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(D) Given that $z_1$ and $z_2$ are roots of $z^2 + az + b = 0$,we have $z_1 + z_2 = -a$ and $z_1 z_2 = b$.Since $OA = OB$ and the angle between them i

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$4$

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(D) Given that $z_1$ and $z_2$ are roots of $z^2 + az + b = 0$,we have $z_1 + z_2 = -a$ and $z_1 z_2 = b$.Since $OA = OB$ and the angle between them is $\alpha$,we can write $z_2 = z_1 e^{i \alpha}$ (assuming $B$ is obtained by rotating $A$ by $\alpha$ counter-clockwise).Then $\frac{z_2}{z_1} = e^{i \alpha}$.Now,consider $\frac{a^2}{b} = \frac{(z_1 + z_2)^2}{z_1 z_2} = \frac{z_1^2 + z_2^2 + 2z_1 z_2}{z_1 z_2} = \frac{z_1}{z_2} + \frac{z_2}{z_1} + 2$.Substituting $\frac{z_2}{z_1} = e^{i \alpha}$,we get $\frac{a^2}{b} = e^{-i \alpha} + e^{i \alpha} + 2 = 2 \cos \alpha + 2$.Using the identity $1 + \cos \alpha = 2 \cos^2 \frac{\alpha}{2}$,we have $\frac{a^2}{b} = 2(1 + \cos \alpha) = 2(2 \cos^2 \frac{\alpha}{2}) = 4 \cos^2 \frac{\alpha}{2}$.Thus,$a^2 = 4b \cos^2 \frac{\alpha}{2}$.Comparing this with $a^2 = \lambda b \cos^2 \frac{\alpha}{2}$,we get $\lambda = 4$.

Let $z_1$ and $z_2$ be two complex numbers and roots of the equation $z^2 + az + b = 0$. If $O$ is the origin such that $OA = OB$ and $a^2 = \lambda b \cos^2 \frac{\alpha}{2}$,where $\alpha$ is the angle $\angle AOB$,then $\lambda$ is equal to:

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