Review Exam III
Complex Analysis
Underlined Propositions or Theorems: Proofs May Be Asked for on Exam
Chapter 3.3
Bi-Linear, Linear Fractional, Moebius Transformations
az + b
, ad − bc ≠ 0
cz + d
ad − bc
Differentiability S '( z ) =
(cz + d ) 2
Properties
S : £ ∞ → £ ∞ , S is 1-1, S is onto, pole of S is -d/c, zero of S is -b/a
− dz + b
dz − b
T ( z ) = S −1 ( z ) =
=
, S conformal on £ \ {− d / c} , Fixed Points,
cz − a
− ca + a
Definition S ( z ) =
Uniqueness
Special Cases: Translations, Dilations, Rotations, Inversion
Every Bi-Linear Transformation can be written as a composition of Translations, Dilations, Rotations, and the
Inversion
Mapping Properties of Special Cases: “Circles” to “Circles”
Geometry of Images
Cross-Ratio
Definition: S ( z ) = ( z , z 2 , z 3 , z 4 ) unique Bi-Linear Transformation such that S ( z 2 ) = 1, S ( z 3 ) = 0 ,
S (z4 ) = ∞
Proposition: Cross-Ratio in Invariant under Bi-Linear Transformations
Properties
Orientation Principle, Symmetry
Joukowski Transformation
Conformal Mappings
Compositions of Standard Functions, Bi-Linear Transformations and Joukowski
Examples of conformal mappings
Problems about constructing conformal mappings
Chapter 4.1
Riemann-Stieltjes Integrals
Definition of function of bounded variation and total variation
Proposition (1.3) If γ :[a , b] → £ is piecewise smooth, then γ is of bounded variation and
b
V (γ ) = ∫ | γ '(t) | dt .
a
Definition of Riemann-Stieltjes Integral
Theorem (1.4) If f :[a , b ] → £ is continuous and if γ :[a , b] → £ is of bounded variation, then the
b
Riemann-Stieltjes integral
∫
a
b
f d γ = ∫ f (t )d γ (t ) exists.
a
(Proof uses Cantor’s Theorem II.3.7).
Proposition 1.7 Let f , g : [a , b ] → £ be continuous, let γ , σ :[ a, b] → £ be of bounded variation and let
α , β ∈ £ . Then,
b
∫ (α f + β g ) dγ
a)
a
b
b
= α ∫ f dγ + β ∫ gd γ
b
∫
a
b
∫
a
b
∫
b) fd (aγ + βσ ) = α f d γ + β gdσ
a
a
a
Proposition Let f :[a , b ] → £ be continuous and let γ :[a , b] → £ be of bounded variation. If
a < t0 < t1 < L < tn = b , then
b
n
tk
∫ fd γ = ∑ ∫
a
fdγ
k =1 t
k −1
Theorem (1.9) If γ :[a , b] → £ is piecewise smooth and f :[a , b ] → £ , then
b
∫
a
b
f d γ = ∫ f (t )γ '(t )dt .
a
Definition for a path γ :[a , b] → £ of trace of γ , {γ } .
b
∫
Definition of rectifiable path γ :[a , b] → £ and length of {γ } = d γ . For γ piece-wise smooth, length of
a
b
{γ } = ∫ | γ '(t) | dt .
a
Definition of line integral: Let γ :[a , b] → £ be rectifiable path and let f :{γ } → £ be continuous, define
line integral
b
b
a
a
∫ f = ∫ ( f o γ )d γ = ∫ f (γ ( t )) d γ (t ) =∫ f (z )dz
γ
Note: if γ piece-wise smooth, then
γ
b
b
b
a
a
a
∫ f = ∫ ( f o γ )d γ = ∫ f (γ ( t ))d γ (t ) =∫ f (γ ( t ))γ '(t ) dt =∫ f (z ) dz
γ
γ
Problems about computing line integrals using the definition
Definition of a change of parameter ϕ
Proposition If ϕ is a change of parameter, i.e., if ϕ :[c , d ] → [a , b ] , ϕ is continuous, strictly increasing and
ϕ is onto, then for γ :[a , b] → £ a rectifiable path and f :{γ } → £ continuous, then
Definition:
∫f
=
γ
∫
f
γ oϕ
(1) a curve as an equivalance class of rectifiable paths;
(2) the trace of a curve is the trace of a representative;
(3) a curve is smooth if some representative is smooth;
(4) a curve is closed if the initial and terminal points on the trace are the same.
Definition for γ :[a , b] → £ a rectifiable path of −γ and of | γ (t) | and definition
∫
γ
b
f ( z ) | dz | = ∫ f (γ ( t ))d | γ | (t )
a
Proposition (1.17) Let γ :[a , b] → £ be a rectifiable path and let f :{γ } → £ be continuous. Then,
a)
∫
−γ
b)
f = −∫ f
γ
| ∫ f |≤ ∫| f || dz |≤ max| f ( z) | V (γ )
γ
z∈{γ }
γ
Theorem (Fundamental of Theorem of Calculus for Line Integrals) Let G be a region and let γ be a rectifiable
path in G with initial and terminal points α and β , resp. If f : G → £ is continuous and if f has a primitive
on G, say F, then
∫f
γ
β
= F ( z) α .
Corollary Let G be a region and let γ be a closed rectifiable path in G . If f : G → £ is continuous and if f
has a primitive on G, say F, then
∫f
=0.
γ
Problems about computing line integrals using the Fund. Thm. of Calc. for Line Integrals
Chapter 4.2
Proposition 2.1 (Leibnitz’s Rule)
Integrals
a)
∫
|w| =1
(w − z )n dw = 0, n = 0,1,2,3,L
b)
n = 2,3,4,5, L
dw
=
0,

n
∫
(
w
−
z
)
| z |≠ 1

| w| =1
c)
 0, | z |> 1
dw
=

∫
w
−
z
2π i, | z | < 1
| w| =1
Cauchy Integral Formula #0 Let f : G → £ be analytic and suppose that B (a , r ) ⊂ G . For z ∈ B (a , r ) ,
f ( z) =
1
f ( w)
dw
2π i | w−∫a|= r w − z
Problems about computing line integrals using the CIF #0
Lemma (2.7) Let γ be a rectifiable curve. Suppose that Fn and F are continuous on{γ } and that
{ Fn } converges uniformly on {γ } to F. Then, lim ∫ Fn = ∫ lim Fn = ∫ F
n →∞
γ
γ
n →∞
γ
Theorem 2.8 Let G be a region and let f : G → £ be analytic. Let B (a , R ) ⊂ G . Then, f has a power
series representation on B (a , R) , say
∞
f ( z ) = ∑ an ( z − a) n
(1)
n =0
where the coefficients
f ( n ) (a )
1
f ( w)
an =
=
dw , for any choice 0 < ρ < R .
n!
2π i | w−∫a| = ρ (w − a)n +1
Furthermore, the radius of convergence of the power series (1) is at least R.
Corollaries (Hypothesis: Let G be a region and let f : G → £ be analytic. Let B (a , R ) ⊂ G . )
a)
b)
the radius of convergence of the power series (1) is equal to dist(a , ∂ G ) , i.e., the distance
(from a) to the nearest singularity of f
f ( n ) (a ) =
n!
f (w)
dw
∫
2π i |w− a|= ρ (w − a) n+1
n! M
.
Rn
c)
Cauchy’s Estimate If | f ( z ) |≤ M on B (a , R) , then | f ( n ) (a) |≤
d)
f has a primitive on B (a , R) , namely F ( z ) = ∑
e)
Proposition 2.15 Suppose γ is a closed rectifiable curve in B (a , R) . Then,
∞
an
( z − a ) n+ 1
n =0 n + 1
∫f
=0
γ
Chapter 4.3
Division Algorthim
Definition: Let G be a region and let f : G → £ be analytic and let f (a) = 0. We say that f has a zero of
order m (multiplicity m) at z = a if
a) there exists g ∈A ( G ) such that (i) f ( z ) = ( z − a ) m g ( z ) and (ii) g (a ) ≠ 0
or alternatively
b) (i) f ( a ) = f '(a ) = f ''(a ) = L = f ( m −1) ( a) = 0 and f
(m)
(a) ≠ 0
Definition: entire function
Liouville’s Theorem If f is a bounded entire function, then f is constant.
Fundamental Theorem of Algebra Every non-constant polynomial with complex coefficients has a root in £
Corollary: Every polynomial with complex coefficients of degree n has exactly n roots (counted according to
multiplicity)
Identity Theorem: Let G be a region. Let f be analytic on G. TFAE
a)
f ≡0
( a ) = 0, n = 0,1,2,3,L
b)
there exists a point a ∈ G such that f
c)
Z f = { z ∈ G : f ( z ) = 0} has a limit point in G.
Corollaries:
a)
(n )
Let g , h ∈A ( G ) and g ( z ) = h ( z ) for z ∈ S ⊂ G . If S has a limit point in G, then
g ≡h.
b)
Let G be a region. Let f be analytic on G. Suppose that f is not identically 0 on G. If
a ∈ G and f (a) = 0, then there exists an integer m such that f has a zero at z = a of
multiplicity m.
c)
Isolated Zeros. Let G be a region. Let f be analytic on G. Suppose that f is not identically 0
on G. If a ∈ G and f (a) = 0, then there exists an R > 0 such that on B (a , R ) \ {a} we have
f (z) ≠ 0 .
d)
Let G be a region. Let f be analytic on G. Suppose that f is not identically 0 on G. Let K
be a compact subset of G. Then, f has at most a finite number of zeros on K. Further, f has
at most a countable number of zeros on G.
Maximum Modulus Theorem Let G be a region. Let f be analytic on G. If there exists a point a ∈ G such
that | f (a ) |≥| f ( z ) | for all z ∈ G , then f is constant on G.
Extension Let G be a region. Let f be analytic on G. If there exists a point a ∈ G and r > 0 such
that | f (a ) |≥| f ( z ) | for all z ∈ B ( a , r ) , then f is constant on G.
Chapter 4.4
Proposition: Let γ be a closed rectifiable curve and let a ∉ {γ } . Then,
Definition: Winding of γ wrt to a (Index of γ wrt to a) n(γ , a) =
1
dz
is an integer.
∫
2π i γ z − a
1
dz
, for γ a closed rectifiable
∫
2π i γ z − a
curve and a ∉ {γ } .
Interpretation: 2π n ( γ , a ) represents the total change in arg(γ (t ) − a ) as γ (t ) parametrizes the curve {γ } ,
i.e., n (γ ,a ) represents the total number of times γ winds around a.
Theorem Let γ be a closed rectifiable curve. Then, n (γ ,a ) is constant on components of £ \ {γ } . Also,
n (γ , a ) = 0 on the unbounded component of £ \ {γ }