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Notes for Math 185: Complex Analysis
Lecture 1: Logistics and Motivation 4
Math 185: Complex Analysis Spring 2021
Lecture 1: Logistics and Motivation
Lecturer: Di Fang 19 January Aditya Sengupta
Note: LATEX format adapted from template for lecture notes from CS 267, Applications of Parallel Comput- ing, UC Berkeley EECS department.
Why take complex analysis? Because you need it to get a math degree. That’s boring.
Let’s try again: why is complex analysis useful?
- It makes things complete: C is algebraically complete and R isn’t.
- Even if you’re fine only dealing with real numbers, complex analysis can still help you! Suppose you’re integrating
−∞
dx 1+x^2. This is a real integral, and we know its antiderivative is arctan^ x. However, if you change it to
−∞
dx 1+x^4 , we don’t know the antiderivative and the old method breaks down.^ We can further ruin our lives by changing the integrand to x
2 1+x^4. There’s no closed-form antiderivative for these. Complex analysis lets us do these integrals easily!
- You might think “okay, but why are we just studying better integration techniques? I can do that numerically”. In fact, there are other applications: machine learning/AI and quantum computing depend heavily on complex analysis. Machine learning has often been helped by Fourier analysis, which is based on complex analysis; quantum mechanics works entirely in the complex world.
- It’s unique and beautiful by itself!
This semester, we’re going to relearn calculus (as we’ve done a few times already): functions, limits, continuity (chapters 1-2), derivatives (chapter 3), integrals (chapter 4), and series (chapter 5). However, we’ll also cover residues (chapter 6-7), which are unique to complex analysis.
Consider two cases: a function f : R → R (over real numbers), and a function f : C → C (over complex numbers).
True-false questions:
- If f is differentiable everywhere, f is infinitely differentiable everywhere. False: f (x) = x|x| at x = 0 (and in fact it’s possible to make a function that is differentiable everywhere but second-differentiable nowhere.) But True for complex numbers.
- If f is smooth (infinitely differentiable) everywhere, then its Taylor series is equal to itself. False:
consider f (x) =
e−^1 /x
2 x 6 = 0 0 x = 0
. f is infinitely differentiable at x = 0 (the derivative is 0), and so the
Taylor series at x = 0 is just the constant function 0. But True for complex numbers.
- If f is differentiable everywhere and bounded, then f must be a constant. False: this is bullshit (direct quote from Di) due to cos x, sin x, but it’s True for complex numbers (Liouville’s theorem).
We’ll start math by reviewing complex numbers. My fridge is getting delivered so this bit might be sparse, but we know what complex numbers are. C is isomorphic to R^2 under the isomorphism (x, y) ↔ x + iy. If z = x + iy, we’ll denote Re z = x and Im z = y.
For the sake of rigor, we’ll review some basic properties and definitions:
- For z 1 , z 2 ∈ C, z 1 = z 2 ⇐⇒ Re z 1 = Re z 2 and Im z 1 = Im z 2.
- Summation works like you would expect: z 1 + z 2 = (x 1 + x 2 ) + i(y 1 + y 2 ).
- The product: z 1 z 2 = (x 1 + iy 1 )(x 2 + iy 2 ) = (x 1 x 2 − y 1 y 2 ) + i(x 1 y 2 + x 2 y 1 ).
We continue (over me dealing with the fridge) to show C is a field.
Lecture 2: Complex Numbers, Definitions 7
Properties The properties are below
- ¯z¯ = z
- |¯z| = |z|
- z = ¯z ⇐⇒ z ∈ R
- z 1 ±¯ z 2 = ¯z 1 ± z¯ 2
- z 1 ¯z 2 = ¯z 1 z¯ 2
- Re z = 12 (z + ¯z), Im z = (^21) i (z − z¯)
- |z|^2 = z z¯ = ¯zz
- |z 1 z 2 | = |z 1 ||z 2 |. This is nontrivial to show:
Proof. |z 1 z 2 |^2 = z 1 z 2 z¯ 1 z¯ 2 = z 1 z¯ 1 z 2 z¯ 2 = |z 1 |^2 |z 2 |^2
Let’s try to re-prove the Triangle Inequality using property 7 above.
WTS: |z 1 + z 2 |^2 ≤ (|z 1 | + |z 2 |)^2
Proof. |z 1 + z 2 |^2 = (z 1 + z 2 )( ¯z 1 + ¯z 2 ) = |z 1 |^2 + |z 2 |^2 + z 1 z¯ 2 + z 2 z¯ 1.
The cross terms look similar: we claim (and can easily show by property 5 and the definition of the complex conjugate above) that they are each other’s complex conjugate. Therefore there is no imaginary part, and we get
|z 1 + z 2 |^2 = |z 1 |^2 + |z 2 |^2 + 2 Re(z 1 z¯ 2 ). (2.1)
Further, we can say that 2 Re(z 1 z¯ 2 ) ≤ 2 |z 1 z¯ 2 |. Finally, using properties 8 and 2, we get
|z 1 + z 2 |^2 ≤ |z 1 |^2 + |z 2 |^2 + 2|z 1 ||z 2 |, (2.2)
which is precisely the right hand side.
2.3 Exponential forms
We know that complex numbers are isomorphic to R^2. We’ve been representing these vectors in R^2 in Cartesian coordinates: is there a polar representation? Yes!
We know that polar coordinates are related to Cartesian coordinates by x = r cos θ, y = r sin θ, and so we get z = r(cos θ + i sin θ) = reiθ^ by Euler’s formula. This is the exponential form of complex numbers.
Remark 2.2. For z = 0, θ is undefined.
Remark 2.3. We call θ the argument of z. θ is not unique, as cos and sin are periodic functions: for any valid θ we could shift it by 2 πn for any n ∈ Z and still have a valid θ. We denote by arg z the set of all arguments. Further, we denote by Argz the principal argument, i.e. the unique argument lying in (−π, π].
Lecture 2: Complex Numbers, Definitions 8
Example 2.2. Arg 1 = 0, Arg(−1) = π, Arg i = π 2 , Arg(−i) = − π 2
Properties
- eiθ^1 eiθ^2 = ei(θ^1 +θ^2 ). This is nontrivial: remember we need to prove everything again for C!
Proof.
eiθ^1 eiθ^2 = (cos θ 1 + i sin θ 1 )(cos θ 2 + i sin θ 2 ) (2.3) = (cos θ 1 cos θ 2 − sin θ 1 sin θ 2 ) + i(cos θ 1 sin θ 2 + sin θ 1 cos θ 2 ) (2.4) = cos(θ 1 + θ 2 ) + i sin(θ 1 + θ 2 ) (2.5) = ei(θ^1 +θ^2 )^ (2.6)
- z 1 z 2 = r 1 r 2 ei(θ^1 +θ^2 )
- z z^12 = r r^12 ei(θ^1 −θ^2 )
- (^1) z = (^1) r e−iθ
- arg(z 1 z 2 ) = arg z 1 + arg z 2 , arg z z^12 = arg z 1 − arg z 2. Note that this does not hold if we consider Arg instead of arg. For example, if z 1 = − 1 , z 2 = i, then Arg z 1 = π 2 , Arg z 2 = π, Arg(z 1 z 2 ) = − π 2 6 = 32 π.
2.4 de Moivre’s formula
Lemma 2.4 (de Moivre’s formula).
(cos θ + i sin θ)n^ = cos nθ + i sin nθ, n ∈ Z (2.7)
Proof.
(cos θ + i sin θ)n^ = (eiθ^ )n^ = einθ^ = cos nθ + i sin nθ. (2.8)
Example 2.3. With n = 2, we get the double-angle formulas,
(cos θ + i sin θ) = (cos^2 θ − sin^2 θ) + i(2 sin θ cos θ) = cos(2θ) + i sin(2θ) (2.9)
−16 = 16eiπ^ (2.14) (−16)^1 /^4 = (16)^1 /^4 eiπ/^4 ei^2 πk/n, k = 0, 1 , 2 , 3. (2.15)
For concreteness, plug in for k: we get
2(1 + i),
2(−1 + i),
2(− 1 − i),
2(1 − i).
Lecture 3: “Topology dictionary”, functions and mappings 11
Math 185: Complex Analysis Spring 2021
Lecture 3: “Topology dictionary”, functions and mappings
Lecturer: Di Fang 26 January Aditya Sengupta
Warm-up question: if z 1 = z 2 where z 1 = r 1 eiθ^1 , z 2 = r 2 eiθ^2 is it necessarily the case that r 1 = r 2 and θ 1 = θ 2? Not necessarily: θs are only unique up to 2π rotations.
Today we’ll start describing the geometric structure of C using theory from topology. If we have a number line over the reals, intervals and comparisons are a useful tool for making some sense of it: this is bigger than that, or this number and that one define an interval and you can reason about everything in between them. But in complex numbers, we don’t have these tools. How do we build them?
3.1 Neighborhoods, interiors, exteriors, boundaries
Our first basic tool will be an −neighborhood (nbhd).
Definition 3.1. For some point zo, an −nbhd of zo is the set of all z ∈ C such that |z − zo| < ; the set of all points within a circle of radius centered at zo.
We denote this by B(zo) (the B is for ball, to allow for higher dimensions.)
Definition 3.2. A deleted or punctured neighborhood is a neighborhood with the center removed.
B′ (zo) = {z ∈ C | 0 < |z − zo| < } (3.1)
Now we define the interior and exterior of a set. It’s tempting to let this just be z ∈ S, z 6 ∈ S, but this doesn’t consider points on the boundary: do you put them in S or not? What does it mean to be on the boundary anyway? The following definitions will leverage -neighborhoods to explain this.
Definition 3.3. z is an interior point of a set S if there exists some such that B(z) ⊂ S.
Definition 3.4. z is an exterior point of a set S if there exists some such that B(z) ∩ S = ∅.
Definition 3.5. z is a boundary point of a set S if it is neither an interior nor an exterior point. More rigorously, for any > 0 , the neighborhood B(z) ∩ S 6 = ∅ (there is some point in S) and also B(z) ∩ Sc^6 = ∅ (there is some point not in S).
We denote these by int S, ext S, ∂S. The interior of S is also denoted So.
Example 3.1. Let S be the unit circle: S = B 1 (0) = {z ∈ C | |z| < 1 }.
The interior, exterior, and boundary can be defined based on this:
int S = {z | |z| < 1 } ext S = {z | |z| > 1 }∂S = {z | |z| = 1}. (3.2)
Lecture 3: “Topology dictionary”, functions and mappings 13
Definition 3.9. A set S ⊆ C is a domain if S is open and connected.
Remark 3.2. This doesn’t necessarily mean it’s the domain of a particular function yet.
Definition 3.10. S is a region if S \ ∂S is a domain. (check this)
Definition 3.11. S is bounded if ∃R > 0 such that S ⊂ BR(0).
Example 3.4. Consider the set {z | Im(1/z) > 1 }. Using the Cartesian form, we see that
x + iy
x − iy x^2 + y^2
so
Im
z
y x^2 + y^2
Completing the square shows us that this is equivalent to
x^2 +
y +
This defines a circle. This set is a domain and a neighborhood of the point − 12 i.
Definition 3.12. zo is called an accumulation point or a limit point of a set S ⊆ C if all B′(zo) (all punctured neighborhoods of zo) satisfy B′(zo) ∩ S 6 = ∅.
Less formally, there is always an element of the set arbitrarily close to an accumulation point. The set of accumulation points is often denoted S′.
Example 3.5. Let S = { 1 −n i| n = 1, 2 ,... }. The accumulation point of S is 0: however close you get to the origin, any neighborhood around the origin will have a point in the set (this follows almost exactly from the limit definition of convergence of 1/n → 0.)
Example 3.6. S = {(−1)n^ | n = 1, 2 ,... } has no accumulation points as it only consists of − 1 , 1.
Remark 3.3. S¯ = S ∪ ∂S = S ∪ S′, but ∂S 6 = S′^ in general. This is proved, e.g. in Rudin, but we’ll just take it as a fact.
Lecture 3: “Topology dictionary”, functions and mappings 14
3.3 Functions
Consider S ⊆ C and a function f : S → C. We can describe this as a rule z → f (z) or x + iy → u + iv.
Example 3.7.
f (z) = z^2 (3.6) f (x, y) = (x^2 − y^2 ) + 2ixy (3.7) u = x^2 − y^2 , v = 2xy (3.8)
Example 3.8.
f (z) = |z|^2 (3.9) f (x, y) = x^2 + y^2 (3.10) u = x^2 + y^2 , v = 0. (3.11)
This is a “real-valued” function because v is identically zero.
Example 3.9. We still have polynomials, i.e. functions of the form p(z) =
∑n k=0 akz
k, and we can still take ratios of polynomials to get rational functions.
Remark 3.4. An interesting type of function is the linear fractional transformation, of the form azcz++db.
We consider a generalization of functions that may not just return a single number: multi-valued functions.
Example 3.10. arg z is multi-valued, as it is only unique up to 2π and so we can add factors of 2kπ to get as many values as we like for this.
v =
2 y
y^2 + c 1 right − 2 y
y^2 + c 1 left
On the right, y ↑ =⇒ v ↑ and on the left, y ↑ =⇒ v ↓.
If we had a horizontal line, we could do the same:
v = 2xy = c 1 =⇒ y =
c 2 2 x
and u = x^2 − 4 cx^22. Thus for x > 0, x ↑ =⇒ u ↑ and x < 0, x ↑ =⇒ u ↓.
Lecture 4: Limits and continuity 17
Math 185: Complex Analysis Spring 2021
Lecture 4: Limits and continuity
Lecturer: Di Fang 28 January Aditya Sengupta
4.1 Warmup
True/False.
- A subset of C is either open or closed. (False: C or ∅ are both, and it’s possible to be neither.)
- Since S¯ = S ∪ ∂S = S ∪ S′, we have ∂S = S′. (False: look at {(−1)n^ | n ∈ Z}. There are no accumulation points, but the boundary is the set itself.)
- S is closed if and only if it contains all of its accumulation points. (True: if S is closed, S = S¯ = S ∪S′).
4.2 Limits
Definition 4.1. lim z→zo = ωo if ∀ > 0 , ∃δ > 0 such that 0 < |z − zo| < δ =⇒ |f (z) − ωo| < .
It is important that 0 < |z − zo| (why?) This statement is similar to real analysis, except instead of absolute value of the difference, we have the modulus. This is a similar statement but should be treated as a 2D thing. Therefore a complex limit is a significantly stronger condition than a real one.
We can equivalently say ∀ > 0 , ∃δ > 0 s. t. z ∈ B δ′ (zo) =⇒ f (z) ∈ B(ωo). This can be interpreted as “there exists a punctured δ-ball centered at zo such that its image under f is a subset of a punctured -ball centered at f (zo)”.
Theorem 4.1. If lim z→zo f (z) exists, it is unique.
The proof of this is exactly the same as in real analysis.
Example 4.1. Let f (z) = iz 2 in |z| < 1. We want to show that lim z→ 1 f (z) = i 2.
Proof. ∀ > 0 ∃δ > 0 such that 0 < |z − 1 | < δ implies
∣f^ (z)^ −^
i 2
∣ =^
|z − 1 | 2
δ 2
so we take δ = 2.
Lecture 4: Limits and continuity 19
(b) products work: lim z→zo f (z)g(z) = ω 1 ω 2
(c) quotients work: if ω 2 6 = 0, then lim z→zo
f (z) g(z) =^
ω 1 ω 2
We can combine the first two properties to show that polynomials have well-defined limits that we can find just by plugging in the point. lim z→zo P (z) = P (zo)
4.4 Limits involving infinity
In real analysis, we had ±∞: two infinities for the two directions. In our case, however, we can go in infinite directions, so do we have infinite infinities? (something something John Green).
We’ll introduce a notion of infinity that is direction-independent. We refer to the complex plane with infinity as the “extended complex plane”, C ∪ {∞}. Imagine this as the surface of a sphere.
Consider a sphere whose projection on the complex plane is the unit circle. For any point in the complex plane z, consider the line between the north pole of the sphere at (0, 0 , 1) and the point z = x + iy = (x, y, 0). This will intersect with the sphere at some point. This defines a mapping from the complex plane to the Riemann sphere(conventionally, we also say the origin maps to the south pole). Under this mapping, we say infinity maps to the north pole (in all directions.)
Definition 4.2. A neighborhood of ∞, BR(∞), is the set |z| > R.
This would be a circle around the North Pole on the Riemann sphere.
Definition 4.3. The three types of limits involving infinity are defined as follows:
- lim z→zo f (z) = ∞ if ∀R > 0 , ∃δ > 0 s. t. z ∈ B δ′ (zo) =⇒ f (z) 3 BR(∞).
- lim z→∞ f (z) = ωo if ∀ > 0 , ∃R > 0 s. t. z ∈ BR(∞) =⇒ f (z) ∈ B(ωo).
- lim z→∞ f (z) = ∞ if ∀R > 0 , ∃r > 0 s. t. z ∈ Br (∞) =⇒ f (z) ∈ BR(∞).
Theorem 4.2. 1. lim z→zo f (z) = ∞ if lim z→zo
1 f (z) = 0.
- lim z→∞ f (z) = ωo if lim z→ 0
f (
z
= ωo.
- lim z→∞ f (z) = ∞ if lim z→ 0
1 f (^) ( (^) z^1 ) = 0.
Example 4.3. Suppose we want to show that lim z→∞
1 z+1 = 0. It suffices to show that lim z→ 0 11 z +^
which is equivalent to lim z→ 0
z 1+z = 0.
4.5 Continuity
Definition 4.4. f is continuous at zo if f (zo) is defined, lim z→zo f (z) is defined, and lim z→zo f (z) = f (zo). Note
that the third condition contains the first two.
More precisely, f : S → C is continuous at zo if for all > 0 there exists δ > 0 such that ∀z ∈ S, |z − zo| < δ =⇒ |f (z) − f (zo)| < . More compactly, f (Bδ (zo)) ⊂ B(f (zo)).
Definition 4.5. f is continuous in a region R if it is continuous at each point in R.
Recall that if f, g are continuous at zo, then so are f + g, f g, and fg provided that g 6 = 0. We also find that every polynomial is continuous in the entire plane.
Theorem 4.3. Let f : A → B, g : B → C for A, B, C ∈ C. If f is continuous at zo and g i continuous at f (zo), then g ◦ f : A → C, z → g(f (z)) is continuous at zo.
We can prove this roughly using the definition twice. For any there exists some α such that continuity holds for g, and for that α there exists some δ such that it holds for f.
Theorem 4.4. If f is continuous at zo and f (zo) 6 = 0, then f 6 = 0 in some neighborhood of zo.
Proof. Let = |f^ (z 2 o )|> 0, since f (zo) 6 = 0. Then by continuity, ∃δ > 0 s. t. |z − zo| < δ =⇒ |f (z) − f (zo)| < |f (zo)|
We show this by contradiction. Suppose there exists a point such that f (z) = 0. Then we would have it
cancel out, and we would have |f (zo)| < |f^ (z 2 o )|, a contradiction.
