[Math] Finding the image of a line or circle under a Möbius transformation

Question

Consider an extended Möbius transformation $f : \hat{\mathbb C} \to \hat{\mathbb C}$ where $\hat{\mathbb C} = \mathbb C \, \cup \, \{\infty\}$ such that

$f(z) := \begin{cases} \frac{az+b}{cz+d} \iff z \in \mathbb C \, \backslash \{ -\frac{d}c\} \\ \infty \iff z = -\frac{d}c \\ \frac{a}c \iff z = \infty\end{cases}$

where $a, b, c, d \in \mathbb C$ so that $f$ is bijective. Let's restrict the domain to either a line or a circle in $\mathbb C$.

As known, every line or circle in $\mathbb C$ maps to a line or circle in $\mathbb C$ under $f$. To find what the image is, consider we evaluate $f$ at $z = z_0, z_1, z_2$ where $z_0, z_1, z_2$ are different complex numbers in the domain (as $f$ is bijective, we know that $z_0, z_1, z_2$ will map to different points as well). Define $x_k := \mathcal{Re}(z_k)$ and $y_k := \mathcal{Im}(z_k)$ We can test if

$\begin{cases}\alpha x_0 + \beta y_0 = \gamma \\ \alpha x_1 + \beta y_1 = \gamma \\ \alpha x_2 + \beta y_2 = \gamma \\ \end{cases}$

has any (non-trivial) solution. If not, we know that the image is a circle and we instead get to solve

$\begin{cases} (x_0 – \alpha)^2 + (y_0 – \beta)^2 = \gamma^2 \\ (x_1 – \alpha)^2 + (y_1 – \beta)^2 = \gamma^2 \\ (x_2 – \alpha)^2 + (y_2 – \beta)^2 = \gamma^2\end{cases}$

which has the general solution

$\begin{cases} \alpha = \frac{x_0^2 \left(-y_1\right)+x_0^2 y_2+x_1^2 y_0-x_2^2 y_0+x_2^2 y_1-x_1^2 y_2+y_0 y_1^2-y_0 y_2^2+y_1 y_2^2-y_0^2 y_1+y_0^2 y_2-y_1^2 y_2}{2 \left(x_1 y_0-x_2 y_0-x_0 y_1+x_2 y_1+x_0 y_2-x_1 y_2\right)} \\ \beta = \frac{-x_0 y_1^2+x_0 y_2^2+x_1 y_0^2-x_2 y_0^2+x_2 y_1^2-x_1 y_2^2+x_1 x_0^2-x_2 x_0^2-x_1^2 x_0+x_2^2 x_0-x_1 x_2^2+x_1^2 x_2}{2 \left(x_1 y_0-x_2 y_0-x_0 y_1+x_2 y_1+x_0 y_2-x_1 y_2\right)} \\ \gamma = i \frac{\sqrt{x_1^2-2 x_2 x_1+x_2^2+y_1^2+y_2^2-2 y_1 y_2} \sqrt{2 x_0^2 y_0^2+x_0^2 y_1^2+x_0^2 y_2^2-2 x_0^2 y_0 y_1-2 x_0^2 y_0 y_2-2 x_1 x_0 y_0^2-2 x_2 x_0 y_0^2-2 x_2 x_0 y_1^2-2 x_1 x_0 y_2^2+4 x_2 x_0 y_0 y_1+4 x_1 x_0 y_0 y_2+x_1^2 y_0^2+x_2^2 y_0^2+x_2^2 y_1^2+x_1^2 y_2^2-2 x_2^2 y_0 y_1-2 x_1^2 y_0 y_2+x_0^4-2 x_1 x_0^3-2 x_2 x_0^3+x_1^2 x_0^2+x_2^2 x_0^2+4 x_1 x_2 x_0^2-2 x_1 x_2^2 x_0-2 x_1^2 x_2 x_0+x_1^2 x_2^2+y_0^4+y_0^2 y_1^2+y_0^2 y_2^2+y_1^2 y_2^2-2 y_0 y_1 y_2^2-2 y_0^3 y_1-2 y_0^3 y_2-2 y_0 y_1^2 y_2+4 y_0^2 y_1 y_2}}{\sqrt{-4 x_0^2 y_1^2-4 x_0^2 y_2^2+8 x_0^2 y_1 y_2+8 x_2 x_0 y_1^2+8 x_1 x_0 y_2^2+8 x_1 x_0 y_0 y_1-8 x_2 x_0 y_0 y_1-8 x_1 x_0 y_0 y_2+8 x_2 x_0 y_0 y_2-8 x_1 x_0 y_1 y_2-8 x_2 x_0 y_1 y_2-4 x_1^2 y_0^2-4 x_2^2 y_0^2+8 x_1 x_2 y_0^2-4 x_2^2 y_1^2-4 x_1^2 y_2^2+8 x_2^2 y_0 y_1-8 x_1 x_2 y_0 y_1+8 x_1^2 y_0 y_2-8 x_1 x_2 y_0 y_2+8 x_1 x_2 y_1 y_2}}
\end{cases}$

As you see, this is not something you want to do by hand. Is there any more profound way to see what the image is?

Answer 1

For a first overview,

• a line/circle maps to a line iff its image contains $\infty$, i.e., if the origial object contains $-\frac dc$.
• if the original is a line, the image will pass through $\frac ac$
• points closer to $-\frac dc$ are mapped to points further from $\frac ac$

This suggests the following method

• A line that passes through $-\frac dc$ maps to a line through $\frac ac$. Lines are easy, just pick another point (not $-\frac dc$, not $\infty$) and see where it maps; or find the direction via the differential
• A line that does not pass through $-\frac dc$ maps to a circle through $\frac ac$. Find the point on the line that is closest to $-\frac dc$; its image will be furthest from $\frac ac$, i.e., these two points form a diameter - their midpoint is the center of the circle
• A circle through $-\frac dc$ will map to a line. Find the point opposite to $-\frac dc$ and map it. The line through its image and perpendicular to the direction to $\frac ac$ is our image line
• A circle not through $-\frac dc$ will map to a circle. The line through $-\frac dc$ and the midpoint intersects the circle in two points. Their image is a diameter of the image circle.