Question

Let $z_1$ and $z_2$ be two root of the equation $z^2+az+b=0$ , $z$ being complex 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)

$a^2=3b$

According to what we learnt in quadratic solutions, since a & b are real numbers:

Sum of roots = $-a=z_1+z_2=|z_1+z_2|$

Product of roots = $b=z_1{z}_2=\left|z_1{z}_2\right|$

Per complex numbers, ${\left|z\right|}^2=z.\overline{z}$

As ${0,z}_1,z_2$ form an equilateral $△$,

$\left|z_1\right|=\left|z_2\right|=|z_2-z_1|$

$\therefore {\left|z_1\right|}^2={\left|z_2\right|}^2$

As shown above,

$\left|z_2\right|=|z_2-z_1|$

Squaring both sides we get

${\left|z_2\right|}^2={|z_2-z_1|}^2$

Since $\overline{z_2-z_1}=\overline{z_2}-\overline{z_1}$, we get

$z_2.\overline{z_2}=(z_2-z_1)\left(\overline{z_2-z_1}\right)$

$\Rightarrow z_2.\overline{z_2}=(z_2-z_1)(\overline{z_2}-\overline{z_1})$

$\Rightarrow z_2.\overline{z_2}=z_2.\overline{z_2}+z_1.\overline{z_1}-(z_1\overline{z_2}+\overline{z_1}{z}_2)$

${\Rightarrow \left|z_1\right|}^2=z_1\overline{z_2}+\overline{z_1}{z}_2$ --- (i)

As shown above

$|z_1+z_2|=-a$

Squaring both sides we get

$(z_1+z_2)+(\overline{z_1}+\overline{z_2})=a^2$

${\Rightarrow \left|z_1\right|}^2+{\left|z_2\right|}^2+z_1\overline{z_2}+\overline{z_1}{z}_2=a^2$

${\Rightarrow 2\left|z_1\right|}^2+z_1\overline{z_2}+\overline{z_1}{z}_2=a^2$ --- (ii)

${\Rightarrow 2\left|z_1\right|}^2+{\left|z_1\right|}^2=a^2$

${\Rightarrow 3\left|z_1\right|}^2=a^2$

${\Rightarrow \left|z_1\right|}^2=\frac{a^2}{3}={\left|z_2\right|}^2$

As shown above,

$\left|z_1{z}_2\right|=b$

Squaring both sides we get

${\left|z_1{z}_2\right|}^2=b^2$

$\Rightarrow \left(z_1{z}_2\right)(\overline{z_1}.\overline{z_2})=b^2$

$\Rightarrow {\left|z_1\right|}^2.{\left|z_2\right|}^2=b^2$

$\Rightarrow \frac{a^2}{3}.\frac{a^2}{3}=b^2$

$\Rightarrow a^4=9b^2$

$\therefore a^2=3b$

Hence the answer is C