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

Concept: We can represent geometric rotations of complex numbers in the Argand plane using Euler's exponential form ($Z = R e^{i\phi}$). If two complex numbers $Z_1$ and $Z_2$ share the same distance from the origin ($OA = OB$) and are separated by a rotation angle of $\alpha$, they are related by the rotation identity: \[ Z_2 = Z_1 \cdot e^{i\alpha} \quad \text{or} \quad Z_1 = Z_2 \cdot e^{-i\alpha} \] Step 1: Establish the root-coefficient relationships using Vieta's formulas.
For the given quadratic equation $Z^2 + pZ + q = 0$, Vieta's identities map the roots directly to the complex coefficients: \[ Z_1 + Z_2 = -p \quad \Rightarrow \quad p^2 = (Z_1 + Z_2)^2 \] \[ Z_1 \cdot Z_2 = q \]

Step 2: Construct the primary fractional ratio expression.
Form the fractional ratio requested by the problem using our root equations: \[ \frac{p^2}{q} = \frac{(Z_1 + Z_2)^2}{Z_1 Z_2} \] Expand the binomial numerator expression and split the individual terms: \[ \frac{p^2}{q} = \frac{Z_1^2 + Z_2^2 + 2Z_1 Z_2}{Z_1 Z_2} = \frac{Z_1}{Z_2} + \frac{Z_2}{Z_1} + 2 \]

Step 3: Substitute the geometric rotation exponential expressions.
Using our geometric rotation condition, the ratio of the two complex numbers can be expressed as an exponential phase factor: \[ \frac{Z_2}{Z_1} = e^{i\alpha} \quad \text{and} \quad \frac{Z_1}{Z_2} = e^{-i\alpha} \] Substitute these phase factors back into our ratio equation: \[ \frac{p^2}{q} = e^{-i\alpha} + e^{i\alpha} + 2 \] Apply Euler's identity to combine the exponential terms into a standard cosine function ($e^{i\alpha} + e^{-i\alpha} = 2\cos\alpha$): \[ \frac{p^2}{q} = 2\cos\alpha + 2 = 2(1 + \cos\alpha) \]

Step 4: Simplify using double-angle trigonometric identities.
Recall the standard documentation identity: $1 + \cos\alpha = 2\cos^2\left(\frac{\alpha}{2}\right)$. Substitute this into the equation: \[ \frac{p^2}{q} = 2 \cdot \left( 2\cos^2\frac{\alpha}{2} \right) = 4\cos^2\frac{\alpha}{2} \] Now multiply both sides of the equation by the reciprocal secant term, noting that $\sec^2\theta = \frac{1}{\cos^2\theta}$: \[ \frac{p^2}{q} \cdot \sec^2\frac{\alpha}{2} = 4\cos^2\frac{\alpha}{2} \cdot \left( \frac{1}{\cos^2\frac{\alpha}{2}} \right) = 4 \] This evaluates to exactly 4, matching option (C).