---
tag: 3DJ46V21KH
kind: example
name: Joukowsky transform
msc: [30C20, 76B10]
illustrates: [0YYT36R4YD]
uses: [0DG0Z72CQD, 0VHNKP1N3K, 0FF6CAQT2Y]
sources:
  - url: https://en.wikipedia.org/wiki/Joukowsky_transform
same_as:
  wikidata: Q1794703
  wikipedia: Joukowsky transform
---

## Example

The Joukowsky transform is the map $J(z) = z + \dfrac{1}{z}$ on $\C \setminus \set{0}$.
It is [[holomorphic|0DG0Z72CQD]] with $J'(z) = 1 - z^{-2}$, which vanishes only at $z = \pm 1$.
So $J$ is a [[conformal map|0VHNKP1N3K]] near each point other than $0$ and $\pm 1$.

Circles go to ellipses.
For $z = r e^{i\theta}$ with $r > 1$,
$$J(z) = \left(r + \frac{1}{r}\right) \cos \theta + i \left(r - \frac{1}{r}\right) \sin \theta,$$
so the circle $\abs{z} = r$ goes onto the [[ellipse|0FF6CAQT2Y]] with semi-axes $r + 1/r$ and $r - 1/r$ and foci $\pm 2$.
For $r = 1$ one gets $J(e^{i\theta}) = 2 \cos \theta$, so the unit circle is folded onto the segment $[-2, 2]$.

The exterior of the unit disc is mapped bijectively onto $\C \setminus [-2, 2]$.
Check of injectivity: $J(z) - J(w) = (z - w)(1 - \frac{1}{zw})$, and $zw \neq 1$ when $\abs{z} > 1$ and $\abs{w} > 1$.
Check of surjectivity: for $c \in \C$, the two roots of $z^2 - cz + 1 = 0$ have product $1$.
If neither root has modulus greater than $1$, both have modulus $1$, and then $c = 2\cos\theta$ lies in $[-2, 2]$.

## Remarks

Circles that pass through $-1$ or $1$ and contain the other point go to curves with one sharp edge, the Joukowsky airfoils.
This is why the map is used to model the flow of air around a wing.
Some authors use $\frac{1}{2}(z + 1/z)$ instead, which only rescales the picture by $\frac{1}{2}$.

## Explanations

### L1 human

The Joukowsky map adds a number and its reciprocal.
Far from the origin $1/z$ is tiny, so the map is almost the identity there.
Near the unit circle the two terms are of the same size, and the circle is squashed flat onto a segment.
Circles a bit larger than the unit circle become ellipses, and suitably shifted circles become wing shapes.