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Q.

The condition for z1, z2 and origin  forms an equilateral triangle is 

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a

z12+z22z1 z2=0

b

z12z22z1 z2=0

c

z12z22+z1 z2=0

d

z12+z22+z1 z2=0

answer is A.

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Detailed Solution

By the property equilateral  Δle

Z12+Z22+Z32Z1Z2Z2Z3Z3Z2=0

Z3=0Z12+Z22Z1Z2=0

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