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Question

Let the complex numbers $$z_1, z_2 $$ and $$ z_3$$ be the vertices of an equilateral triangle. Let $$z_0$$ be the circumcentre of the triangle, then $$z^2_1 + z^2_2 + z^2_3 =$$


A
z20
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B
z20
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C
3z20
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D
3z20
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Solution

The correct option is C $$3z^2_0$$
Theorem - the triangle whose vertices are
the points repressed by the 
compels numbers $$ z_{1},z_{2},z_{3} $$ on the
argent plane is equilateral if
$$ \dfrac{1}{z_{1}-z_{2}}+\dfrac{1}{z_{2}-z_{3}}+\dfrac{1}{z_{3}-z_{1}} = 0 $$
i.e. iff $$ z_{1}^{2}+z_{2}^{2}+z_{3}^{2} = z_{1}z_{2}+z_{2}z_{3}+z_{3}z_{1} ...(A) $$
Circumcentre and centroid of equilateral 
triangle coincides.
Thus, $$ z_{0} = \dfrac{1}{2}(z_{1}+z_{2}+z_{3}) $$
$$ \Rightarrow 3z_{0} = z_{1}+z_{2}+z_{3} $$
$$ \Rightarrow (3z_{0})^{2} = (z_{1}+z_{2}+z_{3})^{2} $$
$$ = z_{1}^{2}+z_{2}^{2}+z_{3}^{2}+2(z_{1}z_{2}+z_{2}z_{3}+z_{3}z_{1})...(B) $$
As the triangle is equilateral 
$$ z_{1}z_{2}+z_{2}z_{3}+z_{3}z_{1} = z_{1}^{2}+z_{2}^{2}+z_{3}^{2}...(2) $$
$$ 97_{0}^{2} = 3(z_{1}^{2}+z_{2}^{2}+z_{3}^{2}) $$ [from (1) and (2)]
$$ 3z_{0}^{2} = z_{1}^{2}+z_{2}^{2}+z_{3}^{2} $$
So, option (c) is correct.


1087250_1033719_ans_db76c1cb62c642f9abcdf1f3085b1d3c.png

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