###### Theorem 3.3.1

Any general linear transformation extended to the domain \({\mathbb{C}}^+\) fixes \(\infty\text{.}\)

Consider again inversion about the circle \(C\) given by \(|z - z_0| = r\text{,}\) and observe that points close to \(z_0\) get mapped to points in the plane far away from from \(z_0\text{.}\) In fact, a sequence of points in \(\mathbb{C}\) whose limit is \(z_0\) will be inverted to a sequence of points whose magnitudes go to \(\infty\text{.}\) Conversely, any sequence of points in \(\mathbb{C}\) having magnitudes marching off to \(\infty\) will be inverted to a sequence of points whose limit is \(z_0\text{.}\)

With this in mind, we define a new point called the point at infinity, denoted \(\infty.\) Adjoin this new point to the plane to get the extended plane, denoted as \({\mathbb{C}}^+\text{.}\) Then, one may extend inversion in the circle \(C\) to include the points \(z_0\) and \(\infty\text{.}\) In particular, inversion of \(\mathbb{C}^+\) in the circle \(C\) centered at \(z_0\) with radius \(r\text{,}\) \(i_C: \mathbb{C}^+ \to \mathbb{C}^+\text{,}\) is given by

\begin{equation*}
i_C(z) =
\begin{cases}\frac{r^2}{(\overline{z-z_0})} + z_0 \amp \text{ if \(z \neq z_0,\infty\); } \\
\infty \amp \text{ if \(z=z_0\); } \\
z_0 \amp \text{ if \(z = \infty\). }
\end{cases}
\end{equation*}

Viewing inversion as a transformation of the extended plane, we define \(z_0\) and \(\infty\) to be symmetric points with respect to the circle of inversion.

The space \({\mathbb{C}}^+\) will be the canvas on which we do all of our geometry, and it is important to begin to think of \(\infty\) as “one of the gang,” just another point to consider. All of our translations, dilations, and rotations can be redefined to include the point \(\infty\text{.}\)

So where is \(\infty\) in \({\mathbb{C}}^+\text{?}\) You approach \(\infty\) as you proceed in either direction along any line in the complex plane. More generally, if \(\{z_n\}\) is a sequence of complex numbers such that \(|z_n| \to \infty\) as \(n \to \infty\text{,}\) then we say \(\displaystyle\lim_{n\to\infty} z_n = \infty.\) By convention, we assume \(\infty\) is on every line in the extended plane, and reflection across any line fixes \(\infty\text{.}\)

Any general linear transformation extended to the domain \({\mathbb{C}}^+\) fixes \(\infty\text{.}\)

If \(T(z) = a z + b\) where \(a\) and \(b\) are complex constants with \(a \neq 0\text{,}\) then by limit methods from calculus, as \(|z_n| \to \infty\text{,}\) \(|a z_n + b| \to \infty\) as well. Thus, \(T(\infty) = \infty\text{.}\)

So, with new domain \({\mathbb{C}}^+\text{,}\) we modify our fixed point count for the basic transformations:

The translation \(T_b\) of \({\mathbb{C}}^+\) fixes one point (\(\infty\)).

The rotation about the origin \(R_\theta\) of \({\mathbb{C}}^+\) fixes 2 points (0 and \(\infty\)).

The dilation \(T(z) = kz\) of \({\mathbb{C}}^+\) fixes 2 points, (0 and \(\infty\)).

The reflection \(r_L(z)\) of \({\mathbb{C}}^+\) about line \(L\) fixes all points on \(L\) (which now includes \(\infty\)).

The following function is a transformation of \({\mathbb{C}}^+\)

\begin{equation*}
T(z) = \frac{i+1}{z+2i},
\end{equation*}

a fact we prove in the next section. For now, we ask where \(T\) sends \(\infty\text{,}\) and which point gets sent to \(\infty\text{.}\)

We tackle the second question first. The input that gets sent to \(\infty\) is the complex number that makes the denominator 0. Thus, \(T(-2i) = \infty.\)

To answer the first question, take your favorite sequence that marches off to \(\infty\text{,}\) for example, \(1, 2, 3,\ldots\text{.}\) The image of this sequence, \(T(1), T(2),\) \(T(3),\ldots\) consists of complex fractions in which the numerator is constant, but the denominator grows unbounded in magnitude along the horizontal line \(\text{Im}(z) = 2\text{.}\) Thus, the quotient tends to 0, and \(T(\infty) = 0.\)

As a second example, you can check that if

\begin{equation*}
T(z) =
\frac{iz+(3i+1)}{2iz+1},
\end{equation*}

then \(T(i/2) = \infty\) and \(T(\infty) = 1/2\text{.}\)

We emphasize that the following key results of the previous section extend to \(\mathbb{C}^+\) as well:

- There exists a unique cline through any three distinct points in \(\mathbb{C}^+\text{.}\) (If one of the given points in Theorem 3.2.4 is \(\infty\text{,}\) the unique cline is the line through the other two points.)
- Theorem 3.2.8 applies to all points \(z\) not on \(C\text{,}\) including \(z = z_0\) or \(\infty\text{.}\)
- Inversion about a cline preserves angle magnitudes at all points in \(\mathbb{C}^+\) (we discuss this below).
- Inversion preserves symmetry points for all points in \(\mathbb{C}^+\) (Theorem 3.2.12 holds if \(p\) or \(q\) is \(\infty\)).
- Theorem 3.2.16 now holds for all clines that do not intersect, including concentric circles. If the circles are concentric, the points symmetric to both of them are \(\infty\) and the common center.

We close this section with a look at stereographic projection. By identifying the extended plane with a sphere, this map offers a very useful way for us to think about the point \(\infty\text{.}\)

The unit 2-sphere, denoted \(\mathbb{S}^2\text{,}\) consists of all the points in 3-space that are one unit from the origin. That is,

\begin{equation*}
\mathbb{S}^2 = \{(a,b,c) \in \mathbb{R}^3 ~|~ a^2+b^2+c^2=1\}.
\end{equation*}

We will usually refer to the unit 2-sphere as simply “the sphere.” Stereographic projection of the sphere onto the extended plane is defined as follows. Let \(N = (0,0,1)\) denote the north pole on the sphere. For any point \(P \neq N\) on the sphere, \(\phi(P)\) is the point on the ray \(\overrightarrow{NP}\) that lives in the \(xy\)-plane. See Figure 3.3.4 for the image of a typical point \(P\) of the sphere.

The stereographic projection map \(\phi\) can be described algebraically. The line through \(N = (0,0,1)\) and \(P = (a,b,c)\) has directional vector \(\overrightarrow{NP}=\langle a,b,c-1\rangle\text{,}\) so the line equation can be expressed as

\begin{equation*}
{\vec r}(t) = \langle 0,0,1\rangle + t\langle a,b,c-1\rangle.
\end{equation*}

This line intersects the \(xy\)-plane when its \(z\) coordinate is zero. This occurs when \(t = \frac{1}{1-c}\text{,}\) which corresponds to the point \((\frac{a}{1-c}, \frac{b}{1-c},0)\text{.}\)

Thus, for a point \((a,b,c)\) on the sphere with \(c \neq 1\text{,}\) stereographic projection \(\phi:\mathbb{S}^2 \to \mathbb{C}^+\) is given by

\begin{equation*}
\phi((a,b,c)) = \frac{a}{1-c}+\frac{b}{1-c}i.
\end{equation*}

Where does \(\phi\) send the north pole? To \(\infty\text{,}\) of course. A sequence of points on \(\mathbb{S}^2\) that approaches \(N\) will have image points in \(\mathbb{C}\) with magnitudes that approach \(\infty\text{.}\)

If we think of \(\infty\) as just another point in \(\mathbb{C}^+\text{,}\) it makes sense to ask about angles at this point. For instance, any two lines intersect at \(\infty\text{,}\) and it makes sense to ask about the angle of intersection at \(\infty\text{.}\) We can be guided in answering this question by stereographic projection, thanks to the following theorem.

Stereographic projection preserves angles. That is, if two curves on the surface of the sphere intersect at angle \(\theta\text{,}\) then their image curves in \(\mathbb{C}^+\) also intersect at angle \(\theta\text{.}\)

Thus, if two curves in \(\mathbb{C}^+\) intersect at \(\infty\) we may define the angle at which they intersect to equal the angle at which their pre-image curves under stereographic projection intersect. The angle at which two parallel lines intersect at \(\infty\) is 0. Furthermore, if two lines intersect at a finite point \(p\) as well as at \(\infty\text{,}\) the angle at which they intersect at \(\infty\) equals the negative of the angle at which they intersect at \(p\text{.}\) As a consequence, we may say that inversion about a circle preserves angle magnitudes at all points in \(\mathbb{C}^+\text{.}\)

In each case find \(T(\infty)\) and the input \(z_0\) such that \(T(z_0) = \infty\text{.}\)

a. \(T(z) = (3 - z)/(2z + i)\text{.}\)

b. \(T(z) = (z + 1)/e^{i\pi/4}\text{.}\)

c. \(T(z) = (az + b)/(cz + d)\text{.}\)

Suppose \(D\) is a circle of Apollonius of \(p\) and \(q\text{.}\) Prove that \(p\) and \(q\) are symmetric with respect to \(D\text{.}\) Hint: Recall the circle \(C\) in the proof of Theorem 3.2.14. Show that \(p\) and \(q\) get sent to points that are symmetric with respect to \(i_C(D)\text{.}\)

Determine the inverse stereographic projection function \(\phi^{-1}:\mathbb{C}^+ \to \mathbb{S}^2\text{.}\) In particular, show that for \(z = x + yi \neq \infty\text{,}\)

\begin{equation*}
\phi^{-1}(x,y) = \bigg(\frac{2x}{x^2+y^2+1},\frac{2y}{x^2+y^2+1},\frac{x^2+y^2-1}{x^2+y^2+1}\bigg).
\end{equation*}

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