# 7. Cauchy s integral theorem and Cauchy s integral formula

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1 7. Cauchy s integral theorem and Cauchy s integral formula 7.. Independence of the path of integration Theorem 6.3. can be rewritten in the following form: Theorem 7. : Let D be a domain in C and suppose that f C(D). Suppose further that F (z) is a continuous antiderivative of f(z) through D D. Let z 0 and z T be distinct points in D. Then the integral zt z 0 f(z)dz does not depend on the path of integration, e.g., for every smooth contour γ D which start at z 0 and terminates at z T, we have zt z 0 f(z)dz = f(z)dz = F (z T ) F (z 0 ). γ Theorem 7.. is called Theorem on the depend of the path of integration. From this theorem we get the following obvious consequence: Corollary 7.2. : Under the conditions on f of Theorem 7.., let γ be a smooth closed contour which lies entirely in D. Then f(z)dz = 0 γ Our coming considerations are based on the following theorem: Theorem 7.3. Let D be a domain in C and f C(D). Then the following statements are equivalent: () f has a continuous antiderivative in D; (2) for every loop γ lying in D. We call such contours loops. γ f(z)dz = 0

2 (3) The integral z2 f(z)dz z is independent of the path of integration; e.g., if γ and γ 2 share the same initial and terminal points, then f(z)dz = f(z)dz. γ γ 2 Proof: Since the implications ) 2) and 2) 3) already established (Theorem 7. and Theorem 7.2), we will concentrate on the proof of 3) ). Select an arbitrary point z 0 D and let z D. Set F (z) := z z 0 f(z)dz. We claim that F (z) is an antiderivative of f in D. Before, we notice that the integral is well defined - because of the connectedness of the domain D there ix a contour which combines z 0 and z. We shall show that F (z + ) F (z) f(z), 0. Indeed, z+ F (z + ) F (z) (w)dw z =, where we integrate along a segment lying completely in the domain. Regarding Theorem 6.4, we may write z+ F (z + ) F (z) f(z) = (f(w) f(z)dw z = f(w) f(z) [z,z+] 0 as 0. Thus F (z) = f(z). This concludes the proof. 2

3 7.2. Continuous deformations of loops. Definition: The loop γ is said to be continuously deformable to the loop γ 2 in the domain D, if there exists a function z(s, t), (s, t) ([0, ] [0, ]) that satisfies the conditions:. z(s, t) C 2 ([0, ] [0, ]); 2. For each fixed s [0, ] the function z(s, t) parametrizes a loop in D; 3. The function z(0, t) parametrizes γ ; 4. The function z(, t) parametrizes γ 2. Example: THE function z(s, t) := ( + s)e 2πit, 0 s, t deforms continuously the circle C 0 () into the circle C 0 (2) Deformation Invariance Theorem. We first recall the definition of a simply connected domain. Definition: Any domain D in the complex plane C possessing the property that every loop in D can be continuously deformed in D to a point is called simply connected. ℵ. For example, any disk D a (r), r > 0 is a simply connected domain. Now we are in position to prove the Deformation Invariance Theorem. Theorem 7.3. Let D be a domain in C and suppose that f A(D). If γ, γ 2 are continuously deformable into each other closed curves, then f(z)dz = f(z)dz. γ γ 2 Proof: Fix s [0, ] and set := z(s, t), t [0, ]. We shall show that the function I(s) := f(z)dz equals a constant. Indeed, z(s, t) f(z)dz = f(z(s, t)) dt. t Look at the derivative of I(s); we have I z(s, t) (s) = f(z(s, t)) dt = t 3

4 = [ f(z(s, t)) dt On the other hand, t) (f(z(s, t)) z(s, ) = dt s z(s, t) z(s, t) s t f(z(s, t)) dt + f(z(s, t)) 2 z(s, t) ]dt. s t z(s, t) z(s, t) + f(z(s, t)) 2 z(s, t) t s t s. The theorem by Weierstrass about the independence of second order derivatives of the order of differentiation guarantees that di(s) ds = 0 t) [f(z(s, t)) z(s, ]dt = dt s z(s, t) z(s, t) = f(z(s, )) (s, ) f(z(s, 0)) (s, 0). s s As we know, the curves are closed which means that for every s [0, ] z(s, 0) = z(s, ). Thus I(s) = f(z)dz = f(z)dz. γ γ 2 Cauchy s integral theorem An easy consequence of Theorem 7.3. is the following, familiarly known as Cauchy s integral theorem. Theorem 7.4.If D is a simply connected domain, f A(D) and is any loop in D, then f(z)dz = 0. Proof: The proof follows immediately from the fact that each closed curve in D can be shrunk to a point. We conclude the following Theorem 7.5. Let D be a domain in C and f A(D) C(D). Set D :=. Then f(z)dz = 0. 4

5 Proof: Without losing the generality, we may assume that all components of are smooth curves. It D is simply connected, then we are done. Assume that D is double connected and let = 2. The domain is positively orientated with respect to ; let be the positive component (clockwise) and 2 the negative (counterclockwise) ( = ( 2 ).) Without loosing the generality we suppose that and 2 are continuously deformable into each other, and by Theorem 7.3. f(z)dz = f(z)dz. 2 () On the other hand f(z)dz = f(z)dz + f(z)dz = f(z)dz f(z)dz = Joining (), we arrive at the statement. The Cauchy s integral theorem indicates the intimate relation between simply connectedness and existence of a continuous antiderivative. Theorem 7.6. Let D be simply connected in C and f A(D). Then f possesses a continuous antiderivative and its contour integral does not depend on the path of integration. The proof follows from Theorem Cauchy s integral formula Theorem 7.7. Let D be a domain in C, := D and f A(D) C(D). Then, for every point a D the representation f(a) = f(z) dz (2) 2πi z a holds. Proof: Take r sufficiently small (e.g. D a (r) D) and consider z a =r (the circle is traversed once in the positive direction). We have f(z) 2πi z a dz = 2π f(a + re iθ ) ire iθ dθ. 2πi re iθ z a =r 5 0 f(z) z a dz.

6 Letting now r 0 we obtain that 2πi z a =r f(z)dz = f(a). To complete the proof, we apply Theorem 7.5. with respect to the function f(z) and to the domain D \ D z a a(r). As an application, we provide the mean value theorem for harmonic functions. Theorem 7.7. Let h be harmonic in the disk D a (R), R > 0. Then h(a) = 2π h(a + Re iθ )dθ. 2π 0 Proof: : We recall that the real and the imaginary components of an analytic function are complex conjugate harmonic functions. Let f A(D a (R)) be such that h(z) := Rf(z). Denote the imaginary component by k(z). Using (2), we get f(f) = h(z) + ik(z), z K a (R). h(a) + ik(a) = 2πi C a(r) h(ζ) + ik(ζ) dζ. ζ a Hence, h(z) = 2π h(a + Re iθ ) ire iθ dθ. 2πi 0 Re iθ The statement follows after completing the needed cancellations. Exercises:. Prove that C a(ρ) { dz 0, m (z a) = m 2πi, m = 6

7 2. Prove that C 0 (ρ) { dz 0, a > ρ (z a) = 2πi, a < ρ 3. Which of the following domains are simply connected? a) {z, Im z < }; b) {z, < z < 2}; c){z, z < }; d) {z, z > }; e) {z, z < } \ {z, 0 < Re z < }. 3. Calculate + z dz, 2 with S being the interval [, + i]. 4. Show that if f(z) is of the form S f(z) = n k=0 A k z k + g(z), where g(z) is analytic outside C 0 (), then f(z)dz = 2πiA. z = (By definition, :=, C z = C 0 () 0() traversed once in positive direction.) 5. Let P be a polynomial of degree 2, such that all zeros lie in D (R), R > 0. Show that z =R dz = 0. P (z) Hint Apply Theorem 7.5. with respect to the annulus {z, R < z < R + r} and then let r increase to infinity. 7

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