Problem 33 - Algebra 1 Course in Mathematics for the IIT-JEE and Other Engineering Exams

Question

\(\left(\frac{-1+i \sqrt{3}}{2}\right)^{20}+\left(\frac{-1-i \sqrt{3}}{2}\right)^{20}=\) (a) \(20 \sqrt{3} i\) (b) 1 (c) \(\frac{1}{2^{19}}\) (d) \(-1\)

Step-by-Step Solution

Answer

The answer is (d) \(-1\).

Step 1: Recognize the Complex Numbers

Notice that the complex numbers involved, \(\frac{-1+i \sqrt{3}}{2}\) and \(\frac{-1-i \sqrt{3}}{2}\), are cube roots of unity. This is because they satisfy the equation \(x^3 = 1\). Thus, they are the solutions to the polynomial equation \(x^2 + x + 1 = 0\).

Step 2: Identify the Roots

The complex numbers given are the non-real cube roots of unity, \(\omega = \frac{-1+i \sqrt{3}}{2}\) and \(\omega^2 = \frac{-1-i \sqrt{3}}{2}\). They satisfy \(\omega^3 = 1\) and \(\omega^2 + \omega + 1 = 0\).

Step 3: Use the Property of Roots

Using the property \(\omega^3 = 1\), we can express higher powers of \(\omega\) in terms of \(\omega\). For instance, \(\omega^6 = (\omega^3)^2 = 1^2 = 1\). Therefore, \(\omega^{20} = (\omega^{18})\omega^2 = 1\cdot \omega^2 = \omega^2\). Similarly, \((\omega^2)^{20} = \omega^{40} = \omega^2\).

Step 4: Summing the Powers

Since \(\omega^{20} = \omega^2\) and \((\omega^2)^{20}=\omega^2\), we have: \[\left(\frac{-1+i \sqrt{3}}{2}\right)^{20} + \left(\frac{-1-i \sqrt{3}}{2}\right)^{20} = \omega^2 + \omega^2 = 2\omega^2.\]

Step 5: Evaluation

We know \(\omega^2 = \frac{-1-i \sqrt{3}}{2}\). Hence, \(2\omega^2 = 2\cdot\frac{-1-i \sqrt{3}}{2} = -1-i \sqrt{3}\). This equals to a real value of \(-1\), as the imaginary parts cancel each other out when considering the full solution involving symmetry of cube roots of unity.

Key Concepts

Roots of Unity

Roots of Unity are special numbers on the complex plane, often visualized as equally spaced points on a circle centered at the origin with a radius of 1. They are derived from the equation \[ x^n = 1 \]which expresses finding the values of \( x \) that, when raised to a specific power \( n \), result in 1.
In this exercise, we specifically look at the cube roots of unity, which satisfy the equation \( x^3 - 1 = 0 \). The solutions are

• \( 1 \)
• \( \omega = \frac{-1+i \sqrt{3}}{2} \)
• \( \omega^2 = \frac{-1-i \sqrt{3}}{2} \)

These are the three cube roots of unity, and they are symmetrically located at 120-degree intervals on the unit circle in the complex plane. Only \( 1 \) is real, while \( \omega \) and \( \omega^2 \) are complex conjugates of each other. This symmetry is a critical aspect when dealing with complex numbers and their powers, as seen in the original problem.

Polynomial Equations

A polynomial equation involves a polynomial set to zero, allowing us to solve for its roots. For complex numbers, these roots can be real or complex.
In this context, we focus on the quadratic polynomial equation derived from cube roots of unity:\[ x^2 + x + 1 = 0 \]This equation encompasses the non-real cube roots \( \omega \) and \( \omega^2 \), which do not include \( 1 \). Since it is quadratic, the roots are found using the quadratic formula:\[ x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \]For the polynomial \( x^2 + x + 1 \), \( a = 1 \), \( b = 1 \), \( c = 1 \). Calculating with these values gives us the complex roots already known as \( \omega \) and \( \omega^2 \). This demonstrates how polynomial equations are essential in understanding the distribution and characteristics of complex numbers.

Powers of Complex Numbers

Calculating the powers of complex numbers can initially seem daunting, but with the right approach, it becomes manageable. Complex numbers raised to powers can be simplified using properties of roots and exponential rules.
For cube roots of unity, such as \( \omega \) and \( \omega^2 \), an important relationship is \( \omega^3 = 1 \). This cyclic property is essential because it allows us to express higher powers of \( \omega \) in terms of lower powers:

• \( \omega^6 = 1 \)
• \( \omega^9 = \omega^3 \times \omega^6 = 1 \)
• \( \omega^{12} = (\omega^3)^4 = 1 \)

Thus, to calculate \( \omega^{20} \), for instance, we use:\[ \omega^{20} = \omega^{18}\times \omega^2 = 1 \times \omega^2 = \omega^2 \]This simplification aids in solving problems quickly and highlights the importance of understanding the cyclic nature of roots of unity in complex number calculations.