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Faye Jackson MATH 596 - TOC
Faye Jackson
- MATH Notes on - December 22, (Complex Analysis)
- I. Introduction and Administration. C o n t e n t s
- II. The Basics.
- II.1. Motivation and Recollections
- II.2. Solving Polynomial Equations
- II.3. Topology Time
- II.4. The Riemann Sphere
- II.5. Examples of Functions
- III. Complex Differentiation
- IV. Complex Integration!
- IV.1. Review of prerequisites.
- IV.2. Defining Complex Integrals
- IV.3. The Complex Fundamental Theorem of Calculus
- IV.4. Cauchy’s Theorem
- IV.5. Liouville’s Theorem
- IV.6. Morera’s Theorem
- IV.7. Goursat’s Theorem
- V. Series!
- V.1. Very Quick Review
- V.2. Power Series
- V.3. Zeros of Analytic Functions
- V.4. The Open Mapping Theorem
- V.5. Analytic Continuation
- VI. Laurent Series
- VI.1. Laurent Decomposition
- VI.2. Isolated Singularities
- VI.3. Singularities at ∞
- VI.4. Partial Fractions decompositions
- VII. Residue Calculus
- VII.1. The Residue Theorem.
- VII.2. Integrals of Rational Functions Faye Jackson MATH 596 - TOC
- VII.3. Integrals of Trig Functions
- VII.4. Integrands with Branch Cuts
- VII.5. Fractional Residues
- VII.6. Principal Values
- VII.7. Jordan’s Lemma
- VII.8. Exterior Domains
- VII.9. Logarithmic Integral
- VII.10. Rouch´e’s Theorem
- VII.11. Hurwitz’s Theorem
- VII.12. Winding Numbers
- VIII. Schwarz Lemma and Hyperbolic Geometry
- VIII.1. Conformal self-maps of D
- VIII.2. Hyperbolic Geometry
- IX. Riemann Mapping Theorem.
- IX.1. Arzel`a-Ascoli and the Proof
- IX.2. Mandelbrot Set Things
- X. Infinite Products.
- A Syllabus
- References
Faye Jackson August 30th, 2022 MATH 596 - II.
Because C = R[i] there is a Galois automorphism for this field extension z = x + iy 7 → z := x − iy called complex conjugation which fixes R. We know z 1 + z 2 = z 1 + z 2 z 1 z 2 = z 1 · z 2.
We then define the norm squared of z to be the multiplication of all its Galois conjugates (works for any finite Galois extension), that is |z|^2 := z · z = x^2 + y^2 ∈ R≥ 0.
Compatibility of |z| =
q |z|^2 with the Euclidean metric on R^2. This along with the fact that C ∼= R^2 as a vector space justifies the identification of C with R^2 If z ̸= 0, z = x + iy then z |z| =^
x |z| +^ i
y |z| lies on the unit circle S^1. Thus we can write this complex number as z |z| = cos^ θ^ +^ i^ sin^ θ for some θ ∈ R. Please note θ is uniquely defined only up to adding integer multiples of 2π and only when z ̸= 0. This number θ is called the argument of z, denoted arg(z). We let eiθ^ := cos θ + i sin θ so that z = |z| eiθ^ in these “polar” coordinates. The nice thing about polar coordinates is that if z = reiα, w = ρeiβ^ then zw = rρ · ei(α+β). Thus |zw| = r 1 r 2.
II.2. Solving Polynomial Equations
A critical feature of C is that it is algebraically closed. In other words, we have Theorem II.2.1 (The Fundamental Theorem of Algebra) Every nonconstant polynomial in C[X] has a root in C. This is the historical origin of the complex numbers. In the 16th century, Cardano and others were solving cubic equations over R which have solutions in R! However, their calculations/algorithms included complex numbers which cancelled in the end to give real solutions. Example II.2. Fix a ∈ C. We must solve z^2 = a. If a = 0 there is one solution, z = 0. If a ̸= 0, write a = reiθ^ , then we may take ω = eiθ/^2 , −ω = ei(θ/2+π)^ so that (±ω)^2 = eiθ^. Then ±ω√r are two roots of a, and neither can be “preferred”
Faye Jackson August 30th, 2022 MATH 596 - II.
Proposition II.2.2 (see HW) There is no continuous (topology!) function f : C → C such that (f (z))^2 = z for all z ∈ C. In other words, there is no continuous choice of a square root.
Proof. HW.
Definition II.2. Let n ∈ N and consider the equation zn^ = 1. The solutions of this equation are called the roots of unity of order n.
Proof there are n n-th roots of unity. Note if zn^ = 1, then |z|n^ = 1, so |z| = 1. Note zn^ − 1 has polynomial derivative nzn−^1 , whose only solution is 0, so no roots are repeated. Explicitly, we have solutions zj = eiθj^ where
θj =^2 πjn
for j ∈ Z. Up to integer multiples of 2π, there are n such arguments, and n such zj.
A similar arguments shows that any nonzero complex number z has n different n-th roots that are exactly spaced around the circle of radius |z|^1 /n.
II.3. Topology Time
C is a metric space as d(z, w) = |z − w|. Refresher on the induced topology is
- If a ∈ C and r > 0 then B(a, r) = {z ∈ C | |z − a| < r} is the open ball centered at a of radius r.
- We call a set U ⊆ C open provided that for every z ∈ U there exists an rz > 0 such that B(z, rz ) ⊆ U.
- A subset A ⊆ C is closed provided that C \ A is open. This is equivalent to the statement that for every convergent sequence zn → z with zn ∈ A, we have z ∈ A (that is the set of limit points A′^ ⊆ A). A′^ := {ω ∈ C | ω is a limit point of A} := {ω ∈ C | There exists {an}n∈N ⊆ A converging to ω}.
- We say a sequence an ∈ C converges to ℓ ∈ C provided that for all ε > 0 there exists an N ∈ N so that for n ≥ N we have |an − ℓ| < ε. We know that C is Cauchy complete as it is essentially the same as R^2. If f : C → C and a ∈ C we say that f is continuous at a provided that for all ε > 0 there exists a δ > 0 such that if |z − a| < δ then |f (z) − f (a)| < ε. We may also take limits to infinity and negative infinity in real analysis. In complex analysis, we only have ONE direction of infinity. For f : C → C let ℓ ∈ C. We define (^) zlim→∞ f (z) = ℓ to mean for all ε > 0 there exists an M ∈ R such that if |z| > M then |f (z) − ℓ| < ε.
II.4. The Riemann Sphere
We add a single point to C, called the point at infinity and denoted ∞ to get Cb := C ∪ {∞}.
Faye Jackson September 1st, 2022 MATH 596 - III.
III. Complex Differentiation
Let U ⊆ C be open, f : U → C, we’re going to define holomorphic functions (in Gamelin [Gam03], this is “analytic”) Definition III.0. The function f is holomorphic at z 0 ∈ U provided that
hlim→ 0 f^ (z^0 +^ h h)^ −^ f^ (z^0 ) exists, and in that case we call that limit the derivative f ′(z 0 ). The function f is holomorphic on U provided that it’s holomorphic at all points inside U. If C ⊆ U is closed, then we say f : C → C is holomorphic on C provided that there is an open set containing C on which f is holomorphic. f is said to be entire provided that f is holomorphic on all of C. Proposition III.0. If f, g : U → C are holomorphic at some z 0 ∈ U then (1) f + g is holomorphic, (f + g)′^ = f ′^ + g′. (2) f g is holomorphic, (f g)′^ = f ′g + f g′. (3) If g(z 0 ) ̸= 0, then f /g is holomorphic at z 0 and (^) f g
(z 0 ) = f^
′(z 0 )g(z 0 ) − f (z 0 )g′(z 0 ) (g(z 0 ))^2 (4) If f : Ω → U is holomorphic at z 0 , g : U → C is holomorphic at f (z 0 ), g ◦ f is holomorphic at z 0 and (g ◦ f )′(z 0 ) = g′(f (z 0 ))f ′(z 0 )
Proof. Same as in R! Just manipulating limits with each other. Example III.0. Polynomials are entire! The proof is now easy. Constants and the identity map are both entire (exercise), and polynomials are sums/products of these. Question: How can we tell if a function is holomorphic at a given point z 0 ∈ U? In the case of complex differenitation, the derivative is a complex number... Consider f : C → C, we can view this as a map F : R^2 → R^2 , and then the derivative of F is a linear transformation with standard basis on R^2 , namely ∂F ∂x 1 ∂F ∂y 1 ∂F 2 ∂x ∂F 2 ∂y
Faye Jackson September 6th, 2022 MATH 596 - III.
Proposition III.0.2 (Cauchy-Riemann Equations) Writing f = u + iv which is holomorphic at some z 0 = x 0 + iy 0 , we have ∂u ∂x =^
∂v ∂y
∂u ∂y =^ −^
∂v ∂x in other words ∂f ∂x =^ −i
∂f ∂y.
Proof. Consider the limit
hlim→ 0 f^ (z^0 +^ h h)^ −^ f^ (z^0 ) Write h = h 1 + ih 2 , and approach along real/imaginary axes
f ′(z 0 ) = (^) hlim 1 →^0
f (z 0 + h 1 ) − f (z 0 ) h 1 =^
∂f ∂x =^
∂u ∂x +^ i
∂v ∂x. Similarly
f ′(z 0 ) = (^) hlim 2 →^0
f (z 0 + ih 2 ) − f (z 0 ) ih 2 =^ −i
∂f ∂y =^
∂v ∂y −^ i
∂u ∂y. Equating these gives the Cauchy-Riemann equations above. Remark III.0. If f is complex differentiable at z 0 ∈ U , then it is continuous at z 0. Write
f (z 0 + h) − f (z 0 ) = h
(^) f (z 0 +^ h)^ −^ f^ (z 0 ) h
and take the limit as h → 0. Note:
|f ′(z 0 )|^2 =
(^) ∂u ∂x +^ i
∂v ∂x
(^) ∂u ∂x
(^) ∂v ∂x
= ∂u∂x · ∂u∂y − ∂v∂x · ∂u∂y
which is the determinant of the Jacobian when we view this as a map R^2 → R^2. Gamelin’s definition requires that f ′(z 0 ) exists and also that f ′(z) is continuous at z 0. Later we will show that the derivative of a holomorphic function (at z)) is also holomorphic at z 0 , which will give us lots of extra stuff If we assume this, then the functions u, v in f = u + iv will have continuous partial derivatives of every order, and so the mixed partials will agree... this is useful to keep in mind. Stuff:
- HW 2 (A due tomorrow, B due next week)
- HW 1B due tonight
Faye Jackson September 6th, 2022 MATH 596 - III.
∆v := ∂
(^2) v ∂x^2 +^
∂^2 v ∂y^2 = 0. Definition III.0. A smooth function u : R^2 → R that satisfies Laplace’s equation
∆u := ∂
(^2) u ∂x^2 +^
∂^2 u ∂y^2 = 0 is said to be harmonic The real and imaginary parts of a holomorphic function are thus harmonic. Definition III.0. If two harmonic functions u, v : R^2 → R satisfy Cauchy-Riemann equations, then v is said to be “the” harmonic conjugate of u (unique up to an additive constant). Example III.0. Let u = x^2 − y^2. This is harmonic. Can we build a harmonic conjugate? Well we know ∂v ∂y =^
∂u ∂x = 2x^
∂v ∂x =^ −^
∂u ∂y = 2y. We’re led to consider v = 2xy + const. Building the function f = u + iv yields f (z) = (x^2 − y^2 ) + i(2xy + const) = z^2 + i · const. Formal + Helpful: Consider f (x, y) = u(x, y) + iv(x, y), z = x + iy, z = x − iy. Then x = 12 (z + z), y = − 2 i (z − z). We want to change variables from (x, y) to (z, z). We define new operators
∂f ∂z =
(^) ∂f ∂x −^ i
∂f ∂y
∂f ∂z =
(^) ∂f ∂x +^ i
∂f ∂y
To say that f is holomorphic (aka u, v satisfy the Cauchy-Riemann equations) is exactly to say ∂f∂z = 0, and this gives ∂f∂z = f ′. If f is holomorphic, then is 1/f holomorphic? Yes, provided that f is nonzero. Rational functions! Let R(z) = P (z)/Q(z) where P, Q are polynomials and P, Q have no common roots. The zeros of Q are called poles of R. We extend R to a function Rb : Cb → Cb by taking Rb(z) = ∞ for z a pole of R. We could also consider
R^ b(∞) = lim z→ 0 R(z).
It is nicer to use a related function R 1 (z) = R(1/z), and define Rb(∞) = Rb 1 (0). Note that R(1/z) is a rational function
R(z) = a^0 +^ a^1 z^ +^ · · ·^ +^ amz
n b 0 + b 1 z + · · · + bmzm R 1 (z) = zm−n
(^) a 0 zn^ +^ a 1 zn−^1 +^ · · ·^ +^ an b 0 zm^ + b 1 zm−^1 + · · · + bm
Faye Jackson September 8th, 2022 MATH 596 - III.
If m > n, R(z) has a zero of order m − n at ∞, define Rb(∞) = 0. If m < n, the point at infinity is a pole of order n − m so Rb(∞) = ∞. If m = n, then Rb(∞) = a bmn ̸= 0, ∞. Consider R(z) = zz^2 +57− 53 i. The zeros of Rb are ±
− 57 i, and the poles are z = 53, ∞. Fact: The total number of zeros of a rational function is equal to max (n, m) which is also equal to the number of poles wher we count with multiplicity. Find it in your book! Definition III.0. The degree of a rational function R(z) = P (z)/Q(z) is max(deg P, deg Q). This will agree with the topological degree, which you might know about. Definition III.0.5 (M¨oius tranformations) A M¨obius transformation is a rational function of degree 1. M¨obius transformations are in fact the automorphisms (bijective, holomorphic, with holomorphic inverse) of Cb. To think about defining whether a function f : bC → Cb is holomorphic at ∞, consider testing if f (1/z) : Cb → Cb is holomorphic at 0. Example III.0. When if f (z) = azcz++bd a M¨obius tranformation? Maybe we should think about if
det a^ b c d
= ad − bc ̸= 0.
We say a M¨obius transformation g is affine provided that g(∞) = ∞, and we can then express g(z) = αz+β for some α, β ∈ C. The affine group is then
{z 7 → αz + β | α ̸= 0, β ∈ C} = Aut(C). Stuff:
- HW 2B due September 13th by 10PM.
- For 6, to say LM is C-linear means there exists α ∈ C so that the following diagram commutes R^2 R^2
C C
z αz
LM
- For 9, the set U ⊆ C which is the domain of f : U → C should be connected.
To say f is holomorphic at a point, we will always mean f is holomorphic on a neighborhood of z 0. We’re headed to Gamelin, II.4-7, Ahlfors 3.2-3.3. Now back to M¨obius transformations If we have a M¨obius transformation f (z) = azcz++db we note that f (∞) = ac , c ̸= 0 f (∞) = ∞, c = 0
f −^1 (∞) = −c d, c ̸= 0 f −^1 (∞) = ∞, c = 0.
Faye Jackson September 8th, 2022 MATH 596 - III.
Algebraically, M¨ob is a group with respect to the binary operation of composition. We can think of this as a matrix group via GL 2 (C). Namely via the map a b c d
7 → fA(z) = azcz ++ d b.
The determinant being nonzero corresponds to ad − bc ̸= 0, which we require. One can check this is a surjective homomorphism. The kernel is
ker(GL 2 (C) → M¨ob) =
λ 1 0 0 1
| λ ∈ C \ { 0 }
which may be easily checked. We often call the quotient of GL 2 (C) by this kernel the “projective general lienar group”
PGL 2 (C) := GL 2 (C)
λ 1 0 0 1
| λ ∈ C \ { 0 }
∼= M¨ob
We could normalize our matrices to have determinant one...
SL 2 (C) = {A ∈ GL 2 (C) | det(A) = 1}
There is then a homomorphism SL 2 (C) → M¨ob with kernel
. This gives us
PSL 2 (C) := SL 2 (C)
∼= M¨ob.
This is in some sense 3-dimensional, as we have four variables and one condition, ad − bc = 1. There are three fundamental types of M¨obius transformations (1) Linear, z 7 → αz where α ∈ C \ { 0 }. (2) Translation, z 7 → z + β for some β ∈ C. (3) Inversion, z 7 → (^1) z. Theorem III.0. We have that (i) The group M¨ob is generated by translation, linear maps, and inversion. (ii) The action of M¨ob on Cb is “simply 3-transitive” i.e. for any two triples of distinct points (p 1 , p 2 , p 3 ), (q 1 , q 2 , q 3 ) on Cb there exists a unique M¨obius transformation taking pj to qj. (iii) The action of M¨ob on Cb preserves circles.
Proof of (ii). For existence, it suffices to show that any triple (p 1 , p 2 , p 3 ) can be sent to (0, 1 , ∞), then take
f (z) = ((pp^2 −^ p^3 )(z^ −^ p^1 ) 2 −^ p 1 )(z^ −^ p 3 )^
Faye Jackson September 13th, 2022 MATH 596 - III.
Caution: Breaking the rules a bit if one of the pi is ∞... but just adjust and change formula a bit. Namely one of these three formulas p 2 − p 3 z − p 3 , p^1 =^ ∞^
z − p 1 z − p 3 , p^2 =^ ∞^ z^7 →^
z − p 1 p 2 − p 1 , p^3 =^ ∞.
To prove uniqueness, suppose g ∈ M¨ob that sends (p 1 , p 2 , p 3 ) 7 → (0, 1 , ∞). We must examine f ◦ g−^1. This is a M¨obius transformation fixing 0, 1 , ∞. Check that the only such map is the identity. Something cool: Suppose p 1 , p 2 , p 3 , p 4 are distinct points in C that lie on a circle Γ ⊆ C. The M¨obius transformation f (z) = (^) z−^1 p 1 sends the circle Γ to a line L = f (Γ). Let qk = f (pk), k = 2, 3 , 4. Choose the ordering so on the circle p 3 is between p 2 , p 4. Then q 3 is between q 2 , q 4. This gives that
|q 2 − q 4 | = |q 2 − q 3 | + |q 3 − q 4 |
plug in qk = (^) pk^1 − 1 and simplyify to get
|p 1 − p 3 | · |p 2 − p 4 | = |p 1 − p 2 | · |p 3 − p 4 | + |p 1 − p 4 | · |p 2 − p 3 |
Theorem III.0.5 (Ptolemy’s Theorem) A quadrilateral can be inscribed in a circle if and only if the sum of products of lengths of opposite edges is equal to the product of the lengths of the diagonals.
Stuff:
- Office Hours Wednesday 10:30-12 EH3855.
- HW 2B due tonight 10pm
- HW 3A, 3B is the next round. For 3A, possibly look at Gamelin for inspiration.
- Walk: Carol + snowcones! Back to M¨obius transformations. Recall that M¨ob := {f : bC → Cb | f (z) = azcz ++ d b, ad − bc ̸= 0}.
This is a group with respect to function composition, we saw an isomorphism with familiar groups earlier. Furthermore we have
M¨ob ⊆ Aut(Cb) := {g : Cb → Cb | g is a biholomorphism}.
Faye Jackson September 13th, 2022 MATH 596 - III.
Theorem III.0. Consider these neat properties (i) If f ∈ M¨ob is the unique element sending (p 1 , p 2 , p 4 ) 7 → (0, 1 , ∞) then [p 1 , p 2 , p 3 , p 4 ] = f (p 3 ). In particular, [p 1 , p 2 , p 3 , p 4 ] takes values in Cb \ { 0 , 1 , ∞}. (ii) Two quadruples (p 1 , p 2 , p 3 , p 4 ) and (q 1 , q 2 , q 3 , q 4 ) can be sent to each other by M¨obius transfor- mations if and only if [p 1 , p 2 , p 3 , p 4 ] = [q 1 , q 2 , q 3 , q 4 ]. (iii) If f : Cb → Cb is a homeomorphism that preserves cross ratios of ALL quadruples, then f is a M¨obius transformation. (iv) Four points p 1 , p 2 , p 3 , p 4 lie on the same circle in C if and only if [p 1 , p 2 , p 3 , p 4 ] ∈ R. Back to holomorphic discussion! Recall III.0. f : U → C, z 0 ∈ U , we say that f is holomorphic at z 0 means f is holomorphic on a neighborhood of z 0 , that is for any z within that neighborhood the limit
f ′(z) := lim h→ 0 f^ (z^ +^ h h)^ −^ f^ (z)∈ C exists. We showed that if f = u + iv then
∂x +^ i^ ∂ ∂y
f = 0
- f satisfies Cauchy-Riemann equations on U ∂u ∂x =^
∂v ∂y
∂u ∂y =^ −^
∂v ∂x.
- For ∆ := (^) ∂x∂^22 + (^) ∂y∂^22 we have ∆u, ∆v = 0 under regularity assumptions on u, v (which will be unnecessary later). This means u, v are harmonic, and in fact they are harmonic conjugates. Theorem III.0. Let u, v : U → R have contiuous first order partials on U and satisfy the Cauchy-Riemann equations on U. Then f = u + iv is holomorphic on U.
Proof. We will use Taylor’s Theorem for real variables. Consider some point z 0 = (x 0 , y 0 ) ∈ U , and consider some small h = (h 1 , h 2 ). We see that
u(x 0 + h, y 0 + k) − u(x 0 , y 0 ) = ∂u∂x (x 0 ,y 0 )
· h 1 + ∂u∂y (x 0 ,y 0 )
· h 2 + ε 1
and
v(x 0 + h, y 0 + k) − v(x 0 , y 0 ) = ∂v∂x (x 0 ,y 0 )
· h 1 + ∂v∂y (x 0 ,y 0 )
· h 2 + ε 2
Faye Jackson September 13th, 2022 MATH 596 - III.
where ε 1 , ε 2 tend to 0 more rapidly than h + ik in the sense that ε 1 h 1 + ih 2 ,^
ε 2 h 1 + ih 2 →^0 ⇐⇒^
|ε 1 |^2 h^21 + h^22 ,^
|ε 2 |^2 h^21 + h^22 →^0 as h = h 1 + ih 2 → 0. Using the Cauchy-Riemann equations, we have that
f (z 0 + h) − f (z 0 ) = ∂f∂x (x 0 ,y 0 )
· h + ε 1 + iε 2. (^) hlim→ 0 f^ (z^0 +^ h h)^ −^ f^ (z^0 )= ∂f∂x (x 0 ,y 0 )
Thus f ′(z 0 ) exists. Using this converse, one may show the exponential exp : C → C is holomorphic. Recall the definition below exp : C → C x + iy 7 → ex^ · eiy = ex^ cos y + i sin y.
Exercise III.0. Show that exp is holomorphic by showing it satisfies the Cauchy-Riemann. For the logarithm, we need to use the inverse function theorem. This gets an upgrade in the setting of complex analysis! Theorem III.0. Suppose f is holomorphic on the open set Ω ⊆ C and f ′(z 0 ) ̸= 0 for z 0 ∈ Ω. Then there exists a neighborhood U containing z 0 on which
- f is injective.
- The image V := f (U ) is open in C.
- The inverse f −^1 : V → U is holomorphic on V and satisfies f −^1
(f (z)) = (^) f ′^1 (z).
Proof. The first two bullet points come from analysis of real variables by using the identification R^2 ∼= C. Take g = f −^1 , and let w = f (z), w 1 = f (z 1 ). Then we want to show g′(w 1 ) exists. Consider
wlim→w 1 g(w w)^ −−^ gw(w^1 ) 1 = lim z→z 1 f (z^ z)^ −−^ zf^1 (z 1 ) = (^) f ′(^1 z 1 )^
We must show Log(z) : C \ R≤ 0 → C is holomorphic, where Log(z) = log |z| + i Arg(z).
Well, consider exp : R × (−π, π) → C \ R≤ 0 and use the inverse function theorem... Note exp is already bijective on this domain.
Faye Jackson September 15th, 2022 MATH 596 - IV.
- Due dates for 4A/4B to be decided
- There exists Math Club (Constructing R!)
- There exists Math S^1 (Thursdays 6:30pm-8pm, starts next week 9/22)
- There is a 40 mile walk on Saturday October 1st!
- There is Super Saturdays (starts 10/8, 9:30am-12pm)
- Math Mental Health Hour Sunday (2-3pm), EH 1866
- Bagels on Sunday 10am-11:30am
- U(M) Undergrad Mathematics Seminar EH 3096 Recall III.0. If f = u + iv, we know if f : U → C is holomorphic, then u is harmonic on U , that is ∆u = 0 and u has continuous first and second order partials. Faye’s question: Given u : U → R^2 harmonic, does there exist a harmonic conjugate? No! Take u(z) = Log |z|, which is harmonic on C \ { 0 }. This does not have a harmonic conjugate on C \ { 0 }, but does have a harmonic conjugate on C \ R≤ 0 , namely Arg (z). Then Log (z) = log |z| + i Arg (z) is harmonic on C \ R≤ 0. What’s the difference in domains? C \ { 0 } is not simply connected, while C \ (−∞, 0] is simply connected. Proposition III.0. If u : U → R is harmonic on U and U is simply connected, then a harmonic conjugate exists. For Gamelin, he constructs a harmonic conjugate p57 on rectangles (and will eventually do star-shaped regions). Back to conformal maps: We saw that if f ′(z 0 ) ̸= 0, then f maps orthogonal curves then z 0 to orthogonal curves at f (z 0 ). Definition III.0. The map f : U → V is conformal on U provided that (1) f is conformal at all points z 0 ∈ U. (2) f is bijective. Example III.0. exp : C → C satisfies (1) but not (2). We can also consider z 7 → z^2 as a map C \ { 0 } → C \ { 0 }.
IV. Complex Integration!
Chapter 3/III in Gamelin.
IV.1. Review of prerequisites
Definition IV.1. A path in the plane is a continuous function γ : [a, b] → C, and we say it is a path from γ(a) to γ(b). A path γ is simple provided that γ (^) [a,b) is injective. The path γ is closed provided that γ(a) = γ(b). All paths γ have an orientation, γ(a) is the initial point and γ(b) is the end point.
Faye Jackson September 15th, 2022 MATH 596 - IV.
A path is called smooth if it is smooth as a function. If we have paths γ : [0, 1] → C from A ∈ C to B ∈ C and δ : [0, 1] → C from B to C ∈ C, we can construct a path
(γ ∗ δ)(t) =
γ(2t) if 0 ≤ t ≤ 1 / 2 δ(2t − 1) if 1/ 2 ≤ t ≤ 1 from A to C. This is called the concatenation. A piecewise smooth path is a concatenation of smooth paths. A curve is a smooth path or piecewise smooth. Let γ be a path in C from A to B and let P (x, y), Q(x, y) be continuous complex-valued functions on the image of γ. Break up the image of γ into pieces (xi, yi) and form the sum X P (xj , yj )(xj+1 − xj ) +
X
Q(xj , yj )(yj+1 − yj ).
where we require γ(tj ) = (xj , yj ) where a = t 0 < t 1 < · · · < tn = b. Definition IV.1. If these sums have a limit as distance between points (xj , yj ) → 0 then we define the limit to be the line integral of P dx + Q dy along γ, denoted Z γ
P dx + Q dy.
More precisely, let γ(t) = (x(t), y(t)) with a ≤ t ≤ b. Suppose tj ∈ [a, b] satisfies γ(tj ) = (xj , yj ) with a ≤ t 0 < t 1 < · · · < tn = b. Apply the Mean Value Theorem to find points t∗ j ∈ [tj , tj+1] so that x(tj+1) − x(tj ) = x′(t∗ j )(tj+1 − tj ). Likewise for y. Plugging into the above sums this gives X P (x(tj ), y(tj ))x′(t∗ j )(tj+1 − tj ) +
X
Q(x(tj ), y(tj ))y′(t∗ j )(tj+1 − tj )
X
(P (x(tj ), y(tj ))x′(t∗ j ) + Q(x(tj ), y(tj ))y′(t∗ j ))(tj+1 − tj ).
As tj+1 − tj go to zero we have this is equal to Z γ
P dx + Q dy =
Z (^) b a
P (x(t), y(t))x′(t) + Q(x(t), y(t))y′(t) dt.
Theorem IV.1.1 (Green’s Theorem) Consider some region Ω ⊆ C which is a connected bounded open set whose boundary consists of a finite # of disjoint piecewise smooth curves. Let P, Q be continuously differentiable on Ω ∪ ∂Ω, then Z ∂D
P dx + Q dy =
x D
∂Q
∂x −^
∂P
∂y
dx dy.
