A) \[-2,\,1-i\sqrt{3}\]
B) \[2,\,1+i\sqrt{3}\]
C) \[1+i\sqrt{3},-2\]
D) None of these
Correct Answer: A
Solution :
One of the number must be a conjugate of \[{{z}_{1}}=1+i\sqrt{3}\,\,i.e.\,{{z}_{2}}=1-i\sqrt{3}\] or \[{{z}_{3}}={{z}_{1}}{{e}^{i2\pi /3}}\]and \[{{z}_{2}}={{z}_{1}}{{e}^{-i2\pi /3}}\] \[{{z}_{3}}=(1+i\sqrt{3})\left[ \cos \left( \frac{2\pi }{3} \right)+i\sin \frac{2\pi }{3} \right]=-2\] Aliter: Obviously \[|z|=2\] is a circle with centre \[O(0,\,0)\] and radius 2. Therefore, \[OA=OB=OC\] and this is satisfied by (a) because two vertices of any triangle cannot be same.You need to login to perform this action.
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