Q.

Let the complex numbers z1,z2,z3  be the vertices of an equilateral triangle. Let z0  be the circum centre of the triangle. Then z12+z22+z32=

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a

2z02

b

3z02

c

4z02

d

4z03

answer is B.

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Detailed Solution

Since z1, z2, z3 are vertices of an equilateral triangle, we have                          z12+z22+z32=z1z2+z2z3+z3z1→1 we have z0=circumcenter=centroid=z1+z2+z33 ⇒z1+z2+z3=3z0 ⇒z1+z2+z32=9z02 ⇒ z12+z22+z32+2z1z2+z2z3+z3z1=9z02 ⇒3z12+z22+z32=9z02      by 1
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