Cyclotomic Polynomials & General N-th Roots of Complex Numbers

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Cyclotomic Polynomials

The n-th cyclotomic polynomial $\Phi_n(x)$ is the minimal polynomial whose roots are exactly the primitive n-th roots of unity: $\Phi_n(x) = \prod_{\substack{1 \leq k \leq n \\ \gcd(k,n) = 1}} \left(x - e^{\frac{2\pi i k}{n}}\right)$. This polynomial has integer coefficients and is irreducible over $\mathbb{Q}$. The degree $\deg(\Phi_n)(x)$ is $\deg(\Phi_n) = \phi(n)$ where $\phi(n)$ is Euler’s totient function.

$\Phi_n(x)$ is the polynomial created by multiplying $(x - \text{root})$ for only the primitive roots. Because there are $\phi(n)$ primitive roots, the degree of $\Phi_n(x)$ is exactly $\phi(n)$

Example ($n=3$): Roots of $x^3 - 1 = 0$ are $1, e^{i2\pi/3}, e^{i4\pi/3}$.

• 1 is not primitive.
• $e^{i2\pi/3}$ and $e^{i4\pi/3}$ are primitive. So, $\Phi_3(x) = (x - e^{i2\pi/3})(x - e^{i4\pi/3})$. Expanding this gives $x^2 + x + 1$.

The Fundamental Relation. The product of Cyclotomic polynomials for all divisors d of n equals $x^n - 1$. In other words, the factorization of $x^n -1$ is $x^n - 1 = \prod_{d \mid n} \Phi_d(x)$ where $\Phi_d(x)$ is the d-th cyclotomic polynomial.

Proof.

1. The polynomial $x^n - 1$ has exactly $n$ roots, namely the n-th roots of unity: $\zeta_n^k=e^{2\pi ik/n}, k=0, 1, \dots, n-1.$ We can write it or group the roots as: $x^n - 1 = \prod_{k=1}^n (x - z_k)$.
2. Every root $z$ has a well-defined “order” $d$. The order is the smallest positive integer d such that $z^d = 1$. This order d must be a divisor of n.
   Proof. Since $z^n = 1$ and d is the smallest positive integer d such that $z^d = 1$, we can write n = qd + r with 0 ≤ r < d. Then, $1 = z^n = z^{qd+r} = (z^d)^q\cdot z^r = 1^q\cdot z^r = z^r$. Since d is minimal, we must have r = 0. Hence, d divides n.
   Explicit Formula for the Order: $\zeta_n^k=e^{2\pi ik/n}, k=0, 1, \dots, n-1;$, the order is $d = \frac{n}{gcd(n, k)}$.
3. If a root $z$ has order $d$, it means $z$ is a primitive $d$-th root of unity. Therefore, it is a root of $\Phi_d(x)$.
   Note. In particular, z is a primitive n-th root of unity exactly when gcd(k, n) = 1.
   

   By definition, the cyclotomic polynomial $\Phi_d(x)$ is the monic polynomial whose roots are precisely the primitive d-th roots of unity.

4. Every root of $x^n - 1$ belongs to exactly one (different orders correspond to disjoint sets of roots) $\Phi_d(x)$ where $d$ divides $n$. Thus, multiplying all $\Phi_d(x)$ together reconstructs $x^n - 1$.

   The product $\prod_{d\mid n}\Phi_d(x)$ is a monic polynomial whose roots are exactly the roots of $x^n-1$, with the same multiplicities (all simple). Both polynomials are monic and have degree n: $\deg \left( \prod_{d\mid n}\Phi_d(x)\right) = \sum_{d\mid n}\phi(d) = n =\deg (x^n-1).$. Two monic polynomials with the same roots and the same degree must be equal.

Computing Cyclotomic Polynomials

Find $\Phi_6(x)$. We typically use the relation: $\Phi_n(x) = \frac{x^n - 1}{\prod_{d \mid n, d < n} \Phi_d(x)}$

Divisors of 6: 1, 2, 3, 6

1. n = 1. Divisors: 1. $x^1 - 1 = \Phi_1(x)$. $\Phi_1(x) = x - 1$.
2. n = 2. Divisors: 1, 2. $x^2 - 1 = \Phi_1(x) \cdot \Phi_2(x)$. $\Phi_2(x) = \frac{x^2 - 1}{\Phi_1(x)} = \frac{(x-1)(x+1)}{x-1} = x+1$
3. n = 3. Divisors: 1, 3. $x^3 - 1 = \Phi_1(x) \cdot \Phi_3(x)$. $\Phi_3(x) = \frac{x^3 - 1}{x - 1} = x^2 + x + 1$
4. n = 4. Divisors: 1, 2, 4. $x^4 - 1 = \Phi_1(x) \Phi_2(x) \Phi_4(x)$. We already know $\Phi_1(x)\Phi_2(x) = x^2 - 1$, so $\Phi_4(x) = \frac{x^4 - 1}{x^2 - 1} = x^2 + 1$.
5. n = 6. Divisors: 1, 2, 3, 6. $x^6 - 1 = \Phi_1(x)\Phi_2(x)\Phi_3(x)\Phi_6(x)$. $\Phi_1(x) \Phi_2(x) \Phi_3(x) = (x-1)(x+1)(x^2+x+1) = (x^2-1)(x^2+x+1)$. $\Phi_6(x) = \frac{x^6 - 1}{(x^2-1)(x^2+x+1)} = \frac{(x^3-1)(x^3+1)}{(x^3-1)(x+1)} = \frac{x^3+1}{x+1} = x^2 - x + 1$

Roots of an Arbitrary Complex Number

Given a complex number w = se^iβ, s = |w| > 0, β = Arg(w), we want to find the solutions of the equation zⁿ = w, where z = re^iθ.

zⁿ = (re^iθ)ⁿ = rⁿ(e^iθ)ⁿ = [De Moivre’s Theorem] rⁿe^inθ

Our goal is to identify all possible z that satisfies: se^iβ = rⁿe^inθ$.

1. Matching moduli: s = rⁿ ⟷[s = |w|, hence s >= 0] $r = \sqrt[n]{s} = |w|^{1/n}$
2. Matching arguments: e^iβ = e^inθ, nθ-β = 2πk ⇒ θ = $\frac{β + 2πk}{n}$ for some k ∈ ℤ

The standard formula for finding the $n$-th roots of a complex number $w$ is $\boxed{z_k = \sqrt[n]{|w|} \cdot e^{i\left(\frac{\text{Arg}(w) + 2\pi k}{n}\right)}, \quad k = 0, 1, \ldots, n-1}$

Alternative Form Using Roots of Unity

In the exponent, we have a fraction $\frac{A + B}{n}$. We can split this into $\frac{A}{n} + \frac{B}{n}$, and we get $z_k = \sqrt[n]{|w|} \cdot e^{i\left(\frac{\text{Arg}(w)}{n} + \frac{2\pi k}{n}\right)}$

Apply the rule of exponents ($x^{a+b} = x^a \cdot x^b$) to split the term into two separate factors: $z_k = \underbrace{\sqrt[n]{|w|} \cdot e^{i\left(\frac{\text{Arg}(w)}{n}\right)}}_{\text{Part A}} \cdot \underbrace{e^{i\left(\frac{2\pi k}{n}\right)}}_{\text{Part B}}$. Next, using the power rule $(x^a)^b = x^{ab}$, we can pull the $k$ outside the parenthesis, $e^{i\left(\frac{2\pi k}{n}\right)} = \left( e^{i\frac{2\pi}{n}} \right)^k$

Define the specific terms:

1. The Principal Root ($\sqrt[n]{w}$). This is Part A. It corresponds to the root where k = 0. The principal nth root of w, denoted $\sqrt[n]{w}$, is defined as: $\sqrt[n]{w} = \sqrt[n]{|w|}e^{i(\frac{Arg(w)}{n})}$ where $\text{Arg}(w) \in (-\pi, \pi]$ is the principal argument.
2. The Principal Root of Unity ($\omega_n = \zeta_n=e^{\frac{2πi}{n}}$ i.e., a complex number satisfying $\omega_n^n = 1$ and $w_n^k \ne 1 \text{ for } 1 ≤ k < n$). This is the base of Part B. It is a specific complex number of length 1.
3. Substitute these new definitions back into the previous equation: $z_k = \left( \sqrt[n]{|w|} \cdot e^{i\frac{\text{Arg}(w)}{n}} \right) \cdot \left( e^{i\frac{2\pi}{n}} \right)^k = \sqrt[n]{w} \cdot (w_n)^k$

Conclusion: $\boxed{z_k = \sqrt[n]{w}\cdot w_n^k}$ where 0 ≤ k ≤ n-1

Definition. The set of all nth roots of a complex number w, denoted $w^{\frac{1}{n}}$, consists of n distinct values. It is given by: $w^{\frac{1}{n}}$ = {$\sqrt[n]{w}, w_n\sqrt[n]{w}, w_n²\sqrt[n]{w}, ..., w_n^{n-1}\sqrt[n]{w}$} where:

• |w| is the magnitude (modulus) and $\text{Arg}(w)$ is the argument (angle) of w, $\text{Arg}(w) ∈ (−π,π].$.
• $w_n=ζ_n=e^{\frac{2πi}{n}}$ is a primitive nth root of unity.
• The index k ranges over integers from 0 to n − 1. Using k = 0, 1, … , n−1 ensures exactly n distinct roots.
• The starting point is $\sqrt[n]{w}$. You find the very first root (the principal root) by taking the $n$-th root of the length ($|w|$) and dividing the original angle by $n$.
• The term $w_n = e^{i \frac{2\pi}{n}}$ is a pure rotation. Multiplying by $w_n$ rotates a number by $\frac{2\pi}{n}$ radians (which is $\frac{360^\circ}{n}$). In other words, these roots are obtained by adding multiples of $\frac{2\pi}{n}$ to the argument of the principal root. Each root corresponds to rotating the principal nth root by $\frac{2\pi}{n}$ radians around the origin. This is why the roots form a regular polygon.

Examples

• Cube Roots of 8.

1. Express 8 in polar form: $8 = 8e^{i \cdot 0}$ (modulus r = 8, argument θ = 0).
2. Apply the formula: $z_k = \sqrt[3]{8} \cdot e^{i\frac{0 + 2\pi k}{3}} = 2e^{i\frac{2\pi k}{3}}$ for k = 0, 1, 2.
3. Calculate for $k = 0, 1, 2$: (i) k = 0, $2e^{i \cdot 0} = 2$, k = 1, $2e^{i\frac{2\pi}{3}} = -1 + i\sqrt{3}$, k = 2, $2e^{i\frac{4\pi}{3}} = -1 - i\sqrt{3}$.
4. The cube roots are 2, $-1 + i\sqrt{3}, -1 - i\sqrt{3}$

• Fifth Roots of $1 + i$.

1. Express $1 + i$ in polar form: $|1 + i| = \sqrt{2}, \quad \text{Arg}(1+i) = \frac{\pi}{4}$, so $1 + i = \sqrt{2}e^{i\frac{\pi}{4}}$
2. Roots formula: $z_k = \sqrt[5]{\sqrt{2}} \cdot e^{i\frac{\frac{\pi}{4} + 2\pi k}{5}} = 2^{1/10} \cdot e^{i\frac{\pi(1 + 8k)}{20}}$, $2^{1/10} = \sqrt[10]{2} \approx 1.072$.
3. Angles for k = 0, 1, 2, 3, 4: k = 0, $2^{1/10}\cdot \frac{\pi}{20}$, k = 1, $2^{1/10}\cdot \frac{9\pi}{20}$, k = 2, $2^{1/10}\cdot \frac{17\pi}{20}$, k = 3, $2^{1/10}\cdot \frac{25\pi}{20} = 2^{1/10}\cdot \frac{5\pi}{4}$, k = 4, $2^{1/10}\cdot \frac{33\pi}{20}$
4. The fifth roots are $2^{1/10}\cdot \frac{\pi}{20}, 2^{1/10}\cdot \frac{9\pi}{20}, 2^{1/10}\cdot \frac{17\pi}{20}, 2^{1/10}\cdot \frac{5\pi}{4}, 2^{1/10}\cdot \frac{33\pi}{20}$