Question

Let \(Z_{1},Z_{2}\) be the roots of the equation \(Z^{2}+pZ+q=0\), where the coefficients \(p\) and \(q\) may be complex numbers and also let \(A,B\) represent \(Z_{1},Z_{2}\) respectively in the complex plane. If \(\angle AOB=\alpha\ne0\) and \(OA=OB\), where \(O\) is the origin, then the value of \(\frac{p^{2}}{q}\sec^{2}\frac{\alpha}{2}\) will be:

• A. $\frac{1}{4}$
• B. $\frac{3}{4}$
• C. 4
• D. 1

Hint:

To solve this quickly during an exam, pick simple values that fit the geometric conditions! Let $Z_1 = 1$ and rotate it by $\alpha = 90^\circ$ ($\pi/2$) to get $Z_2 = i$. The coefficients are $p = -(1+i)$ and $q = 1 \cdot i = i$. Plugging these in gives $p^2/q = (1+i)^2 / i = 2i/i = 2$. Then multiply by $\sec^2(45^\circ) = (\sqrt{2})^2 = 2$, which gives $2 \times 2 = 4$ instantly!

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
In the complex plane, if two points \( Z_1 \) and \( Z_2 \) are at equal distances from the origin (\( OA=OB \)), they can be related via a rotation. Specifically, \( Z_2 = Z_1 e^{i\alpha} \) or \( Z_2 = Z_1 e^{-i\alpha} \). We combine this geometric insight with the root-coefficient relations of the quadratic.
Step 2: Key Formula or Approach:
1. Sum of roots: \( Z_1 + Z_2 = -p \).
2. Product of roots: \( Z_1 Z_2 = q \).
3. Ratio of roots: \( Z_1 / Z_2 = e^{i\alpha} \).
Step 3: Detailed Explanation:
Given \( OA = OB \), we have \( |Z_1| = |Z_2| \).
Let the angle between them be \( \alpha \). Then \( \frac{Z_1}{Z_2} = e^{i\alpha} \) (or \( e^{-i\alpha} \)).
From the quadratic equation:
\[ \frac{p^2}{q} = \frac{(Z_1 + Z_2)^2}{Z_1 Z_2} \]
\[ \frac{p^2}{q} = \frac{Z_1^2 + Z_2^2 + 2Z_1 Z_2}{Z_1 Z_2} = \frac{Z_1}{Z_2} + \frac{Z_2}{Z_1} + 2 \]
Substitute the rotation factor \( e^{i\alpha} \):
\[ \frac{p^2}{q} = e^{i\alpha} + e^{-i\alpha} + 2 \]
Using Euler's identity \( e^{i\theta} + e^{-i\theta} = 2 \cos \theta \):
\[ \frac{p^2}{q} = 2 \cos \alpha + 2 = 2(\cos \alpha + 1) \]
Recall the half-angle formula \( 1 + \cos \alpha = 2 \cos^2 \frac{\alpha}{2} \):
\[ \frac{p^2}{q} = 2(2 \cos^2 \frac{\alpha}{2}) = 4 \cos^2 \frac{\alpha}{2} \]
Now, we find the requested expression:
\[ \frac{p^2}{q} \sec^2 \frac{\alpha}{2} = (4 \cos^2 \frac{\alpha}{2}) \cdot \frac{1}{\cos^2 \frac{\alpha}{2}} = 4 \]
Step 4: Final Answer:
The value of the expression is 4.