Question

Let $ {z_1} $ and $ {z_2} $ be the two roots of the equation $ {z^2} + az + b = 0 $ , $ z $ being complex. Further, assumed that the origin $ {z_1} $ and $ {z_2} $ form an equilateral triangle. Then
A) $ {a^2} = b $
B) $ {a^2} = 2b $
C) $ {a^2} = 3b $
D) $ {a^2} = 4b $

Answer 1

Hint: In this question, the two roots of the equation $ {z^2} + az + b = 0 $ are $ {z_1} $ and $ {z_2} $ also, it is assumed that the origin $ {z_1} $ and $ {z_2} $ form an equilateral triangle by which we know the vertices of an equilateral triangle and apply it in the formula for an equilateral triangle and by simplifying it we will get the solution.

Complete step-by-step answer:
Here, the two roots of the equation $ {z^2} + az + b = 0 $ are $ {z_1} $ and $ {z_2} $ .
The sum of the roots in a quadratic equation is given as the ratio of the negative coefficient of the variable alone and the coefficient of the squared variable present in the equation. Mathematically, for the quadratic equation $ a{x^2} + bx + c = 0 $ the sum of the roots is given as $ \dfrac{{ - b}}{a} $ .
Also, the product of the roots in a quadratic equation is given as the ratio of the constant term and the coefficient of the squared variable present in the equation. Mathematically, for the quadratic equation $ a{x^2} + bx + c = 0 $ the sum of the roots is given as $ \dfrac{c}{a} $ .
Therefore, the sum and the product of the roots be $ {z_1} + {z_2} = - a $ and $ {z_1}{z_2} = b $
Let 0, $ {z_1} $ , $ {z_2} $ are vertices of an equilateral triangle.
We know that, for an equilateral triangle, $ {x_1}^2 + {x_2}^2 + {x_3}^2 = {x_1}{x_2} + {x_2}{x_3} + {x_3}{x_1} $
Let $ {x_1} = 0 $ , $ {x_2} = {z_1} $ and $ {x_3} = {z_2} $
Substituting the values in the formula,
 $
  {0^2} + {z_1}^2 + {z_2}^2 = 0 \times {z_1} + {z_1} \times {z_2} + {z_2} \times 0 \\
   \Rightarrow {z_1}^2 + {z_2}^2 = {z_1}{z_2} \;
  $
Adding $ 2{z_1}{z_2} $ on both sides,
 $ {z_1}^2 + {z_2}^2 + 2{z_1}{z_2} = {z_1}{z_2} + 2{z_1}{z_2} $
Therefore,
 $ {\left( {{z_1} + {z_2}} \right)^2} = 3{z_1}{z_2} $
Now, substituting the values $ {z_1} + {z_2} = - a $ and $ {z_1}{z_2} = b $ , we get,
 $ {a^2} = 3b $
So, the correct answer is “Option C”.

Note: It is important here to note that, we can use the formula $ {x_1}^2 + {x_2}^2 + {x_3}^2 = {x_1}{x_2} + {x_2}{x_3} + {x_3}{x_1} $ , only when the vertices of the equilateral triangle are given. Since, it is an equilateral triangle, we can also determine the solution by $ {z_2} = {z_1}\left( {\cos \dfrac{\pi }{3} + i\sin \dfrac{\pi }{3}} \right) $ .