14.12. SIMPLY CONNECTED REGIONS 367

Theorem 14.11.2 Suppose f is analytic on C, having values in X a Banach space,and that ∥ f (z)∥ is bounded for z ∈ C. Then f is constant.

Proof: It was shown above that if γr is a counter clockwise oriented parametrizationfor the circle of radius r centered at z, then

f ′ (z) =1

2πi

∫γr

f (w)

(w− z)2 dw if |z|< r

and so ∥ f ′ (z)∥≤ 12π

C2πr 1r2 where ∥ f (z)∥<C for all z and this is true for any r so let r→∞

and you can conclude that f ′ (z)= 0 for all z∈C. However, continuing this argument showsthat f (k) (z) = 0 for all z and for each k ≥ 1. Thus the power series for f (z) , which exists

by Theorem 14.8.3, is f (z) = f (0)+∑∞k=1

f (k)(0)k! zk = f (0) .

Alternatively, consider the line segment between z and w and use Problem 26 on Page78 on t→ f (z+ t(w− z)). ■

This leads right away to the shortest proof of the fundamental theorem of algebra.

Theorem 14.11.3 Let p(z) be a non constant polynomial with complex coeffi-cients. Then p(z) = 0 for some z ∈ C. That is, p(z) has a root in C.

Proof: Suppose not. Then 1/p(z) is analytic on C. Also, the leading term dominatesthe others and so 1/p(z) must be bounded. Indeed, lim|z|→∞ (1/ |p(z)|) = 0 and the contin-uous function z→ 1/ |p(z)| achieves a maximum on any bounded closed ball centered at 0.By Liouville’s theorem, this quotient must be constant. However, by assumption, 1/p(z)is not constant. Hence there is a root of p(z). ■

14.12 Simply Connected RegionsThe Riemann sphere is a useful way to present the concept of the extended complex plane.Consider 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. Think of ∞ as a point which, when added in to C gives an“extended complex plane” Ĉ in such a way that ∞ corresponds to θ (∞)≡ (0,0,2).

(0,0,2)

(0,0,1)p

θ(p)

CThen θ

−1 is sometimes called sterographic projection.

Definition 14.12.1 Let Ĉ ≡ C∪{∞} be a metric space as follows.d (z,w)≡ ρ (θz,θw) where ρ is the distance in R3.

Lemma 14.12.2 The above does make Ĉ into a metric space. zn → ∞ is the same aslimn→∞ |zn|= ∞. In fact Ĉ is a compact metric space.

Proof: It is clear that d (z,w)≥ 0 and equals 0 if and only if z = w because θ is one toone. The main issue is the triangle inequality. However, d (z,w)+d (w,u) ≡ ρ (θz,θw)+