Suppose z1,z2,z3 are the vertices of an equilateral triangle inscribed in the circle |z|=2. If z1=1+i√3, then z2 and z3 equal to
A
z2=1+i√3,z3=−2
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B
z2=2−i√3,z3=−1
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C
z2=1−i√3,z3=−2
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D
z2=2+i√3,z3=−1
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Solution
The correct option is Cz2=1−i√3,z3=−2 Since the vertices will appear in conjugate form hence two vertices are (1+√3i),(1−√3i)... (i) Now z21+z22+z23=z1z2+z2z3+z3z1 For an equilateral triangle. Hence 1−3+2√3i+1−3−2√3i+z23=4+z3(2) −4+z23=4+z3(2) z2−2z−8=0 or (z+2)(z−4)=0 z=−2 or z=4 Now Z lies on |z|=2 Hence z3=−2.