Joukowsky transform

Example

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