LemmaStatement: AI-adaptedProof: AI-adaptedverified 2026-09-23 (gpt-6-sol)

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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The binomial theorem over the complex field

Statement

For $z, w \in C$ and $n \in N$ , write $ι_{C} : N \to C$ for the canonical-natural map of The canonical natural $ι (n) = n \cdot 1_{F}$ of a field. Then $(z + w)^{n} = \sum_{k \leq n} ι_{C} ⁣ ((\binom{n}{k})) z^{k} w^{n - k} .$ The conventions and prerequisite facts used below are recorded in $C = R [x] / (x^{2} + 1)$ is a field, every element is uniquely $a + b i$ , and every nonzero element has inverse $(a - b i) / (a^{2} + b^{2})$ , Integer powers in the complex field, The set $[A]^{k}$ of $k$ -element subsets and the binomial coefficient $(\binom{n}{k}) : = ∣ [n]^{k} ∣$ , Pascal's rule $(\binom{n + 1}{k + 1}) = (\binom{n}{k}) + (\binom{n}{k + 1})$ , and the hockey-stick identity $\sum_{i \leq n} (\binom{i}{k}) = (\binom{n + 1}{k + 1})$ , The product $g_{0} g_{1} \dots g_{n - 1}$ of a finite list in a monoid, by recursion, with the empty product ( $n = 0$ ) equal to the identity, Generalised associativity: in a monoid the product of a finite list does not depend on the bracketing, and in a commutative monoid it does not depend on the order of the factors either, The principle of mathematical induction.

Facts & Assumptions

Given: Complex $z, w$ and natural $n$ .

Proof

technique · induction

1.1base

For $n = 0$ the sum has the single index $k = 0$ . Its term is $ι_{C} ((\binom{0}{0})) z^{0} w^{0} = 1$ , since $(\binom{0}{0}) = 1$ and both zeroth powers are $1$ . The left side is $(z + w)^{0} = 1$ as well.

1.2ih

Assume the formula at $n$ .

2.1given

Multiply the formula in step 1.2 by $z + w$ , split the finite initial-segment sums, prove the shift $k \mapsto k + 1$ from the recursive monoid-sum clauses, and group equal powers.

3.1discharge-induction∎

Pascal's rule gives the complex coefficient $ι_{C} ⁣ ((\binom{n + 1}{k}))$ at every index, including the endpoints, so the formula holds at $n + 1$ .

Depends on

Dependency tree · 2 levels

Sources

J. Lebl, Basic Analysis I: Complex Numbers and the Complex Exponential