TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-16

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Complex sine and cosine are unbounded on the complex plane

Statement

Neither $\sin : C \to C$ nor $\cos : C \to C$ is bounded.

Facts & Assumptions

Given: A positive real variable $y$ .

[L1]

For every complex $z$ , $\sinh z = - i \sin (i z)$ and $\cosh z = \cos (i z)$ (The exponential formulas, real restrictions, and trigonometric-hyperbolic dictionary over $C$ ).

[L2]

For real $y$ , the complex exponential equals the real exponential, and $\exp (y) \to + \infty$ while $\exp (- y) \to 0$ as $y \to + \infty$ ( $\exp (z + w) = \exp z \exp w$ , and the complex exponential extends the real exponential, The exponential tends to $+ \infty$ at $+ \infty$ and to $0$ at $- \infty$ ).

[L3]

The definitions give $\sinh y = (\exp y - \exp (- y)) / 2$ and $\cosh y = (\exp y + \exp (- y)) / 2$ (Complex sine, cosine, hyperbolic sine, and hyperbolic cosine from the complex exponential).

Proof

technique · direct

1.1L1algebra

Substituting the real $y$ into [L1] gives $\sin (i y) = i \sinh y$ and $\cos (i y) = \cosh y$ .

1.2L2L3algebra

By [L2] and [L3], both positive real quantities $\sinh y$ and $\cosh y$ tend to $+ \infty$ as $y \to + \infty$ .

2.1step 1.1step 1.2∎

Hence $∣ \sin (i y) ∣ = \sinh y$ and $∣ \cos (i y) ∣ = \cosh y$ are unbounded along the imaginary axis, so both complex functions are unbounded.

Depends on

Used by

FALSE: complex sine and cosine are bounded on the complex plane False statement

Dependency tree · two levels

15 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, §2.3 (standard reference, not scraped)