EI331 Signals and Systems Lecture 26 Bo Jiang John Hopcroft Center - PowerPoint PPT Presentation

EI331 Signals and Systems Lecture 26 Bo Jiang John Hopcroft Center for Computer Science Shanghai Jiao Tong University May 28, 2019

Contents 1. Singularities, Poles and Zeros 2. Residue 3. Behavior at Infinity 4. Evaluation of Definite Integrals Using Residues 1/39

Singularities Recall a point z 0 ∈ C is a singular point or a singularity of a function f if f is not analytic at z 0 . Example. sin z and 1 z have a singularity at z = 0 , since they are z not defined at z = 0 . If f ( z ) is singular at z 0 , but analytic on a deleted open disk 0 < | z − z 0 | < r for some r > 0 , then z 0 is called an isolated singularity of f . 1 Example. 1 z have an isolated singularity at z = 0 . z and e 1 Example. z has a singularity at z = 0 . It also has singularities sin 1 at the roots of sin 1 1 z = 0 , or z n = n π , n ∈ Z \ { 0 } . Since z = 0 is the limit z n as | n | → ∞ . Any deleted open disk centered at 0 contains infinitely many z n ’s, so 0 is not an isolated singularity. 2/39

Classification of Isolated Singularities At an isolated singularity z 0 , f is analytic on a deleted open disk 0 < | z − z 0 | < r , so it has a Laurent series expansion ∞ ∞ ∞ c n ( z − z 0 ) n = c n ( z − z 0 ) n + � � � c − n ( z − z 0 ) − n f ( z ) = n = −∞ n = 0 n = 1 n = 0 c n ( z − z 0 ) n is the regular or analytic part of f φ ( z ) = � ∞ n = 1 c − n ( z − z 0 ) − n is the singular or principal part of f ψ ( z ) = � ∞ • If ψ = 0 , then z 0 is called a removable singularity of f • If ψ has finitely many terms, then z 0 is called a pole of f . ◮ If m ∈ N is the largest integer s.t. c − m � = 0 but c − n = 0 for n > m , then z 0 is a pole of order m . ◮ A pole of order 1 is called a simple pole. • If ψ has infinitely many terms, then z 0 is called an essential singularity of f . 3/39

Removable Singularity At a removable singularity z 0 , the Laurent series of f reduces to a power series ∞ � c n ( z − z 0 ) n , f ( z ) = 0 < | z − z 0 | < r n = 0 The power series defines an analytic function F ( z ) on B ( z 0 , r ) s.t. F ( z ) = f ( z ) when z � = z 0 If we define f ( z 0 ) = lim z → z 0 f ( z ) = c 0 , then f = F is analytic at z 0 , whence the name “removable”. Example. z = 0 is a removable singularity of sin z z , since ∞ ∞ ( − 1 ) n ( − 1 ) n sin z = 1 ( 2 n + 1 )! z 2 n + 1 = � � ( 2 n + 1 )! z 2 n z z n = 0 n = 0 We can define sin 0 = 1 . 0 4/39

Pole At a pole of order m , the Laurent series of f is ∞ g ( z ) c n ( z − z 0 ) n = � f ( z ) = ( z − z 0 ) m , 0 < | z − z 0 | < r n = − m where ∞ � c n − m ( z − z 0 ) n g ( z ) = n = 0 is analytic on B ( z 0 , r ) and g ( z 0 ) = c − m � = 0 . c − m Note f ( z ) ≈ ( z − z 0 ) m , so lim z → z 0 | f ( z ) | = + ∞ , also written lim z → z 0 f ( z ) = ∞ g ( z ) Conversely, if f ( z ) = ( z − z 0 ) m for some analytic function g with g ( z 0 ) � = 0 , then z 0 is a pole of order m of f . z − 2 Example. f ( z ) = ( z 2 + 1 )( z − 1 ) 3 has a pole of order 3 at z = 1 , two simple poles at z = ± j . 5/39

Essential Singularity Theorem (Weierstrass). If z 0 ∈ C is an essential singularity of f , then given any A ∈ C , there is a sequence z n s.t. z n → z 0 and f ( z n ) → A , as n → ∞ . NB. As a consequence, lim z → z 0 f ( z ) does not exist. Example. f ( z ) = e 1 / z has an essential singularity at z = 0 , since ∞ 1 � n ! z − n f ( z ) = 1 + n = 1 We can verify Weierestrass’s Theorem: Given A ∈ C , log A + j 2 n π → 0 and f ( z n ) = e log A + j 2 n π = A • if A � = 0 , then z n = 1 n → 0 and f ( z n ) = e − n → 0 • if A = 0 , then z n = − 1 6/39

Zeros and Poles Let f be analytic and not identically zero on a domain D . • A point z 0 ∈ D is called a zero of f if f ( z 0 ) = 0 . • z 0 is called a zero of order m , if the Taylor series of f at z 0 is ∞ � c n ( z − z 0 ) n , f ( z ) = c m � = 0 n = m A zero of order 1 is called a simple zero • f ( z ) ≈ c m ( z − z 0 ) m , so z 0 is an isolated zero. • z 0 is a zero of order m of f , iff f ( z ) = ( z − z 0 ) m φ ( z ) , where φ is analytic and φ ( z 0 ) � = 0 . • z 0 is a zero of order m of f , iff f ( n ) ( z 0 ) = 0 for n < m , and f ( m ) ( z 0 ) � = 0 . 1 Theorem. f ( z ) has a zero of order m at z 0 iff f ( z ) has a pole of order m at z 0 . 7/39

Zeros and Poles Example. f ( z ) = z ( z − 1 ) 3 and g ( z ) = 1 1 f ( z ) = z ( z − 1 ) 3 • f has a simple zero at z = 0 , g has a simple pole at z = 0 • f a zero of order 3 at z = 1 , g has a pole of order 3 at z = 1 Example. f ( z ) = e z − 1 has a singularity at z = 0 . z 2 ∞ ∞ z n z n f ( z ) = 1 � � n ! = z 2 ( n + 2 )! n = 1 n = − 1 z = 0 is a pole of order 1, not of order 2! Example. f ( z ) = sinh z has a singularity at z = 0 , z 3 ∞ ∞ z 2 n + 1 z 2 n f ( z ) = 1 � � ( 2 n + 1 )! = z 3 ( 2 n + 3 )! n = 0 n = − 1 z = 0 is a pole of order 2, not of order 3! 8/39

Quotient of Functions Theorem. Suppose f has a zero of order m at z 0 and g has a zero of order n at z 0 . Let h ( z ) = f ( z ) / g ( z ) . • If n > m , h ( z ) has a pole of order n − m at z 0 . • If n < m , h ( z ) has a zero of order m − n at z 0 . • If n = m , h ( z ) is analytic at z 0 and h ( z 0 ) � = 0 . Proof. f and g take the following form, f ( z ) = ( z − z 0 ) m f 1 ( z ) , g ( z ) = ( z − z 0 ) n g 1 ( z ) where f 1 ( z ) and g 1 ( z ) are analytic and nonzero at z 0 . Thus h ( z ) = ( z − z 0 ) n − m h 1 ( z ) where h 1 ( z ) = f 1 ( z ) / g 1 ( z ) is analytic and nonzero at z 0 . Example. e z − 1 has a simple zero at z = 0 , and z 2 has a zero of order 2 at z = 0 , so e z − 1 has a pole of order 1 at z = 0 . z 2 9/39

Contents 1. Singularities, Poles and Zeros 2. Residue 3. Behavior at Infinity 4. Evaluation of Definite Integrals Using Residues 10/39

Residue Suppose f has an isolated singularity at z 0 ∈ C , and f is analytic on 0 < | z − z 0 | < R with Laurent series expansion, ∞ � c n ( z − z 0 ) n f ( z ) = n = −∞ Let C be the positively oriented circle | z − z 0 | = r ∈ ( 0 , R ) . Term-by-term integration yields ∞ ∞ � � � � ( z − z 0 ) n dz = f ( z ) dz = c n j 2 πδ [ n + 1 ] = j 2 π c − 1 c n C C n = −∞ n = −∞ The residue of f at z 0 ∈ C is � 1 Res( f , z 0 ) = f ( z ) dz = c − 1 j 2 π C 11/39

Residue Example. f ( z ) = sin z has a removable singularity at z = 0 , z ∞ ( 2 n + 1 )! z 2 n = ( − 1 ) n sin z = � ⇒ Res( f , 0 ) = 0 z n = 0 In general, Res( f , z 0 ) = 0 if z 0 ∈ C is a removable singularity. Example. e 1 / z has an essential singularity at 0 . ∞ e 1 / z = n ! z − n = 1 ⇒ Res( e − z , 0 ) = 1 � n = 0 1 Example. f ( z ) = z ( 1 − z ) has a simple pole at z = 0 , ∞ ∞ z n = z n on 0 < | z | < 1 = f ( z ) = 1 � � ⇒ Res( f , 0 ) = 1 z n = 0 n = − 1 ∞ z − n on 1 < | z | < ∞ with Warning. f ( z ) = − 1 1 � 1 − 1 / z = − z 2 n = 2 c − 1 = 0 , but Res( f , 0 ) � = 0 , since the definition or residue requires using Laurent series on 0 < | z − z 0 | < r 12/39

Residue at Simple Poles If z 0 is a simple pole, then f ( z ) = g ( z ) , z − z 0 where g ( z ) is analytic and nonzero at z 0 . By Cauchy’s Theorem 1 � g ( z ) Res( f , z 0 ) = dz = g ( z 0 ) = lim z → z 0 ( z − z 0 ) f ( z ) j 2 π z − z 0 C If f ( z ) = φ ( z ) ψ ( z ) , where φ and ψ are analytic at z 0 , φ ( z 0 ) � = 0 , and ψ has a simple zero at z 0 , then Res( f , z 0 ) = φ ( z 0 ) ψ ′ ( z 0 ) since φ ( z ) = φ ( z 0 ) Res( f , z 0 ) = lim z → z 0 ( z − z 0 ) f ( z ) = lim ψ ( z ) − ψ ( z 0 ) ψ ′ ( z 0 ) z → z 0 z − z 0 13/39

Residue at Simple Poles 1 Example. f ( z ) = z ( 1 − z ) has simple poles at z = 0 and z = 1 , Res( f , 0 ) = lim z → 0 zf ( z ) = 1 , Res( f , 1 ) = lim z → 1 ( z − 1 ) f ( z ) = − 1 Recall the partial fraction expansion of f ( z ) takes the form f ( z ) = A B z + z − 1 where A = lim z → 0 zf ( z ) = Res( f , 0 ) , B = lim z → 1 ( z − 1 ) f ( z ) = Res( f , 1 ) 1 Example. f ( z ) = sin z has simple poles at z = k π , k ∈ Z . Note f ( z ) = φ ( z ) ψ ( z ) , where φ ( z ) = 1 is analytic and nonzero at z = k π , and ψ ( z ) = sin z has a simple zero at z = k π . Thus 1 1 cos z = ( − 1 ) k Res( f , k π ) = lim (sin z ) ′ = lim z → k π z → k π 14/39

Residue at Higher Order Poles If z 0 is a pole of order m , then g ( z ) f ( z ) = ( z − z 0 ) m where g ( z ) is analytic and nonzero at z 0 . By Cauchy’s Integral Formula ( z − z 0 ) m dz = g ( m − 1 ) ( z 0 ) � g ( z ) 1 Res( f , z 0 ) = j 2 π ( m − 1 )! C For n ≥ m , the product rule for differentiation yields d n − 1 = d n − 1 = ( n − 1 )! � � dz n − 1 [( z − z 0 ) n f ( z )] dz n − 1 [( z − z 0 ) n − m g ( z )] ( m − 1 )! g ( m − 1 ) ( z 0 ) � � � � z = z 0 z = z 0 so d n − 1 1 dz n − 1 [( z − z 0 ) n f ( z )] , Res( f , z 0 ) = lim n ≥ m ( n − 1 )! z → z 0 15/39