33 Cauchy Integral Formula

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1 33 AUHY INTEGRAL FORMULA October 27, 2006 PROOF Use the theorem to write f ()d + 1 f ()d + 1 f ()d = 0 2 f ()d = f ()d f ()d This is the deformation principle; if you can continuously deform 1 to 2, without crossing points where f is not analytic, then the value of f ()d is preserved. EXAMPLE Suppose is a simple, closed contour which contains the origin. Then d = 2πi. Indeed, we know this is true for the unit circle, thus for any circle, and in fact for any simple closed contour. 33 auchy Integral Formula We start with a slight extension of auchy s theorem. LEMMA Let be a simple closed contour. Suppose that f is analytic on an inside, except at a point 0 which satisfies lim ( 0 ) f () = 0. 0 Then f ()d = 0. PROOF Let r be the contour which wraps around the circle of radius r centered at 0 exactly once in the clockwise direction. By the deformation principle, if r is sufficiently small (i.e., if r is in the interior of ), then f ()d = r f ()d. Given any ɛ > 0, there exists a δ > 0 so that if 0 < δ, then ( 0 ) f () < ɛ, and f () < ɛ 0. If necessary, shrink δ so that N δ( 0 ) is in the interior of. If 0 < r < δ, then f ()d = f ()d r f () d r ɛ < d r 0 = ɛ d r r = ɛ r 2πr Lecture 27 = 2πɛ. olorado State University 49 M419: Introduction to omplex Variables

2 33 AUHY INTEGRAL FORMULA October 27, 2006 We have shown that f ()d < 2πɛ for all ɛ, so that f ()d = 0. We can use this to prove the auchy integral formula. THEOREM Suppose f is analytic everywhere inside and on a simple closed positive contour. If 0 is any point interior to, then f ( 0 ) = 1 2πi f () 0 d. PROOF Since f is analytic at 0, we have lim 0 ( 0) f () f ( 0) 0 = 0. By the slightly souped-up auchy s theorem, we have But we can rewrite this integral as f () f ( 0 ) 0 d = 0. f () f ( 0 ) f () f ( d = d 0 ) d f () 1 d = f ( 0 ) d 0 0 = f ( 0 )2πi. APPLIATIONS We can often use this to evaluate contour integrals. Here are couple of problems from the book: Let be the positively oriented boundary of the square whose sides lie along the lines x = ±2 and y = ±2. Let correspond to the square x = ±4, y = ±4. ompute the following four integrals: The point with all these is to identify each of them as g() 0, and thus realie them as g( 0 ): a. exp( ) (π/2) d. Use f () = exp( ); analytic everywhere, and in particular on and inside. Then the integral in question is I a = f () π/2 d = 2πi f (π/2) = 2πi exp( π/2) olorado State University 50 M419: Introduction to omplex Variables

3 33 AUHY INTEGRAL FORMULA October 27, 2006 b. 2+1 d d = 1 2 ( 1 2 ) = 1 2πi f ( 1/2) where f () = 2 = πi/2 c. d. Inside the contour, there s just a single root of the denominator, namely, = So d = f () d where f () = f () d = 2πi f (0) = 2πi/9 d d. For this one, there are three poles to worry about. Let 1 be a small loop around 3i; 2 is a small loop around 0; and 3 a small loop around 3i. Then by the deformation principle, d = j d = j d /(( + 3i) d 1 3i cos(3i) = 2πi 3i(3i + 3i) cos( 3i) 3 d = 2πi + 9 3i( 3i 3i) = 2πi/9 2 For a particular contour, we retrieve Gauss s mean value theorem, as follows. LEMMA Suppose that f is analytic on and in a circle of radius R around 0. Then f ( 0 ) = 1 2π 2π 0 f ( 0 + R exp(it))dt. olorado State University 51 M419: Introduction to omplex Variables

4 33 AUHY INTEGRAL FORMULA October 27, 2006 REMARK This is a continuous analogue of something we did for homework, for polynomials. PROOF Let be a contour which wraps around the circle of radius R around 0 exactly once in the counterclockwise direction. On one hand, we have: f ( 0 ) = 1 f () 2πi ( 0 ) d On the other hand, this is = 1 2π 2πi 0 = 1 2π 2πi = 1 2π 0 2π 0 f ((t)) (t)dt f ( 0 + R exp(it)) ( 0 + R exp(it)) 0 (ir exp(it))dt f ( 0 + R exp(it))dt. As a consequence of auchy s theorem, we can prove the following expression for the derivative of f. THEOREM Suppose f is analytic in a domain containing. Let be a simple closed contour in D which goes around exactly once. Then f () = 1 2πi (w ) 2 dw. PROOF Once you know what you re trying to prove, this isn t too hard. Let M be the maximum value of on. Let m be the minimum distance from to a point on, and let m = m /2. Suppose that < m. Then by the auchy integral theorem, we have f ( + ) f () 1 2πi (w ) 2 dw = 1 2πi (w ( + )) dw 1 2πi (w ) dw 1 2πi (w ) 2 dw = 1 2πi (w ( + ))(w ) 2 dw Then take absolute values, etc. For each w, w m > m and w ( + ) > m, so f ( + ) f () 1 2πi (w ) 2 dw M 2π m 3 dw = M 2πm 3. olorado State University 52 M419: Introduction to omplex Variables

5 33 AUHY INTEGRAL FORMULA October 27, 2006 Now, start taking ever-smaller... Lecture 28 In fact, we can use this to prove that derivatives of all orders exist, as follows. THEOREM function Suppose that f is continuous on and inside the simple closed contour. Then each is analytic in D, with derivative F n () = (w ) n dw F n() = nf n+1 (). PROOF We omit the proof that F 1 is continuous. At least if f is analytic, we have already shown that F 1 is analytic, and its derivative is F 2. We now proceed by induction, and suppose that the theorem is true for all functions f and all orders less than n. Fix a value, and construct the function g(w) = g (w) = (w ). For any complex number α, let G n (α) = = g(w) (w α) n dw (w α) n (w ). Suppose that N r () is any neighborhood of whose closure is contained inside. If w, and if + is inside N r (), then w is bounded away from ero. Since f is continuous, this shows that G n ( + ) is bounded on < r. Then we have identities F n () = G n 1 () ( ) 1 F n ( + ) G n 1 ( + ) = (w ( + )) n 1 (w ( + )) n 1 dw (w ) ( ) = (w )(w ( + )) n dw = G n ( + ). Moreover, since g(w) is continuous on and inside the contour (away from ), we may assume that G n 1 is differentiable, with derivative (n 1)G n (). We now show that F n is continuous at, and then differentiable. For the former, we have F n ( + ) F n () = G n 1 ( + ) G n 1 () + G n ( + ) olorado State University 53 M419: Introduction to omplex Variables

6 34 BOUNDEDNESS AND THE MAXIMUM MODULUS PRINIPLE October 27, 2006 By the continuity of G n 1, lim (F n( + ) F n ()) = lim (G n 1( + ) G n 1 ()) + lim G n( + ) But G n is bounded, so = 0, as desired. To show differentiability, use: F n ( + ) F n () G lim = n 1 ( + ) G n 1 () lim + G n ( + ) 0 0 = (n 1)G n () + G n () since G n is now continuous! = ng n () = nf n+1 (). In summary, we have the following representation of all derivatives of an analytic function: f (n) () = n! dw. 2πi (w ) n+1 Lecture 29 [Morera] Suppose f is continuous on D. If for every closed contour in D we have f ()d = 0, then f is analytic in D. THEOREM PROOF By hypothesis and the auchy-goursat theorem, f admits an antiderivative, F. But if F has one derivative (namely, f ), then it has derivatives of all orders. In particular f = F must have a derivative. 34 Boundedness and the Maximum Modulus Principle an we ever fit the entire complex plane into a finite circle? If we re willing to use a function which is merely continuous, this is no problem; use f (x) = 1 +. In this section we ll prove Liouville s theorem, which says that this is impossible if f is analytic. olorado State University 54 M419: Introduction to omplex Variables