The simplest conformal maps beyond $az + b$ are quotients of linear functions. They are conformal everywhere they are defined, they can be undone, and they have one striking geometric property: they never bend a circle into anything but a circle or a line.
Key idea
A Möbius transformation is $T(z) = \dfrac{az + b}{cz + d}$ with $ad - bc \ne 0$. When $c \ne 0$, $T$ sends $-\frac{d}{c}$ to $\infty$ and $\infty$ to $\frac{a}{c}$; its inverse is $T^{-1}(w) = \dfrac{dw - b}{-cw + a}$. Every Möbius map sends circles and lines to circles and lines: the image of a circle or line is a line exactly when it passes through $-\frac{d}{c}$.
Why it works
$T'(z) = \frac{ad - bc}{(cz + d)^2} \ne 0$, so $T$ is conformal wherever it is defined. Divide out: $T(z) = \frac{a}{c} + \frac{bc - ad}{c}\cdot\frac{1}{cz + d}$, so $T$ is a chain of a linear map, the inversion $\frac{1}{z}$, and another linear map. Linear maps send circles to circles and lines to lines. For $\frac{1}{z}$, a circle or line has an equation $A|z|^2 + B\operatorname{Re} z + C\operatorname{Im} z + D = 0$; replacing $z$ by $\frac{1}{w}$ and multiplying by $|w|^2$ gives $A + B\operatorname{Re} w - C\operatorname{Im} w + D|w|^2 = 0$, the same kind of equation. It is a line exactly when $D = 0$, that is, when the original passes through $0$, the point inversion sends to $\infty$.
Worked examples
Worked example: Special points
For $T(z) = \dfrac{2z + i}{z - 3}$, find $T(0)$, the point sent to $\infty$, and $T(\infty)$.
- $T(0) = \frac{i}{-3} = -\frac{i}{3}$.
- The bottom is $0$ at $z = 3$: $T(3) = \infty$.
- For large $|z|$, $T(z) \approx \frac{2z}{z} = 2$: $T(\infty) = 2$.
Answer: $-\frac{i}{3}$, $3$, $2$
Worked example: The inverse
For $T(z) = \dfrac{z + 1}{z - 1}$, find the $z$ with $T(z) = 2i$.
- Solve $\frac{z + 1}{z - 1} = w$: $z + 1 = wz - w$, so $z = \frac{w + 1}{w - 1}$.
- With $w = 2i$: $z = \frac{1 + 2i}{-1 + 2i} = \frac{(1 + 2i)(-1 - 2i)}{5} = \frac{3 - 4i}{5}$.
- Check: $T\left(\frac{3 - 4i}{5}\right) = \frac{(8 - 4i)/5}{(-2 - 4i)/5} = \frac{8 - 4i}{-2 - 4i} = 2i$.
Answer: $\frac{3}{5} - \frac{4}{5}i$
Worked example: Fixed points
Find the fixed points of $T(z) = \dfrac{3z - 4}{z - 1}$.
- $T(z) = z$ means $3z - 4 = z^2 - z$, so $z^2 - 4z + 4 = 0$.
- $(z - 2)^2 = 0$: a single (double) fixed point $z = 2$.
Answer: $z = 2$
Composing two Möbius maps gives another one, and the coefficients compose like $2 \times 2$ matrices: $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$. The inverse map corresponds to the inverse matrix (up to a scalar), which is where the formula $\frac{dw - b}{-cw + a}$ comes from.
Common mistake
Expecting every circle to map to a circle. Möbius maps send circles to circles or lines. If a circle passes through $-\frac{d}{c}$, one of its points goes to $\infty$, and a "circle through $\infty$" is a line. Check: plug $-\frac{d}{c}$ into the circle's equation. For $T(z) = \frac{z}{z - 1}$, the unit circle passes through $1$, so its image is a line.