Consider again inversion about the circleCgiven by|z−z0|=r, and observe

that points close toz0 get mapped to points in the plane far away from from

z0. In fact, a sequence of points inC whose limit isz0 will be inverted to a

sequence of points whose magnitudes go to ∞. Conversely, any sequence of points inChaving magnitudes marching o to∞will be inverted to a sequence

of points whose limit isz0.

With this in mind, we dene a new point called the point at innity, denoted ∞. Adjoin this new point to the plane to get the extended plane, denoted asC+. Then, one may extend inversion in the circleC to include the

pointsz0 and∞. In particular, inversion ofC+ in the circle Ccentered atz0

with radiusr,iC :C+→C+, is given by

iC(z) = r2 (z−z0)+z0 ifz6=z0,∞; ∞ ifz=z0; z0 ifz=∞.

Viewing inversion as a transformation of the extended plane, we denez0

and∞to be symmetric points with respect to the circle of inversion.

The space C+ will be the canvas on which we do all of our geometry, and

it is important to begin to think of∞as one of the gang, just another point to consider. All of our translations, dilations, and rotations can be redened to include the point∞.

So where is ∞in_{C}+_{? You approach}_{∞}_{as you proceed in either direction}

along any line in the complex plane. More generally, if{zn} is a sequence of

complex numbers such that|zn| → ∞asn→ ∞, then we say lim

n→∞zn =∞.By

convention, we assume∞is on every line in the extended plane, and reection across any line xes∞.

Theorem 3.3.1. Any general linear transformation extended to the domain

C+ xes ∞.

Proof. If T(z) =az+b where a andb are complex constants with a6= 0,

then by limit methods from calculus, as |zn| → ∞, |azn+b| → ∞ as well.

Thus,T(∞) =∞.

So, with new domain C+, we modify our xed point count for the basic

transformations:

The translation Tb ofC+ xes one point (∞).

The rotation about the originRθofC+ xes 2 points (0 and∞).

The dilation T(z) =kz of_{C}+ _{xes 2 points, (0 and}_{∞}_{).}

The reectionrL(z)ofC+about lineLxes all points onL(which now

3.3. The Extended Plane 45

Example 3.3.2: Some transformations not xing ∞. The following function is a transformation ofC+

T(z) = i+ 1

z+ 2i,

a fact we prove in the next section. For now, we ask whereT sends∞, and which point gets sent to∞.

We tackle the second question rst. The input that gets sent to ∞ is the complex number that makes the denominator 0. Thus, T(−2i) =∞.

To answer the rst question, take your favorite sequence that marches o to ∞, for example, 1,2,3, . . .. The image of this sequence, T(1), T(2), T(3), . . . consists of complex fractions in which the numerator is constant, but the denominator grows unbounded in magnitude along the horizontal line Im(z) = 2. Thus, the quotient

tends to 0, andT(∞) = 0.

As a second example, you can check that if T(z) = iz+ (3i+ 1)

2iz+ 1 ,

thenT(i/2) =∞andT(∞) = 1/2.

We emphasize that the following key results of the previous section extend to

C+ as well:

There exists a unique cline through any three distinct points in C+. (If

one of the given points in Theorem 3.2.4 is ∞, the unique cline is the line through the other two points.)

Theorem 3.2.8 applies to all pointsz not onC, includingz=z0 or∞.

Inversion about a cline preserves angle magnitudes at all points in C+

(we discuss this below).

Inversion preserves symmetry points for all points inC+(Theorem 3.2.12

holds ifporqis∞).

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 ∞and the common center.

Stereographic Projection We close this section with a look at stereo- graphic projection. By identifying the extended plane with a sphere, this map oers a very useful way for us to think about the point∞.

Denition 3.3.3. The unit 2-sphere, denotedS2, consists of all the points

in 3-space that are one unit from the origin. That is,

S2={(a, b, c)∈R3 |a2+b2+c2= 1}.

We will usually refer to the unit 2-sphere as simply the sphere. Stereo- graphic projection of the sphere onto the extended plane is dened as follows. LetN = (0,0,1)denote the north pole on the sphere. For any point P 6=N on the sphere,φ(P)is the point on the ray−−→N P that lives in thexy-plane. See Figure 3.3.4 for the image of a typical pointP of the sphere.

x y z N P = (a, b, c) φ(P)

Figure 3.3.4: Stereographic projection.

The stereographic projection mapφcan be described algebraically. The line throughN = (0,0,1)andP = (a, b, c)has directional vector−−→N P =ha, b, c−1i, so the line equation can be expressed as

~r(t) =h0,0,1i+tha, b, c−1i.

This line intersects thexy-plane when itszcoordinate is zero. This occurs
whent=_{1−}1_{c}, which corresponds to the point(_{1−}a_{c},_{1−}b_{c},0).

Thus, for a point(a, b, c)on the sphere withc6= 1, stereographic projection

φ:S2→C+ is given by

φ((a, b, c)) = a 1−c +

b

1−ci.

Where does φsend the north pole? To∞, of course. A sequence of points on S2 that approaches N will have image points in C with magnitudes that

approach∞.

Angles at ∞ If we think of ∞as just another point inC+, it makes sense

to ask about angles at this point. For instance, any two lines intersect at∞, and it makes sense to ask about the angle of intersection at ∞. We can be guided in answering this question by stereographic projection, thanks to the following theorem.

Theorem 3.3.5. Stereographic projection preserves angles. That is, if two curves on the surface of the sphere intersect at angleθ, then their image curves inC+ also intersect at angleθ.

Thus, if two curves in C+ intersect at ∞ we may dene 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 ∞ is 0. Furthermore, if two lines intersect at a nite point p as well as at ∞, the angle at which they intersect at ∞ equals the negative of the angle at which they intersect atp. As a consequence, we may say that inversion about a circle preserves angle magnitudes at all points inC+.

Exercises

1. In each case ndT(∞)and the inputz0 such thatT(z0) =∞.

a. T(z) = (3−z)/(2z+i).

b. T(z) = (z+ 1)/eiπ/4_{.}