X.9 Revisited, Cauchy s Formula for Contours
|
|
- Kelley Eaton
- 5 years ago
- Views:
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.
. Then g is holomorphic and bounded in U. So z 0 is a removable singularity of g. Since f(z) = w 0 + 1
Now we describe the behavior of f near an isolated singularity of each kind. We will always assume that z 0 is a singularity of f, and f is holomorphic on D(z 0, r) \ {z 0 }. Theorem 4.2.. z 0 is a removable
More informationChapter 4: Open mapping theorem, removable singularities
Chapter 4: Open mapping theorem, removable singularities Course 44, 2003 04 February 9, 2004 Theorem 4. (Laurent expansion) Let f : G C be analytic on an open G C be open that contains a nonempty annulus
More informationPart IB Complex Analysis
Part IB Complex Analysis Theorems Based on lectures by I. Smith Notes taken by Dexter Chua Lent 2016 These notes are not endorsed by the lecturers, and I have modified them (often significantly) after
More informationMATH 566 LECTURE NOTES 4: ISOLATED SINGULARITIES AND THE RESIDUE THEOREM
MATH 566 LECTURE NOTES 4: ISOLATED SINGULARITIES AND THE RESIDUE THEOREM TSOGTGEREL GANTUMUR 1. Functions holomorphic on an annulus Let A = D R \D r be an annulus centered at 0 with 0 < r < R
More informationMATH 722, COMPLEX ANALYSIS, SPRING 2009 PART 5
MATH 722, COMPLEX ANALYSIS, SPRING 2009 PART 5.. The Arzela-Ascoli Theorem.. The Riemann mapping theorem Let X be a metric space, and let F be a family of continuous complex-valued functions on X. We have
More informationLECTURE-15 : LOGARITHMS AND COMPLEX POWERS
LECTURE-5 : LOGARITHMS AND COMPLEX POWERS VED V. DATAR The purpose of this lecture is twofold - first, to characterize domains on which a holomorphic logarithm can be defined, and second, to show that
More informationComplex Analysis. Chapter V. Singularities V.3. The Argument Principle Proofs of Theorems. August 8, () Complex Analysis August 8, / 7
Complex Analysis Chapter V. Singularities V.3. The Argument Principle Proofs of Theorems August 8, 2017 () Complex Analysis August 8, 2017 1 / 7 Table of contents 1 Theorem V.3.4. Argument Principle 2
More informationTHE RESIDUE THEOREM. f(z) dz = 2πi res z=z0 f(z). C
THE RESIDUE THEOREM ontents 1. The Residue Formula 1 2. Applications and corollaries of the residue formula 2 3. ontour integration over more general curves 5 4. Defining the logarithm 7 Now that we have
More informationCONSEQUENCES OF POWER SERIES REPRESENTATION
CONSEQUENCES OF POWER SERIES REPRESENTATION 1. The Uniqueness Theorem Theorem 1.1 (Uniqueness). Let Ω C be a region, and consider two analytic functions f, g : Ω C. Suppose that S is a subset of Ω that
More informationMORE CONSEQUENCES OF CAUCHY S THEOREM
MOE CONSEQUENCES OF CAUCHY S THEOEM Contents. The Mean Value Property and the Maximum-Modulus Principle 2. Morera s Theorem and some applications 3 3. The Schwarz eflection Principle 6 We have stated Cauchy
More information9. Series representation for analytic functions
9. Series representation for analytic functions 9.. Power series. Definition: A power series is the formal expression S(z) := c n (z a) n, a, c i, i =,,, fixed, z C. () The n.th partial sum S n (z) is
More informationMTH 3102 Complex Variables Final Exam May 1, :30pm-5:30pm, Skurla Hall, Room 106
Name (Last name, First name): MTH 32 Complex Variables Final Exam May, 27 3:3pm-5:3pm, Skurla Hall, Room 6 Exam Instructions: You have hour & 5 minutes to complete the exam. There are a total of problems.
More informationNATIONAL UNIVERSITY OF SINGAPORE Department of Mathematics MA4247 Complex Analysis II Lecture Notes Part II
NATIONAL UNIVERSITY OF SINGAPORE Department of Mathematics MA4247 Complex Analysis II Lecture Notes Part II Chapter 2 Further properties of analytic functions 21 Local/Global behavior of analytic functions;
More informationFINAL EXAM MATH 220A, UCSD, AUTUMN 14. You have three hours.
FINAL EXAM MATH 220A, UCSD, AUTUMN 4 You have three hours. Problem Points Score There are 6 problems, and the total number of points is 00. Show all your work. Please make your work as clear and easy to
More informationarxiv:math.cv/ v1 23 Dec 2003
EXPONENTIAL GELFOND-KHOVANSKII FORMULA IN DIMENSION ONE arxiv:math.cv/0312433 v1 23 Dec 2003 EVGENIA SOPRUNOVA Abstract. Gelfond and Khovanskii found a formula for the sum of the values of a Laurent polynomial
More informationProblem Set 5. 2 n k. Then a nk (x) = 1+( 1)k
Problem Set 5 1. (Folland 2.43) For x [, 1), let 1 a n (x)2 n (a n (x) = or 1) be the base-2 expansion of x. (If x is a dyadic rational, choose the expansion such that a n (x) = for large n.) Then the
More informationTheorem [Mean Value Theorem for Harmonic Functions] Let u be harmonic on D(z 0, R). Then for any r (0, R), u(z 0 ) = 1 z z 0 r
2. A harmonic conjugate always exists locally: if u is a harmonic function in an open set U, then for any disk D(z 0, r) U, there is f, which is analytic in D(z 0, r) and satisfies that Re f u. Since such
More informationComplex Analysis Problems
Complex Analysis Problems transcribed from the originals by William J. DeMeo October 2, 2008 Contents 99 November 2 2 2 200 November 26 4 3 2006 November 3 6 4 2007 April 6 7 5 2007 November 6 8 99 NOVEMBER
More informationPart IB. Further Analysis. Year
Year 2004 2003 2002 2001 10 2004 2/I/4E Let τ be the topology on N consisting of the empty set and all sets X N such that N \ X is finite. Let σ be the usual topology on R, and let ρ be the topology on
More informationMath 220A - Fall Final Exam Solutions
Math 22A - Fall 216 - Final Exam Solutions Problem 1. Let f be an entire function and let n 2. Show that there exists an entire function g with g n = f if and only if the orders of all zeroes of f are
More informationFUNCTIONAL ANALYSIS LECTURE NOTES: COMPACT SETS AND FINITE-DIMENSIONAL SPACES. 1. Compact Sets
FUNCTIONAL ANALYSIS LECTURE NOTES: COMPACT SETS AND FINITE-DIMENSIONAL SPACES CHRISTOPHER HEIL 1. Compact Sets Definition 1.1 (Compact and Totally Bounded Sets). Let X be a metric space, and let E X be
More information2.31 Definition By an open cover of a set E in a metric space X we mean a collection {G α } of open subsets of X such that E α G α.
Chapter 2. Basic Topology. 2.3 Compact Sets. 2.31 Definition By an open cover of a set E in a metric space X we mean a collection {G α } of open subsets of X such that E α G α. 2.32 Definition A subset
More informationLECTURE-13 : GENERALIZED CAUCHY S THEOREM
LECTURE-3 : GENERALIZED CAUCHY S THEOREM VED V. DATAR The aim of this lecture to prove a general form of Cauchy s theorem applicable to multiply connected domains. We end with computations of some real
More informationHomework #2 Solutions Due: September 5, for all n N n 3 = n2 (n + 1) 2 4
Do the following exercises from the text: Chapter (Section 3):, 1, 17(a)-(b), 3 Prove that 1 3 + 3 + + n 3 n (n + 1) for all n N Proof The proof is by induction on n For n N, let S(n) be the statement
More informationMetric Spaces and Topology
Chapter 2 Metric Spaces and Topology From an engineering perspective, the most important way to construct a topology on a set is to define the topology in terms of a metric on the set. This approach underlies
More information106 CHAPTER 3. TOPOLOGY OF THE REAL LINE. 2. The set of limit points of a set S is denoted L (S)
106 CHAPTER 3. TOPOLOGY OF THE REAL LINE 3.3 Limit Points 3.3.1 Main Definitions Intuitively speaking, a limit point of a set S in a space X is a point of X which can be approximated by points of S other
More informationQualifying Exam Complex Analysis (Math 530) January 2019
Qualifying Exam Complex Analysis (Math 53) January 219 1. Let D be a domain. A function f : D C is antiholomorphic if for every z D the limit f(z + h) f(z) lim h h exists. Write f(z) = f(x + iy) = u(x,
More information13 Maximum Modulus Principle
3 Maximum Modulus Principle Theorem 3. (maximum modulus principle). If f is non-constant and analytic on an open connected set Ω, then there is no point z 0 Ω such that f(z) f(z 0 ) for all z Ω. Remark
More informationRIEMANN MAPPING THEOREM
RIEMANN MAPPING THEOREM VED V. DATAR Recall that two domains are called conformally equivalent if there exists a holomorphic bijection from one to the other. This automatically implies that there is an
More informationSolutions to practice problems for the final
Solutions to practice problems for the final Holomorphicity, Cauchy-Riemann equations, and Cauchy-Goursat theorem 1. (a) Show that there is a holomorphic function on Ω = {z z > 2} whose derivative is z
More informationf (n) (z 0 ) Theorem [Morera s Theorem] Suppose f is continuous on a domain U, and satisfies that for any closed curve γ in U, γ
Remarks. 1. So far we have seen that holomorphic is equivalent to analytic. Thus, if f is complex differentiable in an open set, then it is infinitely many times complex differentiable in that set. This
More informationPart IB. Complex Analysis. Year
Part IB Complex Analysis Year 2018 2017 2016 2015 2014 2013 2012 2011 2010 2009 2008 2007 2006 2005 2018 Paper 1, Section I 2A Complex Analysis or Complex Methods 7 (a) Show that w = log(z) is a conformal
More informationRiemann sphere and rational maps
Chapter 3 Riemann sphere and rational maps 3.1 Riemann sphere It is sometimes convenient, and fruitful, to work with holomorphic (or in general continuous) functions on a compact space. However, we wish
More informationCOMPLEX ANALYSIS Spring 2014
COMPLEX ANALYSIS Spring 2014 1 Preliminaries Homotopical topics Our textbook slides over a little problem when discussing homotopy. The standard definition of homotopy is for not necessarily piecewise
More informationPOWER SERIES AND ANALYTIC CONTINUATION
POWER SERIES AND ANALYTIC CONTINUATION 1. Analytic functions Definition 1.1. A function f : Ω C C is complex-analytic if for each z 0 Ω there exists a power series f z0 (z) := a n (z z 0 ) n which converges
More information4.6 Montel's Theorem. Robert Oeckl CA NOTES 7 17/11/2009 1
Robert Oeckl CA NOTES 7 17/11/2009 1 4.6 Montel's Theorem Let X be a topological space. We denote by C(X) the set of complex valued continuous functions on X. Denition 4.26. A topological space is called
More informationINTRODUCTION TO REAL ANALYTIC GEOMETRY
INTRODUCTION TO REAL ANALYTIC GEOMETRY KRZYSZTOF KURDYKA 1. Analytic functions in several variables 1.1. Summable families. Let (E, ) be a normed space over the field R or C, dim E
More informationCourse 214 Basic Properties of Holomorphic Functions Second Semester 2008
Course 214 Basic Properties of Holomorphic Functions Second Semester 2008 David R. Wilkins Copyright c David R. Wilkins 1989 2008 Contents 7 Basic Properties of Holomorphic Functions 72 7.1 Taylor s Theorem
More informationMASTERS EXAMINATION IN MATHEMATICS SOLUTIONS
MASTERS EXAMINATION IN MATHEMATICS PURE MATHEMATICS OPTION SPRING 010 SOLUTIONS Algebra A1. Let F be a finite field. Prove that F [x] contains infinitely many prime ideals. Solution: The ring F [x] of
More informationLECTURE 15: COMPLETENESS AND CONVEXITY
LECTURE 15: COMPLETENESS AND CONVEXITY 1. The Hopf-Rinow Theorem Recall that a Riemannian manifold (M, g) is called geodesically complete if the maximal defining interval of any geodesic is R. On the other
More informationMath 220A Complex Analysis Solutions to Homework #2 Prof: Lei Ni TA: Kevin McGown
Math 220A Complex Analysis Solutions to Homework #2 Prof: Lei Ni TA: Kevin McGown Conway, Page 14, Problem 11. Parts of what follows are adapted from the text Modular Functions and Dirichlet Series in
More informationMath 213br HW 1 solutions
Math 213br HW 1 solutions February 26, 2014 Problem 1 Let P (x) be a polynomial of degree d > 1 with P (x) > 0 for all x 0. For what values of α R does the integral I(α) = 0 x α P (x) dx converge? Give
More information1. The COMPLEX PLANE AND ELEMENTARY FUNCTIONS: Complex numbers; stereographic projection; simple and multiple connectivity, elementary functions.
Complex Analysis Qualifying Examination 1 The COMPLEX PLANE AND ELEMENTARY FUNCTIONS: Complex numbers; stereographic projection; simple and multiple connectivity, elementary functions 2 ANALYTIC FUNCTIONS:
More informationMATH FINAL SOLUTION
MATH 185-4 FINAL SOLUTION 1. (8 points) Determine whether the following statements are true of false, no justification is required. (1) (1 point) Let D be a domain and let u,v : D R be two harmonic functions,
More informationComplex Analysis Important Concepts
Complex Analysis Important Concepts Travis Askham April 1, 2012 Contents 1 Complex Differentiation 2 1.1 Definition and Characterization.............................. 2 1.2 Examples..........................................
More informationComplex Analysis Qualifying Exam Solutions
Complex Analysis Qualifying Exam Solutions May, 04 Part.. Let log z be the principal branch of the logarithm defined on G = {z C z (, 0]}. Show that if t > 0, then the equation log z = t has exactly one
More informationReview of complex analysis in one variable
CHAPTER 130 Review of complex analysis in one variable This gives a brief review of some of the basic results in complex analysis. In particular, it outlines the background in single variable complex analysis
More informationTopic 4 Notes Jeremy Orloff
Topic 4 Notes Jeremy Orloff 4 auchy s integral formula 4. Introduction auchy s theorem is a big theorem which we will use almost daily from here on out. Right away it will reveal a number of interesting
More information1 Lecture 4: Set topology on metric spaces, 8/17/2012
Summer Jump-Start Program for Analysis, 01 Song-Ying Li 1 Lecture : Set topology on metric spaces, 8/17/01 Definition 1.1. Let (X, d) be a metric space; E is a subset of X. Then: (i) x E is an interior
More informationMath 421, Homework #6 Solutions. (1) Let E R n Show that = (E c ) o, i.e. the complement of the closure is the interior of the complement.
Math 421, Homework #6 Solutions (1) Let E R n Show that (Ē) c = (E c ) o, i.e. the complement of the closure is the interior of the complement. 1 Proof. Before giving the proof we recall characterizations
More informationExam 2 extra practice problems
Exam 2 extra practice problems (1) If (X, d) is connected and f : X R is a continuous function such that f(x) = 1 for all x X, show that f must be constant. Solution: Since f(x) = 1 for every x X, either
More informationHartogs Theorem: separate analyticity implies joint Paul Garrett garrett/
(February 9, 25) Hartogs Theorem: separate analyticity implies joint Paul Garrett garrett@math.umn.edu http://www.math.umn.edu/ garrett/ (The present proof of this old result roughly follows the proof
More informationSolutions to Complex Analysis Prelims Ben Strasser
Solutions to Complex Analysis Prelims Ben Strasser In preparation for the complex analysis prelim, I typed up solutions to some old exams. This document includes complete solutions to both exams in 23,
More informationMath 421, Homework #7 Solutions. We can then us the triangle inequality to find for k N that (x k + y k ) (L + M) = (x k L) + (y k M) x k L + y k M
Math 421, Homework #7 Solutions (1) Let {x k } and {y k } be convergent sequences in R n, and assume that lim k x k = L and that lim k y k = M. Prove directly from definition 9.1 (i.e. don t use Theorem
More informationNotes on Complex Analysis
Michael Papadimitrakis Notes on Complex Analysis Department of Mathematics University of Crete Contents The complex plane.. The complex plane...................................2 Argument and polar representation.........................
More informationSet, functions and Euclidean space. Seungjin Han
Set, functions and Euclidean space Seungjin Han September, 2018 1 Some Basics LOGIC A is necessary for B : If B holds, then A holds. B A A B is the contraposition of B A. A is sufficient for B: If A holds,
More informationTheorem Let J and f be as in the previous theorem. Then for any w 0 Int(J), f(z) (z w 0 ) n+1
(w) Second, since lim z w z w z w δ. Thus, i r δ, then z w =r (w) z w = (w), there exist δ, M > 0 such that (w) z w M i dz ML({ z w = r}) = M2πr, which tends to 0 as r 0. This shows that g = 2πi(w), which
More information= F (b) F (a) F (x i ) F (x i+1 ). a x 0 x 1 x n b i
Real Analysis Problem 1. If F : R R is a monotone function, show that F T V ([a,b]) = F (b) F (a) for any interval [a, b], and that F has bounded variation on R if and only if it is bounded. Here F T V
More informationComplex Analysis, Stein and Shakarchi Meromorphic Functions and the Logarithm
Complex Analysis, Stein and Shakarchi Chapter 3 Meromorphic Functions and the Logarithm Yung-Hsiang Huang 217.11.5 Exercises 1. From the identity sin πz = eiπz e iπz 2i, it s easy to show its zeros are
More informationSelected topics in Complex Analysis MATH Winter (draft)
Selected topics in Complex Analysis MATH 428 - Winter 2015 - (draft) François Monard March 23, 2015 1 2 Lecture 1-01/05 - Recalls from 427 Algebraic manipulations of complex numbers. See for instance [Taylor,
More informationTHE WEIERSTRASS PREPARATION THEOREM AND SOME APPLICATIONS
THE WEIERSTRASS PREPARATION THEOREM AND SOME APPLICATIONS XUAN LI Abstract. In this paper we revisit the Weierstrass preparation theorem, which describes how to represent a holomorphic function of several
More informationDepartment of Mathematics, University of California, Berkeley. GRADUATE PRELIMINARY EXAMINATION, Part A Fall Semester 2014
Department of Mathematics, University of California, Berkeley YOUR 1 OR 2 DIGIT EXAM NUMBER GRADUATE PRELIMINARY EXAMINATION, Part A Fall Semester 2014 1. Please write your 1- or 2-digit exam number on
More informationINTEGRATION WORKSHOP 2004 COMPLEX ANALYSIS EXERCISES
INTEGRATION WORKSHOP 2004 COMPLEX ANALYSIS EXERCISES PHILIP FOTH 1. Cauchy s Formula and Cauchy s Theorem 1. Suppose that γ is a piecewise smooth positively ( counterclockwise ) oriented simple closed
More informationSynopsis of Complex Analysis. Ryan D. Reece
Synopsis of Complex Analysis Ryan D. Reece December 7, 2006 Chapter Complex Numbers. The Parts of a Complex Number A complex number, z, is an ordered pair of real numbers similar to the points in the real
More informationImmerse Metric Space Homework
Immerse Metric Space Homework (Exercises -2). In R n, define d(x, y) = x y +... + x n y n. Show that d is a metric that induces the usual topology. Sketch the basis elements when n = 2. Solution: Steps
More informationMATH8811: COMPLEX ANALYSIS
MATH8811: COMPLEX ANALYSIS DAWEI CHEN Contents 1. Classical Topics 2 1.1. Complex numbers 2 1.2. Differentiability 2 1.3. Cauchy-Riemann Equations 3 1.4. The Riemann Sphere 4 1.5. Möbius transformations
More informationThe Residue Theorem. Integration Methods over Closed Curves for Functions with Singularities
The Residue Theorem Integration Methods over losed urves for Functions with Singularities We have shown that if f(z) is analytic inside and on a closed curve, then f(z)dz = 0. We have also seen examples
More informationA NICE PROOF OF FARKAS LEMMA
A NICE PROOF OF FARKAS LEMMA DANIEL VICTOR TAUSK Abstract. The goal of this short note is to present a nice proof of Farkas Lemma which states that if C is the convex cone spanned by a finite set and if
More informationF (z) =f(z). f(z) = a n (z z 0 ) n. F (z) = a n (z z 0 ) n
6 Chapter 2. CAUCHY S THEOREM AND ITS APPLICATIONS Theorem 5.6 (Schwarz reflection principle) Suppose that f is a holomorphic function in Ω + that extends continuously to I and such that f is real-valued
More informationTHIRD SEMESTER M. Sc. DEGREE (MATHEMATICS) EXAMINATION (CUSS PG 2010) MODEL QUESTION PAPER MT3C11: COMPLEX ANALYSIS
THIRD SEMESTER M. Sc. DEGREE (MATHEMATICS) EXAMINATION (CUSS PG 2010) MODEL QUESTION PAPER MT3C11: COMPLEX ANALYSIS TIME:3 HOURS Maximum weightage:36 PART A (Short Answer Type Question 1-14) Answer All
More informationRiemann Mapping Theorem (4/10-4/15)
Math 752 Spring 2015 Riemann Mapping Theorem (4/10-4/15) Definition 1. A class F of continuous functions defined on an open set G is called a normal family if every sequence of elements in F contains a
More informationCauchy Integral Formula Consequences
Cauchy Integral Formula Consequences Monday, October 28, 2013 1:59 PM Homework 3 due November 15, 2013 at 5 PM. Last time we derived Cauchy's Integral Formula, which we will present in somewhat generalized
More informationContinuity. Matt Rosenzweig
Continuity Matt Rosenzweig Contents 1 Continuity 1 1.1 Rudin Chapter 4 Exercises........................................ 1 1.1.1 Exercise 1............................................. 1 1.1.2 Exercise
More informationFundamental Properties of Holomorphic Functions
Complex Analysis Contents Chapter 1. Fundamental Properties of Holomorphic Functions 5 1. Basic definitions 5 2. Integration and Integral formulas 6 3. Some consequences of the integral formulas 8 Chapter
More informationCOMPLETELY INVARIANT JULIA SETS OF POLYNOMIAL SEMIGROUPS
Series Logo Volume 00, Number 00, Xxxx 19xx COMPLETELY INVARIANT JULIA SETS OF POLYNOMIAL SEMIGROUPS RICH STANKEWITZ Abstract. Let G be a semigroup of rational functions of degree at least two, under composition
More informationMATH SPRING UC BERKELEY
MATH 85 - SPRING 205 - UC BERKELEY JASON MURPHY Abstract. These are notes for Math 85 taught in the Spring of 205 at UC Berkeley. c 205 Jason Murphy - All Rights Reserved Contents. Course outline 2 2.
More informationM17 MAT25-21 HOMEWORK 6
M17 MAT25-21 HOMEWORK 6 DUE 10:00AM WEDNESDAY SEPTEMBER 13TH 1. To Hand In Double Series. The exercises in this section will guide you to complete the proof of the following theorem: Theorem 1: Absolute
More informationv( x) u( y) dy for any r > 0, B r ( x) Ω, or equivalently u( w) ds for any r > 0, B r ( x) Ω, or ( not really) equivalently if v exists, v 0.
Sep. 26 The Perron Method In this lecture we show that one can show existence of solutions using maximum principle alone.. The Perron method. Recall in the last lecture we have shown the existence of solutions
More informationComplex Analysis Slide 9: Power Series
Complex Analysis Slide 9: Power Series MA201 Mathematics III Department of Mathematics IIT Guwahati August 2015 Complex Analysis Slide 9: Power Series 1 / 37 Learning Outcome of this Lecture We learn Sequence
More informationb 0 + b 1 z b d z d
I. Introduction Definition 1. For z C, a rational function of degree d is any with a d, b d not both equal to 0. R(z) = P (z) Q(z) = a 0 + a 1 z +... + a d z d b 0 + b 1 z +... + b d z d It is exactly
More informationTHE REAL NUMBERS Chapter #4
FOUNDATIONS OF ANALYSIS FALL 2008 TRUE/FALSE QUESTIONS THE REAL NUMBERS Chapter #4 (1) Every element in a field has a multiplicative inverse. (2) In a field the additive inverse of 1 is 0. (3) In a field
More informationQuasi-conformal maps and Beltrami equation
Chapter 7 Quasi-conformal maps and Beltrami equation 7. Linear distortion Assume that f(x + iy) =u(x + iy)+iv(x + iy) be a (real) linear map from C C that is orientation preserving. Let z = x + iy and
More informationNATIONAL UNIVERSITY OF SINGAPORE. Department of Mathematics. MA4247 Complex Analysis II. Lecture Notes Part IV
NATIONAL UNIVERSITY OF SINGAPORE Department of Mathematics MA4247 Complex Analysis II Lecture Notes Part IV Chapter 2. Further properties of analytic/holomorphic functions (continued) 2.4. The Open mapping
More informationare Banach algebras. f(x)g(x) max Example 7.4. Similarly, A = L and A = l with the pointwise multiplication
7. Banach algebras Definition 7.1. A is called a Banach algebra (with unit) if: (1) A is a Banach space; (2) There is a multiplication A A A that has the following properties: (xy)z = x(yz), (x + y)z =
More informationMA30056: Complex Analysis. Exercise Sheet 7: Applications and Sequences of Complex Functions
MA30056: Complex Analysis Exercise Sheet 7: Applications and Sequences of Complex Functions Please hand solutions in at the lecture on Monday 6th March..) Prove Gauss Fundamental Theorem of Algebra. Hint:
More informationComplex Analysis. Travis Dirle. December 4, 2016
Complex Analysis 2 Complex Analysis Travis Dirle December 4, 2016 2 Contents 1 Complex Numbers and Functions 1 2 Power Series 3 3 Analytic Functions 7 4 Logarithms and Branches 13 5 Complex Integration
More informationAn Introduction to Complex Analysis and Geometry John P. D Angelo, Pure and Applied Undergraduate Texts Volume 12, American Mathematical Society, 2010
An Introduction to Complex Analysis and Geometry John P. D Angelo, Pure and Applied Undergraduate Texts Volume 12, American Mathematical Society, 2010 John P. D Angelo, Univ. of Illinois, Urbana IL 61801.
More informationCOMPLEX ANALYSIS Spring 2014
COMPLEX ANALYSIS Spring 24 Homework 4 Solutions Exercise Do and hand in exercise, Chapter 3, p. 4. Solution. The exercise states: Show that if a
More informationHW 4 SOLUTIONS. , x + x x 1 ) 2
HW 4 SOLUTIONS The Way of Analysis p. 98: 1.) Suppose that A is open. Show that A minus a finite set is still open. This follows by induction as long as A minus one point x is still open. To see that A
More informationMA3111S COMPLEX ANALYSIS I
MA3111S COMPLEX ANALYSIS I 1. The Algebra of Complex Numbers A complex number is an expression of the form a + ib, where a and b are real numbers. a is called the real part of a + ib and b the imaginary
More informationA RAPID INTRODUCTION TO COMPLEX ANALYSIS
A RAPID INTRODUCTION TO COMPLEX ANALYSIS AKHIL MATHEW ABSTRACT. These notes give a rapid introduction to some of the basic results in complex analysis, assuming familiarity from the reader with Stokes
More informationReal Analysis. Joe Patten August 12, 2018
Real Analysis Joe Patten August 12, 2018 1 Relations and Functions 1.1 Relations A (binary) relation, R, from set A to set B is a subset of A B. Since R is a subset of A B, it is a set of ordered pairs.
More informationMA424, S13 HW #6: Homework Problems 1. Answer the following, showing all work clearly and neatly. ONLY EXACT VALUES WILL BE ACCEPTED.
MA424, S13 HW #6: 44-47 Homework Problems 1 Answer the following, showing all work clearly and neatly. ONLY EXACT VALUES WILL BE ACCEPTED. NOTATION: Recall that C r (z) is the positively oriented circle
More informationCLASS NOTES FOR APRIL 14, 2000
CLASS NOTES FOR APRIL 14, 2000 Announcement: Section 1.2, Questions 3,5 have been deferred from Assignment 1 to Assignment 2. Section 1.4, Question 5 has been dropped entirely. 1. Review of Wednesday class
More information7 Asymptotics for Meromorphic Functions
Lecture G jacques@ucsd.edu 7 Asymptotics for Meromorphic Functions Hadamard s Theorem gives a broad description of the exponential growth of coefficients in power series, but the notion of exponential
More informationSolutions for Problem Set #5 due October 17, 2003 Dustin Cartwright and Dylan Thurston
Solutions for Problem Set #5 due October 17, 23 Dustin Cartwright and Dylan Thurston 1 (B&N 6.5) Suppose an analytic function f agrees with tan x, x 1. Show that f(z) = i has no solution. Could f be entire?
More informationLecture 3. Econ August 12
Lecture 3 Econ 2001 2015 August 12 Lecture 3 Outline 1 Metric and Metric Spaces 2 Norm and Normed Spaces 3 Sequences and Subsequences 4 Convergence 5 Monotone and Bounded Sequences Announcements: - Friday
More informationSOLUTIONS OF VARIATIONS, PRACTICE TEST 4
SOLUTIONS OF VARIATIONS, PRATIE TEST 4 5-. onsider the following system of linear equations over the real numbers, where x, y and z are variables and b is a real constant. x + y + z = 0 x + 4y + 3z = 0
More informationNOTES ON RIEMANN S ZETA FUNCTION. Γ(z) = t z 1 e t dt
NOTES ON RIEMANN S ZETA FUNCTION DRAGAN MILIČIĆ. Gamma function.. Definition of the Gamma function. The integral Γz = t z e t dt is well-defined and defines a holomorphic function in the right half-plane
More informationComplex Analysis review notes for weeks 1-6
Complex Analysis review notes for weeks -6 Peter Milley Semester 2, 2007 In what follows, unless stated otherwise a domain is a connected open set. Generally we do not include the boundary of the set,
More information