Zeros of analytic functions Lecture 14 Zeros of analytic functions
Zeros of analytic functions Suppose that f : D → C is analytic on an open set D ⊂ C . A point z 0 ∈ D is called zero of f if f ( z 0 ) = 0. The z 0 is a zero of multiplicity/order m if there is an analytic function g : D → C such that f ( z ) = ( z − z 0 ) m g ( z ) , g ( z 0 ) � = 0 . In this case f ( z 0 ) = f ′ ( z 0 ) = f ′′ ( z 0 ) = · · · = f ( m − 1) ( z 0 ) = 0 but f m ( z 0 ) � = 0. Understanding of multiplicity via Taylor’s series: If f is analytic function in D , then f has a Taylor series expansion around z 0 ∞ f n ( z 0 ) � ( z − z 0 ) n , f ( z ) = | z − z 0 | < R . n ! n =0 If f has a zero of order m at z 0 then ∞ f n ( z 0 ) f ( z ) = ( z − z 0 ) m � ( z − z 0 ) n − m n ! n = m f n ( z 0 ) ( z − z 0 ) n − m . and define g ( z ) = � ∞ n = m n ! Lecture 14 Zeros of analytic functions
Zeros of analytic functions Suppose that f : D → C is analytic on an open set D ⊂ C . A point z 0 ∈ D is called zero of f if f ( z 0 ) = 0. The z 0 is a zero of multiplicity/order m if there is an analytic function g : D → C such that f ( z ) = ( z − z 0 ) m g ( z ) , g ( z 0 ) � = 0 . In this case f ( z 0 ) = f ′ ( z 0 ) = f ′′ ( z 0 ) = · · · = f ( m − 1) ( z 0 ) = 0 but f m ( z 0 ) � = 0. Understanding of multiplicity via Taylor’s series: If f is analytic function in D , then f has a Taylor series expansion around z 0 ∞ f n ( z 0 ) � ( z − z 0 ) n , f ( z ) = | z − z 0 | < R . n ! n =0 If f has a zero of order m at z 0 then ∞ f n ( z 0 ) f ( z ) = ( z − z 0 ) m � ( z − z 0 ) n − m n ! n = m f n ( z 0 ) ( z − z 0 ) n − m . and define g ( z ) = � ∞ n = m n ! Lecture 14 Zeros of analytic functions
Zeros of analytic functions Suppose that f : D → C is analytic on an open set D ⊂ C . A point z 0 ∈ D is called zero of f if f ( z 0 ) = 0. The z 0 is a zero of multiplicity/order m if there is an analytic function g : D → C such that f ( z ) = ( z − z 0 ) m g ( z ) , g ( z 0 ) � = 0 . In this case f ( z 0 ) = f ′ ( z 0 ) = f ′′ ( z 0 ) = · · · = f ( m − 1) ( z 0 ) = 0 but f m ( z 0 ) � = 0. Understanding of multiplicity via Taylor’s series: If f is analytic function in D , then f has a Taylor series expansion around z 0 ∞ f n ( z 0 ) � ( z − z 0 ) n , f ( z ) = | z − z 0 | < R . n ! n =0 If f has a zero of order m at z 0 then ∞ f n ( z 0 ) f ( z ) = ( z − z 0 ) m � ( z − z 0 ) n − m n ! n = m f n ( z 0 ) ( z − z 0 ) n − m . and define g ( z ) = � ∞ n = m n ! Lecture 14 Zeros of analytic functions
Zeros of analytic functions Suppose that f : D → C is analytic on an open set D ⊂ C . A point z 0 ∈ D is called zero of f if f ( z 0 ) = 0. The z 0 is a zero of multiplicity/order m if there is an analytic function g : D → C such that f ( z ) = ( z − z 0 ) m g ( z ) , g ( z 0 ) � = 0 . In this case f ( z 0 ) = f ′ ( z 0 ) = f ′′ ( z 0 ) = · · · = f ( m − 1) ( z 0 ) = 0 but f m ( z 0 ) � = 0. Understanding of multiplicity via Taylor’s series: If f is analytic function in D , then f has a Taylor series expansion around z 0 ∞ f n ( z 0 ) � ( z − z 0 ) n , f ( z ) = | z − z 0 | < R . n ! n =0 If f has a zero of order m at z 0 then ∞ f n ( z 0 ) f ( z ) = ( z − z 0 ) m � ( z − z 0 ) n − m n ! n = m f n ( z 0 ) ( z − z 0 ) n − m . and define g ( z ) = � ∞ n = m n ! Lecture 14 Zeros of analytic functions
Zeros of analytic functions Suppose that f : D → C is analytic on an open set D ⊂ C . A point z 0 ∈ D is called zero of f if f ( z 0 ) = 0. The z 0 is a zero of multiplicity/order m if there is an analytic function g : D → C such that f ( z ) = ( z − z 0 ) m g ( z ) , g ( z 0 ) � = 0 . In this case f ( z 0 ) = f ′ ( z 0 ) = f ′′ ( z 0 ) = · · · = f ( m − 1) ( z 0 ) = 0 but f m ( z 0 ) � = 0. Understanding of multiplicity via Taylor’s series: If f is analytic function in D , then f has a Taylor series expansion around z 0 ∞ f n ( z 0 ) � ( z − z 0 ) n , f ( z ) = | z − z 0 | < R . n ! n =0 If f has a zero of order m at z 0 then ∞ f n ( z 0 ) f ( z ) = ( z − z 0 ) m � ( z − z 0 ) n − m n ! n = m f n ( z 0 ) ( z − z 0 ) n − m . and define g ( z ) = � ∞ n = m n ! Lecture 14 Zeros of analytic functions
Zeros of analytic functions Suppose that f : D → C is analytic on an open set D ⊂ C . A point z 0 ∈ D is called zero of f if f ( z 0 ) = 0. The z 0 is a zero of multiplicity/order m if there is an analytic function g : D → C such that f ( z ) = ( z − z 0 ) m g ( z ) , g ( z 0 ) � = 0 . In this case f ( z 0 ) = f ′ ( z 0 ) = f ′′ ( z 0 ) = · · · = f ( m − 1) ( z 0 ) = 0 but f m ( z 0 ) � = 0. Understanding of multiplicity via Taylor’s series: If f is analytic function in D , then f has a Taylor series expansion around z 0 ∞ f n ( z 0 ) � ( z − z 0 ) n , f ( z ) = | z − z 0 | < R . n ! n =0 If f has a zero of order m at z 0 then ∞ f n ( z 0 ) f ( z ) = ( z − z 0 ) m � ( z − z 0 ) n − m n ! n = m f n ( z 0 ) ( z − z 0 ) n − m . and define g ( z ) = � ∞ n = m n ! Lecture 14 Zeros of analytic functions
Zeros of analytic functions Zeros of a non-constant analytic function are isolated: If f : D → C is non-constant and analytic at z 0 ∈ D with f ( z 0 ) = 0, then there is an R > 0 such that f ( z ) � = 0 for z ∈ B ( z 0 , R ) \ { z 0 } . Proof. Assume that f has a zero at z 0 of order m . Then f ( z ) = ( z − z 0 ) m g ( z ) where g ( z ) is analytic and g ( z 0 ) � = 0 . For ǫ = | g ( z 0 ) | > 0 , we can find a δ > 0 2 such that | g ( z ) − g ( z 0 ) | < | g ( z 0 ) | , 2 whenever | z − z 0 | < δ (as g is continuous at z 0 ). Therefore whenever | z − z 0 | < δ, we have 0 < | g ( z 0 ) | < | g ( z ) | < 3 | g ( z 0 ) | . Take R = δ. 2 2 Lecture 14 Zeros of analytic functions
Zeros of analytic functions Zeros of a non-constant analytic function are isolated: If f : D → C is non-constant and analytic at z 0 ∈ D with f ( z 0 ) = 0, then there is an R > 0 such that f ( z ) � = 0 for z ∈ B ( z 0 , R ) \ { z 0 } . Proof. Assume that f has a zero at z 0 of order m . Then f ( z ) = ( z − z 0 ) m g ( z ) where g ( z ) is analytic and g ( z 0 ) � = 0 . For ǫ = | g ( z 0 ) | > 0 , we can find a δ > 0 2 such that | g ( z ) − g ( z 0 ) | < | g ( z 0 ) | , 2 whenever | z − z 0 | < δ (as g is continuous at z 0 ). Therefore whenever | z − z 0 | < δ, we have 0 < | g ( z 0 ) | < | g ( z ) | < 3 | g ( z 0 ) | . Take R = δ. 2 2 Lecture 14 Zeros of analytic functions
Zeros of analytic functions Zeros of a non-constant analytic function are isolated: If f : D → C is non-constant and analytic at z 0 ∈ D with f ( z 0 ) = 0, then there is an R > 0 such that f ( z ) � = 0 for z ∈ B ( z 0 , R ) \ { z 0 } . Proof. Assume that f has a zero at z 0 of order m . Then f ( z ) = ( z − z 0 ) m g ( z ) where g ( z ) is analytic and g ( z 0 ) � = 0 . For ǫ = | g ( z 0 ) | > 0 , we can find a δ > 0 2 such that | g ( z ) − g ( z 0 ) | < | g ( z 0 ) | , 2 whenever | z − z 0 | < δ (as g is continuous at z 0 ). Therefore whenever | z − z 0 | < δ, we have 0 < | g ( z 0 ) | < | g ( z ) | < 3 | g ( z 0 ) | . Take R = δ. 2 2 Lecture 14 Zeros of analytic functions
Zeros of analytic functions Identity Theorem: Let D ⊂ C be a domain and f : D → C is analytic. If there exists an infinite sequence { z k } ⊂ D , such that f ( z k ) = 0 , ∀ k ∈ N and z k → z 0 ∈ D , f ( z ) = 0 for all z ∈ D . Proof. Case I: If D = { z ∈ C : | z − z 0 | < r } then ∞ � a n ( z − z 0 ) n , for all z ∈ D . f ( z ) = n =0 To show f ≡ 0 on D it is enough to show f n ( z 0 ) = 0 for all n . If possible assume that f n ( z 0 ) � = 0 for some n > 0. Let n 0 be the smallest positive integer such that f n 0 ( z 0 ) � = 0. Then ∞ a n ( z − z 0 ) n = ( z − z 0 ) n 0 g ( z ) , � f ( z ) = n = n 0 where g ( z 0 ) = a n 0 � = 0. Since g is continuous at z 0 , there exist ǫ > 0 such that g ( z ) � = 0 for all z ∈ B ( z 0 , ǫ ). By hypothesis there exists some k such that z 0 � = z k ∈ B ( z 0 , ǫ ) and f ( z k ) = 0 . This forces g ( z k ) = 0 which is a contradiction. Lecture 14 Zeros of analytic functions
Zeros of analytic functions Identity Theorem: Let D ⊂ C be a domain and f : D → C is analytic. If there exists an infinite sequence { z k } ⊂ D , such that f ( z k ) = 0 , ∀ k ∈ N and z k → z 0 ∈ D , f ( z ) = 0 for all z ∈ D . Proof. Case I: If D = { z ∈ C : | z − z 0 | < r } then ∞ � a n ( z − z 0 ) n , for all z ∈ D . f ( z ) = n =0 To show f ≡ 0 on D it is enough to show f n ( z 0 ) = 0 for all n . If possible assume that f n ( z 0 ) � = 0 for some n > 0. Let n 0 be the smallest positive integer such that f n 0 ( z 0 ) � = 0. Then ∞ a n ( z − z 0 ) n = ( z − z 0 ) n 0 g ( z ) , � f ( z ) = n = n 0 where g ( z 0 ) = a n 0 � = 0. Since g is continuous at z 0 , there exist ǫ > 0 such that g ( z ) � = 0 for all z ∈ B ( z 0 , ǫ ). By hypothesis there exists some k such that z 0 � = z k ∈ B ( z 0 , ǫ ) and f ( z k ) = 0 . This forces g ( z k ) = 0 which is a contradiction. Lecture 14 Zeros of analytic functions
Recommend
More recommend
