A) \[{{a}^{2}}=b\]
B) \[{{a}^{2}}=2b\]
C) \[{{a}^{2}}=3b\]
D) \[{{a}^{2}}=4b\]
Correct Answer: C
Solution :
If \[{{z}_{1}},{{z}_{2}}\] and \[{{z}_{3}}\] are the vertices of an equilateral triangle. Then, \[z_{1}^{2}+z_{2}^{2}+z_{3}^{2}={{z}_{1}}{{z}_{2}}+{{z}_{2}}{{z}_{3}}+{{z}_{3}}{{z}_{1}}\]. Since, origin, \[{{z}_{1}}\] and \[{{z}_{2}}\] are the vertices of an equilateral triangle, then \[z_{1}^{2}+z_{2}^{2}={{z}_{1}}{{z}_{2}}\] \[\Rightarrow \] \[{{({{z}_{1}}-{{z}_{2}})}^{2}}=3{{z}_{1}}{{z}_{2}}\] ... (i) Again, \[{{z}_{1}},{{z}_{2}}\] are the roots of the equation \[{{z}^{2}}+az+b=0\] Then, \[{{z}_{1}}+{{z}_{2}}=-a\] and \[{{z}_{1}}{{z}_{2}}=b\] On putting these values in Eq. (i), we get \[{{(-a)}^{2}}=3b,\Rightarrow {{a}^{2}}=3b\]
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