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

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The zeros of complex sine are the integer multiples of pi, and the zeros of complex cosine are the odd half-integer multiples of pi

Statement

For $z \in C$ , $\sin z = 0 ⟺ z = k π for some k \in Z,$ and $\cos z = 0 ⟺ z = (k + \frac{1}{2}) π for some k \in Z .$

Facts & Assumptions

Given: A complex number $z$ .

[L1]

The definitions are $\sin z = (\exp (i z) - \exp (- i z)) / (2 i)$ and $\cos z = (\exp (i z) + \exp (- i z)) / 2$ (Complex sine, cosine, hyperbolic sine, and hyperbolic cosine from the complex exponential).

[L2]

The exponential satisfies $\exp (u + v) = \exp u \exp v$ , has kernel $2 π i Z$ , and $\exp u = \exp v$ exactly when $u - v \in 2 π i Z$ ( $\exp (z + w) = \exp z \exp w$ , and the complex exponential extends the real exponential, $\ker (\exp) = 2 π i Z$ , and $\exp z = \exp w$ exactly when $z - w \in 2 π i Z$ ).

[L3]

Euler's identity gives $\exp (i π) = - 1$ ( $\exp (x + i y) = e^{x} (\cos y + i \sin y)$ , $∣ \exp (x + i y) ∣ = e^{x}$ , and $e^{i π} + 1 = 0$ ).

Proof

technique · direct

1.1L1L2algebra

By [L1] and multiplication by the nonzero $\exp (i z)$ , $\sin z = 0$ is equivalent to $\exp (2 i z) = 1$ . By [L2], this is equivalent to $2 i z = 2 π i k$ for some integer $k$ , hence to $z = k π$ .

1.2L1L2L3algebra

Similarly, $\cos z = 0$ is equivalent to $\exp (2 i z) = - 1 = \exp (i π)$ by [L3]. By [L2], $2 i z - i π = 2 π i k$ , so $z = (k + 1 / 2) π$ .

2.1step 1.1step 1.2∎

Reversing each algebraic equivalence proves both converses, so the displayed descriptions are exact.

Depends on

Used by

The Basel sum is pi squared over six by a residue computation Corollary
Agreement accumulating only at the boundary does not force a holomorphic identity Counterexample
Cosecant residues sum an alternating rational series over the integers Theorem
Cotangent residues sum a rational function over the integers Theorem
The Mittag-Leffler expansion of pi cotangent Theorem
The Weierstrass product for sine Theorem

Dependency tree · two levels

20 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)