TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck 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

1.1L1L2L3L5

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.

2.1step 1.1algebra

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.

3.1step 2.1L4∎

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.

Depends on

Used by

Complex sine, cosine, hyperbolic sine, and hyperbolic cosine are entire with their standard derivatives Corollary
sin z-z has a zero of order three at the origin Example
The sine product recovers the Basel sum Example
FALSE: an entire function bounded on the real axis is constant False statement
The Weierstrass product for sine Theorem

Dependency tree · two levels

19 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

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