Question

Let $z_1$ and $z_2$ be roots of the equation $z^2+pz+q=0,$ where the coefficients $p$ and $q$ may be complex numbers. Let $A$ and $B$ represents $z_1$ and $z_2$ in the complex plane. If $\angle AOB=\alpha \ne 0$ and $OA=OB,$ where $O$ is the origin, then $p^2=k{\cos }^2\frac{\alpha }{2},$ where $k=$

• A. $q$
• B. $2q$
• C. $4q$
• D. None of these

Answer 1

The correct option is C $4q$

We have, $z_1+z_2=-p$ and $z_1{z}_2=q$ We know that $\frac{z_1}{z_2}=\frac{\left|z_1\right|}{\left|z_2\right|}(\cos \alpha +i\sin \alpha )$

Since $\left|z_1\right|=\left|z_2\right|(\because OA=OB)$

we get $\frac{z_1}{z_2}=\frac{\cos \alpha +i\sin \alpha }{1}$

Applying Componendo and Dividendo, we get

$\frac{z_1+z_2}{z_1+z_2}=\frac{\cos \alpha +i\sin \alpha +1}{\cos \alpha +i\sin \alpha -1}$

$=\frac{2{\cos }^2\frac{\alpha }{2}+2i\sin \frac{\alpha }{2}\cos \frac{\alpha }{2}}{-2{\sin }^2\frac{\alpha }{2}+2i\sin \frac{\alpha }{2}\cos \frac{\alpha }{2}}$

$=\frac{2\cos \frac{\alpha }{2}[\cos \frac{\alpha }{2}+i\sin \frac{\alpha }{2}]}{2i\sin \frac{\alpha }{2}[\cos \frac{\alpha }{2}+i\sin \frac{\alpha }{2}]}=-i\cot \frac{\alpha }{2}$

$\Rightarrow \frac{p}{z_1-z_2}=i\cot \frac{\alpha }{2}$

Squaring we obtain

$\frac{p^2}{{(z_1-z_2)}^2-4z_1{z}_2}=-{\cot }^2\frac{\alpha }{2}$

$\Rightarrow \frac{p^2}{p^2-4q}=-{\cot }^2\frac{\alpha }{2}$

$\Rightarrow p^2=-p^2{\cot }^2\frac{a}{2}+4q{\cot }^2\frac{\alpha }{2}$

$\Rightarrow p^2(1+{\cot }^2\frac{\alpha }{2})=4q{\cot }^2\frac{\alpha }{2}$

$\Rightarrow p^2{\csc }^2\frac{\alpha }{2}=4q{\csc }^2\frac{\alpha }{2}$

$\Rightarrow p^2=4q{\cos }^2\frac{\alpha }{2}$

$\therefore k=4q$