Section3.3The Extended Plane

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 $\infty\text{.}$

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 $\infty$).

Example3.3.2Some transformations not fixing $\infty$

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.
Stereographic Projection

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 $\infty\text{.}$

Definition3.3.3

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 $\infty\text{.}$

Angles at $\infty$

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.

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 $\mathbb{C}^+\text{.}$

SubsectionExercises

1

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

2

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 $i_C(D)\text{.}$

3

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*}