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Power maps are biholomorphisms on sectors of width less than $2 π / n$

Statement

Let $n \geq 1$ be an integer, let $α < β$ be real numbers with $β - α < 2 π / n$ , and define

$S : = {r e^{i θ} : r > 0, α < θ < β}, T : = {ρ e^{i ϕ} : ρ > 0, n α < ϕ < n β} .$

Then the power map $f (z) = z^{n}$ is a biholomorphism from $S$ onto $T$ .

Facts & Assumptions

Given: The integer $n \geq 1$ and the sectors $S, T$ above.

[F1]

On the slit plane the principal logarithm is holomorphic and satisfies $\exp (Log z) = z$ (The principal logarithm is the normalised holomorphic branch on the slit plane).

[F2]

If $L$ is a holomorphic logarithm branch on a domain, then $z_{L}^{α} : = \exp (α L (z))$ defines the branch power (Complex powers defined from a holomorphic logarithm branch).

[F3]

For integer exponents, branch powers agree with ordinary powers (Branch-defined complex powers agree with integer powers).

[F4]

The complex exponential is entire (The complex exponential is entire and its complex derivative is itself).

[F5]

A map is biholomorphic when it is bijective, holomorphic, and has holomorphic inverse (Biholomorphic maps between complex domains).

Proof

technique · direct

1.1

Put $γ = (α + β) / 2$ and $δ = (β - α) / 2$ , so $δ < π / n \leq π$ ; if $z \in S$ then $e^{- i γ} z$ has argument in $(- δ, δ) \subset (- π, π)$ , so $L (z) : = Log (e^{- i γ} z) + i γ$ is a holomorphic logarithm branch on $S$ by [F1].

F1givenconstruct

2.1

By [F2] and [F3], $f (z) = z^{n} = \exp (n L (z))$ on $S$ , so $f$ is holomorphic there; moreover $\arg f (z) = n \arg z$ with $n \arg z \in (n α, n β)$ , hence $f [S] \subseteq T$ .

F2F3F4step 1.1algebra

2.2

If $w \in T$ , then $e^{- i n γ} w$ has argument in $(- n δ, n δ) \subset (- π, π)$ , so $M (w) : = Log (e^{- i n γ} w) + i n γ$ is a holomorphic logarithm branch on $T$ by [F1]; define $g (w) : = \exp (M (w) / n)$ . Then $g$ is holomorphic on $T$ , $\arg g (w) \in (α, β)$ , and $g [T] \subseteq S$ .

F1F4step 1.1constructalgebra

3.1

By [F2] and [F3], $g (w)^{n} = \exp (M (w)) = w$ for $w \in T$ , while for $z \in S$ one has $g (f (z)) = \exp (n L (z) / n) = z$ because $n L (z)$ has imaginary part in $(n α, n β)$ and so is the chosen branch value $M (f (z))$ . Therefore $g = f^{- 1}$ , and [F5] makes $f$ biholomorphic from $S$ onto $T$ .

F2F3F5step 2.1step 2.2algebra∎

Depends on

Used by

A power map sends a sector to a half-plane Example

Dependency tree · two levels

37 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

Lars V. Ahlfors, Complex Analysis, 3rd ed., Ch. 3 §4.2 (standard reference, not scraped)
Elias M. Stein and Rami Shakarchi, Complex Analysis, Ch. 8 §1.2 (standard reference, not scraped)

Power maps are biholomorphisms on sectors of width less than 2π/n2π/n