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The n-th roots of unity are the complex solution to the equation: xn = 1. This equation asks for all complex numbers z such that raising z to the power n yields 1. This can be approached using Euler’s formula, which connects complex exponentials with trigonometric functions.
In the real number system, the only solution to this equation is x = 1, since 1 is the only real number that, when raised to any power, results in 1.
However, in the complex domain, there are n distinct solutions to this equation, distributed symmetrically on the unit circle (the set of points in the complex plane at a distance of 1 from the origin). These roots are known as the n-th roots of unity.
The number 1 can be represented using Euler’s formula by choosing θ = 0: 1 = $e^{i·0} = cos(0)+ isin(0) = 1 + i·0$. The complex equation zn = 1 can be rewritten using exponential form: $z^n = e^{i·0} = e^{i·2πk}$ for any integer k because adding multiples of 2π to the angle 0 does not change the value of eiθ due to the periodicity of sine and cosine functions.
$z^n = e^{i·2πk}$ ⇒[Solving for z:] $z = e^{\frac{i·2πk}{n}}$ for k = 0, 1, 2, ···, n -1. There are n distinct solutions corresponding to k = 0 through k = n-1. Using Euler’s formula, each root can be expressed as: $z_k = cos(\frac{i·2πk}{n})+isin(\frac{i·2πk}{n})$ where k = 0, 1, 2, ···, n -1.
Conclusion: The n-th roots of unity are the solution to the equation zn = 1, where z is a complex number. These roots are given by the formula: $z_k = e^{\frac{2πik}{n}} = cos(\frac{2πk}{n}) + isin(\frac{2πk}{n})$ where k = 0, 1, 2, ···, n-1 and:
k = 0, $z_0 = e^{\frac{2πi·0}{2}}=e^0=1$. k = 1, $z_1 = w_2 = e^{\frac{2πi·1}{2}}=e^{πi} = cos(π)+isin(π) = -1 + i·0 = -1.$
The square roots of unity are 1 and −1, lying on opposite ends of the real axis on the unit circle.
n = 3, k = 0, $z_0 = e^{\frac{2πi·0}{3}}=e^0=1, z_1 = w_3 = e^{\frac{2πi}{3}} = cos(\frac{2πi}{3}i)+isin(\frac{2πi}{3}i), w_3² = e^{\frac{4πi}{3}}$
k = 0, $z_0 = e^{\frac{2πi·0}{4}} = e^0 = 1$. k = 1, $z_1 = w_3 = e^{\frac{2πi·1}{4}} = e^{\frac{πi}{2}} = cos(\frac{π}{2})+isin(\frac{π}{2}) = 0 + i·1 = i$. k = 2, $z_2 = w_3² = e^{\frac{2πi·2}{4}}=e^{πi} = cos(π)+isin(π) = -1 + i·0 = -1$. k = 3, $z_3 = w_3³ = e^{\frac{2πi·3}{4}} = e^{\frac{3πi}{2}} = cos(\frac{3π}{2})+isin(\frac{3π}{2}) = 0 + i·(-1) = -i$
The fourth roots of unity are 1, i, −1, and -i, which are positioned at equal angles of π⁄2 radians around the unit circle.
In the complex plane, the n-th roots of unity are arranged as the vertices of a regular n-sided polygon inscribed in the unit circle. Each vertex corresponds to one of the roots, and the angle between any two consecutive roots is $\frac{2π}{n}$ radians (Refer to Figure iv for a visual representation and aid in understanding it).
Properties of N-th Roots of Unity:
This property holds because the roots are symmetrically distributed around the origin in the complex plane, so their vector sum cancels out.
In modular arithmetic, the integers coprime (relatively prime) to n from the set { 0 , 1 , … , n − 1 } of n non-negative integers form a group under multiplication modulo n, called the multiplicative group of integers modulo n. This group is usually denoted as $\mathbb{Z}/n\mathbb{Z}$.
The set of complex numbers {$1,ζ_n,ζ_n^2, \ldots ,ζ_n^{n−1 }$} where $ζ_n = w_n = e^{\frac{2\pi i}{n} }$ and $\zeta_n^n = 1$ is cyclic, with $ζ_n$ as a generator.
There is an isomorphism between $\mathbb{Z}/n\mathbb{Z}$ and the group of nth roots of unity. Let $ϕ:(Z/nZ)^*→{ζ_n^k∣k∈Z}$ be defined by ϕ(a) = $\zeta_n^a$, e.g., $\phi(ab) = \zeta_n^{ab} = \zeta_n^a \cdot \zeta_n^b = \phi(a) \cdot \phi(b)$ (homomorphism)
Given a complex number w = seiβ, β = Arg(w), we are looking at the solution set of the equation zn = w, where z = reiθ.
zn = (reiθ)n = rn(eiθ)n = [De Moivre’s Theorem] rneinθ
Our goal is to identify all possible z that satisfies: seiβ = rneinθ ⟷ s = rn, eiβ = einθ ⟷[s = |w|, hence s >= 0] $r = \sqrt{s}, nθ-β = 2πk ⇒ θ = \frac{β + 2πk}{n}$ for some k ∈ ℤ
If w = zn, z ∈ $\sqrt[\frac{1}{n}]{w}$ (n roots of w)
We have already study that for this equation to hole, z = $\sqrt[n]{s}e^{i(\frac{β + 2πk}{n})}$ for some k ∈ ℤ
Next, considering that $e^{i(\frac{β + 2πk}{n})} = e^{i(\frac{β}{in})}e^{i(\frac{2πk}{n})}$ = [De Moivre’s Theorem] $e^{i(\frac{β}{n})}(e^{i(\frac{2π}{n})})^k$ = [Notation, the principal root of unity, $w_n=ζ_n=e^{\frac{2πi}{n}}$] $e^{i(\frac{Arg(w)}{n})}w_n^k$
z = $\sqrt[n]{s}e^{i(\frac{β + 2πk}{n})} = \sqrt[n]{|w|}e^{i(\frac{Arg(w)}{n})}w_n^k$ where 0 ≤ k ≤ n-1
Definition. The principal nth root of w can be given by: $\sqrt[n]{w} = \sqrt[n]{|w|}e^{i(\frac{Arg(w)}{n})}$ where $w^{\frac{1}{n}} = ${$\sqrt[n]{w}, w_n\sqrt[n]{w}, w_n²\sqrt[n]{w}, ..., w_n^{n-1}\sqrt[n]{w}$} is the set of nth roots of a complex number w = $|w| e^{i \text{Arg}(w)}$, where |w| is the magnitude (modulus) and $\text{Arg}(w)$ is the argument (angle) of w.
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.