Alphabeta Math
TheoremStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-08-02
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exp⁡(z+w)=exp⁡z exp⁡w, and the complex exponential extends the real exponential

Facts & Assumptions

Given: Complex z,w and real x.

[L1]

The Cauchy product of two absolutely convergent complex series converges absolutely to the product of their sums says that the product has nth coefficient ∑k≤nakbn−k.

[L3]

The binomial theorem over the complex field gives (z+w)n=∑k≤nιC ⁣((nk))zkwn−k, where ιC is its canonical-natural map.

[L4]

The complex exponential series converges absolutely for every complex argument states that ∑zn/n! converges absolutely for every complex z.

Proof

technique · direct
1.1

By [L4], the two exponential series converge absolutely, so [L1] makes their product the Cauchy product.

L1L4
2.1

Its degree-n coefficient is ∑k≤nzkwn−k/(k!(n−k)!). By [L2], after applying the canonical-natural map into C, each summand is ιC ⁣((nk))zkwn−k/n!, and [L3] turns their sum into (z+w)n/n!.

L2L3step 1.1
3.1

The resulting series is the defining series of exp⁡(z+w). When z=x+0i, every term is the corresponding real term in the definition of ex, so the two values agree.

step 2.1∎

Depends on

Used by

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Dependency tree · two levels

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Sources