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