Problem 95 - Algebra and Trigonometry

Question

Products of Roots of Unity Find the product of the three cube roots of 1 (see Exercise 93\() .\) Do the same for the fourth, fifth, sixth, and eighth roots of \(1 .\) What do you think is the product of the \(n\) th roots of \(1,\) for any \(n ?\)

Step-by-Step Solution

Answer

The product of the \(n\)th roots of unity is always \(1\).

Step 1: Understand Roots of Unity

The cube roots of unity are the solutions to the equation \(z^3 = 1\). They are \(1, \omega, \omega^2\) where \(\omega = e^{2\pi i / 3}\) and \(\omega^2 = e^{-2\pi i / 3}\).

Step 2: Calculate Product of Cube Roots

To find the product of the cube roots of 1, multiply all roots: \(1 \times \omega \times \omega^2\). Using the property \(\omega^3 = 1\) and \(\omega^2 = 1/\omega\), we find the product is \(1\).

Step 3: Fourth Roots of Unity

The fourth roots of unity are the solutions \(z^4 = 1\), resulting in \(1, i, -1, -i\). The product is \(1 \times i \times -1 \times -i = 1\).

Step 4: Fifth Roots of Unity

The fifth roots of unity are solutions to \(z^5 = 1\), which are \(1, \zeta, \zeta^2, \zeta^3, \zeta^4\), where \(\zeta = e^{2\pi i / 5}\). The product is 1 because it forms a geometric sequence and \(z^n = 1\).

Step 5: Sixth Roots of Unity

The sixth roots of unity are the solutions to \(z^6 = 1\), which are \(1, \omega, \omega^2, -1, \omega^3, \omega^4\). The product is \(1\), as negative terms cancel and others form a full cycle.

Step 6: Eighth Roots of Unity

The eighth roots of unity solve \(z^8 = 1\), with roots \(1, \omega, \omega^2, \ldots, \omega^7\). The product of these roots circles back to \(1\) by symmetry and property of unity.

Step 7: Generalize for nth Roots

The \(n\)th roots of unity are the solutions to \(z^n = 1\) and include \(1, \zeta, \zeta^2, ..., \zeta^{n-1}\), where \(\zeta = e^{2\pi i / n}\). The product forms a complete cycle back to 1 due to symmetry.

Key Concepts

Cube Roots of Unity

The cube roots of unity are the numbers that solve the equation \( z^3 = 1 \). These roots can be represented as \( 1, \omega, \omega^2 \). Here, \( \omega = e^{2\pi i / 3} \) and \( \omega^2 = e^{-2\pi i / 3} \), which are complex numbers. These numbers divide the complex unit circle into three equal parts. When we multiply these cube roots together, we find that the product is \( 1 \).

This is due to a special property: \( \omega^3 = 1 \) and subsequently \( (\omega)(\omega^2) = 1 \). The cube roots of unity are notable for their symmetrical placement on the complex plane. This symmetry makes calculations like their products predictable and particularly elegant.

Nth Roots of Unity

For the general case of the \( n \)th roots of unity, these are the solutions to the equation \( z^n = 1 \). Such roots can be expressed as \( 1, \zeta, \zeta^2, \ldots, \zeta^{n-1} \), where \( \zeta = e^{2\pi i / n} \) and is known as a primitive \( n \)th root of unity. These roots are spaced evenly around the unit circle in the complex plane.

Just like with cube roots, these roots form a complete circle of symmetry and always include 1. A crucial property of the \( n \)th roots of unity is that their product equals \( 1 \). This is because multiplying all these roots together leads back to unity, as each pair of complex conjugate roots cancels out to 1, and the remaining products also revolve symmetrically back to 1.

Product of Roots of Unity

One of the fascinating aspects of roots of unity is their ability to multiply into a simple and elegant result. The product of all \( n \)th roots of unity is always \( 1 \). This stems from the symmetrical nature of their positions on the unit circle in the complex plane.

• Each root can be paired with its conjugate, balancing out to unity.
• The uniform distribution across the circle ensures all angular components sum up to integer multiples of \( 2\pi \), leading back to \(1\).

The special property that the product of roots loops back to \(1\) illustrates the beauty and balance inherent in mathematical symmetry. This principle holds not only for cube roots but for all \( n \)th roots regardless of \( n \), affirming the orderly structure of complex numbers.