Let Z_1 and Z_2 be two roots 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 :

Complex Numbers PYQ — JEE Main 2003 · Shift 0 · Q98 with Solution

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Correct answer: D

By Vieta's formulas,

\qquad Z_1Z_2=b.$$ 2. Since the points $0, Z_1, Z_2$ form an equilateral triangle, the distances from the origin to the two roots are equal, and the distance between the roots is also the same. So, $$|Z_1|=|Z_2|=|Z_1-Z_2|.$$ Let $$Z_2=\omega Z_1,$$ where $|\omega|=1$ and the angle between $Z_1$ and $Z_2$ is $\pm 60^\circ$. Thus, $$\omega=e^{\pm i\pi/3}.$$ 3. Now compute the sum and product of roots: $$Z_1+Z_2=Z_1(1+\omega),$$ $$Z_1Z_2=\omega Z_1^2.$$ Hence, $$a^2=(Z_1+Z_2)^2=Z_1^2(1+\omega)^2,$$ $$b=\omega Z_1^2.$$ Therefore, $$\frac{a^2}{b}=\frac{Z_1^2(1+\omega)^2}{\omega Z_1^2}=\frac{(1+\omega)^2}{\omega}.$$ 4. Take $\omega=e^{i\pi/3}=\frac12+i\frac{\sqrt3}{2}$. Then, $$1+\omega=\frac32+i\frac{\sqrt3}{2}.$$ But a simpler identity is: $$1+e^{i\theta}=2\cos\frac{\theta}{2}\,e^{i\theta/2}.$$ For $\theta=\frac{\pi}{3}$, $$1+\omega=2\cos\frac{\pi}{6}\,e^{i\pi/6}=\sqrt3\,e^{i\pi/6}.$$ So, $$(1+\omega)^2=3e^{i\pi/3}=3\omega.$$ Thus, $$\frac{a^2}{b}=\frac{3\omega}{\omega}=3.$$ Therefore, $$a^2=3b.$$ The same result comes for $\omega=e^{-i\pi/3}$ as well. 5. Hence the correct option is $$\boxed{a^2=3b}.$$

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Correct answer: D

By Vieta's formulas,

\qquad Z_1Z_2=b.$$ 2. Since the points $0, Z_1, Z_2$ form an equilateral triangle, the distances from the origin to the two roots are equal, and the distance between the roots is also the same. So, $$|Z_1|=|Z_2|=|Z_1-Z_2|.$$ Let $$Z_2=\omega Z_1,$$ where $|\omega|=1$ and the angle between $Z_1$ and $Z_2$ is $\pm 60^\circ$. Thus, $$\omega=e^{\pm i\pi/3}.$$ 3. Now compute the sum and product of roots: $$Z_1+Z_2=Z_1(1+\omega),$$ $$Z_1Z_2=\omega Z_1^2.$$ Hence, $$a^2=(Z_1+Z_2)^2=Z_1^2(1+\omega)^2,$$ $$b=\omega Z_1^2.$$ Therefore, $$\frac{a^2}{b}=\frac{Z_1^2(1+\omega)^2}{\omega Z_1^2}=\frac{(1+\omega)^2}{\omega}.$$ 4. Take $\omega=e^{i\pi/3}=\frac12+i\frac{\sqrt3}{2}$. Then, $$1+\omega=\frac32+i\frac{\sqrt3}{2}.$$ But a simpler identity is: $$1+e^{i\theta}=2\cos\frac{\theta}{2}\,e^{i\theta/2}.$$ For $\theta=\frac{\pi}{3}$, $$1+\omega=2\cos\frac{\pi}{6}\,e^{i\pi/6}=\sqrt3\,e^{i\pi/6}.$$ So, $$(1+\omega)^2=3e^{i\pi/3}=3\omega.$$ Thus, $$\frac{a^2}{b}=\frac{3\omega}{\omega}=3.$$ Therefore, $$a^2=3b.$$ The same result comes for $\omega=e^{-i\pi/3}$ as well. 5. Hence the correct option is $$\boxed{a^2=3b}.$$