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

In answer to the particular example you have given in your comments, and to illustrate the approach you could take, without using the general solution or considering the image of three particular points, you have $$w=\frac {2z}{2+z}\implies z=\frac{2w}{2-w}=\frac{(2u+2iv)(2-u+iv)}{(2-u-iv)(2-u+iv)}$$

Hence $$x+iy=\frac{(4u-2u^2-2v^2)+i(4v)}{(2-u)^2+v^2}$$

So the image of the line $x=y$ is the circle $$u^2+v^2-2u+2v=0$$

Answer 2

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.

Answer 3

Notice that $f$ is an automorphism on $\widehat{\mathbb{C}}$, and in particular it is bijective. Check that when $c=0$, $f(\infty)=\infty$ and when $c\neq 0$ we have $f(-d/c)=\infty$.

Straight lines always contain the point an infinity, and classic circles never do. Hence, to determine whether an object (line or circle) is mapped to a line or a circle, it suffices to determine for each case $(c=0$ and $c\neq0)$ whether the corresponding preimage of infinity belongs to the object.

Now that you know what your image looks like, just take a few distinct point in the object (two if your image is a line, three if it's a circle), determine their image under $f$, and plug them into the corresponding equation for line/circle to obtain your answer.

Answer 4

The equation of a circle in $\mathbb{C}$ has a simple form which is expressed just using complex variables, namely $|z-p|^2 = t$ where $p$ is the center and $t$ is the squared radius which is positive. Using properties of the complex conjugate this has the alternate form $z \bar z - \bar p z - p \bar z + p \bar p = t$.

The equation of a line in $\mathbb{C}$ has a similarly simple form.

Use these forms, substitute $z=f^{-1}(w)$ which is a fractional linear transformation, and rearrange terms to get a $w$ equation in one of the same two forms.