Cauchy s Theorem (rigorous) In this lecture, we will study a rigorous proof of Cauchy s Theorem. We start by considering the case of a triangle.

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1 Cauchy s Theorem (rigorous) In this lecture, we will study a rigorous proof of Cauchy s Theorem. We start by considering the case of a triangle. Given a certain complex-valued analytic function f(z), for any triangle P within the domain of definition of f(z), we denote (always counter-clockwise orientation): I(P ) = P f(z) dz. Theorem 1 Let T be any triangle in C such that f(z) is analytic in a neighborhood of T. Then I(T ) = 0. First observation: For any triangle P, P 1 dz = z dz = 0. (1) This is proved by observing that in both cases we have obvious primitive functions (z and z 2 /2), and then applying the Fundamental Theorem of Calculus. P

2 The next step is a subdivision procedure which goes as follows: We first divide R into four congruent subtriangles T 1, T 2, T 3, T 4 and observe that I(T ) = I(T 1 ) + I(T 2 ) + I(T 3 ) + I(T 4 ). This is simply because all inner boundaries cancel out. The triangle inequality now gives I(T ) I(T 1 ) + I(T 2 ) + I(T 3 ) + I(T 4 ), from which we conclude that there is at least one subtriangle (which we can denote by T (1) ) which has the property that I(T (1) ) 1 I(T ). 4

3 We can now go on and split T (1) up into four equally large subtriangle and choose one of these (which we call T (2) ) with the property that I(T (2) ) 1 4 I(T (1) ) 1 42 I(T ). Continuing in this way we obtain a decreasing sequence of triangles T.T (1), T (2),..., T (k),... such that I(T (k) ) 1 4k I(T ), (2) and also the intersection of all T (k) consists of exactly one point z 0. This last statement follows from compactness and the fact that the diameters tend to zero. In fact, if we denote the diameter of T by D and the perimeter of T by p, then it is easily seen that diam(t (k) ) = D 2 k and per(t (k) ) = p 2 k. Since f(z) is complex differentiable at z 0, we can (given any ɛ > 0) find a δ > 0 so that f(z) f(z 0 ) f (z 0 ) z z 0 < ɛ for z z 0 < δ, or equivalently f(z) f(z 0 ) f (z 0 )(z z 0 ) < ɛ z z 0. (3)

4 Choose k so large that diam(t (k) ) = D 2 k < δ. Then inequality (3) is true on T (k) and we get I(T (k) = T f(z) dz (k) = (4) = f(z) f(z 0) f (z 0 )(z z 0 ) dz T (k) (by the first observation (1)) sup T (k) f(z) f(z 0 ) f (z 0 )(z z 0 ) per(t (k) ) ɛ diam(t (k) ) per(t (k) ) < ɛ D 2 k p 2 k = ɛdp 4 k Combining (2)) and (4)) we now obtain 1 4 k I(T ) I(T (k) ) ɛdp 4 k, or after multiplying by 4 k : I(T ) ɛdp. D and p are fixed constants, and since ɛ can be chosen arbitrarily small, we conclude that I(T ) = 0 and hence also I(T ) = 0. The proof in the case of a triangle is complete. We now proceed to the general case. Recall that a closed curve Γ is called null-homotopic

5 in a domain Ω if there exists a continuous map H : [0, 1] [0, 1] Ω with the following properties: 1. z(t) = H(1, t), 0 t 1 parametrizes Γ. 2. H(0, t) = z 0 for 0 t 1 and some z 0 Ω. 3. H(s, 0) = H(s, 1) for all 0 s 1. Theorem 2 (General Cauchy Theorem) Let f(z) be analytic in a domain Ω, and let Γ be a closed null-homotopic contour in Ω. Then f(z) dz = 0. (5) Γ Theorem 3 In particular (5) holds for all Γ and analytic f(z) if Ω is simply connected. Theorem 3 follows immediately from Theorem 2. The idea of the proof of Theorem 2 is to use the homotopy-map H to construct a subdivision of the interior of Γ into triangles where we can apply Theorem 1. There is however a slight technical problem related to the fact that Γ is not in general piecewise linear. To deal with this is not difficult if we start with the correct definition of a line-integral (i.e. not

6 referring to a parametrization but rather starting from the curve itself). In fact, in this case the following proposition becomes trivial (see Chap. 4.3 in Saff-Snider): Proposition 1 For every ɛ > 0, let Γ ɛ be a piecewise linear contour which is ɛ-close to Γ in the sense that 1. Every corner of Γ ɛ lies on Γ. 2. The maximal length d ɛ of the line-segments in Γ ɛ is less than ɛ. Then lim ɛ 0 Γ ɛ f(z) dz = Γ f(z) dz. Hence, it is enough to construct for each ɛ > 0 a Γ ɛ so that Cauchy s Theorem is valid for Γ ɛ. To prove Theorem 2 for Γ ɛ, we first note that K = H([0, 1] [0, 1]) Ω is compact (the continuous image of a compact set is compact). This implies that the infimum ɛ 0 of the distance between points in K and C \ Ω is strictly positive (see Appendix). In the following, we can assume that 0 < ɛ < ɛ 0. By uniform continuity, there is a δ > 0 such that if (s, t), (s, t ) [0, 1] [0, 1] satisfy

7 (s, t) (s, t ) < δ, then H(s, t) H(s, t ) < ɛ (see Appendix). We can now construct a subdivision of [0, 1] [0.1] into triangles with diameter less then δ as in the picture. We denote the set of these triangles by {S k } k=1. We can not use the images of these S k directly to construct the triangular subdivision in Ω. Rather, for each S k we let T k be the triangle spanned by the images under H of the three corners in S k. Note that some triangles may intersect or degenerate into lines or points. This does not effect any of the arguments below. It is important to observe that, although T k is not necessarily a subset of K, it still lies within

8 Ω. This is because the corners belong to K and hence by construction every point in T k lies within distance ɛ < ɛ 0 from K. To construct Γ ɛ and to prove that the integral of f(z) over Γ ɛ is zero, we proceed as follows: By Theorem 1, the integral of f(z) over each T k is zero. Let Γ ɛ be the closed (oriented) contour made up of all edges between corners corresponding to points with s = 1. Then 0 = N k=1 T k f(z) dz = Γ ɛ f(z) dz. In fact, when we integrate along all the boundaries of all the T k :s, then every inner edge will appear twice and the integrations in different directions will cancel out by the usual argument in vector-analysis. Similarly, the edges corresponding t = 0 and t = 1 cancel out against each other by the property 3 in the definition of homotopy above. Also, all edges along the boundary corresponding to s = 0 contribute with zero, since they are all equal to the point z 0 by property 2 in the definition of homotopy. So the only edges which remain are the ones corresponding to s = 1, which together make up Γ ɛ. By construction, all corners of Γ ɛ lie on Γ and the length of each line-segment is less than ɛ. This completes the proof.