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

Statement

Let

$E : = {z \in C : ∣ z ∣ > 1}, Ω : = C ∖ [- 1, 1],$

and define the Joukowski map by

$J (z) : = \frac{1}{2} (z + \frac{1}{z}) (z \in E) .$

Then $J : E \to Ω$ is a biholomorphism.

Facts & Assumptions

Given: The exterior disc $E$ , the slit-complement $Ω$ , and the map $J$ above.

[F1]

For $n = 2$ , the slit-plane root branch $R_{2} (t) = \exp (Log t / 2)$ is a biholomorphism from $C ∖ {x \in R : x \leq 0}$ onto the right half-plane ${s \in C : Re s > 0}$ , with inverse $s \mapsto s^{2}$ (A slit-plane root branch biholomorphically parametrizes a sector).

Proof

technique · direct

1.1F1givenconstructalgebra

For $w \in Ω$ , put $m (w) : = (w - 1) / (w + 1)$ . If $m (w)$ were a nonpositive real number, then $w = (1 + m (w)) / (1 - m (w))$ would lie in $[- 1, 1]$ , contradicting $w \in Ω$ ; also $m (w) \neq 1$ because $(w - 1) / (w + 1) = 1$ has no finite solution. Hence $m$ maps $Ω$ holomorphically into the slit plane of [F1]. Define $s (w) : = R_{2} (m (w))$ , so $Re s (w) > 0$ for every $w \in Ω$ .

2.1F1step 1.1algebra

Define $H (w) : = (1 + s (w)) / (1 - s (w))$ on $Ω$ . Since $s (w) \neq 1$ , this is holomorphic there; using $s (w)^{2} = m (w)$ gives $J (H (w)) = \frac{1}{2} (\frac{1 + s}{1 - s} + \frac{1 - s}{1 + s}) = \frac{1 + s^{2}}{1 - s^{2}} = \frac{1 + m (w)}{1 - m (w)} = w$ , and $∣ 1 + s (w) ∣^{2} - ∣ 1 - s (w) ∣^{2} = 4 Re s (w) > 0$ yields $∣ H (w) ∣ > 1$ , so $H (Ω) \subseteq E$ .

3.1F1step 2.1algebra

For $z \in E$ , put $s : = (z - 1) / (z + 1)$ . Then $Re s = (∣ z ∣^{2} - 1) / ∣ z + 1 ∣^{2} > 0$ , so $s$ lies in the right half-plane, and a direct calculation gives $(J (z) - 1) / (J (z) + 1) = ((z - 1) / (z + 1))^{2} = s^{2}$ . By [F1], the right-half-plane inverse of squaring is exactly $R_{2}$ , so $R_{2} ((J (z) - 1) / (J (z) + 1)) = s$ , and substituting this into the definition of $H$ yields $H (J (z)) = (1 + s) / (1 - s) = z$ .

4.1step 2.1step 3.1∎

Steps 2.1 and 3.1 show that $J$ and $H$ are holomorphic two-sided inverses between $E$ and $Ω$ . Therefore $J : E \to Ω$ is a biholomorphism.

Depends on

Used by

Dependency tree · two levels

24 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)

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