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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