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Lecture Note for Math 220B Complex Analysis of One Variable

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Lecture Note for Math 220B Complex Analysis of One Variable Lecture Note for Math 220B Complex Analysis of One Variable Song-Ying Li University of California, Irvine Contents 1 The Residue theorem applied to real integrals 3 1.1 Several types of real integrals . 3 1.2 The Logarithm function . 11 1.3 Real integrals involving ln x ................... 13 1.4 Homework 1 . 17 2 The Zero Set of Holomorphic Functions 18 2.1 The Argument Principle and its Applications . 18 2.2 The Open Mapping theorem . 20 2.3 Rouch´eTheorems . 20 2.4 Applications and Examples . 21 2.5 Hurwitz's Theorem and Applications . 23 2.6 Examples/Applications . 24 2.7 Homework 2 . 25 3 The Geometry of Holomorphic Mappings 27 3.1 The Maximum Modulus Theorem . 27 3.2 Schwarz Lemma . 29 3.2.1 Schwarz-Pick Lemma . 29 3.3 Homework 3 . 32 3.4 Conformal and proper holomorphic function maps . 33 3.5 Automorphism groups . 35 3.6 M¨obiustransformations . 37 1 3.7 Cross ratio . 38 3.8 Properties of the Cross Ratio . 39 3.8.1 Symmetric points . 39 3.9 Construction of conformal maps . 41 3.10 Homework 4 . 44 3.11 Midterm Review . 46 3.12 Normal families . 48 3.12.1 Examples for normal families . 50 3.13 The Riemann Mapping Theorem . 52 3.13.1 Existence of the Riemann map . 52 3.13.2 Proof of Riemann mapping theorem . 54 3.14 Homework 5 . 55 3.15 The Reflection Principle . 57 3.16 Homework 6 . 58 3.17 Singular Points and Regular Points . 59 4 Infinite Products 60 4.1 Basic properties of infinite products . 61 4.2 Examples . 63 4.3 Infinite Products and Factorization Factors . 64 4.4 Weierstrass Factorization Theorem . 66 4.5 Application to Singular Points . 68 4.6 Mittag-Leffler’s Theorem . 69 4.7 Homework 7 . 71 List of Figures 1 The nice contour . 5 2 The semi-circular contour with a bump . 8 3 Another common contour . 9 4 The contour that works well with ln(x) . 14 5 Conformal Map 1 . 42 6 Upper half disc to unit disc . 43 7 Region between two circles to unit disc . 44 8 conformal map 4 . 45 Acknowledgement: I would like to thank J. N. Treuer for reading through the first draft and making some revisions. 2 1 The Residue theorem applied to real inte- grals We are going to apply the Residue theorem to evaluate integrals of real- valued functions over subsets of IR. We will call such integrals real integrals. 1.1 Several types of real integrals I) Let R(x; y) be a rational function in x and y. How does one evaluate Z 2π R(cos t; sin t)dt? 0 Solution. Let z = eit; t 2 [0; 2π): Then z + 1 z2 + 1 z − 1 z2 − 1 dz cos t = z = ; sin t = z = and dt = : 2 2z 2i 2iz iz Therefore, Z 2π Z z2 + 1 z2 − 1 1 R(cos t; sin t)dt = R( ; ) dz: 0 jzj=1 2z 2iz iz We can apply the Residue theorem to evaluate the right hand side. EXAMPLE 1 For a > 1, evaluate the integral Z 2π 1 dt 0 a + cos t Solution. Let z =: eit Then 1 1 cos t = (z + ); dz = izdt 2 z 3 Therefore, Z 2π 1 Z 1 1 dt = 1 dz 0 a + cos t jzj=1 a + 2 (z + 1=z) iz 2 Z 1 = dz i jzj=1 2az + z2 + 1 1 p = 4πRes ; −a + a2 − 1 z2 + 2az + 1 4π = p 2(−a + a2 − 1) + 2a 2π = p : a2 − 1 II) How does one evaluate Z 1 R(x)dx? −∞ THEOREM 1.1 Let Pm(x) and Qn(x) be polynomials of degree m and n respectively and (P ;Q ) = 1. Let R(x) = Pm(x) . Suppose also that Q (x) 6= m n Qn(x) n 0 for all x 2 IR and n − m ≥ 2. Then ` Z 1 X R(x)dx = 2πi( Res(R; zk)) −∞ k=1 where z1; ··· ; z` are the zeros of Qn not counting multiplicity in the upper 2 half plane IR+. Proof. Let fz1; :::; z`g be the zeros stated in the theorem's hypothesis and choose r >> 1 such that fz1; ··· ; z`g ⊂ D(0; r). Apply the Residue theorem 2 to R(z) on Dr = fz 2 IR+ : jzj < rg. Then ` Z r Z X R(x) = − R(z)dz + 2πi Res(R; zk) −r Cr k=1 iθ where Cr = fz = re : 0 ≤ θ ≤ πg. Since n ≥ m + 2, we have Z lim R(z)dz = 0: r!1 Cr The theorem is proved. 4 Cr Dr −r r Figure 1: The nice contour EXAMPLE 2 Evaluate the integral Z 1 1 dx: 0 (1 + x2)2 Solution. Let f(z) = 1=(z2 + 1)2. Then Z 1 1 1 Z 1 1 dx = dt 0 (1 + x2)2 2 −∞ (1 + x2)2 −2 = πiRes(f; i) = πi 3 (z + i) z=i 2πi π = − = (2i)3 4 III) How does one evaluate Z 1 f(x)eixdx? −∞ 2 Lemma 1.2 Let f(z) be meromorphic in IR+ such that lim f(reiθ) = 0 r!1 2 uniformly for θ 2 (0; π). Let fzk : 1 ≤ k ≤ mg be zero set of f in IR+. Then Z lim f(z)eizdz = 0 R!1 CR where CR = fz = x + iy 2 C: jzj = R; y > 0g 5 iθ Proof. Let MR = maxfjf(Re )j : θ 2 [0; π]g. Then Z Z π iz iθ −R sin θ f(z)e dz ≤ jf(Re )je Rdθ CR 0 Z π −R sin θ ≤ MR e Rdθ 0 Z π=2 −R sin θ = 2MR e Rdθ 0 Z π=2 −2Rθ/π ≤ 2MR e Rdθ 0 Z 1 −t ≤ πMR e dt 0 = πMR ! 0 as R ! 1 π where the third inequality follows because sin(θ) is concave on [0; 2 ]; hence 2θ π sin(θ) > π on [0; 2 ]: 2 THEOREM 1.3 Let f(z) be meromorphic in IR+ such that lim f(reiθ) = 0 r!1 2 uniformly for θ 2 (0; π). Let fzk : 1 ≤ k ≤ mg be zero set of f in IR+.Then Z 1 m ix X iz f(x)e dx = 2πi Res(fe ; zk) −∞ k=1 2 Proof. Let DR = fz 2 IR+ : jzj < Rg. By the Residue theorem, one has Z R Z Z ix iz iz X iz f(x)e + f(z)e dz = f(z)e dz = 2πi Res(f(z)e ; zk) −R CR @DR zk2DR Let R ! 1. By the previous lemma, Z 1 m ix X iz f(x)e = 2πi Res(f(z)e ; zk) −∞ k=1 The proof is complete. EXAMPLE 3 Evaluate the following integral: Z 1 cos x dx 0 1 + x2 6 Solution. Since Z 1 cos x 1 Z 1 cos x dx = dx 0 1 + x2 2 −∞ 1 + x2 1 Z 1 eix = Re dx 2 −∞ 1 + x2 eiz = Re πiRes( ; i) 1 + z2 π = : 2e EXAMPLE 4 Evaluate the following integral: Z 1 sin x dx 0 x Solution.We would like to write Z 1 sin x 1 Z 1 sin x 1 Z 1 eix dx = dx = Im dx: 0 x 2 −∞ x 2 −∞ x However, the last equality may cause trouble because the last integral does not converge. So, instead, we write Z 1 sin x h Z −r sin(x) Z R sin x i dx = lim dx + dx −∞ x R!1;r!0 −R x r x Let 2 Dr;R = fz 2 IR+ : jzj < R; jzj > rg and @Dr;R = CR [ [−R; −r] [ (−Cr) [ [r; R]: Thus Z 1 sin x h Z −r eix Z R eix i dx = lim Im dx + dx −∞ x R!1;r!0 −R x r x h Z eiz Z eiz Z eiz i = lim Im dz − dz + dz R!1;r!0 @Dr;R z CR z Cr z Z eiz = 0 + 0 + lim Im dz r!0 Cr z Z π eri cos θ−r sin θ = lim Im ireiθdθ r!0 0 reiθ = π 7 CR Dr;R −Cr -R -r r R Figure 2: The semi-circular contour with a bump eiz where the third equality follows because z is holomorphic in a neighborhood of Dr;R and lemma 1.2. Therefore, Z 1 sin x π dx = : 0 x 2 EXAMPLE 5 Evaluate the following integral: Z 1 sin2 x dx: 0 x2 Solution. 2 2 Z 1 sin x sin x 1 Z 1 2 sin x cos x dx = − + dx 0 x2 x 0 0 x Z 1 sin(2x) = dx 0 x Z 1 sin(2x) = d(2x) 0 2x π = : 2 EXAMPLE 6 Evaluate the following integral: Z 1 1 − cos x dx 0 x2 8 Solution. Z 1 1 − cos x 1 − cos x 1 Z 1 sin x π dx = − + = : 0 x2 x 0 0 x 2 IV) Choosing special integral path. EXAMPLE 7 Evaluate the following integral: Z 1 1 dx: 0 1 + x3 Solution. We know that (e2πi=3)3 = 1. Choose iθ DR = fz = jzje : 0 < θ < 2π=3; jzj < Rg Then for R > 2; Re2πi=3 2π 3 R Figure 3: Another common contour Z 1 1 iπ=3 2πi 3 dz = 2πiRes( 3 ; e ) = 2πi=3 @DR 1 + z 1 + z 3e On the other hand, Z Z R Z Z 0 1 1 1 1 2πi=3 3 dz = 3 dx + 3 dz + 3 e dx @DR 1 + z 0 1 + x CR 1 + z R 1 + x Z 1 1 ! (1 − e2πi=3) dx 0 1 + x3 9 as R ! 1.
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