Question

Let $z_{1}and{z}_2$ be the roots of the equation $z^2+az+b=0,z$ is a complex number. Assume that origin, $z_{1}and{z}_2$form an equilateral triangle, then

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

Answer 1

The correct option is B

$a^2=3b$

Explanation for the correct answer:

Calculating the value of $a^2$:

Given, $z_{1}and{z}_2$ are the roots of the given quadratic equation $z^2+az+b=0$.

Therefore, we can say that:

$z_1+z_2=-a,z_1{z}_2=b$

Writing in the polar form:

$$\begin{gathered} z_2=z_1{e}^{i\frac{\mathrm{\pi }}{3}} \\ \Rightarrow z_2=z_1\left(\cos \left(\frac{\mathrm{\pi }}{3}\right)+i\sin \left(\frac{\mathrm{\pi }}{3}\right)\right)\mathbf{[}\mathbf{\because }{\mathbf{e}}^{\mathbf{i}\mathbf{\theta }}\mathbf{=}\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{\theta }\mathbf{+}\mathbf{i}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{\theta }\mathbf{]} \\ \Rightarrow z_2=z_1\left(\frac{1}{2}+\frac{i\sqrt{3}}{2}\right)\left[\because \sin \frac{\pi }{3}=\frac{\sqrt{3}}{2},\cos \frac{\pi }{3}=\frac{1}{2}\right] \\ \Rightarrow 2z_2-z_1=\sqrt{3}i{z}_1 \\ \Rightarrow {\left(2z_2-z_1\right)}^2=-3{z_1}^2 \end{gathered}$$

Thus,

$$\begin{gathered} {z_1}^2+{4z_2}^2-4z_1{z}_2={-3z_1}^2 \\ {\Rightarrow 4z_1}^2+{4z_2}^2=4z_1{z}_2 \\ \Rightarrow a^2-2b=b\left[\because {(z_1+z_2)}^2={z_1}^2+{z_2}^2+2z_1{z}_2\right] \\ \therefore a^2=3b \end{gathered}$$

Hence, the correct answer is option (B).