Kenneth Kuttler

1760 CHAPTER 55. COMPLEX MAPPINGS

55.5.4 A Brief Review

First recall the definition of the metric on Ĉ. For convenience it is listed here again. Con-sider the unit sphere, S2 given by (z−1)2 + y2 + x2 = 1. Define a map from the complexplane to the surface of this sphere as follows. Extend a line from the point, p in the complexplane to the point (0,0,2) on the top of this sphere and let θ (p) denote the point of thissphere which the line intersects. Define θ (∞)≡ (0,0,2).

(0,0,2)

(0,0,1)p

θ(p)

CThen θ

−1 is sometimes called sterographic projection. The mapping θ is clearly con-tinuous because it takes converging sequences, to converging sequences. Furthermore, it isclear that θ

−1 is also continuous. In terms of the extended complex plane, Ĉ, a sequence,zn converges to ∞ if and only if θzn converges to (0,0,2) and a sequence, zn converges toz ∈ C if and only if θ (zn)→ θ (z) .

In fact this makes it easy to define a metric on Ĉ.

Definition 55.5.9 Let z,w ∈ Ĉ. Then let d (x,y) ≡ |θ (z)−θ (w)| where this last distanceis the usual distance measured in R3.

Theorem 55.5.10(Ĉ,d

)is a compact, hence complete metric space.

Proof: Suppose {zn} is a sequence in Ĉ. This means {θ (zn)} is a sequence in S2

which is compact. Therefore, there exists a subsequence,{

θznk

}and a point, z ∈ S2 such

that θznk → θz in S2 which implies immediately that d(znk ,z

)→ 0. A compact metric

space must be complete.Also recall the interesting fact that meromorphic functions are continuous with values

in Ĉ which is reviewed here for convenience. It came from the theory of classification ofisolated singularities.

Theorem 55.5.11 Let Ω be an open subset of C and let f : Ω→ Ĉ be meromorphic. Thenf is continuous with respect to the metric, d on Ĉ.

Proof: Let zn → z where z ∈ Ω. Then if z is a pole, it follows from Theorem 51.7.11that

d ( f (zn) ,∞)≡ d ( f (zn) , f (z))→ 0.

If z is not a pole, then f (zn)→ f (z) in C which implies

|θ ( f (zn))−θ ( f (z))|= d ( f (zn) , f (z))→ 0.

1760 CHAPTER 55. COMPLEX MAPPINGS55.5.4 A Brief ReviewFirst recall the definition of the metric on C. For convenience it is listed here again. Con-sider the unit sphere, S* given by (z— 1) +y* +x? = 1. Define a map from the complexplane to the surface of this sphere as follows. Extend a line from the point, p in the complexplane to the point (0,0,2) on the top of this sphere and let 6 (p) denote the point of thissphere which the line intersects. Define 0 (c°) = (0,0, 2).(0,0,2)Then 97! is sometimes called sterographic projection. The mapping @ is clearly con-tinuous because it takes converging sequences, to converging sequences. Furthermore, it isclear that @~! is also continuous. In terms of the extended complex plane, C, a sequence,Zn converges to ce if and only if 0z, converges to (0,0,2) and a sequence, z, converges toz€ C if and only if 6 (z,) + 6(z).In fact this makes it easy to define a metric on ¢.Definition 55.5.9 Let z,w € C. Then let d(x,y) = |@ (z) — @(w)| where this last distanceis the usual distance measured in R?.Theorem 55.5.10 (C.a) is a compact, hence complete metric space.Proof: Suppose {z,} is a sequence in C. This means {6 (z,)} is a sequence in S?which is compact. Therefore, there exists a subsequence, { OZn, } and a point, z € S? suchthat @z,, — Oz in S* which implies immediately that d (Zing +Z) — 0. A compact metricspace must be complete._Also recall the interesting fact that meromorphic functions are continuous with valuesin C which is reviewed here for convenience. It came from the theory of classification ofisolated singularities.Theorem 55.5.11 Let Q be an open subset of Cand let f : Q— C be meromorphic. Thenf is continuous with respect to the metric, d on C.Proof: Let z, — z where z € Q. Then if z is a pole, it follows from Theorem 51.7.11thatd(f (Zn) °°) =4(F (Zn), f(z) 49.If zis not a pole, then f (z,) + f (z) in C which implies19 (f (Zn)) — 9 (F (z))| =4(F (zn), F(z) > 9.