What is conformal map?

Example 18.2 (Möbius transformations)

For $\bigl(\begin{smallmatrix}a & b\\ c & d\end{smallmatrix}\bigr) \in GL_2(\C)$, the Möbius transformation $z \mapsto \frac{az + b}{cz + d}$ is conformal from $\C\setminus\{-d/c\}$ onto $\C\setminus\{a/c\}$ (inverse of the same type, from the inverse matrix; composition corresponds to matrix product). The Cayley map

maps $\mathbb H$ conformally onto $\mathbb D$: indeed $\abs{z - \iu} < \abs{z + \iu}$ exactly when $z$ is closer to $\iu$ than to $-\iu$, i.e. $\operatorname{Im}z > 0$; the inverse is $w \mapsto \iu\frac{1 + w}{1 - w}$. Möbius maps send the family circles-and-lines to itself (Exercise 18.1).

Example 18.3 (The Joukowski map)

Beyond Möbius, the most useful conformal map of classical applied mathematics is

On the exterior $\Omega = \{\abs z > 1\}$ of the unit disc, $J$ is injective: $J(z) = J(w)$ gives $(z - w)(1 - \frac1{zw}) = 0$ and $\abs{zw} > 1$. Its derivative $J'(z) = \frac12(1 - z^{-2})$ vanishes only at $z = \pm1$, on the boundary: $J$ is a conformal equivalence from $\Omega$ onto its image, which is $\C\setminus\intcc{-1}1$ — the unit circle itself is folded two-to-one onto the segment ($J(\eu^{\iu\theta}) = \cos\theta$). Thus the exterior of a segment, a slit plane with no smooth boundary, is conformally the exterior of a disc: corners are no obstacle to conformal equivalence, only to boundary smoothness. Images of circles through $\pm1$ but off-center are airfoil-shaped curves, and composing $J$ with Möbius maps transported the flow past a cylinder — computable by hand — to the flow past a wing: for the first half of the twentieth century, this example was aerodynamics. It is also the door to Chebyshev: $J$ conjugates $z \mapsto z^n$ to the Chebyshev polynomial $T_n$ (Problem 13.1, Part V), since $J(z^n) = \cos(n\theta)$ when $z = \eu^{\iu\theta}$.