By Robert B. Ash, W.P. Novinger

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Be any sequence in Ω × Ω converging to (w, z) ∈ Ω × Ω. (zn ) → If w = z, then eventually wn = zn , and by continuity of f , g(wn , zn ) = f (wwnn)−f −zn f (w)−f (z) w−z = g(w, z). However, if w = z, then 1 wn −zn g(wn , zn ) = f (zn ) [zn ,wn ] f (τ ) dτ if wn = zn if wn = zn . In either case, the continuity of f at z implies that g(wn , zn ) → f (z) = g(z, z). Finally, the function z → g(w, z) is continuous on Ω and analytic on Ω \ {w} (because f is analytic on Ω). 13). 3 CHAPTER 3.

8. Characterize the entire functions f, g such that f 2 + g 2 = 1. ) 9. Let f and g be continuous mappings of the connected set S into C \ {0}. 3. CAUCHY’S THEOREM 9 k = 0, 1, . . Hence if f (s0 ) = g(s0 ) for some s0 ∈ S, then f ≡ g. (b) Show that C \ {0} cannot be replaced by C in the hypothesis. 3 Cauchy’s Theorem This section is devoted to a discussion of the global (or homology) version of Cauchy’s theorem. The elementary proof to be presented below is due to John Dixon, and appeared in Proc.

We conclude that Re(f (z)/f (z0 )) = 1 on D(z0 , δ). But if |w| = Re w = c, then w = c, hence f (z) = f (z0 ) on D(z0 , δ). By the identity theorem, f is constant on Ω. (b) If λ = +∞ there is nothing to prove, so assume λ < +∞. If |f (z0 )| = λ for some z0 ∈ Ω, then f is constant on Ω by (a). (c) If λ is deﬁned as in (b), then there is a sequence {zn } in Ω such that |f (zn )| → λ. But since Ω is bounded, there is a subsequence {znj } that converges to a limit z0 . If z0 ∈ Ω, then |f (z0 )| = λ, hence f is constant by (b).