0 and qo = bo, q-1 = bl, q-2 = b2, . , q-,,, = bm; likewise bm+1 = bm+2 = . = 0 by the absence of additional negative powers in (2-6). Thus g(z) is a polynomial and f(z) is rational. D. REVIEW OF COMPLEX ANALYSIS 30 COROLLARY 1 A nonconstant rational function w = f(z) of degree n assumes every value a on the w sphere for exactly n values of z on the z sphere (using conventions of multiplicity of root and order of pole). PROOF We start with formulas (2-3) for f(z) = Pl(z)/P2(z), and we make the simplifying assumption that f (oo) = oo.
1-S Illustration for remarks on homology. Let us view a very familiar situation in the light of topological concepts by referring to Fig. 1-3. Here a function f (z) is analytic in some region (R* (not marked in the diagram), but 63 is a region whose closure lies interior to 63*. Then we would say (1-26) f, f(z) dz + f,, f(z) dz + fe f(z) dz = 0 For proof, we would divide 63 into 631 and 632 (as shown in Fig. 1-3b), and in this manner we would take dz = fx=f(z) dz = 0 Then we would show that the chain 6C1 + 6C2 is equivalent to e1 + e2 + e3 (by the cancellation of retraced portions in usual elementary fashion).