Problem 109 - A Complete Resource Book in Mathematics for JEE Main

Question

Let \(z_{1}\) and \(z_{2}\) be 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, then \(p^{2}=k \cos ^{2} \frac{\alpha}{2}\), where \(k=\) (A) \(q\) (B) \(2 q\) (C) \(4 q\) (D) None of these

Step-by-Step Solution

Answer

(C) 4q

Step 1: Identify conditions from geometry

Since \(A\) and \(B\) in the complex plane represent the roots \(z_1\) and \(z_2\) and form an angle \(\angle AOB = \alpha\) while satisfying \(OA = OB\), the two vectors \(OA\) and \(OB\) are equal in length. This implies that \(z_1\) and \(z_2\) lie on the same circle with center \(O\). Also, \(z_1\) and \(z_2\) are symmetric with respect to a line that makes an angle of \(\frac{\alpha}{2}\) with the x-axis.

Step 2: Relate roots to coefficients using Vieta's formulas

For a quadratic equation \(z^2 + pz + q = 0\), Vieta's formulas tell us that the sum of roots \(z_1 + z_2 = -p\) and the product of roots \(z_1 z_2 = q\).

Step 3: Represent the roots using polar form

Since the roots are equidistant from the origin and form an angle \( \alpha \), assume \( z_1 = r \text{cis} \frac{\alpha}{2} \) and \( z_2 = r \text{cis} \left(-\frac{\alpha}{2}\right) \) where \( r = |z_1| = |z_2| \).

Step 4: Calculate \(z_1 + z_2\) and equate to Vieta's sum

\(z_1 + z_2 = r \left( \text{cis} \frac{\alpha}{2} + \text{cis} \left(-\frac{\alpha}{2}\right) \right) = 2r \cos \frac{\alpha}{2}\). From Vieta's, we have \( z_1 + z_2 = -p \). Thus, \(-p = 2r \cos \frac{\alpha}{2}\).

Step 5: Solve for \(r\) using the sum of coefficients

From Step 4, \( r = \frac{p}{2 \cos \frac{\alpha}{2}} \).

Step 6: Calculate \(z_1 z_2\) and equate to Vieta's product

The product \(z_1 z_2 = r^2 [\text{cis} \frac{\alpha}{2} \times \text{cis} \left(-\frac{\alpha}{2}\right)] = r^2 \cdot 1 = r^2\). From Vieta's, we have \(z_1 z_2 = q\). Thus, \(q = r^2\).

Step 7: Substitute \(r^2\) in terms of \(p\) and \(q\)

Substitute \(r = \frac{p}{2 \cos \frac{\alpha}{2}}\) into \(r^2 = q\): \[ \left(\frac{p}{2 \cos \frac{\alpha}{2}}\right)^2 = q \]This simplifies to \[ p^2 = 4q \cos^2 \frac{\alpha}{2} \].

Step 8: Identify the value of \(k\)

Comparing the result \( p^2 = 4q \cos^2 \frac{\alpha}{2} \) with \( p^2 = k \cos^2 \frac{\alpha}{2} \), we find that \(k = 4q\).

Key Concepts

Quadratic Equations

Quadratic equations are a central concept within mathematics, often introduced early in algebra. A quadratic equation is a second-degree polynomial expressed generally as \( ax^2 + bx + c = 0 \). Here, \( a \), \( b \), and \( c \) are coefficients, and \( x \) represents the variable.

In our problem, the quadratic equation is \( z^2 + pz + q = 0 \), with \( z \) being a complex number, and the coefficients \( p \) and \( q \) can also be complex. Thus, this generalizes the classic quadratic equation to the set of complex numbers, which requires a deeper understanding of complex solutions.

Solving quadratic equations involves finding the values of \( z \) that satisfy the equation. One powerful approach is using the quadratic formula:
\[ z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
In the realm of complex numbers, the discriminant \( b^2 - 4ac \) can take negative values, yielding complex roots that can provide insight into their geometric properties on the complex plane.

Vieta's Formulas

Vieta's formulas provide a fascinating link between the coefficients of a polynomial and its roots, allowing us to express the sum and product of its roots in terms of its coefficients. For a quadratic equation of the form \( z^2 + pz + q = 0 \), these formulas are defined as:

• Sum of roots: \( z_1 + z_2 = -p \)
• Product of roots: \( z_1 z_2 = q \)

In our problem, Vieta's provides the key relationships needed to connect the geometric configuration of the roots \( z_1 \) and \( z_2 \) with the coefficients \( p \) and \( q \). By using the sum of roots, we find an expression involving the angle \( \alpha \) that these roots form with the origin, and equate this with \( -p \). Similarly, by considering the product of roots, we establish a direct link to \( q \), allowing us to solve for complex coefficients effectively.

Polar Form Representation

The polar form is a useful representation for complex numbers that emphasizes their geometric aspects. A complex number \( z \) can be expressed as \( z = r \text{cis} \theta \), where \( r \) is the magnitude (distance from the origin), and \( \theta \) is the argument (the angle it makes with the positive real axis).

In this problem, since \( z_1 \) and \( z_2 \) are roots satisfying \( \angle AOB = \alpha \) and \( OA = OB \), they equidistantly lie on a circle centered at the origin. Thus, each root can be represented in polar form as:

• \( z_1 = r \text{cis} \frac{\alpha}{2} \)
• \( z_2 = r \text{cis} \left(-\frac{\alpha}{2}\right) \)

This representation simplifies computations involving angles and distances, making it particularly useful when dealing with geometric problems in the complex plane.

Roots of Complex Equations

Understanding the roots of complex equations requires familiarity with both algebraic and geometric properties. In complex equations, roots are not always real numbers and frequently lie in the complex plane, expressed as \( a + bi \) where \( i = \sqrt{-1} \).

For our problem, the roots \( z_1 \) and \( z_2 \) exhibit symmetric properties due to their geometric configuration \( \angle AOB = \alpha \) with equal radii from the origin. This symmetry plays a critical role in simplifying the relationships between the roots and coefficients.

By using the roots' polar form, we compute the sum and product of the roots, linking them back to the coefficients through Vieta's formulas. This approach enables the translation of geometric conditions, such as the angle and equidistance criteria, into algebraic equations. The conclusion, \( p^2 = 4q \cos^2 \frac{\alpha}{2} \), highlights how such an interplay of geometric intuition and algebraic manipulation can yield deep insights into the nature of complex roots.