Math 752 Spring 2015

Class 2/11

1 Winding numbers

Theorem 1 (Cauchy’s formula). G ⊆ C open, f : G → C analytic, γ closed path in G. If n(γ; w) = 0 for all w in the complement of G, then for a ∈ G\{γ} 1 Z f(z) n(γ; a)f(a) = dz. 2πi γ z − a Remark 1. The condition n(γ; w) = 0 is crucial since it prohibits γ winding about holes outside the domain of f. R Proof. By definition of winding number we need to show that f(z) γ(w − −1 R −1 z) dw = γ f(w)(w − z) dw for z ’inside’ γ. We consider their difference and show that it defines a bounded entire (which by Liouville’s theorem is constant). If n(γ; z) = 0 for z ∈ G (make a sketch of examples to understand this R −1 condition), then f(z) γ(w − z) dw = 0. Hence for such z

Z f(w) − f(z) Z f(w) dw = dw. (1) γ w − z γ w − z We note that the integral on the left defines an in all f(w)−f(z) of G (since for every w the function z 7→ w−z can be continued to an analytic function by setting its value to f 0(z) at z = w), and the integral on the right defines an analytic function on the complement of γ (differentiation may be performed under the integral sign), in particular on the complement of G. Let H = {w ∈ C : n(γ, w) = 0}. Consider (R f(w)−f(z) γ w−z dw if z ∈ G, F (z) = R f(w) γ w−z dw if z ∈ H.

It follows from (1) that F is well-defined (This is an issue since G ∩ H 6= ∅), and it is entire (note that C = H ∪ G). Since γ is closed, it is compact, and the right hand side goes uniformly to zero as |z| → ∞. Hence f is bounded, and by Liouville’s theorem constant. The constant can only be zero by considering lim|z|→∞ f(z). We get in particular Z 1 Z f(w) f(z) dw = dw γ w − z γ w − z for all z ∈ G, which is the claimed identity.

This formula extends to sums over closed . The winding number P condition can be relaxed to n(γk; w) = 0 for every w in the complement of the union of the paths. The substitution f(z) = g(z)(z − a) in this theorem gives then

Theorem 2. Let G be an open subset of the and f : G → C be P analytic. If γ1, ..., γn are closed rectifiable curves in G with k n(γk; w) = 0 for all w the complement of G, then n X Z f(w)dw = 0. k=1 γk Proof. This will be a midterm question.

The integral formulas may be differentiated under the integral sign.

Theorem 3 (Morera’s theorem). G region, f : G → C continuous so that R for each triangular path Γ in G the identity Γ f = 0 holds. Then f is analytic in G. Proof. May assume that G is a disk with center a. Let [a, z] be the line segment connecting a and z. Define F by Z F (z) = f(w)dw. [a,z] We will show that F 0 = f, and since f is continuous, this will imply that F is analytic. But then also F 0 = f is analytic. The assumed independence of triangular paths implies Z F (z) − F (z0) = f(w)dw [z0,z] and continuity of f implies that

F (z) − F (z0) − f(z0) ≤ sup{|f(w) − f(z0)| : w ∈ [z0, z]} → 0 z − z0 as z → z0. 2 and simple connectivity

We need to investigate under which conditions on the paths γi (and/or G) P the identity n(γi; w) = 0 for w∈ / G is satisfied.

Definition 1. Let γ0, γ1 : [0, 1] → G be two closed paths. Then γ0 is homotopic to γ1 if there exists continuous Γ : [0, 1] × [0, 1] → G so that

Γ(., 0) = γ0, Γ(., 1) = γ1, and s 7→ Γ(s, t) is closed for every t. Notation: γ0 ∼ γ1. Note that ∼ defines an equivalence relation.