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
B
z20
C
3z20
D
3z20

Solution

The correct option is C $$3z^2_0$$Theorem - the triangle whose vertices arethe points repressed by the compels numbers $$z_{1},z_{2},z_{3}$$ on theargent 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.Maths

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