Complex Numbers 4 Question 14
14.
Let $z _1$ and $z _2$ be the roots of the equation $z^{2}+p z+q=0$, where the coefficients $p$ and $q$ may be complex numbers. Let $A$ and $B$ represent $z _1$ and $z _2$ in the complex plane. If $\angle A O B=\alpha \neq 0$ and $O A=O B$, where $O$ is the origin prove that $p^{2}=4 q \cos ^{2} \frac{\alpha}{2}$.
(1997, 5M)
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Solution:
- Since, $z _1+z _2=-p$ and $z _1 z _2=q$
Now,
$ \frac{z _1}{z _2}=\frac{\left|z _1\right|}{\left|z _2\right|}(\cos \alpha+i \sin \alpha) $
$\Rightarrow \frac{z _1}{z _2}=\frac{\cos \alpha+i \sin \alpha}{1}$
${\left[\because\left|z _1\right|=\left|z _2\right|\right]}$
Applying componendo and dividendo, we get
$ \begin{aligned} & \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}(\alpha / 2)+2 i \sin (\alpha / 2) \cos (\alpha / 2)}{-2 \sin ^{2}(\alpha / 2)+2 i \sin (\alpha / 2) \cos (\alpha / 2)} \\ & =\frac{2 \cos (\alpha / 2)[\cos (\alpha / 2)+i \sin (\alpha / 2)]}{2 i \sin (\alpha / 2)[\cos (\alpha / 2)+i \sin (\alpha / 2)]} \\ & =\frac{\cot (\alpha / 2)}{i}=-i \cot \alpha / 2 \Rightarrow \frac{-p}{z _1-z _2}=-i \cot (\alpha / 2) \end{aligned} $
On squaring both sides, we get $\frac{p^{2}}{\left(z _1-z _2\right)^{2}}=-\cot ^{2}(\alpha / 2)$
$ \begin{array}{lc} \Rightarrow & \frac{p^{2}}{\left(z _1+z _2\right)^{2}-4 z _1 z _2}=-\cot ^{2}(\alpha / 2) \\ \Rightarrow & \frac{p^{2}}{p^{2}-4 q}=-\cot ^{2}(\alpha / 2) \\ \Rightarrow & p^{2}=-p^{2} \cot ^{2}(\alpha / 2)+4 q \cot ^{2}(\alpha / 2) \\ \Rightarrow & p^{2}\left(1+\cot ^{2} \alpha / 2\right)=4 q \cot ^{2}(\alpha / 2) \\ \Rightarrow & p^{2} \operatorname{cosec}^{2}(\alpha / 2)=4 q \cot ^{2}(\alpha / 2) \\ \Rightarrow & p^{2}=4 q \cos ^{2} \alpha / 2 \end{array} $
