Review Exam II Complex Analysis Underlined Definitions: May be asked for on exam Underlined Propositions or Theorems: Proofs may be asked for on exam Chapter 5.3 Definition. Let G be a region. A function f is meromorphic on G if . . . Argument Principle Let G be a region and let f be meromorphic on G with poles { p1 , p2 , { z1 , z 2 , , z n } , counted according to multiplicity. Let , pm } and zeros γ be a closed rectifiable curve in G which does not pass through any of the points p j and z k and let γ ≈ 0 in G. Then, n m 1 f ′( z ) = − dz n ( ; z ) n(γ ; p j ) . γ ∑ ∑ k 2π i ∫γ f ( z ) k =1 j =1 Corollary Let G be a region and let f be meromorphic on G with poles { p1 , p2 , , pm } and zeros γ be a simple closed rectifiable curve in G which is positively oriented which does not pass through any of the points p j and z k and let γ ≈ 0 in G. Let { z1 , z 2 , , z n } , counted according to multiplicity. Let int(γ ) denote the bounded component of { p1 , p2 , \ {γ } . Let Pf ,γ and Z f ,γ denote the cardinality of , pm } ∩ int(γ ) and { z1 , z 2 , , z n } ∩ int(γ ) , respectively . Then, 1 f ′( z ) dz = Z f ,γ − Pf ,γ . 2π i ∫γ f ( z ) Corollary Let G be a region and let f be analytic on G with zeros { z1 , z 2 , , z n } , counted according to multiplicity. Let γ be a simple closed rectifiable curve in G which is positively oriented which does not pass through any of the points z k and let γ ≈ 0 in G. Let int(γ ) denote the bounded component of Z f ,γ denote the cardinality of { z1 , z 2 , , z n } ∩ int(γ ) . Then, \ {γ } . Let 1 f ′( z ) dz = Z f ,γ . ∫ 2π i γ f ( z ) Problems about counting the number of zeros of a given function in a given region using Argument Principle Rouche’s Theorem (Ver #1) Let G be a region and let f and g be mereomorphic on G. Let B ( a , r ) ⊂ G such that f and g have no zeros or poles on C ( a , r ) . Let Z f , Z g , Pf , Pg denote the cardinality of the zeros of f and g and the cardinality of the poles of f and g inside C ( a , r ) , respectively. If | f ( z ) + g ( z ) |<| f ( z ) | + | g ( z ) | on C ( a , r ) , then Z f − Pf = Z g − Pg . Corollary Let G be a region and let f and g be analytic on G. Let B ( a , r ) ⊂ G such that f and g have no zeros on C ( a , r ) . Let Z f , Z g denote the cardinality of the zeros of f and g inside C ( a , r ) , respectively. If | f ( z ) + g ( z ) |<| f ( z ) | + | g ( z ) | on C ( a , r ) , then Z f = Z g . Rouche’s Theorem (Ver #2) Let G be a region and let f and g be mereomorphic on G. Let B ( a , r ) ⊂ G such that f and g have no zeros or poles on C ( a , r ) . Let Z f , Z f + g , Pf , Pf + g denote the cardinality of the zeros of f and g and the cardinality of the poles of f and f+g inside C ( a , r ) , respectively. If 0 < | g ( z ) | < | f ( z ) | < ∞ on C ( a , r ) , then Z f − Pf = Z f + g − Pf + g . Corollary Let G be a region and let f and g be analytic on G. Let B ( a , r ) ⊂ G such that f and g have no zeros on C ( a , r ) . Let Z f , Z g denote the cardinality of the zeros of f and g inside C ( a , r ) , respectively. If 0 < | g ( z ) | < | f ( z ) | < ∞ on C ( a , r ) , then Z f = Z f + g . Problems about counting the number of zeros of a given function in a given region using Rouche’s Theorem Chapter 6.1 Maximum Modulus Theorem (Ver #1) Let G be a region and let f ∈ A (G ) . If there exists a ∈ G such that | f ( a ) |≥| f ( z ) | for all z ∈ G , then f is constant. Maximum Modulus Theorem (Ver #2) Let G be a bounded region and let f ∈ A (G ) ∩ C (G ) . Then, max | f ( z ) |= max | f ( z ) | . z ∈G z ∈ ∂G Definition. Let G be a region and let f : G → . Let a ∈ ∂G or a = ∞ . Then, lim sup f ( z ) = . . . z→a Definition. Let G be a region. Then, ∂ ∞ G = . . . Maximum Modulus Theorem (Ver #3) Let G be a region and let f ∈ A (G ) . Suppose there exists a constant M such that lim sup | f ( z ) | ≤ M for all a ∈ ∂ ∞ G . Then, | f ( z ) | ≤ M for all z ∈ G . z→a Supplement Maximum Modulus Definition (for Supplement). Let D = { z : | z | < 1} and for 0 < r ≤ 1 let Dr = { z : | z | < r} Theorem For f ∈ A ( D ) let a) M ( r , f ) = max | f ( z ) | b) A( r , f ) = max Re f ( z ) | z| = r c) 1 I p (r , f ) = 2π | z| = r π ∫π | f (re iθ ) | p dθ − Then, (i) M ( r , f ) is a strictly increasing function of r on (0,1) unless f is constant. (i) A( r , f ) is a strictly increasing function of r on (0,1) unless f is constant. (i) I p ( r , f ) is a strictly increasing function of r on (0,1) unless f is constant, for p ∈ Parseval’s Identity Let f ∈ A ( D ) , f ( z ) = ∞ ∑a z n=0 1 I 2 (r , f ) = 2π π ∞ −π n=0 n n . For 0 < r < 1 , iθ 2 2 2n ∫ | f (re ) | dθ = ∑ | an | r Chapter 6.2 Let D denote the open unit disk (centered at 0) and let Dr denote open disk of radius r (centered at 0). Schwarz’s Lemma (Ver #1) Let f ∈ A ( D ) such that (a) | f ( z ) | ≤ 1 for all z ∈ D and (b) f (0) = 0 . Then, (i) | f ( z ) | ≤| z | for all z ∈ D and (ii) | f ′(0) | ≤ 1 . Moreover, equality occurs in (i) or (ii) if and only if f ( z ) = ζ z for some ζ , | ζ |= 1 . Schwarz’s Lemma (Ver #2) Let f ∈ A ( D ) such that (a) f ( D ) ⊂ D and (b) f (0) = 0 . Then, (i) f ( Dr ) ⊂ Dr and (ii) | f ′(0) | ≤ 1 . Moreover, equality occurs in (i) or (ii) if and only if f ( z ) = ζ z for some ζ ∈ ∂D . z−a . Then, ϕ a ∈ A ( D ), ϕ a : D → D and, ϕ a is one-to-one and onto. 1 − az Further, for | z |= 1 , | ϕ a ( z ) |= 1 and ϕ a′ (0) = 1− | a |2 , ϕ a′ ( a ) = (1− | a |2 ) −1 . Proposition. For a ∈ D let ϕa ( z) = 1− | α |2 Proposition Let f ∈ A ( D ), f : D → D . For a ∈ D let f ( a ) = α . Then, | f ′( a ) | ≤ . Equality 1− | a |2 occurs if and only if f ( z ) = ϕ −α (ζϕ a ( z )) for some ζ ∈ ∂D . Theorem Let f ∈ A ( D ), f : D → D such that f is one-to-one and onto. Then, there exists a ∈ D and ζ ∈ ∂D such that f = ζϕ a . Supplement Subordination Definition (for Supplement). Let D = { z : | z | < 1} and for 0 < r ≤ 1 let Dr = { z : | z | < r} Definition Let f , g ∈ A ( D ) . Then, f is subordinate to g ( f ≺ g ) if . . . Theorem. Let f ≺ g , f ( z ) = ∞ ∑ an z n , g ( z ) = n=0 i) ii) iii) iv) v) ∑|a k =0 vii) ∑b n=0 n z n . Then, f (0) = g (0), | f ′(0) |≤| g ′(0) | f (D) ⊂ g (D) f ( Dr ) ⊂ g ( Dr ) M (r , f ) ≤ M (r , g ) I p (r , f ) ≤ I p (r , g ) m vi) ∞ m | ≤ ∑ | bk |2 , m = 0, 1, 2, 3, 2 k k =0 max (1− | z |2 ) | f ′( z ) |≤ max (1− | z |2 ) | g ′( z ) | z ∈ Dr z ∈ Dr Proposition. Let f , g ∈ A ( D ) . Suppose (a) f (0) = g (0) , (b) f ( D ) ⊂ g ( D ) , (c) g is one-to-one. Then, f ≺ g. Example. Let P = { f ∈ A ( D ) : Re f ( z ) > 0 for all z ∈ D , f (0) = 1} . Let p ( z ) = f ∈P implies f ≺ p . 1+ z . Then, 1− z Chapter 10.1 Definition. Let G be a region and let f : G → . f is harmonic on G if . . . Definition. Let G be a region and let f : G → . f satisfies the Mean Value Property (MVP) on G if . . . Proposition. Let G be a region. Let f be harmonic on G. Then, f satisfies the MVP on G. Maximum Principle (Ver. #1) Let G be a region and let u : G → R satisfy the MVP on G. If there exists a ∈ G such that u ( a ) ≥ u ( z ) for all z ∈ G , then u is constant. Maximum Principle (Ver. #2) Let G be a region and let u , v : G → be bounded functions satisfying the MVP on G. Suppose for each a ∈ ∂ ∞ G that lim sup u ( z ) ≤ lim sup v ( z ) , then either z→a z→a u ( z ) < v ( z ) for all z ∈ G or else u ≡ v . Minimum Principle (Ver. #1) Let G be a region and let u : G → R satisfy the MVP on G. If there exists a ∈ G such that u ( a ) ≤ u ( z ) for all z ∈ G , then u is constant. Chapter 10.2 Definition. The Poisson kernel Pr (θ ) = . . . 1 + reiθ 1− r2 eiθ + r Proposition Pr (θ ) = Re = = Re iθ 1 − reiθ 1 − 2r cos(θ ) + r 2 e −r Proposition. The Poisson kernel satisfies (i) 1 2π π ∫π P (θ ) dθ = 1 , r 0 < r <1 − (ii) Pr (θ ) > 0 , Pr (θ ) = Pr ( −θ ) (iii) Pr (θ ) < Pr (δ ) for 0 < δ <| θ |< π , i.e., Pr (θ ) is strictly decreasing on (0, π ) (iv) lim Pr (δ ) = 0 for each δ , 0 < δ ≤ π r → 1− Theorem Let f ∈ C ( ∂D ) , f : ∂D → (i) (ii) . Then, there exists u ∈ C ( D ) ∩ Har ( D ) such that u (e iθ ) = f (e iθ ) u is unique 1 Corollary Let u ∈ C ( D ) ∩ Har ( D ) . Then, u ( re ) = 2π iθ Corollary Let h ∈ C (C ( a , R )) , h : C ( a , R ) → w ( z ) = h ( z ) on C ( a , R ) . π ∫π P (θ − t )u (e r it ) dt , 0 < r < 1 − . Then, there exists w ∈ C ( B ( a , R )) ∩ Har ( B ( a , R )) such that Corollary Let u ∈ C ( B ( a , R )) ∩ Har ( B ( a , R )) . Then, u ( a + reiθ ) = 1 2π π ∫π P r/R (θ − t )u ( a + reit ) dt , 0 < r < R − Harnack’s Inequality Let u ∈ C ( B ( a , R )) ∩ Har ( B ( a , R )) , u ( z ) ≥ 0 . Then, R−r R+r u ( a ) ≤ u ( a + reiθ ) ≤ u (a) , 0 ≤ r < R R+r R−r Chapter 7.1 Definition. Let G be a region and let ( Ω , d ) be a complete metric space. Then, C (G , Ω ) = . . . Proposition. Let G be a region. Then there exists a sequence of subsets { K n } of G such that (i) K n ⊂⊂ G (ii) K n ⊂ int( K n +1 ) ∞ (iii) ∪K n =G n =1 (iv) K ⊂⊂ G implies K ⊂ K n for some n ∈ Lemma If ( S , d ) is a metric space, then ( S , σ ) is a metric space, where σ ( s, t ) = d ( s, t ) . A set O is open 1 + d ( s, t ) in ( S , d ) if and only if O is open in ( S , σ ) . Definition. For K ⊂⊂ G and f , g ∈C (G , Ω ) , let ρ ( f , g ) = K , σ K ( f , g) = , Bρ K ( f , δ ) = . Definition. For { K n } a compact exhaustion of a region G and for f , g ∈ C (G , Ω ) let ρ ( f , g ) = . . . Proposition. (C (G , Ω ), ρ ) is a metric space. Lemma 1.7 (i) Given ε > 0 there exists δ > 0 and K ⊂⊂ G such that for f , g ∈ C (G , Ω ) ρK ( f , g) < δ ⇒ ρ ( f , g) < ε (ii) Given δ > 0 and K ⊂⊂ G there exists ε > 0 such that ρ ( f , g) < ε ⇒ ρK ( f , g ) < δ Lemma 1.10 (i) A set O ⊂ C (G , Ω ) is open if and only if for each f ∈ O there exists δ > 0 and K ⊂⊂ G such that O ⊃ Bρ K ( f , δ ) (ii) A sequence { f n } ⊂ C (G , Ω ) converges to f (in the ρ metric) if and only if for each K ⊂⊂ G { f n } converges to f in the ρ K metric. Proposition (C (G , Ω ), ρ ) is a complete metric space. Definition. A set F ⊂ C (G , Ω ) is normal . . . Proposition. A set F ⊂ C (G , Ω ) is normal if and only if F is compact. Proposition. A set F ⊂ C (G , Ω ) is normal if and only if for each δ > 0 and K ⊂⊂ G there exist functions f1 , f 2 , , f n ∈ F such that F ⊂ n ∪ Bρ k =1 K ( fk ,δ ) . Definition. A set F ⊂ C (G , Ω ) is equicontinuous at a point z 0 ∈ G if . . . Definition. A set F ⊂ C (G , Ω ) is equicontinuous on a set E ⊂ G if . . . Proposition. Suppose a set F ⊂ C (G , Ω ) is equicontinuous at each point of G. Then, F is equicontinuous on each K ⊂⊂ G . Arzela-Ascoli Theorem A set F ⊂ C (G , Ω ) is normal if and only if (a) for each z ∈ G the orbit of z under F, i.e., { f ( z ) : f ∈ F } , has compact closure and (b) F ⊂ C (G , Ω ) is equicontinuous at each point of G. Chapter 7.2 Theorem Let G be a region and let A ( D ) denote the set of analytic functions on G. Then, 1'. Let f ∈ C (G , Ω ) . If f is a limit point of A ( D ) , then f ∈ A ( D ) . 1. Let { f n } ⊂ A ( D ) and let f ∈ C (G , Ω ) . If f n → f (in C (G , Ω ) ), then f ∈ A ( D ) . 2. Let { f n } ⊂ A ( D ) and let f ∈ C (G , Ω ) . If f n → f (in C (G , Ω ) ), then f n ( k ) → f C (G , Ω ) ) for each k > 0. Corollary. A ( D ) is closed in C (G , Ω ) Corollary. A ( D ) is a complete metric space Corollary. Let { f n } ⊂ A ( D ) and let f ∈ C (G , Ω ) . If ∞ f ( k ) ( z ) = ∑ f n( k ) ( z ) . n =1 ∞ ∑f n =1 n ( z ) → f ( z ) (in C (G , Ω ) ), then (k ) (in