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A slit-plane root branch biholomorphically parametrizes a sector

Statement

Let $n \geq 1$ be an integer, let

$S : = C ∖ {x \in R : x \leq 0}, V_{n} : = {r e^{i θ} : r > 0, - π / n < θ < π / n},$

and define

$R_{n} (z) : = \exp ⁣ (\frac{Log z}{n}) (z \in S) .$

Then $R_{n}$ is a biholomorphism from $S$ onto $V_{n}$ , and its inverse is the power map $w \mapsto w^{n}$ on $V_{n}$ .

Facts & Assumptions

Given: The integer $n \geq 1$ , the slit plane $S$ , and the map $R_{n}$ above.

[F1]

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

[F2]

Branch powers are defined by $z_{L}^{α} = \exp (α L (z))$ (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]

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

Proof

technique · direct

1.1

Since $Log$ is holomorphic on $S$ by [F1], the map $R_{n} (z) = \exp (Log z / n)$ is holomorphic on $S$ ; if $z = r e^{i θ}$ with $- π < θ < π$ , then $\arg R_{n} (z) = θ / n \in (- π / n, π / n)$ , so $R_{n} [S] \subseteq V_{n}$ .

F1givenalgebra

2.1

If $w \in V_{n}$ , write $w = r e^{i θ}$ with $r > 0$ and $- π / n < θ < π / n$ . Then $w^{n} = r^{n} e^{i n θ}$ lies in $S$ because $n θ \in (- π, π)$ , and the principal logarithm of $w^{n}$ is $Log (w^{n}) = \log (r^{n}) + i n θ = n (\log r + i θ) .$ Therefore $R_{n} (w^{n}) = \exp ⁣ (\frac{Log (w^{n})}{n}) = \exp (\log r + i θ) = w .$

F1step 1.1algebra

3.1

For $z \in S$ , [F2] with [F3] gives $(R_{n} (z))^{n} = \exp (Log z) = z$ by [F1], so the inverse of $R_{n}$ on $V_{n}$ is $w \mapsto w^{n}$ . Therefore [F4] makes $R_{n} : S \to V_{n}$ biholomorphic.

F1F2F3F4step 2.1algebra∎

Depends on

Used by

The Joukowski map is a biholomorphism from the exterior disc onto ℂ∖[-1,1] Theorem

Dependency tree · two levels

35 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)