De Moivre’s Theorem, Roots of Unity & Trigonometric Expansions

1. Statement of De Moivre's Theorem

Abraham de Moivre established one of the foundational bridges between complex analysis and trigonometry: $$\mathbf{(\cos\theta + i\sin\theta)^n = \cos(n\theta) + i\sin(n\theta)}$$ In compact Euler notation, this expresses the exponential property $(e^{i\theta})^n = e^{in\theta}$.

2. Rigorous Proof for Positive Integers ($n \in \mathbb{N}$)

We prove the theorem by the Principle of Mathematical Induction on $n$:
Base Step ($n = 1$): $(\cos\theta + i\sin\theta)^1 = \cos(1\cdot\theta) + i\sin(1\cdot\theta)$, which is trivially true.
Inductive Hypothesis: Assume the identity holds for some positive integer $k \ge 1$: $$(\cos\theta + i\sin\theta)^k = \cos(k\theta) + i\sin(k\theta)$$ Inductive Step: Consider $n = k + 1$: $$(\cos\theta + i\sin\theta)^{k+1} = (\cos\theta + i\sin\theta)^k \cdot (\cos\theta + i\sin\theta)$$ Using the inductive hypothesis: $$= [\cos(k\theta) + i\sin(k\theta)] [\cos\theta + i\sin\theta]$$ Expanding the complex multiplication: $$= [\cos(k\theta)\cos\theta - \sin(k\theta)\sin\theta] + i[\sin(k\theta)\cos\theta + \cos(k\theta)\sin\theta]$$ Applying standard trigonometric angle-addition identities: $$\cos(k\theta)\cos\theta - \sin(k\theta)\sin\theta = \cos((k+1)\theta)$$ $$\sin(k\theta)\cos\theta + \cos(k\theta)\sin\theta = \sin((k+1)\theta)$$ Thus: $$(\cos\theta + i\sin\theta)^{k+1} = \cos((k+1)\theta) + i\sin((k+1)\theta)$$ By induction, the theorem is proved for all $n \in \mathbb{N}$. $\blacksquare$

3. Extension to Negative Integers and Rational Powers

Case $n = 0$: $(\cos\theta + i\sin\theta)^0 = 1 = \cos 0 + i\sin 0$.
Case $n = -m$ where $m \in \mathbb{N}$: $$(\cos\theta + i\sin\theta)^{-m} = \frac{1}{(\cos\theta + i\sin\theta)^m} = \frac{1}{\cos(m\theta) + i\sin(m\theta)}$$ Multiplying numerator and denominator by $\cos(m\theta) - i\sin(m\theta)$: $$= \frac{\cos(m\theta) - i\sin(m\theta)}{\cos^2(m\theta) + \sin^2(m\theta)} = \cos(-m\theta) + i\sin(-m\theta) = \cos(n\theta) + i\sin(n\theta)$$ Rational Powers $n = p/q$ ($q > 0$): One of the values of $(\cos\theta + i\sin\theta)^{p/q}$ is $\cos(p\theta/q) + i\sin(p\theta/q)$, with exactly $q$ distinct complex values given by $\cos\left(\frac{p\theta + 2k\pi}{q}\right) + i\sin\left(\frac{p\theta + 2k\pi}{q}\right)$ for $k = 0, 1, \dots, q-1$.

1. Binomial Expansion Technique

By De Moivre's theorem, $\cos(n\theta) + i\sin(n\theta) = (\cos\theta + i\sin\theta)^n$. Expanding the right-hand side using the Binomial Theorem: $$(\cos\theta + i\sin\theta)^n = \sum_{k=0}^n \binom{n}{k} \cos^{n-k}\theta \, (i\sin\theta)^k$$ Since $i^k$ alternates signs for even powers ($i^0 = 1, i^2 = -1, i^4 = 1$) and yields imaginary units for odd powers ($i^1 = i, i^3 = -i, i^5 = i$), we equate real and imaginary parts.

2. Explicit Formulas

Expansion of $\cos(n\theta)$ (Real Part): $$\mathbf{\cos(n\theta) = \cos^n\theta - \binom{n}{2}\cos^{n-2}\theta\sin^2\theta + \binom{n}{4}\cos^{n-4}\theta\sin^4\theta - \dots}$$ Substituting $\sin^2\theta = 1 - \cos^2\theta$ expresses $\cos(n\theta)$ purely as a polynomial in $\cos\theta$ of degree $n$, matching the celebrated Chebyshev Polynomial of the First Kind $T_n(x)$ where $x = \cos\theta$: $\cos(n\theta) = T_n(\cos\theta)$.

Expansion of $\sin(n\theta)$ (Imaginary Part): $$\mathbf{\sin(n\theta) = \binom{n}{1}\cos^{n-1}\theta\sin\theta - \binom{n}{3}\cos^{n-3}\theta\sin^3\theta + \binom{n}{5}\cos^{n-5}\theta\sin^5\theta - \dots}$$ Dividing by $\sin\theta$ yields the Chebyshev Polynomial of the Second Kind $U_{n-1}(\cos\theta)$.

3. Expansion of $\tan(n\theta)$

Taking the quotient $\frac{\sin(n\theta)}{\cos(n\theta)}$ and dividing both numerator and denominator by $\cos^n\theta$: $$\mathbf{\tan(n\theta) = \frac{\binom{n}{1}\tan\theta - \binom{n}{3}\tan^3\theta + \binom{n}{5}\tan^5\theta - \dots}{1 - \binom{n}{2}\tan^2\theta + \binom{n}{4}\tan^4\theta - \dots}}$$

1. The Reciprocal Exponential Variables

Let $x = e^{i\theta} = \cos\theta + i\sin\theta$. Then $\frac{1}{x} = e^{-i\theta} = \cos\theta - i\sin\theta$.
Adding and subtracting these relations: $$\mathbf{2\cos\theta = x + \frac{1}{x}, \qquad 2i\sin\theta = x - \frac{1}{x}}$$ More generally, for any integer $k \ge 1$: $$\mathbf{2\cos(k\theta) = x^k + \frac{1}{x^k}, \qquad 2i\sin(k\theta) = x^k - \frac{1}{x^k}}$$

2. Systematic Power Reduction for $\cos^n\theta$

Raising $2\cos\theta$ to the $n$-th power and applying the Binomial Theorem: $$(2\cos\theta)^n = \left(x + \frac{1}{x}\right)^n = \sum_{k=0}^n \binom{n}{k} x^{n-k} \left(\frac{1}{x}\right)^k = \sum_{k=0}^n \binom{n}{k} x^{n-2k}$$ Pairing terms symmetrically from the ends: $\binom{n}{k} = \binom{n}{n-k}$, so: $$x^{n-2k} + \frac{1}{x^{n-2k}} = 2\cos((n - 2k)\theta)$$ For example, for $n = 4$: $$2^4 \cos^4\theta = \left(x + \frac{1}{x}\right)^4 = \left(x^4 + \frac{1}{x^4}\right) + 4\left(x^2 + \frac{1}{x^2}\right) + 6$$ $$16\cos^4\theta = 2\cos(4\theta) + 4[2\cos(2\theta)] + 6 \implies \mathbf{\cos^4\theta = \frac{1}{8}[\cos(4\theta) + 4\cos(2\theta) + 3]}$$

3. Systematic Power Reduction for $\sin^n\theta$

Similarly, raising $2i\sin\theta$ to the $n$-th power: $$(2i\sin\theta)^n = \left(x - \frac{1}{x}\right)^n = \sum_{k=0}^n (-1)^k \binom{n}{k} x^{n-2k}$$ For odd $n$, the terms pair into $2i\sin((n-2k)\theta)$. For even $n$, the terms pair into $2\cos((n-2k)\theta)$, providing closed forms essential for calculus and physics integrals!

1. The Fundamental Root Equation

Let $w = R e^{i\phi}$ be a given non-zero complex number, where $R = |w| > 0$ and $\phi = \text{Arg}(w)$. We seek all complex solutions $z = r e^{i\theta}$ to the polynomial equation: $$z^n = w \iff r^n e^{in\theta} = R e^{i\phi}$$ Equating moduli and arguments: $$r^n = R \implies r = \sqrt[n]{R} \in \mathbb{R}^+$$ $$n\theta = \phi + 2k\pi \implies \theta_k = \frac{\phi + 2k\pi}{n}, \quad k \in \mathbb{Z}$$

2. The Exactly $n$ Distinct Complex Roots

As $k$ ranges through $0, 1, 2, \dots, n-1$, we obtain $n$ distinct values of $\theta_k$ within a span of $2\pi$. For $k \ge n$, the arguments differ from previous ones by multiples of $2\pi$, producing the identical complex numbers. Thus, the equation $z^n = w$ possesses exactly $n$ distinct roots: $$\mathbf{z_k = \sqrt[n]{R} \exp\left(i \frac{\phi + 2k\pi}{n}\right) = \sqrt[n]{R} \left[ \cos\left(\frac{\phi + 2k\pi}{n}\right) + i\sin\left(\frac{\phi + 2k\pi}{n}\right) \right]}$$ for $k = 0, 1, 2, \dots, n-1$.

3. Geometric Configuration of Roots

In the Argand plane:

• All $n$ roots have identical modulus $r = \sqrt[n]{R}$, placing them on a circle of radius $\sqrt[n]{R}$ centered at the origin.
• The angular separation between adjacent roots is constantly $\Delta\theta = \frac{2\pi}{n}$.
• The roots form the vertices of a regular $n$-sided polygon inscribed in the circle $|z| = \sqrt[n]{R}$.

1. The Roots of Unity

Setting $w = 1 = 1 \cdot e^{i\cdot 0}$, the solutions to $z^n = 1$ are the $n$-th roots of unity: $$\mathbf{\omega_k = e^{i\frac{2k\pi}{n}} = \cos\left(\frac{2k\pi}{n}\right) + i\sin\left(\frac{2k\pi}{n}\right), \quad k = 0, 1, \dots, n-1}$$ Defining the fundamental root $\omega \equiv \omega_1 = e^{i 2\pi/n}$, the complete set of roots is: $$U_n = \{1, \omega, \omega^2, \omega^3, \dots, \omega^{n-1}\}$$ Under complex multiplication, $(U_n, \cdot)$ forms a finite cyclic group of order $n$ isomorphic to $\mathbb{Z}/n\mathbb{Z}$.

2. The Fundamental Algebraic Identities

• Sum of Roots Identity: Since $z^n - 1 = (z - 1)(z^{n-1} + z^{n-2} + \dots + z + 1) = 0$, for any root $\omega \ne 1$: $$\mathbf{1 + \omega + \omega^2 + \dots + \omega^{n-1} = \sum_{k=0}^{n-1} \omega^k = 0}$$ Geometrically, the centroid of the regular $n$-gon is at the origin $\sum z_k / n = 0$.
• Product of Roots Identity: $$\prod_{k=0}^{n-1} \omega_k = (-1)^{n-1}$$

3. Primitive Roots and Cyclotomic Polynomials

A root $\omega_k$ is a primitive $n$-th root of unity if its multiplicative order is exactly $n$—that is, $\omega_k^m \ne 1$ for all $1 \le m < n$.
Theorem: $\omega_k = e^{i 2k\pi/n}$ is primitive if and only if $\gcd(k, n) = 1$. The number of primitive $n$-th roots of unity is given by Euler's totient function $\phi(n)$.
The $n$-th cyclotomic polynomial $\Phi_n(x)$ is defined as the monic polynomial whose roots are precisely the primitive $n$-th roots of unity: $$\Phi_n(x) \equiv \prod_{\substack{1 \le k \le n \\ \gcd(k, n) = 1}} \left(x - e^{i\frac{2k\pi}{n}}\right)$$ Remarkably, $\Phi_n(x)$ always has integer coefficients and is irreducible over $\mathbb{Q}$, with $x^n - 1 = \prod_{d | n} \Phi_d(x)$.