Question

If $z_1$ and $z_2$ are 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

The correct option is C

$a^2=3b$

Explanation for the correct option.

Find the relation between $a$ and $b$.

It is given that the roots of the equation $z^2+az+b=0$ are $z_1$ and $z_2$.

So, the sum of roots is $-a$, thus $z_1+z_2=-a$.

And the product of roots is $b$, so $z_1{z}_2=b$.

Now, it is given that the origin, $z_1$ and $z_2$ form an equilateral triangle. So, $z_2=z_1{e}^{i\frac{\mathrm{\pi }}{3}}$.

Now, using $e^{i\theta }=\cos \theta +i\sin \theta$ it can be written as:

$$\begin{gathered} z_2=z_1{e}^{i\frac{\mathrm{\pi }}{3}} \\ \Rightarrow z_2=z_1\left(\cos \frac{\mathrm{\pi }}{3}+i\sin \frac{\mathrm{\pi }}{3}\right) \\ \Rightarrow z_2=z_1\left(\frac{1}{2}+i\frac{\sqrt{3}}{2}\right)\left[\cos \frac{\mathrm{\pi }}{3}=\frac{1}{2};\sin \frac{\mathrm{\pi }}{3}=\frac{\sqrt{3}}{2}\right] \\ \Rightarrow 2z_2=z_1+i\sqrt{3}{z}_1 \\ \Rightarrow 2z_2-z_1=i\sqrt{3}{z}_1 \end{gathered}$$

Now, square both sides.

$\begin{array}{rcl}{\left(2z_2-z_1\right)}^2& =& {\left(i\sqrt{3}{z}_1\right)}^2\\ & \Rightarrow & 4{z_2}^2-4z_2{z}_1+{z_1}^2=i^{2}3{z_1}^2\\ & \Rightarrow & 4{z_2}^2-4z_2{z}_1+{z_1}^2=-3{z_1}^2\left[i^2=-1\right]\\ & \Rightarrow & 4{z_2}^2-4z_2{z}_1+{z_1}^2+3{z_1}^2=0\\ & \Rightarrow & 4\left({z_1}^2+{z_2}^2-z_1{z}_2\right)=0\\ & \Rightarrow & {z_1}^2+{z_2}^2=z_1{z}_2\end{array}$

Now, add $2z_1{z}_2$ both sides.

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

So, the relation between $a$ and $b$ for the given condition is $a^2=3b$.

Hence, the correct option is C.