Question

Let $z_1$ and $z_2$ be the roots of a quadratic equation $z^2+pz+q=0$, where $p,q$ are complex numbers. Let A and $B$ represent $z_1$ and $z_2$ in complex plane. If $\angle AOB=\alpha \ne 0$ and $OA=OB$ (where O is origin), then :

• A. $p^2=4q{\cos }^2\frac{\alpha }{2}$
• B. $p^2=2q{\cos }^2\frac{\alpha }{2}$
• C. $p^2=4q{\cos }^2\alpha$
• D. $p^2=2q{\cos }^2\alpha$

Answer 1

The correct option is A $p^2=4q{\cos }^2\frac{\alpha }{2}$

We have,
$z_1+z_2=-p---\left(1\right)$
$z_1{z}_2=q---\left(2\right)$
$\left[\begin{array}{c}\because fora{x}^2+bx+c\\ sumofroots=\frac{-coefficientofx}{coefficientof{x}^2}\\ productsofroots=\frac{constant}{co.of{x}^2}\end{array}\right]$
also given,
$\angle AOB=\alpha$
$\Rightarrow \frac{z_2}{z_1}=e^{i\alpha }$
$\Rightarrow z_2=z_1{e}^{i\alpha }$---(3) $\left[\begin{array}{c}asOA=OB\\ \Rightarrow \left|z_1\right|=\left|z_2\right|\end{array}\right]$
So, Substituting (3) & (2),
$z_1+z_1{e}^{i\alpha }=-p$
$\Rightarrow z_1+{z_1}^2{e}^{i\alpha }=-p$
$\Rightarrow z_1+(1+e^{i\alpha }{)}^2=-p$
$\Rightarrow \frac{9}{e^{i\alpha }}(1+2e^{i\alpha }+e^{i2\alpha })=p^2$ [from (4)]
$\Rightarrow q(e^{-i\alpha }+2+e^{i\alpha })=p^2\left[\begin{array}{c}{e}^{i\alpha }+e^{-i\alpha }=cos\alpha +isin\alpha +cos\alpha -isin\alpha \\ =2cos\alpha \end{array}\right]$
$\Rightarrow q(2+2cos\alpha )=p^2$
$\Rightarrow 4qco{s}^2\frac{\alpha }{2}=p^2$