Problem 29
Question
If \(\omega\) is a cube root of unity, then \((1+\omega-\) \(\left.\omega^{2}\right)\left(1-\omega+\omega^{2}\right)=\) (a) 1 (b) 0 (c) 2 (d) 4
Step-by-Step Solution
Verified Answer
(b) 0
1Step 1: Understanding Cube Roots of Unity
The cube roots of unity are the solutions to the equation \(x^3 = 1\). They can be represented as \(1, \omega, \omega^2\), where \(\omega\) is a primitive cube root of unity. These roots satisfy the property: \(\omega^3 = 1\), \(1 + \omega + \omega^2 = 0\), and \(\omega^2 = \omega + 1\).
2Step 2: Substitution and Simplification
Given the expression \((1 + \omega - \omega^2)(1 - \omega + \omega^2)\), we substitute \(\omega^2 = \omega + 1\) in both parts of the expression.Thus, it becomes \((1 + \omega - (\omega + 1))(1 - \omega + (\omega + 1))\), which simplifies to \(0 \times 1 = 0\).
3Step 3: Verification
Verify by simplifying each term separately: \(1 + \omega - \omega^2\) simplifies to \(1 + \omega - (\omega + 1) = 0\), and \(1 - \omega + \omega^2\) simplifies to \(1 - \omega + (\omega + 1) = 1\). Multiplying these results: \(0 \times 1 = 0\).
Key Concepts
Primitive RootsComplex NumbersPolynomial Equations
Primitive Roots
Primitive roots are crucial in understanding concepts such as roots of unity in mathematics. A primitive root of unity is a number that, when raised to a certain power (the order of the root), equals one. Specifically, for cube roots of unity, this involves solving a polynomial equation where the possible solutions are complex numbers. A primitive cube root of unity is the most basic root from which all other roots of unity can be derived. For cube roots, these are numbers that satisfy the equation:
- \(x^3 = 1\)
- Where the solutions are 1, \(\omega\), and \(\omega^2\).
- \(\omega^3 = 1\)
- \(1 + \omega + \omega^2 = 0\)
Complex Numbers
Complex numbers are an extension of the real numbers, featuring elements that are both real and imaginary components. Within the realm of complex numbers, we often encounter the concept of cube roots of unity. These appear predominantly in the study of polynomial equations. A complex number is generally expressed as \(a + bi\),
- where \(i\) is the imaginary unit \(\sqrt{-1}\).
- \(a\) is the real part, and \(b\) is the imaginary part.
Polynomial Equations
Polynomial equations form the backbone of algebra and involve terms raised to exponent powers and their sums. The degree of the polynomial is dictated by the highest power present in the equation. For cube roots of unity, we deal with third-degree polynomial equations. The general equation for cube roots of unity is: \(x^3 - 1 = 0\). Solving this specific polynomial involves finding the three roots that satisfy the equation. These roots are fundamentally important for forming expressions such as:
- \((1 + \omega - \omega^2)(1 - \omega + \omega^2)\)
- \(\omega^3 = 1\)
- which simplifies other expressions, verifying \(0 \times 1 = 0\).
Other exercises in this chapter
Problem 27
If \(n\) is a positive integer not a multiple of 3 , then \(1+\omega^{n}+\omega^{2 n}=\) (a) 3 (b) 1 (c) 0 (d) none of these
View solution Problem 28
If \(1, \omega, \omega^{2}\) are the three cube roots of unity, then \(\left(3+\omega^{2}+\omega^{4}\right)^{6}=\) (a) 64 (b) 729 (c) 2 (d) 0
View solution Problem 30
One of the cube roots of unity is (a) \(\frac{-1+i \sqrt{3}}{2}\) (b) \(\frac{1+i \sqrt{3}}{2}\) (c) \(\frac{1-i \sqrt{3}}{2}\) (d) \(\frac{\sqrt{3}-i}{2}\)
View solution Problem 31
If \(\omega\) is the cube root of unity, then \((3+5 \omega\) \(\left.+3 \omega^{2}\right)^{2}+\left(3+3 \omega+5 \omega^{2}\right)^{2}=\) (a) 4 (b) 0 (c) \(-4\
View solution