Hadamard s Theorem and Entire Functions of Finite Order For Math 331
|
|
- Gerald Reeves
- 6 years ago
- Views:
Transcription
1 Hadamard s Theorem and Entire Functions of Finite Order For Math 33 Taylor Dupuy Entire functions of finite order Definition.. An entire function f is finite order if and only if ρ, R such that f(z) < exp( z ρ ) whenever z R. The infimum of such ρ is called the order of f and is denoted by ρ = ρ(f). Lemma.2. Let f be a entire function of finite order. were M(f, r) = max z =r f(z). Proof. If f is finite order, ρ(f) = lim R sup r R log log M(f, r), log(r) M(f, r) exp(r ρ ) = log M(f, r) r ρ = log log M(f, r) ρ log(r) So we have lim lim log log M(f, r) ρ R r R log r Example.3. Here are some functions and their orders:. e z, ρ = 2. sin(z), ρ = 3. cos( z), ρ = /2 4. e ez, ρ = 5. e z2, ρ = 2
2 Here is the converse of the Weierstrass product formula. This is the main goal of these notes. Theorem.4 (Hadamard). If f be an entire function of finite order ρ then f can be written as where f(z) = e g(z) z m E d (z/a j ) (.5) a, a 2,... are the zeros of f(z) repeating according to their multiplicity. m = ord z= f(z) g(z) C[z] Furthermore we have deg(g), d ρ. The proof in Ahlfors is a little incomplete and the proof in McMullen uses hyperbolic geometry. The proof in Schlag isn t finished. The proof in Green and Krantz is essentially the same al Ahlfors. We are going to follow McMullen s proof supplementing it with computations from Ahlfors and replacing his hyperbolic geometry arguments with a more down-to-earth estimate. Strategy of Proof.. Prove this for entire function f(z) without zeros: by existence of logarithms we know f(z) = e g(z). We show g(z) is a polynomial (This involves the so-called Borel-Cartheodory inequality) 2. By Weierstrass any entire function can be written as f(z) = e g(z) P (z) where P (z) is a canonical product and g(z) is some function. (a) Study the order of P (z) (this uses Jensen s Formula a formula that relates the size of the zeros of a function to its absolute value) (b) Get bounds on /P (z) (c) Conclude that f(a)/p (z) is entire of finite order as in the first case. 2 Case: functions without zeros Exercise 2.. If f : D R () D R () is a conformal map with f() = then f(z) z. (Hint: consider g(z) = g(z/r)/r and apply Schwarz.) This replaces a hyperbolic geometry argument that McMullen uses an actually uses the Schwarz Lemma, an incarnation of hyperbolic geometry 2
3 Theorem 2.2 (Borel-Caratheodory). Let f be holomorphic on a region containing D R (). For all r < R, M(f, r) 2r R + r M(Re f, R) + R r R r f(). Proof. Let A = M(Re f, R) (since Re f is harmonic this is positive). We break the proof into two cases: f() = and f(). Suppose f() =. The map f takes D R () to the region {z : Re f(z) < A}. The map in figure explains how the map g(z) = Rf(z) f(z) 2A is a conformal map g : D R () D R () with g() =. Lemma (see Exercise 2.) we have g(z) z. This gives By Schwarz s R f(z) f(z) 2A z. If z < r we have which implies R f(z) r f(z) 2A r f(z) + 2Ar f(z) 2r M(Re f, R). R r Suppose f(). Consider g(z) = f(z) f() and apply the previous case (you will find the string of inequalities below) This implies here we used f(z) f() f(z) f() 2r R r max Re(f(w) f()) w R 2r R r ( max Re f(w)) + f(). w R f(z) 2r R r max R + r Re f(w) + w R R r f(), 2r R r + = r+r R r. Theorem 2.3. If f(z) is an entire function of finite order ρ without zeros then f(z) = e g(z) where g(z) is a polynomial of degree ρ (in particular ρ N). 3
4 f D R () x = M(Re(f), R) = A z z/a z R z+ z Re(z) < D R () Figure : The composition of conformal maps used to imply the Borel- Caratheodory theorem. 4
5 Remark 2.4. Theorem 2.3 is equivalent to saying the following: if for all ε > there exists some r n such that Re g(z) < r ρ+ε n whenever z = r n then g(z) is a polynomial of degree at most ρ. The sequence r n is necessary sometimes to avoid weird behavior at particular radii in applications. Proof. By the finite order hypothesis, there exists some R such that f(z) = e Re g(z) e z ρ+ε for z R. This implies Re g(z) z ρ+ε. By Borel- Caratheodory, taking R = 2r we have M(g, r) 2r R + r M(Re g, R) + R r R r g() = 2M(Re g, 2r) + 3 g() 2(2r) ρ+ε + 3 g() = O(r ρ+ε ). By Liouville s Theorem (the souped-up version) g(z) must be a polynomial of degree less than or equal to ρ. 2 3 Jensen s formula To move prove Hadamard s theorem where the entire function f(z) has zeros we need to know something about the growth of the zeros. This is provided by Jensen s Formula: Theorem 3. (Jensen s Formula). Let f(z) be analytic in a region containing D R (), f() and f(z) has no zeros with z = R. We have log f() n log a j R + 2π 2π log f(re iθ ) dθ, (3.2) here a, a 2,..., a n are the zeros of f(z) insider the regions D R (). When looking at this at first, think of the case when f() = so the left hand term vanishes. What this formula does is relate the absolute values of the roots to the absolute value of f(z) on the circle z = R. Proof. Suppose f(z) has no zeros. Then log f(z) is harmonic and log f() = 2π 2π log f(re iθ ) dθ 2 Observe that in the application of Liouville s theorem we just need bounds on M(f, r) for particular radii approaching infinity so the version in the remark holds 5
6 by the mean value theorem. Suppose now f(z) has zeros inside D R (). Let a,..., a n be the zeros of f(z) in D R () where we repeat zeros according to multiplicity. Define g(z) = f(z) n R 2 a j z R(z a j ). This function is analytic, free from zeros inside D R () (see Exercise 3.3). Consequently, log g() = 2π 2π log g(re iθ ) dθ = 2π 2π log f(re iθ ) dθ. Since, g() = f() n R a j, we have and hence log f() + log g() = log f() + R log R a j + 2π n log R a j 2π log f(re iθ ) dθ. Exercise 3.3. The factors B R/a (z) = R2 az R(z a) appearing in the product are called Blashke factors. Here a < R.. Show that B R/a (z) = when z = R 2. Show that the zeros of B R/a (z) lie outside D R (). Exercise 3.4. This exercise extends Jensen s formula to the case when f(z) has zeros on the boundary.. Show that 2π log e iθ dθ = (first show that this integral makes sense) Consider g(z) = f(z)/ r (z Reiθj ) where Re iθ, Re iθ2,..., Re iθr are the zeros of f(z) on the boundary. Show that by applying Jensen s Formula to g(z) we recover Jensen s formular for f(z). 4 Case: canonical products We want to know when canonical products of zeros of an entire function converge. In this section we keep in mind the following criterion: a j h+ < = ( ) z E h converges, a j where E h (z) = ( z) exp(z + z 2 /2 + + z h /h). We will control the left hand side using Jensen s formula. 3 See Ahlfors, Chapter 5, Section 6, Exercise 4 I think, 6
7 Definition 4.. Let {a j } be a sequence of non-zero complex number ordered by size (meaning a j a j+ ). The critical exponent of the sequence is α = inf{β : a j β < }. Definition 4.2. Let {a j } be a sequence of non-zero complex number ordered by size. The counting function of the sequence is N(r) = #{n : a n r}. A basic philosophy is that knowledge about the N(r) is the same as knowledge of r n = a n. Lemma 4.3. For all β > we have a n β = n= Proof. Let a j r j and define r =. We have N(r)βr β dr. (4.4) N(r)βr β dr = = = = rj+ N(r j ) βr β dr r j N(r j )(r β j r β j+ ) (N(r j ) N(r j )r β a j β. j The last line follows from the fact that N(r j ) N(r j ) = #{a k : a k = r j } Lemma 4.5. Let {a j } be a sequence of non-zero complex numbers ordered by size. Let N(r) = #{j : a j < r}. We have α = lim R sup r>r I couldn t make a proof of this work using the p-test. log N(r). (4.6) log r log N(r) Proof. Let α = lim R sup r>r log r. Let α be the critical exponent. We need to show α α and α α. 7
8 We first show α α. We show convergence of a j β whenever β > α. By the definition of lim sup for all ε > there exist some R such that for all r > R we have log(n(r))/ log(r) α + ε. This implies Since and N(r)r β dr converges R N(r) > r α+ε when r R. N(r)r β dr R R N(r)r β dr converges r α+ε β+ dr we have convergence when α + ε β < or α + ε < β. Since ε > was arbitrary this condition turns says and hence {β : β > α } {β : a j β < }, α = inf{β : β > α } inf{β : a j β < } = α We show now α α. We divergence of a j β whenever β < α. We would like to use a similar divergence argument as in the pervious case but we don t get clear lower bounds because of the lim sup in the definition log N(r) lim sup r log r = α we have: ε >, R and a sequence r n with r n R such that N(r n ) > rn α ε. Because r n could be spread out in some weird way, we don t get a simple lower bound on our integral. First, observe that u a j 2u a j β u a j 2u N(2r) N(r) = (2u) β (2u) β. 4 In [McM] he shows divergence of a j β when we have N(r) > r α ε and β < α. 5 We do the opposite of on-the-convergence-exponent-of-zeros-of-entire-functions. Although the person answering is technically giving the wrong direction, this is a good example of where a technically wrong response gives the correct tool to solve a problem. 6 Another post order-of-growth-of-a-counting-function obnoxiously asserts α is the quantity we log(n) should start with, asserts that it is equal to α = α = lim sup n without proof log a n (which is done is say [War] or [Lev96]) and proves α = α making a funny convention for ε. The treatment of Hadamard s Theorem in [BY4] proves α = α and doesn t use α at all. 8
9 This means a j <2 m+ u a j β = a j u m m a j β + j= 2 j u< a j 2 j+ u N(2 j+ u) N(2 j u) (2 j u) β (2 m+ u) β a j β m N(2 j+ u) N(2 j u) = N(2j+ u) N(u) (2 m+ u) β Now we have a lower bound! Define m n and u n from the sequence r n by log 2 (r n ) = m n +, r n 2 mn+ = u n, we get r n = 2 mn+ u n with u n < 2 and a j r n a j β N(r n) N(u n ) r β n N(r n) N(2) r β n N(r n) r β n rα ε n rn β = r α ε β n. Hence the series diverges if α ε > β. Since this is done for arbitrary ε > we have the series diverges whenever α > β. Hence which implies {β : β < α } {β : a j β diverges } α = sup{β : β < α } sup{β : a j β diverges } = α. Exercise 4.7. Show that α = lim sup n log n log a n. log n Proof. Let α = lim sup n log a n and let α = inf{β : j a j β < }. We show that α α and α α. 9
10 Suppose β > α. This implies β > α +ε for some ε >. By the definition of lim sup: ε >, n, n N: n n = log n log a n < α + ε. Fix such an n. This condition is the same as n /(α+ε) < a n. Hence for n n we have a n β. n β α +ε Since β > α + ε the series n a n β converges. Suppose β < α. This implies β < α ε for some ε >. By the definition of lim sup: ε >, n and a sequence n j as j with n j n such that for all j log n j log a nj > α ε. Fix such a sequence n j. This condition is equivalent to a nj < n /(α ε) j. Hence for n j we have n j n= a n β n j a nj β > n j β n α ε j = n β α ε j. Since β < α ε we have β/(α ε) < and the series diverges. Lemma 4.8. Let f be an entire function of finite order. The critical exponent of the non-zero zeros of f(z) is less than the order of f: Proof. log(2)n(r) log M(2r) α(f) ρ(f). = log N(r) log 2 M(2r) log 2 (2) = log N(r) log(r) = α = lim sup r log2 M(2r) log 2 (2) log(2r) log(2) log N(r) log(r) lim sup r = log2 (M(2r))/ log(2r) log 2 (2)/ log(2r) log(2)/ log(2r) log 2 M(2r) log(2r) = ρ. Remark 4.9. We actually have α(f) = ρ(f).
11 Lemma 4.. Let f be an entire function of finite order ρ. Let a, a 2,... be its non-zero zeros. The number d = ρ is the smallest integer making ( E z d converge. a j ) Proof. We have ρ ρ ρ +. Since α < ρ we have that α < ρ +. This implies a j ρ converges and hence the desired canonical product converges. If f(z) is ( an entire function of finite order ρ and zeros a, a 2,... we now know that f(z)/ z m ) E ρ (z/a j ) is an entire function of finite order without zero. It remains to get a bound on the order of this function. To do this, we need a lower bound on E ρ (z/a j ). McMullen and Ahlfors say some things about lower bounds and it looks like McMullen is following Stein and Shakarchi. Here we follow the material before Stein and Shakarchi, Lemma 5.3 to get the bound. Lemma 4.. when z /2.. There exists some C such that 2. There exists some C 2 such that when z /2. Proof. We follow [SS3]. log E h (z) C z h+ log E h (z) C 2 z h. Suppose z /2. This implies log( z) = j zj /j. We have h E h (z) = exp log( z) + z j /j = exp z j /j! = exp(w). j=h+ Since e w e w and w C z h+ we have ( ) E h (z) exp z j /j e C z h+. 2. Suppose z /. We have The proof follows from for some C. h+ E h (z) = z e z+z2 /2+ +z h /h. e z+z2 /2+ +z h /h e z+z2 /2+ z h /h e C z h
12 Lemma 4.2. Let {a j } be a sequence of complex numbers with critical exponent α. Let h = α. For all z sufficiently large and outside j D /r α+ε (a j ) j there exists a constant B = B ε such that E h (z/a j ) exp( B z α+ε ). (Hence there exists a sequence z j such that this inequality holds.) Proof. E h (z/a j ) = z/a j </2 E h (z/a j ) } {{ } case I z/a j /2 E h (z/a j ) } {{ } case II Case I: z/a j < /2 (there are infinitely many such a j since a j ). By the bounding lemma, log E h (z/a j ) c z/a j h+ Observe that a j h = a j α ε a j α+ε h a j 2z α+ε h. Here we used α + ε α < and 2z < a j. These together give z/a j </2 E h (z/a j ) exp c z/a j </2 exp c z h+ = exp = e B z α+ε z/a j h+ z/a j </2 c 2z α+ε 2 h+ z/a j </2 a j α ε 2z α+ε h a j α ε Case II: z/a j /2 (there are finitely many such a j ). We have z/a j /2 E h (z/a j ) z/a j /2 z } {{ } case II.2 z/a j /2 e c2 z/aj h } {{ } case II.2 2
13 Case II.: We have z/a j /2 e c2 z/aj h = exp c 2 z/a j /2 We will now bound the exponent from below: c 2 z/a j h = c 2 z/a j h z/a j /2 z/a j /2 c 2 z α+ε z/a j >/2 = c 2 z α+ε = b 2 z α+ε. z/a j /2 Case II.2: We look at z/a j /2 z a j. z/a j h. ( a j /2) h α ε a j h ( z > a j /2 and h α ε < ) 2 h α ε a j α+ε Since z/a j > /2 we have 2 z > a j which implies there are exactly N(2 z ) terms. From the hypotheses z a j > / a j α+ε we get This implies N(2 z ) log( z/a j ) = > z a j = z a j a j > a j α+ε a j = a j α++ε N(2 z ) N(2 z ) log z/a j (α + + ε) log a j N(2 z ) > (α + + ε) log(2 z ) = (α + + ε) log(2 z )N(2 z ). (α + + ε) log(2 z )(2 z ) α by formula for α b z α+ε 3
14 Combining cases I, II. and II.2 show that for every ε > there exists a constant such that for all z not in j D /r α+ε (a j ) we have j this is equivalent to E h (z/a j ) exp( B z α+ε ) 5 Proof of Hadamard s Factorization Theorem Proof of Hadmard s Theorem. Consider now f(z)/z m P (z) = e g(z) where P (z) = j E ρ (z/a j ). sufficiently large we have For z C \ j D /r α n (a j) and z = r f(z) z m P (z) e rρ+ε r m e Brα+ε = exp(r ρ+ε + Br α+ε m log(r)) e (+B)rρ+ε. This shows that Re g(z) = O(r ρ+ε ) (along particular radii where the lower bound applies) and hence we are in the hypotheses of Theorem 2.3 and Remark 2.4. This proves that g(z) is a polynomial of degree at most ρ. References [BY4] Freiling Buterin and Yurko, Lectures on the theory of entire functions, SM-UDE-779, 24. [Lev96] B Ya Levin, Lectures on entire functions, vol. 5, American Mathematical Soc., 996. [McM] [SS3] [War] McMullen, A concise course in complex analysis. Elias M Stein and Rami Shakarchi, Complex analysis. princeton lectures in analysis, ii, 23. Warner, Zeros, University of Washington. 4
Math 259: Introduction to Analytic Number Theory Functions of finite order: product formula and logarithmic derivative
Math 259: Introduction to Analytic Number Theory Functions of finite order: product formula and logarithmic derivative This chapter is another review of standard material in complex analysis. See for instance
More informationMath 259: Introduction to Analytic Number Theory Functions of finite order: product formula and logarithmic derivative
Math 259: Introduction to Analytic Number Theory Functions of finite order: product formula and logarithmic derivative This chapter is another review of standard material in complex analysis. See for instance
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 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 informationMath 520a - Final take home exam - solutions
Math 52a - Final take home exam - solutions 1. Let f(z) be entire. Prove that f has finite order if and only if f has finite order and that when they have finite order their orders are the same. Solution:
More information3. 4. Uniformly normal families and generalisations
Summer School Normal Families in Complex Analysis Julius-Maximilians-Universität Würzburg May 22 29, 2015 3. 4. Uniformly normal families and generalisations Aimo Hinkkanen University of Illinois at Urbana
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 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 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 informationAccumulation constants of iterated function systems with Bloch target domains
Accumulation constants of iterated function systems with Bloch target domains September 29, 2005 1 Introduction Linda Keen and Nikola Lakic 1 Suppose that we are given a random sequence of holomorphic
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 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 informationMATH 566 LECTURE NOTES 6: NORMAL FAMILIES AND THE THEOREMS OF PICARD
MATH 566 LECTURE NOTES 6: NORMAL FAMILIES AND THE THEOREMS OF PICARD TSOGTGEREL GANTUMUR 1. Introduction Suppose that we want to solve the equation f(z) = β where f is a nonconstant entire function and
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 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 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 informationConsidering our result for the sum and product of analytic functions, this means that for (a 0, a 1,..., a N ) C N+1, the polynomial.
Lecture 3 Usual complex functions MATH-GA 245.00 Complex Variables Polynomials. Construction f : z z is analytic on all of C since its real and imaginary parts satisfy the Cauchy-Riemann relations and
More informationPlaces of Number Fields and Function Fields MATH 681, Spring 2018
Places of Number Fields and Function Fields MATH 681, Spring 2018 From now on we will denote the field Z/pZ for a prime p more compactly by F p. More generally, for q a power of a prime p, F q will denote
More informationComplex Analysis for F2
Institutionen för Matematik KTH Stanislav Smirnov stas@math.kth.se Complex Analysis for F2 Projects September 2002 Suggested projects ask you to prove a few important and difficult theorems in complex
More information8 8 THE RIEMANN MAPPING THEOREM. 8.1 Simply Connected Surfaces
8 8 THE RIEMANN MAPPING THEOREM 8.1 Simply Connected Surfaces Our aim is to prove the Riemann Mapping Theorem which states that every simply connected Riemann surface R is conformally equivalent to D,
More informationMath 425 Fall All About Zero
Math 425 Fall 2005 All About Zero These notes supplement the discussion of zeros of analytic functions presented in 2.4 of our text, pp. 127 128. Throughout: Unless stated otherwise, f is a function analytic
More informationENTRY POTENTIAL THEORY. [ENTRY POTENTIAL THEORY] Authors: Oliver Knill: jan 2003 Literature: T. Ransford, Potential theory in the complex plane.
ENTRY POTENTIAL THEORY [ENTRY POTENTIAL THEORY] Authors: Oliver Knill: jan 2003 Literature: T. Ransford, Potential theory in the complex plane. Analytic [Analytic] Let D C be an open set. A continuous
More informationMA30056: Complex Analysis. Revision: Checklist & Previous Exam Questions I
MA30056: Complex Analysis Revision: Checklist & Previous Exam Questions I Given z C and r > 0, define B r (z) and B r (z). Define what it means for a subset A C to be open/closed. If M A C, when is M said
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 680 Fall A Quantitative Prime Number Theorem I: Zero-Free Regions
Math 68 Fall 4 A Quantitative Prime Number Theorem I: Zero-Free Regions Ultimately, our goal is to prove the following strengthening of the prime number theorem Theorem Improved Prime Number Theorem: There
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 Topic: Singularities
Complex Analysis Topic: Singularities MA201 Mathematics III Department of Mathematics IIT Guwahati August 2015 Complex Analysis Topic: Singularities 1 / 15 Zeroes of Analytic Functions A point z 0 C is
More informationLecture 14 Conformal Mapping. 1 Conformality. 1.1 Preservation of angle. 1.2 Length and area. MATH-GA Complex Variables
Lecture 14 Conformal Mapping MATH-GA 2451.001 Complex Variables 1 Conformality 1.1 Preservation of angle The open mapping theorem tells us that an analytic function such that f (z 0 ) 0 maps a small neighborhood
More informationGROWTH CONDITIONS FOR ENTIRE FUNCTIONS WITH ONLY BOUNDED FATOU COMPONENTS
GROWTH CONDITIONS FOR ENTIRE FUNCTIONS WITH ONLY BOUNDED FATOU COMPONENTS AIMO HINKKANEN AND JOSEPH MILES Abstract Let f be a transcendental entire function of order < 1/2 We denote the maximum and minimum
More informationMath 220A Homework 4 Solutions
Math 220A Homework 4 Solutions Jim Agler 26. (# pg. 73 Conway). Prove the assertion made in Proposition 2. (pg. 68) that g is continuous. Solution. We wish to show that if g : [a, b] [c, d] C is a continuous
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 informationGeometric Complex Analysis. Davoud Cheraghi Imperial College London
Geometric Complex Analysis Davoud Cheraghi Imperial College London May 9, 2017 Introduction The subject of complex variables appears in many areas of mathematics as it has been truly the ancestor of many
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 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 informationComplex Analysis Qual Sheet
Complex Analysis Qual Sheet Robert Won Tricks and traps. traps. Basically all complex analysis qualifying exams are collections of tricks and - Jim Agler Useful facts. e z = 2. sin z = n=0 3. cos z = z
More informationMath Homework 2
Math 73 Homework Due: September 8, 6 Suppose that f is holomorphic in a region Ω, ie an open connected set Prove that in any of the following cases (a) R(f) is constant; (b) I(f) is constant; (c) f is
More informationComplex Analysis, Stein and Shakarchi The Fourier Transform
Complex Analysis, Stein and Shakarchi Chapter 4 The Fourier Transform Yung-Hsiang Huang 2017.11.05 1 Exercises 1. Suppose f L 1 (), and f 0. Show that f 0. emark 1. This proof is observed by Newmann (published
More informationMATH 452. SAMPLE 3 SOLUTIONS May 3, (10 pts) Let f(x + iy) = u(x, y) + iv(x, y) be an analytic function. Show that u(x, y) is harmonic.
MATH 45 SAMPLE 3 SOLUTIONS May 3, 06. (0 pts) Let f(x + iy) = u(x, y) + iv(x, y) be an analytic function. Show that u(x, y) is harmonic. Because f is holomorphic, u and v satisfy the Cauchy-Riemann equations:
More informationOn multiple points of meromorphic functions
On multiple points of meromorphic functions Jim Langley and Dan Shea Abstract We establish lower bounds for the number of zeros of the logarithmic derivative of a meromorphic function of small growth with
More informationf(w) f(a) = 1 2πi w a Proof. There exists a number r such that the disc D(a,r) is contained in I(γ). For any ǫ < r, w a dw
Proof[section] 5. Cauchy integral formula Theorem 5.. Suppose f is holomorphic inside and on a positively oriented curve. Then if a is a point inside, f(a) = w a dw. Proof. There exists a number r such
More informationComplex Analysis Math 147 Winter 2008
Complex Analysis Math 147 Winter 2008 Bernard Russo March 14, 2008 Contents 1 Monday January 7 Course information; complex numbers; Assignment 1 1 1.1 Course information................................
More informationζ(u) z du du Since γ does not pass through z, f is defined and continuous on [a, b]. Furthermore, for all t such that dζ
Lecture 6 Consequences of Cauchy s Theorem MATH-GA 45.00 Complex Variables Cauchy s Integral Formula. Index of a point with respect to a closed curve Let z C, and a piecewise differentiable closed curve
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 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 informationThe result above is known as the Riemann mapping theorem. We will prove it using basic theory of normal families. We start this lecture with that.
Lecture 15 The Riemann mapping theorem Variables MATH-GA 2451.1 Complex The point of this lecture is to prove that the unit disk can be mapped conformally onto any simply connected open set in the plane,
More informationNon-real zeroes of real entire functions and their derivatives
Non-real zeroes of real entire functions and their derivatives Daniel A. Nicks Abstract A real entire function belongs to the Laguerre-Pólya class LP if it is the limit of a sequence of real polynomials
More informationCourse 214 Section 2: Infinite Series Second Semester 2008
Course 214 Section 2: Infinite Series Second Semester 2008 David R. Wilkins Copyright c David R. Wilkins 1989 2008 Contents 2 Infinite Series 25 2.1 The Comparison Test and Ratio Test.............. 26
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 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 informationExercises involving elementary functions
017:11:0:16:4:09 c M K Warby MA3614 Complex variable methods and applications 1 Exercises involving elementary functions 1 This question was in the class test in 016/7 and was worth 5 marks a) Let z +
More informationSchwarz lemma and automorphisms of the disk
Chapter 2 Schwarz lemma and automorphisms of the disk 2.1 Schwarz lemma We denote the disk of radius 1 about 0 by the notation D, that is, D = {z C : z < 1}. Given θ R the rotation of angle θ about 0,
More informationMath 411, Complex Analysis Definitions, Formulas and Theorems Winter y = sinα
Math 411, Complex Analysis Definitions, Formulas and Theorems Winter 014 Trigonometric Functions of Special Angles α, degrees α, radians sin α cos α tan α 0 0 0 1 0 30 π 6 45 π 4 1 3 1 3 1 y = sinα π 90,
More information11 COMPLEX ANALYSIS IN C. 1.1 Holomorphic Functions
11 COMPLEX ANALYSIS IN C 1.1 Holomorphic Functions A domain Ω in the complex plane C is a connected, open subset of C. Let z o Ω and f a map f : Ω C. We say that f is real differentiable at z o if there
More informationExercises for Part 1
MATH200 Complex Analysis. Exercises for Part Exercises for Part The following exercises are provided for you to revise complex numbers. Exercise. Write the following expressions in the form x + iy, x,y
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 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 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 informationWe have been going places in the car of calculus for years, but this analysis course is about how the car actually works.
Analysis I We have been going places in the car of calculus for years, but this analysis course is about how the car actually works. Copier s Message These notes may contain errors. In fact, they almost
More informationF (x) = P [X x[. DF1 F is nondecreasing. DF2 F is right-continuous
7: /4/ TOPIC Distribution functions their inverses This section develops properties of probability distribution functions their inverses Two main topics are the so-called probability integral transformation
More informationGROWTH OF SOLUTIONS TO HIGHER ORDER LINEAR HOMOGENEOUS DIFFERENTIAL EQUATIONS IN ANGULAR DOMAINS
Electronic Journal of Differential Equations, Vol 200(200), No 64, pp 7 ISSN: 072-669 URL: http://ejdemathtxstateedu or http://ejdemathuntedu ftp ejdemathtxstateedu GROWTH OF SOLUTIONS TO HIGHER ORDER
More informationA UNIQUENESS THEOREM FOR MONOGENIC FUNCTIONS
Annales Academiæ Scientiarum Fennicæ Series A. I. Mathematica Volumen 8, 993, 05 6 A UNIQUENESS THEOREM FOR MONOGENIC FUNCTIONS Jörg Winkler Technische Universität Berlin, Fachbereich 3, Mathematik Straße
More informationMATH 185: COMPLEX ANALYSIS FALL 2009/10 PROBLEM SET 9 SOLUTIONS. and g b (z) = eπz/2 1
MATH 85: COMPLEX ANALYSIS FALL 2009/0 PROBLEM SET 9 SOLUTIONS. Consider the functions defined y g a (z) = eiπz/2 e iπz/2 + Show that g a maps the set to D(0, ) while g maps the set and g (z) = eπz/2 e
More information2.4 The Extreme Value Theorem and Some of its Consequences
2.4 The Extreme Value Theorem and Some of its Consequences The Extreme Value Theorem deals with the question of when we can be sure that for a given function f, (1) the values f (x) don t get too big or
More informationComplex Analysis Math 220C Spring 2008
Complex Analysis Math 220C Spring 2008 Bernard Russo June 2, 2008 Contents 1 Monday March 31, 2008 class cancelled due to the Master s travel plans 1 2 Wednesday April 2, 2008 Course information; Riemann
More informationIII.2. Analytic Functions
III.2. Analytic Functions 1 III.2. Analytic Functions Recall. When you hear analytic function, think power series representation! Definition. If G is an open set in C and f : G C, then f is differentiable
More information= 0. Theorem (Maximum Principle) If a holomorphic function f defined on a domain D C takes the maximum max z D f(z), then f is constant.
38 CHAPTER 3. HOLOMORPHIC FUNCTIONS Maximum Principle Maximum Principle was first proved for harmonic functions (i.e., the real part and the imaginary part of holomorphic functions) in Burkhardt s textbook
More informationHere are brief notes about topics covered in class on complex numbers, focusing on what is not covered in the textbook.
Phys374, Spring 2008, Prof. Ted Jacobson Department of Physics, University of Maryland Complex numbers version 5/21/08 Here are brief notes about topics covered in class on complex numbers, focusing on
More informationFunctions of a Complex Variable and Integral Transforms
Functions of a Complex Variable and Integral Transforms Department of Mathematics Zhou Lingjun Textbook Functions of Complex Analysis with Applications to Engineering and Science, 3rd Edition. A. D. Snider
More informationWEIERSTRASS THEOREMS AND RINGS OF HOLOMORPHIC FUNCTIONS
WEIERSTRASS THEOREMS AND RINGS OF HOLOMORPHIC FUNCTIONS YIFEI ZHAO Contents. The Weierstrass factorization theorem 2. The Weierstrass preparation theorem 6 3. The Weierstrass division theorem 8 References
More informationUNIQUE RANGE SETS FOR MEROMORPHIC FUNCTIONS CONSTRUCTED WITHOUT AN INJECTIVITY HYPOTHESIS
UNIQUE RANGE SETS FOR MEROMORPHIC FUNCTIONS CONSTRUCTED WITHOUT AN INJECTIVITY HYPOTHESIS TA THI HOAI AN Abstract. A set is called a unique range set (counting multiplicities) for a particular family of
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 informationOn Bank-Laine functions
Computational Methods and Function Theory Volume 00 0000), No. 0, 000 000 XXYYYZZ On Bank-Laine functions Alastair Fletcher Keywords. Bank-Laine functions, zeros. 2000 MSC. 30D35, 34M05. Abstract. In this
More informationTaylor and Laurent Series
Chapter 4 Taylor and Laurent Series 4.. Taylor Series 4... Taylor Series for Holomorphic Functions. In Real Analysis, the Taylor series of a given function f : R R is given by: f (x + f (x (x x + f (x
More informationMATH COMPLEX ANALYSIS. Contents
MATH 3964 - OMPLEX ANALYSIS ANDREW TULLOH AND GILES GARDAM ontents 1. ontour Integration and auchy s Theorem 2 1.1. Analytic functions 2 1.2. ontour integration 3 1.3. auchy s theorem and extensions 3
More informationChapter 1, Exercise 22
Chapter, Exercise 22 Let N = {,2,3,...} denote the set of positive integers. A subset S N is said to be in arithmetic progression if S = {a,a+d,a+2d,a+3d,...} where a,d N. Here d is called the step of
More informationProperties of Entire Functions
Properties of Entire Functions Generalizing Results to Entire Functions Our main goal is still to show that every entire function can be represented as an everywhere convergent power series in z. So far
More informationON THE GROWTH OF ENTIRE FUNCTIONS WITH ZERO SETS HAVING INFINITE EXPONENT OF CONVERGENCE
Annales Academiæ Scientiarum Fennicæ Mathematica Volumen 7, 00, 69 90 ON THE GROWTH OF ENTIRE FUNCTIONS WITH ZERO SETS HAVING INFINITE EXPONENT OF CONVERGENCE Joseph Miles University of Illinois, Department
More informationSelected Solutions To Problems in Complex Analysis
Selected Solutions To Problems in Complex Analysis E. Chernysh November 3, 6 Contents Page 8 Problem................................... Problem 4................................... Problem 5...................................
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 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 - Final exam - Answers
Complex Analysis - Final exam - Answers Exercise : (0 %) Let r, s R >0. Let f be an analytic function defined on D(0, r) and g be an analytic function defined on D(0, s). Prove that f +g is analytic on
More informationIntroduction to Complex Analysis MSO 202 A
Introduction to Complex Analysis MSO 202 A Sameer Chavan Semester I, 2016-17 Course Structure This course will be conducted in Flipped Classroom Mode. Every Friday evening, 3 to 7 videos of total duration
More informationOn the mean values of an analytic function
ANNALES POLONICI MATHEMATICI LVII.2 (1992) On the mean values of an analytic function by G. S. Srivastava and Sunita Rani (Roorkee) Abstract. Let f(z), z = re iθ, be analytic in the finite disc z < R.
More informationExercises for Part 1
MATH200 Complex Analysis. Exercises for Part Exercises for Part The following exercises are provided for you to revise complex numbers. Exercise. Write the following expressions in the form x+iy, x,y R:
More informationPICARD S THEOREM STEFAN FRIEDL
PICARD S THEOREM STEFAN FRIEDL Abstract. We give a summary for the proof of Picard s Theorem. The proof is for the most part an excerpt of [F]. 1. Introduction Definition. Let U C be an open subset. A
More informationMATH 117 LECTURE NOTES
MATH 117 LECTURE NOTES XIN ZHOU Abstract. This is the set of lecture notes for Math 117 during Fall quarter of 2017 at UC Santa Barbara. The lectures follow closely the textbook [1]. Contents 1. The set
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 informationComplex Analysis Math 185A, Winter 2010 Final: Solutions
Complex Analysis Math 85A, Winter 200 Final: Solutions. [25 pts] The Jacobian of two real-valued functions u(x, y), v(x, y) of (x, y) is defined by the determinant (u, v) J = (x, y) = u x u y v x v y.
More informationz b k P k p k (z), (z a) f (n 1) (a) 2 (n 1)! (z a)n 1 +f n (z)(z a) n, where f n (z) = 1 C
. Representations of Meromorphic Functions There are two natural ways to represent a rational function. One is to express it as a quotient of two polynomials, the other is to use partial fractions. The
More informationProblems for MATH-6300 Complex Analysis
Problems for MATH-63 Complex Analysis Gregor Kovačič December, 7 This list will change as the semester goes on. Please make sure you always have the newest version of it.. Prove the following Theorem For
More informationcarries the circle w 1 onto the circle z R and sends w = 0 to z = a. The function u(s(w)) is harmonic in the unit circle w 1 and we obtain
4. Poisson formula In fact we can write down a formula for the values of u in the interior using only the values on the boundary, in the case when E is a closed disk. First note that (3.5) determines the
More information2. Complex Analytic Functions
2. Complex Analytic Functions John Douglas Moore July 6, 2011 Recall that if A and B are sets, a function f : A B is a rule which assigns to each element a A a unique element f(a) B. In this course, we
More informationSOLUTIONS OF NONHOMOGENEOUS LINEAR DIFFERENTIAL EQUATIONS WITH EXCEPTIONALLY FEW ZEROS
Annales Academiæ Scientiarum Fennicæ Mathematica Volumen 23, 1998, 429 452 SOLUTIONS OF NONHOMOGENEOUS LINEAR DIFFERENTIAL EQUATIONS WITH EXCEPTIONALLY FEW ZEROS Gary G. Gundersen, Enid M. Steinbart, and
More informationGluing linear differential equations
Gluing linear differential equations Alex Eremenko March 6, 2016 In this talk I describe the recent work joint with Walter Bergweiler where we introduce a method of construction of linear differential
More informationModule 6: Deadbeat Response Design Lecture Note 1
Module 6: Deadbeat Response Design Lecture Note 1 1 Design of digital control systems with dead beat response So far we have discussed the design methods which are extensions of continuous time design
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 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 informationMATH5685 Assignment 3
MATH5685 Assignment 3 Due: Wednesday 3 October 1. The open unit disk is denoted D. Q1. Suppose that a n for all n. Show that (1 + a n) converges if and only if a n converges. [Hint: prove that ( N (1 +
More informationAn alternative proof of Mañé s theorem on non-expanding Julia sets
An alternative proof of Mañé s theorem on non-expanding Julia sets Mitsuhiro Shishikura and Tan Lei Abstract We give a proof of the following theorem of Mañé: A forward invariant compact set in the Julia
More informationIntroduction to Series and Sequences Math 121 Calculus II Spring 2015
Introduction to Series and Sequences Math Calculus II Spring 05 The goal. The main purpose of our study of series and sequences is to understand power series. A power series is like a polynomial of infinite
More information