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

${a}^{2}=b$

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B

${a}^{2}=2b$

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C

${a}^{2}=3b$

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D

${a}^{2}=4b$

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Solution

## 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 ${e}^{i\theta }=\mathrm{cos}\theta +i\mathrm{sin}\theta$ it can be written as:${z}_{2}={z}_{1}{e}^{i\frac{\mathrm{\pi }}{3}}\phantom{\rule{0ex}{0ex}}⇒{z}_{2}={z}_{1}\left(\mathrm{cos}\frac{\mathrm{\pi }}{3}+i\mathrm{sin}\frac{\mathrm{\pi }}{3}\right)\phantom{\rule{0ex}{0ex}}⇒{z}_{2}={z}_{1}\left(\frac{1}{2}+i\frac{\sqrt{3}}{2}\right)\left[\mathrm{cos}\frac{\mathrm{\pi }}{3}=\frac{1}{2};\mathrm{sin}\frac{\mathrm{\pi }}{3}=\frac{\sqrt{3}}{2}\right]\phantom{\rule{0ex}{0ex}}⇒2{z}_{2}={z}_{1}+i\sqrt{3}{z}_{1}\phantom{\rule{0ex}{0ex}}⇒2{z}_{2}-{z}_{1}=i\sqrt{3}{z}_{1}$Now, square both sides.$\begin{array}{rcl}{\left(2{z}_{2}-{z}_{1}\right)}^{2}& =& {\left(i\sqrt{3}{z}_{1}\right)}^{2}\\ & ⇒& 4{{z}_{2}}^{2}-4{z}_{2}{z}_{1}+{{z}_{1}}^{2}={i}^{2}3{{z}_{1}}^{2}\\ & ⇒& 4{{z}_{2}}^{2}-4{z}_{2}{z}_{1}+{{z}_{1}}^{2}=-3{{z}_{1}}^{2}\left[{i}^{2}=-1\right]\\ & ⇒& 4{{z}_{2}}^{2}-4{z}_{2}{z}_{1}+{{z}_{1}}^{2}+3{{z}_{1}}^{2}=0\\ & ⇒& 4\left({{z}_{1}}^{2}+{{z}_{2}}^{2}-{z}_{1}{z}_{2}\right)=0\\ & ⇒& {{z}_{1}}^{2}+{{z}_{2}}^{2}={z}_{1}{z}_{2}\end{array}$Now, add $2{z}_{1}{z}_{2}$ both sides.${{z}_{1}}^{2}+{{z}_{2}}^{2}+2{z}_{1}{z}_{2}={z}_{1}{z}_{2}+2{z}_{1}{z}_{2}\phantom{\rule{0ex}{0ex}}⇒{\left({z}_{1}+{z}_{2}\right)}^{2}=3{z}_{1}{z}_{2}\left[{\left(a+b\right)}^{2}={a}^{2}+{b}^{2}+2ab\right]\phantom{\rule{0ex}{0ex}}⇒{a}^{2}=3b\left[{z}_{1}+{z}_{2}=a,{z}_{1}{z}_{2}=b\right]$So, the relation between $a$ and $b$ for the given condition is ${a}^{2}=3b$.Hence, the correct option is C.

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