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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 $m$ th root.

More precisely, if $h$ is nowhere zero and holomorphic on $D (a, r)$ and $m \in N$ satisfies $m \geq 1$ , then there is a holomorphic $q$ on $D (a, r)$ with $q^{m} = h$ . If $ξ^{m} = h (a)$ 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 $m \geq 1$ , and, for the normalized form, a scalar $ξ$ satisfying $ξ^{m} = h (a)$ . 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 $m$ th roots (The $n$ -th roots of a complex number and the $n$ distinct roots of unity for every $n \geq 1$ ).

[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 $z, w$ , $\exp (z + w) = \exp z \exp w$ ( $\exp (z + w) = \exp z \exp w$ , and the complex exponential extends the real exponential).

Proof

technique · direct

1.1

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

2.1

Because $m \geq 1$ , division by $m$ is defined. Put $q : = \exp (L / m)$ . Repeated use of [L2] gives $q^{m} = \exp L = h$ , and $q$ is holomorphic; for $m = 1$ this construction gives $q = h$ .

step 1.1L2givenalgebra

3.1

For the prescribed value, both $q (a)$ and $ξ$ are nonzero and have $m$ th power $h (a)$ . Thus $c : = ξ / q (a)$ satisfies $c^{m} = 1$ , and $\tilde{q} : = c q$ is holomorphic with ${\tilde{q}}^{m} = h$ and $\tilde{q} (a) = ξ$ .

step 1.1step 2.1givenalgebra∎

Depends on

Used by

Local normal form of a nonconstant holomorphic map Theorem

Dependency tree · two levels

34 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

J. Lebl, Guide to Cultivating Complex Analysis, Corollary 4.3.5 (standard reference, not scraped)