X.9 Revisited, Cauchy s Formula for Contours

Transcription

1 X.9 Revisited, Cauchy s Formula for Contours Let G C, G open. Let f be holomorphic on G. Let Γ be a simple contour with range(γ) G and int(γ) G. Then, for all z 0 int(γ), f (z 0 ) = 1 f (z) dz 2πi Γ z z 0 Also, for all integers n 0, f (n) (z 0 ) = n! 2πi Γ f (z) dz (z z 0 ) n+1

2 Proof of the Second Formula Let z 0 int(γ). f (z) Set g(z) = (z z 0 ) n+1 for z G {z 0}. Note that g is holomorphic on G {z 0 }. Because z 0 int(γ), it is not in the range of Γ. By the Residue Theorem g(z) dz = 2πi ind Γ (z 0 ) res z0 (g(z)) Γ For z near z 0, f (z) = k=0 a k (z z 0 ) k with a k = f (k) (z 0 ) k!

3 Proof Continued So, for z near z 0 but not equal to z 0, g(z) = a k (z z 0 ) k n 1 k=0 Recall that the residue of g at z 0 is the coefficient of (z z 0 ) 1 in the Laurent Series for g. Thus res z0 (g) = a n = f (n) (z 0 ) n! Because Γ is a simple contour, its winding numbers are 0 or 1 (by definition). Because z 0 int(γ), its winding number is not zero (by def n of int(γ)). Thus ind Γ (z 0 ) = 1. Thus Γ g(z) dz = 2πi 1 f (n) (z 0 ) n! and f (n) (z 0 ) = n! g(z) dz 2πi Γ

4 X.11, The Argument Principle Let G be an open subset of C. Let Γ be a simple contour with range(γ) G and int(γ) G. Let f be holomorphic on G and nonzero on the range of Γ. Then the number of zeros of f in int(γ), counted according to multiplicity, is equal to 1 f (z) 2πi Γ f (z) dz

5 Proof, Slide 1 Claim: f has finitely many zeros in int(γ). ext(γ) is open. ext(γ) contains a neighborhood of infinity. Because int(γ) range(γ) = C ext(γ), we have H = int(γ) range(γ) both closed and bounded, and hence compact. Suppose that f has infinitely many zeros in int(γ) (and hence also in H). Because H is compact, these zeros have a cluster point z in H. Because H G, and f is holomorphic (and hence continuous) on G, we have f (z) = 0. If z were a zero of finite order, it would be isolated. Since it is not isolated, z is a zero of infinite order for f. Since f is nonzero on the range of Γ, we have z int(γ).

6 Proof, Slide 2 Let D be the component of z int(γ). Because int(γ) is an open set, D is also an open set. With D connected and open, f holomorphic on D, and f having a zero of infinite order in D, f is zero on all of D. Let A = inf S where S = { x 0 : z + x / int(γ) } Along the ray {z + x x 0}, we have z + x ext(γ) for all x sufficiently large. Therefore S is not empty. Since S is not empty, and bounded below by 0, inf S is the greatest lower bound of S (and a real number).

7 Proof, Slide 3 Suppose z + A int(γ). Since int(γ) is open, there is some δ > 0 so that w (z + A) < δ w int(γ) So, for x A < δ, we have (z + x) (z + A) < δ and thus z + x int(γ). Consequently, (A δ, A + δ) S =. By the definition of A, [0, A) S =. So A + δ > A is a lower bound for S, greater than the greatest lower bound. This contradiction proves that z + A / int(γ). Since z + 0 int(γ), this proves that A > 0.

8 Proof, Slide 4 Suppose that z + A ext(γ). Since ext(γ) is open, there is some δ > 0 so that w (z + A) < δ w ext(γ) So, for x A < δ, we have (z + x) (z + A) < δ and thus z + x ext(γ). Consequently, (A δ, A + δ) S. So A δ/2 S but A δ/2 < A. This contradicts A being a lower bound for S. That proves that z + A / ext(γ). By this slide and the previous slide, z + A is in the range of Γ.

9 Proof, Slide 5 By the definition of A, we have z + x int(γ) for x [0, A), and z + x is clearly connected by a line segment in int(γ) to z. So z + x D for x [0, A). With z + A range(γ) G, and f holomorphic on G, we have f continuous at z + A. So f (z + A) = lim x A f (z + x) = lim x A 0 = 0 This contradicts the assumption that f is nonzero on the range of Γ. Thus, f has at most finitely many zeros in int(γ). Let z 1,..., z p be the zeros of f in int(γ), with multiplicities m 1,... m p.

10 Proof, Slide 6 By hypothesis, f is nonzero on range(γ), range(γ) G, G is open and f is holomorphic (hence continuous) on G. At each z range(γ), there is some δ z > 0 such that Let D(z, δ z ) G and 0 / f (D(z, δ z ) G 1 = int(γ) z range(γ) D(z, δ z ) We have G 1 open, G 1 G, f holomorphic on G 1, f nonzero on G 1 {z 1,..., z p }, int(γ) G 1 and range(γ) G 1.

11 Proof, Slide 7 Suppose f has no zeros in int(γ). Then f /f is holomorphic on G 1. By Cauchy s Theorem for contours, Section IX.10, applied to f and G 1, f (z) f (z) dz = 0 Thus Γ 1 f (z) 2πi Γ f (z) dz = 0 which is the correct number of zeros.

12 Proof, Slide 7 Suppose that f does indeed have zeros. Then f /f is holomorphic on G 1 {z 1,..., z p }. By the Residue Theorem applied to f /f and G 1, Γ f (z) dz = 2πi f (z) p ind Γ (z k ) res zk (f /f ) k=1 Because z k int(γ) and Γ is a simple contour, ind Γ (z k ) = 1. By Exercise VIII.12.3, we have res zk (f /f ) = m k. Thus Γ f (z) dz = 2πi f (z) p k=1 m k and 1 f (z) 2πi Γ f (z) dz = p k=1 m k

13 X.12 Rouché s Theorem Let G be an open subset of C. Let K G, K compact. Let f and g be holomorphic on G such that, for all z in the boundary of K, f (z) g(z) < f (z) Then, counting according to multiplicities, f and g have the same number of zeros in the interior of K. The hypotheses imply that both f and g are nonzero on the boundary of K : Consequently, g(z) f (z) (g f )(z) < f (z) f (z) < g(z) f (z) < f (z) and thus g(z) > f (z) f (z) = 0.

14 Proof, Slide 1 Let G 1 = interior(k ) B where B = { z G : f (z) g(z) < f (z) }. Because f and g are holomorphic on G, and hence continuous there, B is an open set. Since interior(k ) is open, G 1 is open (the union of open sets is open). Recall that K is the union of its interior and its boundary. Since its boundary is a subset of B by hypothesis, we have K G 1. By the Separation Lemma, there is a simple contour Γ with range(γ) G and K int(γ) G.

15 Proof, Slide 2 For t [0, 1], let f t = (1 t)f + tg. Let m t be the number of zeros of f t in the interior of K, counted according to multiplicities. Note that f 0 = f and f 1 = g. Also f t = f + t(g f ). On B we have and thus f t is nonzero on B. f t f = t g f g f < f Consequently, all the zeros of f t in G 1 are in interior(k ). Since int(γ) G 1, all the zeros of f t in int(γ) are in interior(k ). Conversely, since K int(γ), every zero of f t in the interior of K is also in int(γ). So No. of zeros in int(γ) = m t

16 Proof, Slide 3 By the Argument Principle, the number of zeroes of f t in int(γ) equals 1 f t (z) 2πi Γ f t (z) dz We will argue later that the integral above is a continuous function of t, 0 t 1. Since the interval [0, 1] is connected and the integral is an integer multiple of 2πi, the integral must be constant. Hence m t is constant as a function of t [0, 1], giving us m 0 = m 1 as desired. Let Γ = p k=1 n kγ k, with each n k a nonzero integer and γ k a piecewise C 1 curve. Set L = p k=1 n k L(γ k ). Note that the range of Γ is the union of the ranges of γ k, 1 k p, by definition. The range of each γ k is compact. Because the union of finitely many compact sets is again compact, the range of Γ is compact.

17 Proof, Slide 4 We next argue that f t (z) is a continuous function of (t, z) with t [0, 1] and z range(γ). Because the range of Γ is a subset of G 1 G, with f and g holomorphic on G, g f and hence g f is continuous on the range of Γ. Because the range of Γ is compact, there is some M such that (g f )(z) M for z in the range of Γ. Let ε > 0. Let s and t be in [0, 1], u and w in range(γ). There is some δ 1 > 0 such that, u w < δ 1 f (u) f (w) < ε/3 There is some δ 2 > 0 such that u w < δ 2 (g f )(u) (g f )(w) < ε/3 Let δ = min { ε/(3(m + 1)), δ 1, δ 2 }.

18 Proof, Slide 5 If s t < δ and u w < δ, then f t (u) f s (w) = f (u) + t(g f )(u) f (w) s(g f )(w) f (u) f (w) + t (g f )(u) (g f )(w) + t s (g f )(w) < ε/3 + ε/3 + (ε/(3m + 3))M < ε The product topology space [0, 1] range(γ) is compact. Since f t (u) is continuous there (in (t, u) simultaneously), f t (u) is continuous there and has a minimum value. Since every f t is nonzero on the range of Γ, this minimum absolute value, say m, is positive.

19 Proof, Slide 6 Let s and t be in [0, 1], and z in the range of Γ. Note that f t (z) f t (z) f s(z) f s (z) = f t (z)f s(z) f t (z)f s(z) f t (z)f s (z) Hence f t (z) f t (z) f s(z) f s (z) f t (z)f s(z) f t (z)f s(z) m 2 Next we simplify the numerator: f t (z)f s (z) f t (z)f s(z) = (s t)(f g fg )(z) We have f, f, g, and g all continuous on the range of Γ, a compact set, and thus f g fg has a maximum there, say M. m 2 ε Let ε > 0. Let δ = ( M. If s t < δ, then for z in + 1)(L + 1) the range of Γ, f t (z) f t (z) f s(z) δ M f s (z) m 2 < ε L + 1

20 Proof, Slide 7 Thus Γ [ f t f ] s (z) dz f t f s ε L + 1 L < ε

21 X.13, The Local Mapping Theorem Then Let G be an open and connected subset of C. Let f be a nonconstant and holomorphic on G. Let z 0 G and set w 0 = f (z 0 ). 1 The function g(z) = f (z) w 0 has a zero of finite order m at z 0. 2 There is some δ 0 > 0 such that, for every δ (0, δ 0 ], there is an ε > 0 such that For all w D (w 0, ε), there are exactly m distinct z in D (z 0, δ) such that f (z) = w. If f (z) = w and z D (z 0, δ), then the function h(u) = f (u) w has a zero of order 1 at z.

22 Intuition For z near z 0, f (z) has some of the properties of w 0 + (z z 0 ) m.

23 Proof, Slide 1 g has a zero at z 0 because g(z 0 ) = f (z 0 ) f (z 0 ) = 0. If this were a zero of infinite order, since G is connected g would be zero on G. Then f (z) = w 0 throughout G, which contradicts the hypothesis that f is nonconstant on G. Therefore g has a zero of finite order m, for some positive integer m. This implies that the zero of g is isolated. There is some λ > 0 so that g(z) = 0 for z D (z 0, λ). If f (z 0 ) = 0, by the continuity of f there is some μ > 0 such f (z) = 0 for z D(z 0, μ). If f (z 0 ) = 0, it is a zero of finite order since [f ] (m 1) (z 0 ) = f (m) (z 0 ) = 0. Since it has a zero of finite order, this zero is isolated. There is some μ > 0 so that f (z) = 0 for z D (z 0, μ).

24 Proof, Slide 2 Let δ 0 = min{λ, μ}. Let δ (0, δ 0 ]. Set K = { z C : z z 0 δ }. Set K equal to the boundary of K (the circle of radius δ centered at z 0 ). Set ε = min{ f (z) w 0 : z K }. We have K compact and a subset of G. Since f is holomorphic on G, g(z) = f (z) w 0 is continuous on G and thus f (z) w 0 is continuous on G. Therefore, the minimum exists at some point in K. However, g(z) = f (z) w 0 is nonzero on K {z 0 }. Therefore ε > 0. Let w D (w 0, ε). For z K, (f (z) w 0 ) (f (z) w) = w w 0 < ε f (z) w 0 By Rouché s Theorem, f (z) w 0 and f (z) w have the same number of zeros in int(k ) = D(z 0, δ), counted according to multiplicity. f (z) w 0 has exactly one zero there, of order m. So the sum of the orders of the zeros of f (z) w is also m.

25 Proof, Slide 3 Since w = w 0, f (z 0 ) = w 0 = w. So the zeros of f (z) w are in D (z 0, δ). There f = 0, and thus each zero of f (z) w has order 1. Since the orders of the zeros of f (z) w add to m, each with order 1, f (z) w has exactly m distinct zeros in D (z 0, δ).

26 X.14, Open Mappings 1 If f is holomorphic and nonconstant on an open and connected G C, then for all open U G, f (U) is open. 2 Let f be holomorphic in a neighborhood of z 0. If f (z 0 ) = 0, then there is some δ > 0 so that f is univalent (one-to-one) on D(z 0, δ). 3 Let f be holomorphic on an open set G C. If f is univalent on G, then f is nonzero on G.

27 Proof of Item 1 To prove the first item, we apply the Local Mapping Theorem, Section X.13. Let U be an open subset of G. Let w 0 f (U). By the def n of f (U), there is some z 0 U such that f (z 0 ) = w 0. Because U is open, there is some λ > 0 such that D(z 0, λ) U. Note that D(z 0, λ) is connected. Claim: f is nonconstant on D(z 0, λ). Proof of Claim: Suppose that f is constant on D(z 0, λ). Then f (z) w 0 = 0 on D(z 0, λ). Because f (z) w 0 is holomorphic on G, with G open and connected, we have f (z) w 0 = 0 on G. This contradicts the hypothesis in Item (1) that f is nonconstant on G.

28 Proof of Item 1, Continued By the Local Mapping Theorem, applied to f on D(z 0, λ), there is some δ 0 > 0 with δ 0 λ such that, for all δ (0, δ 0 ], there is ε > 0 such that, for all w D (w 0, ε) there is some z D (z 0, δ) such that f (z) = w. In symbols, D (w 0, ε) f (D (z 0, δ)) Of course f (z 0 ) = w 0 and z 0 D(z 0, δ). So D(w 0, ε) f (D(z 0, δ)) f (D(z 0, λ) f (U) We ve proved that, for every w 0 f (U), there is some ε > 0 such that D(w 0, ε) f (U). That makes f (U) open.

29 Proof of Item 2 There is some λ > 0 such that f is holomorphic on D(z 0, λ). Let w 0 = f (z 0 ). Note that D(z 0, λ) is connected. Because f (z 0 ) = 0, the function g(z) = f (z) w 0 has a zero of order 1 at z 0. This zero of g is therefore isolated. There is some κ > 0 with κ λ such that f (z) = w 0 on D (z 0, κ). By the Local Mapping Theorem, there δ 0 > 0 with δ 0 λ such that, for all δ (0, δ 0 ] there is some ε > 0 such that, for all w D (w 0, ε) there is exactly one z D(z, δ) such that f (z) = w. Let δ = δ 0 and get ε > 0 for it from the local mapping theorem. Since f is holomorphic on D(z 0, λ) and hence continuous, there is ρ > 0 with ρ λ such that f (D(z 0, ρ)) D(w 0, ε)

30 Proof of Item 2, Continued Let u and v be in B = D(z 0, min{ κ, ρ, δ 0 }) such that f (u) = f (v). Set w = f (u). Suppose w = w 0. We chose κ so that w 0 / f (D (z 0, κ)). Since B D(z 0, κ), we have u = v = z 0. Suppose w = w 0. Then u = z 0 and v = z 0. So u, v B {z 0 } D (z 0, δ 0 ) Also, since B D(z 0, ρ) and u B, we have w = f (u) D(w 0, ε). By our choices of δ 0 and ε, the equation f (z) = w has exactly one solution for z D (z 0, δ 0 ). Therefore u = v.

31 Proof of Item 3 Let f be holomorphic and univalent on an open set G, and suppose that f (z 0 ) = 0 for some z 0 G. There is some λ > 0 such that D(z 0, λ) G. Because f is univalent on G, f is univalent on D(z 0, λ) and hence nonconstant on D(z 0, λ). D(z 0, λ) is connected. By the Local Mapping Theorem applied to f on D(z 0, λ), g(z) = f (z) w 0 has a zero of finite order m at z 0. Since f (z 0 ) = 0, we have m > 1. By the Local Mapping Theorem, there is some δ > 0 with δ λ and there is ε > 0 such that, for all w D(w 0, ε) there are exactly m distinct points z in D (z 0, δ) such that f (z) = w. Thus f is not univalent on D(z 0, δ), a contradiction.

32 X.15, Inverses Theorem: Let f : G C be holomorphic and univalent on G (with G an open subset of C). Then f 1 exists, f 1 : f (G) G with f (G) an open subset of C, and f 1 is holomorphic on f (G). For all w f (G), (f 1 ) (w) = 1 f (f 1 (w)) Note: By Item 3 of Section X.14, f (z) = 0 for z G and thus f (f 1 (w)) = 0. Also, to help remember the formula for (f 1 ), use the chain rule. For w f (G), (f f 1 )(w) = w and thus f (f 1 (w)) (f 1 ) (w) = 1

33 Proof, Slide 1 For any sets A, B, with h : A B being one-to-one (univalent), h 1 exists and h 1 : h(a) A. Claim: For all open U G, we have f (U) open. Let w 0 f (U). There is some z 0 G such that f (z 0 ) = w 0. Because U is open, there is some λ > 0 such that D(z 0, λ) U. Because f is univalent on G, it is univalent on D(z 0, λ) and hence nonconstant on D(z 0, λ). By the first item of Section X.14, with f holomorphic and nonconstant on the connected, open set D(z 0, λ), f (D(z 0, λ)) is open. Since w 0 = f (z 0 ) f (D(z 0, λ)), there is some ε > 0 such that D(w 0, ε) f (D(w 0, δ)) f (U) Since this can be done for any w 0 f (U), that makes f (U) open.

34 Proof, Slide 2 In particular, f (G) is open. Claim: f 1 is continuous on f (G). General Topology Argument: the property that f (U) is open for all open U G = domain(f ) makes f an open mapping. If f is one-to-one, it is a mini-theorem that f is an open mapping if and only if f 1 is continuous.

35 Proof, Slide 3 Second Argument for the claim: Let w 0 f (G). Let z 0 = f 1 (w 0 ). Of course z 0 G. Because G is open there is some λ > 0 such that D(z 0, λ) G. Let ε > 0 be arbitrary. Set ε = min{ε, λ}. Because f is an open mapping, f (D(z 0, ε )) is an open set. Because w 0 = f (z 0 ) is in f (D(z 0, ε )), there is some δ > 0 such that D(w 0, δ) f (D(z 0, ε )) Then f 1 [D(w 0, δ)] f 1 [f (D(z 0, ε ))] = (f 1 f )[D(z 0, ε )] Since f 1 f is the identify function on G, and D(z 0, ε ) G (we chose ε λ), (f 1 f )[D(z 0, ε )] = D(z 0, ε )

36 Proof, Slide 4 Since ε ε, we have f 1 [D(w 0, δ)] D(z 0, ε ) D(z 0, ε) We have satisfied the ε δ definition for f 1 to be continuous at w 0.

37 Proof, Slide 5 We next show that f 1 has a derivative at all points of its domain. Let w 0 f (G) and z 0 = f 1 (w 0 ). We want to prove that f 1 (w) z 0 lim = 1 w w 0 w w 0 f (z 0 ) This will show that (f 1 ) (w 0 ) = 1 f (f 1 (w 0 )).

38 Proof, Slide 6 By the third item of Section X.14, since z 0 G, G is open and f is univalent on G, we have f (z 0 ) = 0. We have f (z) f (z 0 ) lim = f (z 0 ) z z 0 z z 0 The function u 1/u is continuous at f (z 0 ) since f (z 0 ) = 0. Since continuous functions preserve limits (when they are continuous at the value of the limit), we have z z 0 lim z z 0 f (z) f (z 0 ) = 1 f (z 0 ) Let ε > 0 be given. There is some δ > 0 so that 0 < z z 0 < δ z z 0 f (z) f (z 0 ) 1 f (z 0 ) < ε

39 Proof, Slide 7 Because f 1 is continuous at w 0, there is some ρ > 0 such that w w 0 < ρ f 1 (w) z 0 = f 1 (w) f 1 (w 0 ) < δ Suppose that 0 < w w 0 < ρ. Since f is one-to-one, so is f 1. Thus f 1 (w) = f 1 (w 0 ) = z 0. Consequently, we can let z = f 1 (w) in the last statement of the previous slide. We have f 1 (w) z 0 f (f 1 1 (w)) z 0 f (z 0 ) < ε Since f (f 1 )(w) = w, we have proved that 0 < w w 0 < ρ f 1 (w) z 0 1 w z 0 f (z 0 ) < ε Since we can produce ρ > 0 for each ε > 0, we have proved the desired limit.