ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-29

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The Joukowski map sends circles centered at the origin to ellipses

Example

For $r > 1$ , the Joukowski map

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

sends the circle $∣ z ∣ = r$ to the ellipse

$\frac{x^{2}}{{(\frac{r + r^{- 1}}{2})}^{2}} + \frac{y^{2}}{{(\frac{r - r^{- 1}}{2})}^{2}} = 1.$

Facts & Assumptions

Given: A real $r > 1$ and the Joukowski map of The Joukowski map is a biholomorphism from the exterior disc onto $C ∖ [- 1, 1]$ .

[F1]

The Joukowski map is $J (z) = \frac{1}{2} (z + \frac{1}{z})$ (The Joukowski map is a biholomorphism from the exterior disc onto $C ∖ [- 1, 1]$ ).

Verification

1.1

Parameterizing the circle by $z = r e^{i t}$ , $0 \leq t < 2 π$ , [F1] gives $J (r e^{i t}) = \frac{1}{2} (r e^{i t} + r^{- 1} e^{- i t}) = \frac{r + r^{- 1}}{2} \cos t + i \frac{r - r^{- 1}}{2} \sin t$ .

F1givenalgebra

2.1

Writing $x = \frac{r + r^{- 1}}{2} \cos t$ and $y = \frac{r - r^{- 1}}{2} \sin t$ , one gets $\frac{x^{2}}{{(\frac{r + r^{- 1}}{2})}^{2}} + \frac{y^{2}}{{(\frac{r - r^{- 1}}{2})}^{2}} = \cos^{2} t + \sin^{2} t = 1$ , so the image is the stated ellipse.

step 1.1algebra∎

Depends on

The Joukowski map is a biholomorphism from the exterior disc onto $\mathbb{C}\setminus[-1,1]$

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

3 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)