As $z_1,z_2,z_3$ are vertices of equilateral triangle,circumcentre coincides with centroid.
$\Rightarrow \large\frac{1}{3}$$(z_1+z_2+z_3)=0$
$\Rightarrow (z_1+z_2+z_3)=0$
Hence (A) is the correct answer.
answered Apr 9, 2014 by sreemathi.v

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