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

Statement

For all $z, w \in C$ , $\exp (z + w) = \exp z \exp w$ . For real $x$ , the complex value $\exp (x + 0 i)$ equals the published real exponential $e^{x}$ . The conventions and prerequisite facts used below are recorded in The complex exponential by its power series, The complex exponential series converges absolutely for every complex argument, The Cauchy product of two absolutely convergent complex series converges absolutely to the product of their sums, The real exponential function and the number $e$ by a power series, The binomial theorem over the complex field, $(\binom{n}{k}) k! (n - k)! = n!$ for $k \leq n$ ; hence $(\binom{n}{k}) k! = n^{\underline{k}}$ , the quotient $n! / (k! (n - k)!)$ is a natural number, and $(\binom{n}{k}) = (\binom{n}{n - k})$ .

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 $n$ th coefficient $\sum_{k \leq n} a_{k} b_{n - k}$ .

[L2]

$(\binom{n}{k}) k! (n - k)! = n!$ for $k \leq n$ ; hence $(\binom{n}{k}) k! = n^{\underline{k}}$ , the quotient $n! / (k! (n - k)!)$ is a natural number, and $(\binom{n}{k}) = (\binom{n}{n - k})$ gives $(\binom{n}{k}) k! (n - k)! = n!$ for $k \leq n$ .

[L3]

The binomial theorem over the complex field gives $(z + w)^{n} = \sum_{k \leq n} ι_{C} ⁣ ((\binom{n}{k})) z^{k} w^{n - k}$ , where $ι_{C}$ is its canonical-natural map.

[L4]

The complex exponential series converges absolutely for every complex argument states that $\sum z^{n} / n!$ converges absolutely for every complex $z$ .

Proof

technique · direct

1.1L1L4

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

2.1L2L3step 1.1

Its degree- $n$ coefficient is $\sum_{k \leq n} z^{k} w^{n - k} / (k! (n - k)!)$ . By [L2], after applying the canonical-natural map into $C$ , each summand is $ι_{C} ⁣ ((\binom{n}{k})) z^{k} w^{n - k} / n!$ , and [L3] turns their sum into $(z + w)^{n} / n!$ .

3.1step 2.1∎

The resulting series is the defining series of $\exp (z + w)$ . When $z = x + 0 i$ , every term is the corresponding real term in the definition of $e^{x}$ , so the two values agree.

Depends on

Dependency tree · 2 levels

Sources

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