NATIONAL UNIVERSITY OF SINGAPORE Department of Mathematics MA4247 Comple Analsis II Lecture Notes Part I Chapter 1 Preliminar results/review of Comple Analsis I These are more detailed notes for the results studied in MA3111 Comple Analsis I As man of these results are required for this module, ou are epected to know them, especiall the statement of the theorems and also the various formulae like the Cauch integral formula, etc However, ou will not be required in this module to reproduce the proofs of these results (unless the occur in the proofs of theorems covered in this module) Note that some of the notation ma differ from that used in the later parts of these notes 11 Comple Numbers (111) Notation and Basic Results We denote the set of comple numbers b C, that is C = { + i :, R} Remark (i) For z = + i, is called the real part of z, and is called the imaginar part of z We usuall write = Re z, = Im z (ii) (Equalit of two comple numbers) Two comple numbers are equal if and onl if their real parts are equal and their imaginar parts are equal In other words, if z 1 = 1 + i 1 and z 2 = 2 + i 2, then z 1 = z 2 1 = 2 & 1 = 2 (iii) The comple conjugate z of a comple number z = + i is defined b z = i For an z = + i, z 1, z 2 C, we have Re z = = z + z 2 z z, Im z = = 2i 1
2 (iv) The modulus z of a comple number z = + i is the distance of z from the origin in the comple plane B the Pthagoras theorem, we have z = 2 + 2 (v) Geometricall, z 1 z 2 is the distance between z 1 and z 2 in the comple plane (vi) For an z C, we have (vii) For an z 1, z 2 C, we have z z = z 2 z 1 z 2 = z 1 z 2, z 1 z 1 = z 2 z 2 (viii) (Triangle Inequalit) For an z 1, z 2 C, we have z 1 z 2 z 1 ± z 2 z 1 + z 2 (i) For more than two comple numbers, we have z 1 + z 2 + + z m z 1 + z 2 + + z m Eercise: Prove the triangle inequalit (viii) above (112) Polar form, Eponential form (, ) r θ The polar form of a comple number z = + i is given b z = r(cos θ + i sin θ), where r = z, and θ is an argument of z Write e iθ := cos θ + i sin θ
3 Then we have the eponential form of z given b Sometimes, we also write z = re iθ ep(iθ) := e iθ, so that the eponential form ma also be written as z = r ep(iθ) Remark The argument of z is defined onl up to a multiple of 2π The collection of all possible values of the argument of z is denoted b arg z Thus we ma write arg z = θ + 2nπ, n Z, where θ is an one of the values of arg z Note: This ambiguit in the definition of the argument accounts for much of the interesting phenomena which occurs in the stud of comple analsis The general polar form of z is given b z = r[cos(θ + 2nπ) + i sin(θ + 2nπ)], n =, ±1, ±2,, where θ is an argument of z form of z is given b Similarl, the general eponential z = r ep[i(θ + 2nπ)], n =, ±1, ±2, The principal argument of a non-zero z, denoted b Arg z, is the unique value of the argument of z satisfing π < Arg z π One has the following de Moivre s formula: (cos θ + i sin θ) n = cos nθ + i sin nθ, which is valid for an rational number n Application: To find all the n-th roots of a comple number c Write c = r o ep(iθ o ) The problem is equivalent to solving the equation z n = c = r o ep[i(θ o + 2kπ)], k =, ±1, ±2, = z k = {r o ep[i(θ o + 2kπ)]} 1 n, k =, ±1, ±2, = z k = r 1 n o ep [ i ( θ o + 2kπ )], n k =, 1, 2,, n 1 Eercise: (i) Wh is it necessar to consider onl k =, 1,, n 1 in the solution above? (ii) Show that the four 4-th roots of 16 are 2 + 2i, 2 + 2i, 2 2i, 2 2i
4 12 Analtic functions (121) Comple-valued function of a comple variable, limits, continuit Let f(z) be a comple-valued function of a comple variable z = + i Then we can write Write f(z) = u(, ) + iv(, ), where u(, ) and v(, ) are real-valued functions in two real-variables, The function u(, ) is called the real-part of f(z), while v(, ) is called the imaginar part of f(z) And we simpl write u(, ) = Re (f), v(, ) = Im (f) Geometricall, z f(z ) = u + iv f(z) For a function f(z), we write lim z z f(z) = w o if the following condition is satisfied: For an ɛ >, there eists δ = δ(ɛ) > such that f(z) w o < ɛ whenever < z z < δ A comple-valued function f(z) is continuous at z iff lim z z f(z) = f(z ) Remark Write f(z) = u(, ) + iv(, ) Then f(z) is a continuous function u(, ) and v(, ) are continuous functions (122) Infinit For comple numbers, there is onl ONE infinit, It is often convenient to enlarge the comple plane C b adding the element, and the enlarged set Ĉ := C { } is called the etended comple plane
Remark (i) z z (second limit is in the usual sense for real numbers); 1 (ii) lim f(z) = lim z z z z f(z) = ; (iii) lim f(z) = w o lim f ( 1) = wo z s s (123) Differentiations, Cauch-Riemann equations Definition: The derivative of f at z is defined as d dz f(z) z=z = f (z ) := lim z z f(z) f(z ) z z provided that the limit eists If f (z ) eists,we sa that f is differentiable at z Remark Alternativel, we ma define f (z ) := lim z f(z + z) f(z ) z Remark (i) If f(z) is differentiable at z, then f(z) is continuous at z (ii) The usual Power Rule, Addition Rule, Product Rule, Quotient Rule and Chain Rule hold Theorem (Necessar conditions for differentiabilit) If f(z) = u(, ) + iv(, ) is differentiable at z = o + i o, then u and v satisf the Cauch-Riemann equations at ( o, o ), ie, u ( o, o ) = v ( o, o ), u ( o, o ) = v ( o, o ), 5 and f (z ) = u + i v = v i u [In short, f(z) differentiable = u = v, u = v and f (z) = u + iv = v iu ] Theorem (A Sufficient Condition for Differentiabilit) Let f(z) = u(, ) + iv(, ) be defined near the point z = o + i o Suppose that (i) u, v satisf the Cauch-Riemann equations at ( o, o ) ie u = v, u = v at ( o, o ); and
6 (ii) u, u, v, v are continuous at ( o, o ) Then f is differentiable at z [In short, continuous partial derivatives + CR-equations = differentiable] (124) Domains, Analtic functions For z C and r >, we define the ball centered at z and of radius r b B(z, r) := {z C : z z < r} We also use the following notation for the deleted ball (that is, with the centre removed): B (z, r) := B(z, r) \ {z } = {z C : < z z < r} z r A subset U in C is said to be open if for each z U, there eists some r > such that z B(z, r) U (Here the radius r ma depend on the point z) [Roughl, an open set U is a set which contains no boundar points] Definition An open subset S of C is said to be connected if each pair of points z 1, z 2 in S can be joined b a polgonal line, consisting of a finite number of line segments joined end to end, which lie entirel in S (Remark: This is not the usual definition for connectedness More generall, we have the notion of connectedness, path-connectedness, and polgonall path-connectedness For open subsets of C, these are all equivalent) Definition A domain in C is a non-empt, open and connected subset of C
7 eg The annulus U = {z C : 1 < z < 2} is open 1 2 Definition A function f is analtic at a point z if f is differentiable everwhere in some open set U containing z z f(z) Roughl speaking, analtic at z means differentiable everwhere near z Definition A function f is analtic in an open set U if f is differentiable everwhere in U Remark If a function f is differentiable onl at a finite number of (isolated) points, sa z 1, z 2,, z n, then f is nowhere analtic z 2 z n z 1 Theorem Let f(z) be analtic in a domain D If f (z) on D, then f(z) is constant in D Definition An entire function is a function which is analtic in the whole comple plane C (ie differentiable everwhere in C)
8 13 Elementar Functions (131) The eponential function The (comple) eponential function given b e z = n= It can be shown that z n n! = 1 + z + z2 2! + z3 3! + ( z < ) e z = e (cos + i sin ) z = + i C Some properties: (i) When z is real, e z coincides with the real eponential function e, ie, when z = + i, e z = e (ii) e z is entire and d dz (ez ) = e z (iii) e z = e, arg(e z ) = + 2nπ, n =, ±1, ±2, In particular, we have e z for an z C (132) The logarithmic function Definition The (comple) logarithmic function log z is defined as the inverse function of the eponential function e z, ie, One easil sees that w = log z z = e w log z = ln z + i arg z Remark (i) log z is a multi-valued function So it is strictl speaking, not a function under the usual definition (ii) log z is defined for z C \ {} Question: What can we do to tr to obtain a well-defined log function? Definition The principal logarithmic function (or the principal branch of log z) is given b Log z = ln z + i Arg z
9 Remark (i) Log z is a single-valued function defined on C \ {} (ii) Log z is continuous on the cut comple plane C \ (, ] (iii) In fact, Log z is analtic on C \ (, ], and d dz Log z = 1 z z C \ (, ] Question: Can one define other branches of the log function? must the domain satisf for these branches? What (133) The trigonometric functions The (comple) cosine and sine functions are given b cos z = eiz + e iz, 2 sin z = eiz e iz 2i for an z C Remark (i) cos z and sin z are entire functions (Wh?) (ii) When z is real, ie z = + i, we have (iii) We have cos z = cos, sin z = sin d d (cos z) = sin z, (sin z) = cos z, dz dz cos 2 z + sin 2 = 1 (iv) (Eercise) sin z = z = nπ, n Z We ma define the other comple trigonometrical functions b: tan z = sin z cos z, cot z = 1 tan z,
1 Again, we have sec z = 1 cos z, csc z = 1 sin z d dz (tan z) = sec2 z, d d (sec z) = sec z tan z, dz d dz (cot z) = csc2 z, (csc z) = csc z cot z dz Eercise: Determine the domains for each of the other trigonometric functions (134) The hperbolic functions The (comple) hperbolic cosine and sine functions are given b cosh z = ez + e z, sinh z = ez e z 2 2 1 cosh z, sinh z are entire functions 2 When z is real, ie z = + i, cosh z = cosh, sinh z = sinh The other hperbolic functions are given b tanh z = sinh z cosh z, coth z = 1 tanh z, sech z = 1 cosh z, csch z = 1 sinh z Some Properties: One has cosh 2 z sinh 2 z = 1, d d (cosh z) = sinh z, dz (sinh z) = cosh z dz Eercise: Find the relation between the functions sin and sinh, and cos and cosh (135) Comple Eponents z c with z, c C For z, c C (where we regard z as the variable and c is a fied comple constant), with z, one defines z c := e c log z Remark For fied c, z c is a multi-valued function in z B fiing a branch of the log function (equivalentl a branch of the argument), we
ma obtain a well-defined function on suitable subsets of C Note that when c is an integer, then z c is well-defined (136) Inverse Trigonometric/Hperbolic Functions One can define the the inverse trigonometric/hperbolic functions (eg cos 1 z, tanh 1 z,, etc) in terms of log z b solving the appropriate equations Usuall these are multi-valued functions Eample cos 1 z = i log [ z + i(1 z 2 ) 1/2] 11
12 14 Contour Integrals (141) Contours A curve in the comple plane is (parametrized b) a continuous function : [a, b] C That is, if we write (t) = (t)+i(t), then (t) and (t) are continuous on [a, b] Remark (i) is simple if t 1 t 2 = (t 1 ) (t 2 ), ie does not cross itself (ii) is closed if (a) = (b) (iii) is a simple closed curve if it is closed and a < t 1 < t 2 < b = (t 1 ) (t 2 ), ie does not cross itself ecept at the end points Eample (i) The line segment [z 1, z 2 ] One can parametrize the straight line segment joining z 1 to z 2 b (t) = z 1 + t(z 2 z 1 ), t 1 z 2 z 1 (ii) One can parametrize the circle centered at z and of radius R b (t) = z + Re it, t 2π z R
13 A curve : [a, b] C is said to be smooth if (i) (t) = (t) + i (t) eists and is continuous on [a, b]; and (ii) (t) for all t [a, b] Definition The integral of f along a smooth curve is defined b b f(z) dz = f((t)) (t) dt a Definition A contour is a finite sequence { 1, 2,, n } of smooth curves such that the terminal point of k coincides with the initial point of k+1 for 1 k n 1 In this case, we write = 1 + 2 + + n The contour integral of f along a contour is defined to be f(z) dz = f(z) dz + 1 f(z) dz + + 2 f(z) dz n Some basic properties: (i) [f(z) + g(z)] dz = f(z) dz + g(z) dz (ii) z f(z) dz = z f(z) dz (iii) f(z) dz = f(z) dz (iv) the value of f(z) dz remains unchanged under reparametrizaton of (v) (M L-inequalit) Suppose that f is continuous along a contour, and f(z) M z Then where L = length of f(z) dz ML, Note The length of a curve = (t) : [a, b] C ma be calculated b the formula Length of = b a (t) dt
14 (142) Anti-derivatives, Cauch-Goursat Theorem Let f be a continuous function on a domain D A function F such that F (z) = f(z) z D is called an antiderivative of f in D Theorem Let f be continuous on a domain D The following are equivalent: (a) f has an antiderivative in D; (b) for an closed contour in D, f(z) dz = ; (c) the contour integrals of f are independent of paths in D, that is, if z 1, z 2 D and 1, 2 are contours in D joining z 1 to z 2, then 1 f(z) dz = f(z) dz 2 Theorem (Cauch-Goursat Theorem) If a function f is analtic at all points inside and on a simple closed contour, then f(z) dz = Definition A domain D is simpl connected if ever simple closed contour in D encloses onl points in D (ie D has no holes) Remark The interior of a simple closed contour is simpl connected Theorem (Cauch-Goursat Theorem for simpl-connected domains) If f is analtic in a simpl connected domain D, then for ever closed contour in D f(z) dz =
15 Corollar If f is analtic in a simpl connected domain D, then it has an antiderivative in D Eercise: Find an analtic function in a non-simpl connected domain D which does not have an anti-derivative in D (143) Cauch integral formula, Liouville s Theorem A simple closed contour is positivel oriented if the interior domain lies to the left of an observer tracing out the points in order Remark: A circle is positivel oriented if it is traversed in the anticlockwise direction Theorem (Cauch Integral Formula) Let be a positivel oriented simple closed contour and let f be analtic everwhere within and on Then for an z inside, f(z ) = 1 2πi f(z) z z dz z Theorem (Cauch integral formula for derivatives) Let f(z) be analtic everwhere inside and on a positivel oriented simple closed contour C Then for an z inside C and an integer n 1, f (n) (z ) = n! 2πi C f(z) dz (n = 1, 2, 3, ) (z z ) n+1 Corollar If f is analtic in a domain D, then all its derivatives f, f, f, eist and are analtic in D
16 Corollar (Morera s Theorem) If f is continuous on a domain D, and f(z) dz = for ever closed contour in D, then f is analtic in D Corollar (Cauch s inequalit) If f is analtic within and on the circle R centered at z and of radius R, then for n Z +, f (n) (z ) n!m R R n, where M R = ma z R f(z) Let S C A function f : S C is bounded if there eists some K > such that f(z) K for all z S Theorem (Liouville s Theorem) If an entire function f is bounded, then it must be a constant function [In short, entire + bounded = constant] Theorem (The Fundamental Theorem of Algebra) An polnomial p(z) of degree 1 has a zero in C (ie, p(z) = has at least one solution) Idea of proof Suppose p(z) for all z C Then 1/p(z) can be shown to be an entire and bounded function Thus 1/p(z) and hence p(z) are constant functions This contradicts the assumption that p(z) is of degree 1
17 15 Power series, Talor series, Laurent series Definition We sa that a sequence of functions {f n } n=1 converges uniforml to f on D if for an ɛ >, there eists N = N(ɛ) Z + such that f n (z) f(z) < ɛ for all n > N and all z D Note: Here N depends on ɛ but not on z Theorem Let be a contour and let {f n } n=1 be a sequence of continuous functions on If {f n } n=1 converges uniforml to f on, then lim f n (z) dz = lim f n(z) dz = f(z) dz n n Theorem Let {f n } n=1 be a sequence of analtic functions on a domain D If {f n } n=1 converges uniforml to f on D, then f is also analtic on D Moreover, lim f n(z) = f d (z), ie, lim n n dz f n(z) = d ( ) lim dz f n(z) z D n Similar results hold for a series of functions Theorem Given an power series a k (z z ) k, there is an associated number R, R, called the radius of convergence, such that: (i) a k (z z ) k converges absolutel at each point z satisfing z k= z < R, (ii) a k (z z ) k diverges at each z satisfing z z > R, and (iii) k= k= a k (z z ) k converges uniforml on the closed ball B(z, ρ) for k= an ρ satisfing < ρ < R (Here, B(z, ρ) := {z C : z z ρ}) 1 1 Moreover, R =, and also R =, if the limit lim sup a k 1 k a k+1 lim k k a k eits z ρ R
18 Theorem (Convergent power series are analtic functions) Let R be the radius of convergence of a k (z z ) k Then (i) S(z) = k= a k (z z ) k is an analtic function on B(z, R) k= (ii) (Term-b term integration) If is a contour in B(z, R) and g(z) is continuous on, then g(z) a k (z z ) k dz = k= (c) (Term-b-term differentiation) a k k= g(z)(z z ) k dz d dz a k (z z ) k = k= ka k (z z ) k 1 k=1 on B(z, R) Theorem (Talor s Theorem) Suppose f(z) is analtic in B(z, R) Then f (k) (z ) f(z) = (z z ) k ( z z < R) k! k= z z R This power series is called the Talor series of f(z) at z The coefficients f (k) (z ) are called the Talor coefficents k! Remark: (i) When z =, the Talor series at z = is also called the Maclaurin series of f(z) (ii) Roughl speaking, Talor s Theorem sas that analtic functions are equal to their Talor series Together with its preceding theorem, it follows that power series and analtic functions are more or less the same objects Theorem (Uniqueness of Talor series) Let f(z) be an analtic function If f(z) = a k (z z ) k for all z B(z, R) for some R >, k=
then a k (z z ) k is THE Talor series k= k= 19 f (k) (z ) (z z ) k of f(z) at k! z, ie, a k = f (k) (z ) for all k =, 1, 2, k! Consequences: We can use those standard power series to write f(z) in the form f(z) = a n (z z ) n, which will automaticall be the n= Talor series of f(z) at z Some standard power series: e z = sin z = cos z = n= n= z n n! = 1 + z + z2 2! + z3 3! ( 1) n z 2n+1 (2n + 1)! ( 1) n z 2n n= (2n)! + ( z < ) = z z3 3! + z5 5! z7 7! = 1 z2 2! + z4 4! z6 6! 1 1 z = z n = 1 + z + z 2 + z 3 + ( z < 1) n= + ( z < ) + ( z < ) 1 1 + z = ( 1) n z n = 1 z + z 2 z 3 + ( z < 1) n= Theorem (Laurent s Theorem) Suppose f(z) is analtic in the annulus A := {z C : R 1 < z z < R 2 } Then f(z) = where a n (z z ) n + n= a n = 1 2πi n=1 b n (z z ) n (R 1 < z z < R 2 ), (*) f(s) (s z ) n+1 ds and b n = 1 2πi f(s) ds, (s z ) n+1 and is an positivel oriented simple closed contour around z and ling in A R 2 z R 1
2 Remark The epression in (*) is called the Laurent series of f for the annulus R 1 < z z < R 2 The coefficients a n, b n are called the Laurent coefficients Theorem (Uniqueness of Laurent series representation) If an analtic function f in the annulus R 1 < z z < R 2 satisfies f(z) = a n (z z ) n + n= n=1 b n (z z ) n for R 1 < z z < R 2, (1) then the epression in (1) is THE Laurent series of f(z) for R 1 < z z < R 2 Consequence: One can use the standard power series to find Laurent series of certain functions 5z + 14 Eample Find the Laurent series of for the annulus (z + 2)(z + 3) 2 < z < 3 ( 1) n 2 n+2 ( 1) n Answer: z n+1 + 3 n+1 zn n= n=
21 16 Residues and poles (161) Singular points and residues Definition (1) A point z is said to be a singular point of a function f if (i) f is not analtic at z ; but (ii) f is analtic at some point in B(z, ɛ) for all ɛ > (2) A singular point z of f is isolated if there eists R > such that f is analtic in B(z, R) \ {z } Eample (i) f(z) = 1 sin z f(z) has singular points at sin z = z = nπ, n Z (ii) f(z) = Log z is analtic on C \ (, ] Thus, each point in (, ] is a singular point of f(z) None of them is an isolated singular point of f(z) Suppose that f(z) has an isolated singular point at z Then there eists R > such that f is analtic in the punctured ball B(z, R) \ {z } = {z C : < z z < R}, (*)
22 which ma be regarded an annulus z R B Laurent s theorem, we have where f(z) = b n = 1 2πi a n (z z ) n + n= n=1 b n (z z ) n ( < z z < R), f(s) (s z ) n+1 ds, and thus b 1 = 1 f(s) ds 2πi Definition The residue of f(z) at z is given b: Res z=z f(z) = b 1 = coeff of 1 z z in the Laurent series of f at z Theorem (Cauch s Residue Theorem) If is a positivel oriented simple closed contour and f(z) is analtic everwhere inside and on ecept for a finite number of isolated singular points z k (k = 1, 2,, n) inside, then f(z) dz = 2πi n k=1 Res f(z) (*) z=z k
23 (163) Classification of isolated singularities Definition Suppose f(z) has an isolated singular point at z Consider the Laurent series of f(z) at z : f(z) = b n a n (z z ) n + n= n=1 b n (z z ) n ( < z z < R) is called the principal part of the Laurent series of f(z) (z z n=1 ) n at z (i) Removable singular point: If b n = for all n = 1, 2, (ie the principal part vanishes), we sa that z is a removable singular point of f(z) In this case, the Laurent series of f(z) reduces to a power series in z z, ie f(z) = a n (z z ) n n= Res f(z) = z=z ( < z z < R), and (ii) Essential singular point: If b n for infinitel man n (ie the principal part has infinitel man non-zero terms), then we sa that z is an essential singular point of f(z) Note that in this case, some of the b n s ma still be zero (iii) Pole If there eists m Z + such that b m but b n = for all n > m (ie the principal part has finitel man non-zero terms ) so that f(z) = n= a n (z z ) n + b 1 z z + then we sa that z is a pole of order m for f(z) b 2 (z z ) 2 + + b m (z z ) m, We also sa that z is a simple pole of f(z) if m = 1, and we also sa that z is a double pole of f(z) if m = 2 Eercise: Give eamples of functions with removable and essential singularities and poles
24 Behavior of a function near a singular point (a) Behavior of a function near a removable singular point Suppose a function f(z) has a removable singular point at a point z B Laurent s theorem, f(z) = a n (z z ) n ( < z z < R) (*) n= = a + a 1 (z z ) + for some R > (recalling that all b n = ) Note that the RHS of (*) is a convergent power series, and thus it is an analtic (and thus continuous) function on z z < R It follows that one has lim f(z) = lim [a + a 1 (z z ) + ] = a + a 1 + = a z z z z δ z ɛ a f(z) Observe that if one etend f(z) across the point z b letting f(z ) = a, then (*) will hold everwhere on z z < R The (etended) function f(z) is equal to the analtic function a n (z z ) n everwhere on z z < R Eample The function sin z has an isolated singular point at z = z For < z <, Thus sin z z sin z z = 1 z n= (z z3 3! + z5 5! + ) = 1 z2 3! + z4 5! + (*) has a removable singular point at z = Now we define { sin z if z, f(z) = z 1 if z = (**) Then f(z) is equal to the convergent power series 1 z2 3! + z4 5! + at all z C (B (*) and (**), both f(z) and the power series are equal to sin z z for z ; at z =, both are equal to 1) The power series is necessaril analtic at z = Hence f is also analtic at z =
25 (b) Behavior of a function near a pole Proposition Suppose f has a pole of order m at z (m 1) Then (i) R > such that f(z) = φ(z) (z z ) m for < z z < R, (1) where φ(z) is analtic at z and φ(z ) (ii) In particular, lim z z f(z) = z f(z) Proof of Proposition (i) Suppose that f has a pole of order m at z Then there eists R > such that f(z) = a n (z z ) n + b 1 b m + + z z (z z ) m, ( < z z < R), n= where b m Then for < z z < R, (z z ) m f(z) = a n (z z ) n+m + b 1 (z z ) m 1 + + b m 1 (z z ) + b m n= Consider the power series φ(z) = a n (z z ) n+m + b 1 (z z ) m 1 + + b m 1 (z z ) + b m n= for z B(z, R) Then φ(z) converges everwhere on B(z, R)\{z } (being convergent to (z z ) m f(z)) Clearl φ(z) converges at z (Wh?) Thus the radius of convergence of φ(z) is at least R Therefore, the power series φ(z) is convergent on B(z, R), and thus it is an analtic function on B(z, R) So we have (z z ) m f(z) = φ(z) with φ(z ) = b m, and φ(z) is analtic on B(z, R) This proves (i) To prove (ii), we let φ(z) be as in (i) Then 1 lim z z f(z) = lim (z z ) m z z φ(z) Thus, we have lim z z f(z) = = (z z ) m φ(z ) =
26 (c) Behavior of a function near an essential singular point It turns out that a function has complicated behaviors near an essential singular point, which was beond the scope of MA3111 Let me just state (in a rough manner) without proof one result along this direction: Theorem (Casorati-Weierstrass Theorem) Suppose f(z) has an essential singular point at z = z and let w be an comple number Then, for an positive number ε, the inequalit f(z) w < ε is satisfied at some point z in each deleted neighborhood < z z < δ of z Roughl speaking, this sas that for each (small) r >, the image f(b(z, r)) is dense in C Note that this cannot hold if z is a removable singularit or pole (wh?) For more details, see [Churchill, p 249(7th ed) or p 22-23(6th ed)] Eercise: Write up a proof for the Casorati-Weierstrass theorem z f(z) (164) Methods for computing residues Proposition (Method I) Suppose f(z) can be written in the form f(z) = φ(z) z z near z for some function φ(z) analtic at z Then Res f(z) = φ(z ) z=z Remark Such f has either a simple pole or a removable singular point at z
Proposition (Method II) Suppose f(z) can be written in the form f(z) = φ(z) (z z ) m near z for some function φ(z) analtic at z and m 1 Then Res f(z) = φ(m 1) (z ) z=z (m 1)! Remark Such f has a pole of order m or a removable singular point at z Proposition (Method III) If p(z) and q(z) are analtic at z, and q(z) has a simple zero at z, (ie q(z ) = and q (z ) ), then p(z) Res z=z q(z) = p(z ) q (z ) Remark Such p(z) q(z) has either a simple pole at z (if p(z ) ) or a removable singular point at z (if p(z ) = ) Method IV (None of the above) When the above methods all fail, we can still directl compute the residue b 1, b first finding the Laurent series of f(z) at z using standard power series, and then reading off the appropriate Laurent coefficient b 1 Summar of methods of finding residues 27 I II III f(z) Res f(z) z=z φ(z) z z φ(z ) φ(z) φ (m 1) (z ) (z z ) m (m 1)! p(z) q(z) p(z ) q (z ) IV None of above Find Laurent series and read b 1 17 Applications of Residues Some of the tpical applications of the Cauch s residue theorem (and the methods of evaluating residues) are in the evaluation of certain improper integrals of certain functions from (, ) and in the evaluation of certain trigonometric integrals over [, 2π] See [Churchill, Chapter 7] for details We will see later that it also plas an important role in the proof of the argument principle and Rouché s theorem