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Rigidity and rotation of a unit-phasor average

Statement

Let $n \geq 1$ and $z_{1}, \dots, z_{n} \in S^{1} = {z \in C : ∣ z ∣ = 1}$ . Their average $m = n^{- 1} \sum_{j = 1}^{n} z_{j}$ satisfies $∣ m ∣ \leq 1$ , with equality exactly when all $z_{j}$ coincide.

For real $t$ , rotation about the origin through angle $t$ means the map $R_{t} (x + i y) : = (x \cos t - y \sin t) + i (x \sin t + y \cos t)$ for real $x, y$ . A common rotation of a set is its image under one map $R_{t}$ .

For real angles $θ_{j}$ and a positive integer $k$ , put $m_{k} = n^{- 1} \sum_{j = 1}^{n} e^{i k θ_{j}}$ . Then $∣ m_{k} ∣ = 1$ exactly when all $e^{i k θ_{j}}$ coincide. Equivalently, their unit phasors $e^{i θ_{j}}$ lie in one common rotation $e^{i θ_{1}} μ_{k}$ , where $μ_{k} = {e^{2 π i r / k} : r = 0, \dots, k - 1}$ is the set of complex $k$ th roots of unity, meaning the complex numbers $w$ satisfying $w^{k} = 1$ .

If $m \neq 0$ , multiplication by the unit phasor $\overline{m} / ∣ m ∣$ rotates $m$ to $∣ m ∣$ . This expression is asserted only when $m \neq 0$ .

Facts & Assumptions

[F1]

A unit phasor is $e^{i u} = \cos u + i \sin u$ for real $u$ , has modulus one, and $e^{i (u - v)} = e^{i u} \overline{e^{i v}}$ ; it is unchanged by adding an integer multiple of $2 π$ to its angle (Phasors, trigonometric identities and periodic derivatives).

[F2]

In a complex inner product space, Cauchy–Schwarz gives $∣ ⟨ u, v ⟩ ∣ \leq ∥ u ∥ ∥ v ∥$ , with equality exactly when $u, v$ are linearly dependent. An inner product is linear in its first argument, conjugate symmetric and positive definite (Real and complex inner product spaces, with the inner product linear in the first argument, Cauchy–Schwarz: |⟨ u,v⟩|≤‖ u‖‖ v‖, with equality exactly for linearly dependent vectors). Its induced norm is $∥ w ∥ = \sqrt{⟨ w, w ⟩}$ (The norm ‖ v‖=√⟨ v,v⟩ induced by a real or complex inner product). The two-vector list $(u, v)$ is linearly dependent when $a u + b v = 0$ for complex scalars $a, b$ not both zero (Linear independence: a finite list v : n → V is independent when ∑_i<n λᵢ vᵢ = 0_V forces every λᵢ = 0_F, and a subset S ⊆ V is independent when every injective finite list into S is independent).

[F3]

For $w = a + i b$ with real $a, b$ , its modulus is $∣ w ∣ = \sqrt{a^{2} + b^{2}}$ (Real and imaginary parts, complex conjugation, and modulus). Complex modulus is multiplicative, and $w \overline{w} = ∣ w ∣^{2}$ ; it vanishes only at zero (Real and complex modulus product and triangle inequalities, Conjugation is an involutive real-field automorphism, zz̄=|z|², and modulus is definite, multiplicative, and subadditive).

[F4]

The real zeros of sine are precisely the integer multiples of $π$ ; $e^{i π} = - 1$ and the complex exponential satisfies $e^{a + b} = e^{a} e^{b}$ (The zero sets of sine and cosine and the least positive common period 2 pi, exp(x+iy)=eˣ(cos y+i sin y), |exp(x+iy)|=eˣ, and e^iπ+1=0, exp(z+w)=exp z exp w, and the complex exponential extends the real exponential).

[F5]

For an integer and a positive integer divisor, division with remainder gives an integer quotient and a remainder between zero inclusive and the divisor exclusive (Division with remainder in ℤ: for a ∈ ℤ and b > 0 there are unique q, r ∈ ℤ with a = qb + r and 0 ≤ r < b).

[F6]

Every point of the unit circle is $(\cos t, \sin t)$ for some real $t$ ([t↦(cos t,sin t) is a bijection from 0,2π) onto the real unit circle).

[F7]

Complex numbers have unique real coordinates, and their multiplication obeys $(a + i b) (c + i d) = (a c - b d) + i (a d + b c)$ for real $a, b, c, d$ (ℂ=ℝ[x]/(x²+1) is a field, every element is uniquely a+bi, and every nonzero element has inverse (a-bi)/(a²+b²)).

[F8]

The standard coordinate pairing on $C^{n}$ , with coordinate addition and scalar multiplication, is an inner product linear in the first argument (The standard formulas ⟨ x,y⟩=∑_k<nxₖ yₖ on ℝⁿ and ∑_k<nxₖyₖ on ℂⁿ are inner products).

Proof

Given: The unit complex numbers $z_{1}, \dots, z_{n}$ , with $n \geq 1$ ; and, for the root-fibre assertion, real $θ_{1}, \dots, θ_{n}$ and a positive integer $k$ .

Goal: Prove the bound and equality criterion, identify the root fibre, and establish the rotation formula.

1.1F2F3F8algebra

On $C^{n}$ with its coordinate vector-space operations, use the inner product $⟨ u, v ⟩ = \sum_{j} u_{j} \overline{v_{j}}$ from [F8]. Coordinate calculation verifies the inner-product axioms in [F2]: linearity and conjugate symmetry hold termwise, and $⟨ u, u ⟩ = \sum_{j} ∣ u_{j} ∣^{2}$ is nonnegative and vanishes only for $u = 0$ . For $u = (z_{1}, \dots, z_{n})$ and $v = (1, \dots, 1)$ , both norms are $\sqrt{n}$ and $⟨ u, v ⟩ = n m$ . Hence Cauchy–Schwarz gives $∣ m ∣ \leq 1$ . Equality holds exactly when $u$ and $v$ are dependent. A dependence witness in [F2] has $a u + b v = 0$ with $a, b$ not both zero. Since $v \neq 0$ , $a = 0$ would force $b = 0$ , so $a \neq 0$ and $u = - (b / a) v$ . Conversely, $u = c v$ gives the dependence witness $(1, - c)$ . Thus equality is equivalent to $u = c v$ for some complex $c$ , or $z_{1} = \dots = z_{n} = c$ .

2.1F1F4step 1.1algebra

Each $e^{i k θ_{j}}$ has modulus one, so step 1.1 applied to these numbers gives the stated criterion for $∣ m_{k} ∣ = 1$ . By the exponential addition law, $e^{i k θ_{j}} = e^{i k θ_{1}}$ exactly when $(e^{i (θ_{j} - θ_{1})})^{k} = 1$ .

3.1F1F3F4F5F6step 2.1algebra

If $e^{i t}$ is a $k$ th root of unity, then $e^{i k t} = 1$ . Its imaginary part is zero by [F1], so [F4] gives $k t = a π$ for an integer $a$ . Divide $a$ by $2$ : $a = 2 q + s$ with $s \in {0, 1}$ . The exponential addition law and $e^{i π} = - 1$ yield $e^{i a π} = (- 1)^{s}$ , so $s = 0$ . Divide $q$ by $k$ , writing $q = b k + r$ with $0 \leq r < k$ . Then $t = 2 π b + 2 π r / k$ and $e^{i t} = e^{2 π i r / k}$ . Conversely, the addition law and $2 π$ -periodicity show that every displayed $e^{2 π i r / k}$ has $k$ th power one. Any complex $k$ th root $w$ has $∣ w ∣^{k} = 1$ by [F3], hence $∣ w ∣ = 1$ ; [F6] and [F1] write it as $e^{i t}$ . Thus these displayed numbers are exactly all complex $k$ th roots of unity. Combining this with step 2.1 proves the stated membership in $e^{i θ_{1}} μ_{k}$ .

4.1F1F3F6F7step 1.1step 2.1step 3.1algebra∎

For any real $t, x, y$ , [F1] and [F7] give $e^{i t} (x + i y) = (x \cos t - y \sin t) + i (x \sin t + y \cos t) = R_{t} (x + i y)$ . Hence multiplication by $e^{i t}$ is exactly the rotation defined in the Statement; in particular $e^{i θ_{1}} μ_{k} = R_{θ_{1}} (μ_{k})$ , completing the common-rotation interpretation in step 3.1. If $m \neq 0$ , then $∣ m ∣ > 0$ . By [F3], $∣ \overline{m} / ∣ m ∣ ∣ = 1$ and $(\overline{m} / ∣ m ∣) m = ∣ m ∣^{2} / ∣ m ∣ = ∣ m ∣$ . By [F6] and [F1], $\overline{m} / ∣ m ∣ = e^{i t}$ for some real $t$ , so the same coordinate identity shows that this multiplier rotates $m$ to $∣ m ∣$ . Together with steps 1.1, 2.1 and 3.1 this proves all assertions.

Depends on

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