SATHEE: Complex Numbers 4 Question 14

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 $∠ A O B = α \neq 0$ and $O A = O B$ , where $O$ is the origin prove that $p^{2} = 4 q \cos^{2} \frac{α}{2}$ .

(1997, 5M)

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Solution:

Since, $z_{1} + z_{2} = - p$ and $z_{1} z_{2} = q$

Now,

$\frac{z_{1}}{z_{2}} = \frac{| z_{1} |}{| z_{2} |} (\cos α + i \sin α)$

$\Rightarrow \frac{z_{1}}{z_{2}} = \frac{\cos α + i \sin α}{1}$

$[∵ | z_{1} | = | z_{2} |]$

Applying componendo and dividendo, we get

$\begin{aligned} \frac{z_{1} + z_{2}}{z_{1} - z_{2}} = \frac{\cos α + i \sin α + 1}{\cos α + i \sin α - 1} \\ = \frac{2 \cos^{2} (α / 2) + 2 i \sin (α / 2) \cos (α / 2)}{- 2 \sin^{2} (α / 2) + 2 i \sin (α / 2) \cos (α / 2)} \\ = \frac{2 \cos (α / 2) [\cos (α / 2) + i \sin (α / 2)]}{2 i \sin (α / 2) [\cos (α / 2) + i \sin (α / 2)]} \\ = \frac{\cot (α / 2)}{i} = - i \cot α / 2 \Rightarrow \frac{- p}{z_{1} - z_{2}} = - i \cot (α / 2) \end{aligned}$

On squaring both sides, we get $\frac{p^{2}}{{(z_{1} - z_{2})}^{2}} = - \cot^{2} (α / 2)$

$\begin{array}{lc} \Rightarrow & \frac{p^{2}}{{(z_{1} + z_{2})}^{2} - 4 z_{1} z_{2}} = - \cot^{2} (α / 2) \\ \Rightarrow & \frac{p^{2}}{p^{2} - 4 q} = - \cot^{2} (α / 2) \\ \Rightarrow & p^{2} = - p^{2} \cot^{2} (α / 2) + 4 q \cot^{2} (α / 2) \\ \Rightarrow & p^{2} (1 + \cot^{2} α / 2) = 4 q \cot^{2} (α / 2) \\ \Rightarrow & p^{2} {cosec}^{2} (α / 2) = 4 q \cot^{2} (α / 2) \\ \Rightarrow & p^{2} = 4 q \cos^{2} α / 2 \end{array}$

Complex Numbers 4 Question 14

14.

Solution:

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Complex Numbers 4 Question 14

14.

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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