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Question: Let \( {z_1} \) and \( {z_2} \) be the two roots of the equation \( {z^2} + az + b = 0 \) , \( z \) ...

Let z1{z_1} and z2{z_2} be the two roots of the equation z2+az+b=0{z^2} + az + b = 0 , zz being complex. Further, assumed that the origin z1{z_1} and z2{z_2} form an equilateral triangle. Then
A) a2=b{a^2} = b
B) a2=2b{a^2} = 2b
C) a2=3b{a^2} = 3b
D) a2=4b{a^2} = 4b

Explanation

Solution

Hint : In this question, the two roots of the equation z2+az+b=0{z^2} + az + b = 0 are z1{z_1} and z2{z_2} also, it is assumed that the origin z1{z_1} and z2{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 z2+az+b=0{z^2} + az + b = 0 are z1{z_1} and z2{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 ax2+bx+c=0a{x^2} + bx + c = 0 the sum of the roots is given as −ba\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 ax2+bx+c=0a{x^2} + bx + c = 0 the sum of the roots is given as ca\dfrac{c}{a} .
Therefore, the sum and the product of the roots be z1+z2=−a{z_1} + {z_2} = - a and z1z2=b{z_1}{z_2} = b
Let 0, z1{z_1} , z2{z_2} are vertices of an equilateral triangle.
We know that, for an equilateral triangle, x12+x22+x32=x1x2+x2x3+x3x1{x_1}^2 + {x_2}^2 + {x_3}^2 = {x_1}{x_2} + {x_2}{x_3} + {x_3}{x_1}
Let x1=0{x_1} = 0 , x2=z1{x_2} = {z_1} and x3=z2{x_3} = {z_2}
Substituting the values in the formula,
02+z12+z22=0×z1+z1×z2+z2×0 ⇒z12+z22=z1z2   {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 2z1z22{z_1}{z_2} on both sides,
z12+z22+2z1z2=z1z2+2z1z2{z_1}^2 + {z_2}^2 + 2{z_1}{z_2} = {z_1}{z_2} + 2{z_1}{z_2}
Therefore,
(z1+z2)2=3z1z2{\left( {{z_1} + {z_2}} \right)^2} = 3{z_1}{z_2}
Now, substituting the values z1+z2=−a{z_1} + {z_2} = - a and z1z2=b{z_1}{z_2} = b , we get,
a2=3b{a^2} = 3b
So, the correct answer is “Option C”.

Note : It is important here to note that, we can use the formula x12+x22+x32=x1x2+x2x3+x3x1{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 z2=z1(cos⁡π3+isin⁡π3){z_2} = {z_1}\left( {\cos \dfrac{\pi }{3} + i\sin \dfrac{\pi }{3}} \right) .