TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-16

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The exponential definitions of complex sine, cosine, hyperbolic sine, and hyperbolic cosine equal their entire power series

Statement

For every $z \in C$ , $\sin z = \sum_{n \geq 0} \frac{(- 1)^{n} z^{2 n + 1}}{(2 n + 1)!}, \cos z = \sum_{n \geq 0} \frac{(- 1)^{n} z^{2 n}}{(2 n)!},$ $\sinh z = \sum_{n \geq 0} \frac{z^{2 n + 1}}{(2 n + 1)!}, \cosh z = \sum_{n \geq 0} \frac{z^{2 n}}{(2 n)!} .$ All four series have infinite radius.

Facts & Assumptions

Given: A complex number $z$ .

[L1]

The complex exponential is defined by the series $\exp z = \sum_{n \geq 0} z^{n} / n!$ , the cited Definition recording that convergence for every $z \in C$ is discharged elsewhere (The complex exponential by its power series).

[L2]

Sine, cosine, hyperbolic sine, and hyperbolic cosine are the symmetric and antisymmetric exponential combinations displayed in their definition (Complex sine, cosine, hyperbolic sine, and hyperbolic cosine from the complex exponential).

[L3]

Every absolutely convergent complex series may be rearranged without changing its sum (Every absolutely convergent complex series converges, and rearrangements preserve its sum).

[L4]

If $L = {lim sup}_{k \to \infty} ∣ c_{k + 1} ∣^{1 / (k + 1)}$ , Cauchy–Hadamard gives radius $+ \infty$ when $L = 0$ (Cauchy-Hadamard for complex power series, including zero and infinite radius).

[L5]

For every $z \in C$ the series $\sum z^{n} / n!$ converges absolutely (The complex exponential series converges absolutely for every complex argument).

Proof

technique · direct

1.1

Substitute the series [L1] at $z, - z, i z, - i z$ into [L2]. Absolute convergence, which [L5] supplies for every complex argument, allows [L3] to separate the even and odd indices.

L1L2L3L5

2.1

The identities $i^{2 n} = (- 1)^{n}$ and $i^{2 n + 1} = i (- 1)^{n}$ simplify those even and odd parts to the four displayed series.

step 1.1algebra

3.1

Their factorial coefficients have root limsup $0$ : for $n \geq 2$ the factorial satisfies $n! \geq (n / 2)^{⌊ n / 2 ⌋}$ , since at least $⌊ n / 2 ⌋$ of the factors $1, \dots, n$ are at least $n / 2$ , so $(1 / n!)^{1 / n} \leq (2 / n)^{⌊ n / 2 ⌋ / n} \to 0$ . Hence [L4] gives infinite radius. The constant terms are retained in the even series and absent from the odd series.

step 2.1L4∎

Depends on

Used by

Complex sine, cosine, hyperbolic sine, and hyperbolic cosine are entire with their standard derivatives Corollary

Dependency tree · next 3 levels

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Sources

E. Stein and R. Shakarchi, Complex Analysis, Ch. 1 (standard reference, not scraped)