Suppose z1,z2,z3 are the vertices of an equilateral triangle inscribed in the circle |z|=2. If z1=1+i√3, then the other two vertices are
A
1+i√3
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B
1−i√3
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C
2
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D
−2
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Solution
The correct options are B1−i√3 D−2 Given z1=1+i√3=−2ω2 Rotating OA about O by an angle 2π/3, we have z−0−2ω2−0=|z−0||−2ω2−0|e±i2π/3 ⇒z=−2ω2(ω),−2ω2(ω2) ⇒z=−2ω3,−2ω ⇒z=−2,1−i√3