Question

Previous year IIT jee Question
Chapter :- complex number and quadratic equation.​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

$\rm :\longmapsto\:z_1 \: and \: z_2 \: are \: roots \: of \: {z}^{2} + az + b = 0$

We know that,

$\boxed{\red{\sf Product\ of\ the\ zeroes=\frac{Constant}{coefficient\ of\ x^{2}}}}$

$\bf\implies \: z_1z_2 = \dfrac{b}{1} = b$

and

$\boxed{\red{\sf Sum\ of\ the\ zeroes=\frac{-coefficient\ of\ x}{coefficient\ of\ x^{2}}}}$

$\bf\implies \: z_1 + z_2 = - \dfrac{a}{1} = - \: a$

Now,

Further it is given that,

$\rm :\longmapsto\:(0,0), \:z_1 \: ,z_2 \: forms \: vertices \: of \: equilateral \: triangle.$

We know that,

$\red{\rm :\longmapsto \:z_1, z_2,z_3 \: forms \: vertices \: of \: equilateral \: triangle \: then}$

$\red{\rm :\longmapsto\: {z_1}^{2} + {z_2}^{2} + {z_3}^{2} = z_1z_2 + z_2z_3 + z_3z_1 }$

Now, we have

$\rm :\longmapsto\:z_1 = z_1$

$\rm :\longmapsto\:z_2 = z_2$

$\rm :\longmapsto\:z_3 = 0$

So, on substituting these values, we get

${\rm :\longmapsto\: {z_1}^{2} + {z_2}^{2} + {0}^{2} = z_1z_2 + z_2 \times 0+ 0 \times z_1 }$

${\rm :\longmapsto\: {z_1}^{2} + {z_2}^{2} = z_1z_2 }$

${\rm :\longmapsto\: {z_1}^{2} + {z_2}^{2} + 2z_1z_2 = z_1z_2 + 2z_1z_2}$

$\rm :\longmapsto\: {(z_1 + z_2)}^{2} = 3z_1z_2$

$\rm :\longmapsto\: {( - a)}^{2} = 3b$

$\bf\implies \: {a}^{2} = 3b$

Hence, Option (3) is correct

Additional Information :-

$\boxed{ \sf{ \: |z| = |\overline{z}| }}$

$\boxed{ \sf{ \:z \: \overline{z} \: = { |z| }^{2}}}$

$\boxed{ \sf{ \: |z| = 1 \: \rm\implies\:\overline{z} = \frac{1}{z}}}$

$\boxed{ \sf{ \:z = r {e}^{i \theta} = r(cos\theta + i \: sin\theta)}}$