Question

Let $z_1$ and $z_2$ be two roots of the equation $z^2+az+b=0,z$ being a complex number. Further, assume that the origin, $z_1$ and $z_2$ form an equilateral triangle. Then

• A. $a^2=4b$
• B. $a^2=b$
• C. $a^2=2b$
• D. $a^2=3b$

Answer 1

The correct option is D $a^2=3b$

for $z_1,z_2,z_3$ forming the vertices of an equilateral triangle, the condition is,

$z_1^2+z_2^2+z_3^2-z_1{z}_2-z_2{z}_3-z_3{z}_1=0$

Here, the vertices are given as $0,z_1,z_2$

$\therefore 0+z_1^2+z_2^2-0-z_1{z}_2=0$

$\therefore z_1^2+z_2^2-z_1{z}_2=0--\left(1\right)$

given that $z_1,z_2$ are roots of $z^2+az+b=0$

$\therefore z_1^2+a{z}_1+b=0$ & $z_1^2+a{z}_2+b=0$

$\Rightarrow z_1^2+z_2^2+a(z_1+z_2)+2b=0--\left(2\right)$

From (1) & (2)

$z_1^2+z_2^2-z_1{z}_2=z_1^2+z_2^2+a(z_1+z_2)+2b$

$\therefore -2b-z_1{z}_2=a(z_1+z_2)--\left(3\right)$

By theory of equations,

sum of roots $=z_1+z_2=-a$

Product of roots $=z_1{z}_2=b$

substituting in eq(3),

$-2b-b=a(-a)$

$\therefore a^2=-3b$

$\therefore a^2=3b$