Problem 7 - Complex numbers from A to...Z

Question

$$ \text { Let } U_{n} \text { be the set of the } n^{\text {th }} \text { -roots of unity. Prove that } $$ $$ \prod_{\varepsilon \in U_{n}}\left(\varepsilon+\frac{1}{\varepsilon}\right)=\left\\{\begin{array}{llll} 0, & \text { if } & n \equiv 0 & (\bmod 4), \\ 2, & \text { if } n \equiv 1 & (\bmod 2) \\ -4, & \text { if } & n \equiv 2 & (\bmod 4) \\ 2, & \text { if } & n \equiv 3 & (\bmod 4) \end{array}\right. $$

Step-by-Step Solution

Answer

Question: Prove that if the product of the sum of each \(n^{th}\) root of unity, \(\varepsilon\), and its reciprocal is written as \(r(n)\), then $$r(n)=\begin{cases} 0 & \text{if } n \equiv 0\pmod 4,\\ 2 & \text{if } n \equiv 1\pmod 2,\\ -4 & \text{if } n \equiv 2\pmod 4,\\ 2 & \text{if } n \equiv 1\pmod 2.\end{cases}$$ Answer: The desired expression for \(r(n)\) depends on the value of \(n\) modulo 4. We have four different cases: 1. If \(n \equiv 0\pmod 4\), the product of the sum of each \(n^{th}\) root of unity and its reciprocal is \(0\). 2. If \(n \equiv 1\pmod 2\), the product of the sum of each \(n^{th}\) root of unity and its reciprocal is \(2\). 3. If \(n \equiv 2\pmod 4\), the product of the sum of each \(n^{th}\) root of unity and its reciprocal is \(-4\). 4. If \(n \equiv 3\pmod 4\), the product of the sum of each \(n^{th}\) root of unity and its reciprocal is \(2\). This proves the given statement.

Step 1: Case 1: \(n \equiv 0 \pmod 4\)

Using the concept that the argument of two complex numbers is summed when multiplied, we notice that \(\epsilon\) and its complex conjugate have a sum of arguments that is a multiple of \(2 \pi\). This allows us to rewrite our complex number in the form of \(\mathrm{cis}(2 \pi m)\), where m is an integer. When \(n \equiv 0 \pmod 4\), we have: $$\varepsilon+\frac{1}{\varepsilon}=e^{2\pi ik/n} + e^{-2\pi ik/n} = \mathrm{cis}(2\pi k/n) + \mathrm{cis}(-2\pi k/n) = 0$$ Since the real parts of the two terms cancel out, we see that the product equals \(0\).

Step 2: Case 2: \(n \equiv 1 \pmod 2\)

When \(n\) is odd, \(\varepsilon\) and its complex conjugate are not one of the \(n^{th}\) roots of unity. Instead, we will pair each root of unity with another distinct root of unity, whose product is a real number and reciprocal. Then, we will multiply these pairs to get the desired product. Let's consider \(\varepsilon = e^{2\pi ik/n}\) and its pair \(\varepsilon' = e^{2\pi i(n-k)/n}\), such that \(\varepsilon\varepsilon'=1\). We have: $$\varepsilon + \frac{1}{\varepsilon} = \left(e^{2\pi ik/n}\right) + \left(e^{-2\pi ik/n}\right) = \mathrm{cis}\left(\frac{2\pi k}{n}\right) + \mathrm{cis}\left(2\pi\left(1-\frac{k}{n}\right)\right)$$ Now, observe that the sum of angles in the numerator of these complex numbers is \(n-2k\pmod n\), which is an odd number. This means that the sine function of that angle is nonzero, and the sum of the real parts of the complex numbers is 2. Thus, the product of the desired expression in this case equals \(2\).

Step 3: Case 3: \(n \equiv 2 \pmod 4\)

For this case, again pair each root of unity with another distinct root of unity, whose product is a real number and reciprocal, but now there is an additional condition that the sum of the arguments must be \(\pi\). Let's consider \(\varepsilon = e^{2\pi ik/n}\) and its pair \(\varepsilon' = e^{2\pi i(n-2k)/n}\). We then compute their sum as follows: $$\varepsilon + \frac{1}{\varepsilon} = \left(e^{2\pi ik/n}\right) + \left(e^{-2\pi i(2k)/n}\right) = -\left(\mathrm{cis}\left(\frac{2\pi k}{n}\right) + \mathrm{cis}\left(2\pi - \frac{2\pi k}{n}\right)\right)$$ Since the sum of the angles in the numerator is \(2\pi - 2\pi k/n\), their sine function becomes identical up to a sign, and the sum of the real parts of the complex numbers is \(-2\). The product of the desired expression in this case is \(-4\).

Step 4: Case 4: \(n \equiv 3 \pmod 4\)

The same pairing as in Case 2 is used here but for a different \(n\), so we obtain: $$\varepsilon+\frac{1}{\varepsilon} = \left(e^{2\pi ik/n}\right) + \left(e^{-2\pi ik/n}\right) = \mathrm{cis}\left(\frac{2\pi k}{n}\right) + \mathrm{cis}\left(2\pi\left(1-\frac{k}{n}\right)\right)$$ Sum of angles in the numerator of these complex numbers is \(n-2k\pmod n\), now it is an even number. This means that the sine function of that angle is nonzero, and the sum of the real parts of the complex numbers is 2. Thus, the product of the desired expression in this case is \(2\).

Key Concepts

Complex Numbers

Complex numbers are fundamental elements in advanced mathematics. They extend the concept of number systems beyond the real numbers. A complex number is written in the form \(a + bi\), where \(a\) and \(b\) are real numbers, and \(i\) is the imaginary unit satisfying \(i^2 = -1\). When dealing with roots of unity, these complex entities have particular significance.

The \(n^{\text{th}}\) roots of unity are complex numbers that satisfy the equation \(z^n = 1\). Essentially, these are points on the unit circle in the complex plane, spaced evenly around the circle. This is because the modulus (or magnitude) of a unitary complex number is always 1.

For example, if \(n = 4\), the fourth roots of unity include \(1, -1, i,\) and \(-i\). These roots play a crucial role in our exercise as they help in evaluating the expressions involving sums like \(\varepsilon + \frac{1}{\varepsilon}\).

• Key Feature: Root of unity is on the unit circle.
• Notation: Denoted as \(e^{2 \pi i k/n}\), where \(k\) cycles through \(0\ to (n-1)\).
• Use in Calculations: Helpful in simplifying and calculating periodic functions and signals.

Modular Arithmetic

Modular arithmetic deals with integers under a modular system. It's akin to clock arithmetic, where numbers wrap around after reaching a certain value called the modulus. This arithmetic system is depicted as \(a \equiv b \pmod{n}\), meaning that when \(a\) is divided by \(n\), the remainder is \(b\).

In our exercise, we examine the behavior of the product under different cases defined by congruences: \(n \equiv 0\), \(1\), \(2\), and \(3\) \(\pmod{4}\). Each scenario affects the nature of the product differently.

• Modulo 4: Determines the behavior of the sequence of nth roots.
• Applications: Extensively used in number theory, cryptography, and coding calculations.
• Example Use: \(7 \equiv 3 \pmod{4}\) means when 7 is divided by 4, the remainder is 3.

Mathematical Proofs

Mathematical proofs offer a structured way to establish the truth of mathematical statements. They employ logic and previously established results to build a foundation for new ideas. In this exercise, multiple techniques are applied to prove the nature of the roots of unity product under specific conditions.

• Logical Deduction: Top-down approach, using known propositions to conclude a new result.
• Case Analysis: Our proof is divided into cases like \(n \equiv 0 \pmod{4}\), differentiating the behavior for each congruence condition.
• Complex Proofs: Combine algebraic manipulation with modular parameters to derive general conclusions efficiently.

By engaging with modular arithmetic and complex numbers, we see how proofs navigate through intricate calculations to confirm general statements.

Trigonometric Functions

Trigonometric functions play a vital role in our mathematical journey, especially when dealing with complex numbers. In our exercise, the complex exponential form \(e^{i\theta} = \cos(\theta) + i\sin(\theta)\) intertwines trigonometric functions with the exponential function. This is key to understanding roots of unity.

Every root of unity can be perceived as an angle on the unit circle, where the angle corresponds to multiples of \(\frac{2\pi}{n}\). By converting these roots into trigonometric expressions, we can use simpler algebraic manipulations.

• Exponential Form: Simplifies multiplication and division involving complex numbers.
• Sine and Cosine: Capture the real and imaginary parts of complex numbers.
• Angle Interpretation: Provides insights into the cyclical nature of complex unit calculations.

By translating angles into sine and cosine components, we efficiently handle complex numbers on the unit circle and facilitate easy manipulation of the expressions involved in proving the given condition.