Problem 58 - A Complete Resource Book in Mathematics for JEE Main

Question

Common roots of the equations \(z^{3}+2 z^{2}+2 z+1=0\) and \(z^{1985}+z^{100}+1=0\) are (A) \(\omega, \omega^{2}\) (B) \(1, \omega, \omega^{2}\) (C) \(-1, \omega, \omega^{2}\) (D) \(-\omega,-\omega^{2}\)

Step-by-Step Solution

Answer

Option (A) \( \omega, \omega^2 \).

Step 1: Understand the Problem

We are asked to find common roots of two equations: 1. \( z^3 + 2z^2 + 2z + 1 = 0 \) 2. \( z^{1985} + z^{100} + 1 = 0 \)This means we need to find values of \( z \) that satisfy both these equations.

Step 2: Solve the First Equation

We start with \( z^3 + 2z^2 + 2z + 1 = 0 \). To find the roots, use a known root theorem or check for obvious roots such as -1 or cube roots of unity. Since the problem hints at roots involving unity (\( \omega \)), test \( z = \omega \) and \( z = \omega^2 \), where \( \omega \) is a cube root of unity \( \omega = e^{2\pi i/3} \). In fact, both \( \omega \) and \( \omega^2 \) satisfy this equation.

Step 3: Solve the Second Equation

For the second equation \( z^{1985} + z^{100} + 1 = 0 \), substitute the cube roots of unity:The properties of cube roots of unity tell us:\[ \omega^3 = 1 \quad \text{and} \quad \omega^2 + \omega + 1 = 0. \]Therefore, we have the equations:- \( \omega^{1985} = \omega^{3 \times 661 + 2} = \omega^2 \) because \( \omega^3 = 1 \).- \( \omega^{100} = \omegae 1 \).Substituting \( \omega^2 \) and \( \omega \) in, we find both satisfy \( z^{1985} + z^{100} + 1 = 0 \).

Step 4: Compare the Solutions

The common roots which solve both equations are \( \omega \) and \( \omega^2 \). Since these satisfy both equations, no other root like -1 appears in both, further confirming \( \omega \) and \( \omega^2 \) are the correct solutions.

Step 5: Choose the Correct Option

Based on the solutions obtained, the common roots are \( \omega \) and \( \omega^2 \). Therefore, the correct option is (A) \( \omega, \omega^2 \).

Key Concepts

Cube Roots of Unity

Complex numbers can sometimes seem a bit daunting, but when you break them down, they are not so scary. One fascinating part of complex numbers is the "cube roots of unity." They are the solutions to the equation \[ z^3 = 1. \]These solutions are typically written as \( 1 \), \( \omega \), and \( \omega^2 \), where \( \omega = e^{2\pi i/3} \) is a complex number with special symmetry properties.

• \( 1 \) is one of the cube roots, which is straightforward as it satisfies the equation since \( 1^3 = 1 \).
• \( \omega = e^{2\pi i/3} \), a complex number on the unit circle in the complex plane, rotates by 120 degrees.
• \( \omega^2 = e^{4\pi i/3} \) is another root which is simply \( \omega \) squared.

These roots of unity give rise to useful properties:

• They are evenly spaced on the unit circle in the complex plane.
• They satisfy \( \omega^3 = 1 \) and \( \omega^2 + \omega + 1 = 0 \).

Understanding cube roots of unity can make solving polynomial equations much simpler by leveraging their symmetrical and cyclical properties.

Polynomial Equations

Polynomial equations are algebraic expressions set equal to zero and involve a sum of terms featuring different powers of a variable. For instance, the equations like \[ z^3 + 2z^2 + 2z + 1 = 0 \] and \[ z^{1985} + z^{100} + 1 = 0 \] are examples of polynomial equations.

• The degree of a polynomial is the highest power in the expression. In our example, the first equation has degree 3, the second equation has degree 1985, represented by the exponent on \( z \).
• Solving polynomial equations involves finding values of the variable (called roots) that make the equation true, meaning they are solutions.

For the first polynomial, using techniques like factoring or known mathematical theorems can help in deducing these roots. The roots have significant implications in mathematics, such as in understanding the behavior of functions, pioneering fields like calculus, and more advanced areas like differential equations or dynamical systems.
In our exercise, by identifying the presence of cube roots of unity, the task of finding common roots becomes simpler. This shows how algebra interconnects with complex numbers to solve equations effectively.

Algebraic Equations

Algebraic equations are at the heart of many mathematical problems and solutions. These are equations that involve only algebraic terms, such as sums, differences, multiplications, divisions, and integer exponents. Examples include basic equations like \[ x + 3 = 7 \] as well as more challenging ones like \[ z^3 + 2z^2+2z+1 = 0. \]
In algebra, we often encounter the need to find the unknown quantity (or quantities) that satisfy these equations. Here is how we generally approach algebraic equations:

• Identify and isolate the variable, which is the symbol representing an unknown.
• Perform operations on both sides of the equation to simplify or modify it, while maintaining the equality.
• Factor or use identities to simplify higher degree polynomial or algebraic equations.

The significance of solving algebraic equations lies in their universal application, from physics to economics, providing solutions that define functions or model real-world phenomena. In the exercise above, understanding the structure of the two given equations and the notion of common roots using algebraic manipulation aids us in pinpointing the shared solutions, \( \omega \) and \( \omega^2 \). This reveals how algebraic techniques interlock with the rest of mathematics to provide clarity and solutions.