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CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + claude-opus-5[1m])audited 2026-08-24
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Holomorphic roots of a nonvanishing function on a disc

Statement

For every positive natural m, a nowhere-zero holomorphic function on a disc has a holomorphic mth root.

More precisely, if h is nowhere zero and holomorphic on D(a,r) and m∈N satisfies m≥1, then there is a holomorphic q on D(a,r) with qm=h. If ξm=h(a) is a prescribed scalar root, q may be chosen so that q(a)=ξ.

Facts & Assumptions

Given: A disc D(a,r) with r>0, a nowhere-zero holomorphic function h on it, a natural m≥1, and, for the normalized form, a scalar ξ satisfying ξm=h(a). The complex exponential is entire (The complex exponential is entire and its complex derivative is itself), holomorphic compositions obey the complex chain rule (The chain rule for complex derivatives), and every nonzero complex number has exactly m distinct mth roots (The n-th roots of a complex number and the n distinct roots of unity for every n≥1).

[L1]

If h is nowhere zero and holomorphic on a disc, then there is a holomorphic L on that disc with exp⁡L=h (A nonvanishing holomorphic function on a disc has a holomorphic logarithm).

[L2]

For all complex z,w, exp⁡(z+w)=exp⁡zexp⁡w (exp⁡(z+w)=exp⁡z exp⁡w, and the complex exponential extends the real exponential).

Proof

technique · direct
1.1L1

Take from [L1] a holomorphic function L with exp⁡L=h.

2.1step 1.1L2givenalgebra

Because m≥1, division by m is defined. Put q:=exp⁡(L/m). Repeated use of [L2] gives qm=exp⁡L=h, and q is holomorphic; for m=1 this construction gives q=h.

3.1step 1.1step 2.1givenalgebra∎

For the prescribed value, both q(a) and ξ are nonzero and have mth power h(a). Thus c:=ξ/q(a) satisfies cm=1, and q~:=cq is holomorphic with q~m=h and q~(a)=ξ.

Depends on

Used by

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Sources