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Equality in the unit-complex finite-sum bound

Statement

For an integer $n > 0$ and $z_{1}, \dots, z_{n} \in C$ with $∣ z_{i} ∣ = 1$ , one has $∣ \sum_{i = 1}^{n} z_{i} ∣ \leq n$ . Equality holds if and only if all the $z_{i}$ are equal. This includes $n = 1$ .

Facts & Assumptions

Given: $n > 0$ and $∣ z_{i} ∣ = 1$ for $1 \leq i \leq n$ ; natural scalars in inequalities are their real images.

[F1]

The modulus of $a + b i$ is the nonnegative square root of $a^{2} + b^{2}$ (Real and imaginary parts, complex conjugation, and modulus).

[F2]

Real finite sums may be computed in any enumeration (The sum $\sum_{i \in S} a_{i}$ over a finite index set, and its product form).

[F3]

Finite real sums distribute over addition and scaling, and a sum of nonnegative terms vanishes only when every term vanishes (Laws of finite sums and finite products).

[F4]

Complex finite sums are defined in the additive monoid, with empty sum zero (A finite sum in a commutative monoid indexed by an arbitrary finite set).

[F5]

Finite monoid sums can be reindexed, split over disjoint subsets, and summed in either order on a Cartesian product (Finite commutative-monoid sums are invariant under bijective reindexing, split over disjoint unions, and satisfy the finite Fubini rule).

[F6]

Conjugation preserves addition and multiplication; $z \overline{z} = ∣ z ∣^{2}$ , $∣ z ∣ = 0$ exactly when $z = 0$ , and $∣ z w ∣ = ∣ z ∣ ∣ w ∣$ (Conjugation is an involutive real-field automorphism, $z \overline{z} = ∣ z ∣^{2}$ , and modulus is definite, multiplicative, and subadditive).

[F7]

Base and successor steps establish a property for all finite lengths (The principle of mathematical induction).

[F8]

For nonnegative reals, $a \leq b$ if and only if $a^{2} \leq b^{2}$ (Squaring is monotone on the nonnegatives).

Proof

technique · direct

1.1

Put $S = \sum_{i = 1}^{n} z_{i}$ , using F4. For any complex finite lists, distributivity over a finite sum follows by induction: it holds for the empty sum; appending $a$ changes $c A$ to $c (A + a) = c A + c a$ . Likewise conjugation through a finite sum holds at zero and is preserved by $\overline{A + a} = \overline{A} + \overline{a}$ . F7 therefore gives these identities for every finite length.

F4F6F7algebra

2.1

Consequently $∣ S ∣^{2} = S \overline{S} = \sum_{i, j} z_{i} \overline{z_{j}}$ . Split the pairs into $i = j$ , $i < j$ and $j < i$ , and reindex the third part by swapping its coordinates, as permitted by F5. The diagonal sum is $n$ since $z_{i} \overline{z_{i}} = 1$ , giving $∣ S ∣^{2} = n + \sum_{i < j} (z_{i} \overline{z_{j}} + z_{j} \overline{z_{i}})$ .

step 1.1F5F6given

3.1

For each pair $i < j$ , expand $∣ z_{i} - z_{j} ∣^{2} = (z_{i} - z_{j}) (\overline{z_{i}} - \overline{z_{j}}) = 2 - z_{i} \overline{z_{j}} - z_{j} \overline{z_{i}}$ . There are $n (n - 1) / 2$ such pairs: the count is zero at $n = 1$ , and adjoining index $n + 1$ adds the $n$ pairs $(i, n + 1)$ , so the formula is preserved by $n (n - 1) / 2 + n = n (n + 1) / 2$ . Induction gives this count for all positive $n$ . Adding the displayed expansions to step 2.1 cancels each off-diagonal term and yields $\sum_{i < j} ∣ z_{i} - z_{j} ∣^{2} = n (n - 1) - (∣ S ∣^{2} - n) = n^{2} - ∣ S ∣^{2}$ .

step 2.1F6F7F3F4algebra

4.1

Each squared modulus is a nonnegative real by F1 and F6. Use F2 to enumerate the pair set and F3 to conclude $n^{2} - ∣ S ∣^{2} \geq 0$ . Since $∣ S ∣ \geq 0$ and $n > 0$ , F8 now gives $∣ S ∣ \leq n$ .

step 3.1F1F2F3F6F8

4.2

If $∣ S ∣ = n$ , the identity in step 3.1 makes the sum of squared differences zero. F3 forces $∣ z_{i} - z_{j} ∣^{2} = 0$ for every pair, and F6 gives $z_{i} = z_{j}$ . For $n > 1$ , take the pairs $(1, j)$ to conclude every $z_{j} = z_{1}$ . For $n = 1$ that conclusion already holds and the pair sum is empty.

step 3.1F3F6F4

5.1

Conversely, if every $z_{i} = z_{1}$ , repeated addition gives $S = n z_{1}$ . Thus $∣ S ∣ = ∣ n ∣ ∣ z_{1} ∣ = n$ by F6 and the real modulus in F1. At $n = 1$ this reads $∣ z_{1} ∣ = 1$ . This proves both directions of the equality characterization.

F6F1givenstep 4.2∎

Sources

Etingof et al., Lemma 5.4.5 proof, p. 101, uses strictness for unequal unit roots. The pairwise squared-distance calculation supplies that equality argument locally for all unit complex numbers, without complex arguments or trigonometry.

Depends on

Used by

Distinct unit summands need not attain the sum bound Counterexample

Dependency tree · two levels

37 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Etingof et al., Introduction to Representation Theory (standard reference, not scraped)