Question

Let $z_1,z_2$ 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. $a^2=b$
• B. $a^2=2b$
• C. $a^2=3b$
• D. $a^2=4b$

Answer 1

Answer: (C)

$a^2=3b$

Let the roots of the equation be

$z_1=p+iq$ & $z_2=p-iq$

Both roots forming an equilateral triangle with origin means all three are equidistant from one another

$\Rightarrow |z_1|=|z_2|=|z_1-z_2|$

$\Rightarrow z_1^2=|z_1-z_2{|}^2$

$\Rightarrow p^2+q^2=(2q{)}^2$

$\Rightarrow p^2=3q^2$

now, we know that $a=-(z_1+z_2)$

$b=z_1{z}_2$

$a=-2p$

$\Rightarrow a^2=4p^2$

and

$b=p^2+q^2$

On substituting the values of p and q, we get

$\Rightarrow a^2=3b$