Complex Analysis Notes

Math 444 - Fall 2024

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Week 1 Notes

Tentative Schedule

Day Section Topic
Mon, Aug 26 1.1 Definitions and algebraic properties
Wed, Aug 28 1.2 From algebra to geometry and back
Fri, Aug 30 1.3 Geometry of complex numbers

Mon, Aug 26

Today we introduced the complex numbers \mathbb{C} which are a field. We defined the real and imaginary parts of a complex number and also the absolute value and argument of a complex number. We did the following examples.

  1. Find the real and imaginary parts of 13+4i\dfrac{1}{3+4i} by rationalizing the denominator.

  2. Factor the polynomial z2+3iz2z^2 + 3iz - 2.

  3. The polynomial z2+1z^2 + 1 is irreducible over \mathbb{R}, but not over \mathbb{C}. Show this by factoring it over \mathbb{C}.

Later in the course we will use complex analysis to prove one of most important theorems in algebra:

The Fundamental Theorem of Algebra. Every polynomial p(z)p(z) with coefficients in \mathbb{C} factors into a product of linear expressions of the form (az+b)(az + b) where a,ba, b \in \mathbb{C}.

We finished with a discussion of the polar form z=r(cosθ+isinθ)z = r (\cos \theta + i \sin \theta) of a complex number.

Wed, Aug 28

Following Beck et. al., we will define eiθ:=cosθ+isinθ.e^{i \theta} := \cos \theta + i \sin \theta. Later, when we define the complex exponential function, we will revisit this definition. For now, we will do some calculations that suggest the definition is a good one.

  1. Use the angle addition formulas to show that (eiα)(eiβ)=ei(α+β).(e^{i \alpha}) (e^{i \beta}) = e^{i (\alpha + \beta)}.

  2. Simplify 1eiθ\frac{1}{e^{i \theta}}.

  3. Calculate ddθeiθ\dfrac{d}{d \theta} e^{i \theta}. Hint: Use the definition of eiθe^{i \theta} and treat ii like any other constant.

  4. What is (eiθ)n(e^{i \theta})^n?

  5. What is i\sqrt{i}?

  6. Explain why ddθeiθ=ieiθ\dfrac{d}{d \theta} e^{i \theta} = i e^{i \theta} makes sense in the context of the velocity of a point moving counterclockwise along the unit circle.

After introducing eiθe^{i\theta}, we discussed the n-th roots of unity which are the complex numbers zz such that zn=1z^n = 1. They are given by the formula e2πik/n where k{0,1,,n1}.e^{2 \pi i k/ n} \text{ where } k \in \{0, 1, \ldots, n-1\}.

  1. Find the 3rd roots of 8.

  2. Find the 3rd roots of (22+i22)\left(\frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2} \right).

We finished with a discussion of square and cube roots of complex numbers and how they relate to roots of unity. We also defined the complex conjugate of z=a+ibz = a+ib to be z=aib\bar{z} = a - ib.

Fri, Aug 30

We reviewed some of the useful formulas involving complex conjugates including the following:

We also introduced and proved the triangle inequality for complex numbers |z+w||z|+|w|.| z + w | \le |z| + |w|.

  1. Use the triangle inequality to prove the reverse triangle inequality |zw||z||w| | z - w | \ge |z| - |w| by introducing a substitution u=zwu = z - w.

  2. If |zw|ϵ|z - w| \le \epsilon, prove that |1z1w|ϵ|z||w|.\left| \frac{1}{z} - \frac{1}{w} \right| \le \frac{\epsilon}{|z| \, |w|}.

A convex combination of z,wz, w \in \mathbb{C} is any point tz+(1t)w,0t1. t z + (1-t) w, ~ 0 \le t \le 1. The set of all convex combinations of zz and ww is a line segment connecting zz to ww, and the formula for the convex combinations is a parametric formula for the line segment. If you allow any tt \in \mathbb{R}, then you get an affine combination of zz and ww. The set of all affine combinations of zz and ww is line passing through zz and ww.


Week 2 Notes

Tentative Schedule

Day Section Topic
Mon, Sep 2 Labor Day, no class
Wed, Sep 4 1.4 Elementary topology of the plane
Fri, Sep 6 1.4 Elementary topology of the plane - con’d

Wed, Sep 4

Let BR(z)={w:|zw|<R}B_R(z) = \{w \in \mathbb{C}: |z-w| < R\} denote the open disk of radius RR around zz. We used BR(z)B_R(z) to define interior and boundary points for a subset of \mathbb{C}. We proved the following trichotomy result

Theorem. If AA \subseteq \mathbb{C} and xx \in \mathbb{C}, then exactly one of the following holds.

  1. xx is an interior point of AA.
  2. xx is a boundary point of AA.
  3. xx is an interior point of AcA^c (the complement of AA).

We also defined open and closed sets, and proved the following theorem.

Theorem. A set AA \subseteq \mathbb{C} is closed if and only if its complement is open.

  1. If a set is both open and closed, then it can’t have any boundary points, since it includes its boundary (because it is closed) but every point is an interior point (because it is open). Are there any sets with no boundary points?

We also defined bounded sets. We finished by talking about paths which are continuous functions γ:[a,b]\gamma: [a,b] \rightarrow \mathbb{C}. The range of a path is a subset of \mathbb{C} called a curve (note that not all books follow this terminology). We defined the derivative of a path, and discussed how to define a smooth path so that it matches our intuition for a curve with no sharp turns.

  1. Why isn’t the path γ(t)=t2+it3\gamma(t) = t^2 + i t^3 for t[1,1]t \in [-1,1] smooth?

A path γ:[a,b]\gamma: [a,b] \rightarrow \mathbb{C} is simple if γ(t1)γ(t2)\gamma(t_1) \ne \gamma(t_2) for all t1t2t_1 \ne t_2, except possibly at the endpoints t1=at_1 = a and t2=bt_2 = b. Intuitively, a path is simple if it cannot cross itself, except possibly at the endpoints. A path is closed if γ(a)=γ(b)\gamma(a) = \gamma(b).

Fri, Sep 6

A set AA \subseteq \mathbb{C} is path connected if for any w,zAw, z \in A, there is a (continuous) path γ:[a,b]A\gamma: [a,b] \rightarrow A such that γ(a)=w\gamma(a) = w and γ(b)=z\gamma(b) = z. The following theorem seems obvious, but it is actually famously tricky to prove (see https://en.wikipedia.org/wiki/Jordan_curve_theorem).

The Jordan Curve Theorem. The complement of the range of a simple closed curve in \mathbb{C} consists of two disjoint open path connected sets, one of which (the inside) is bounded and the other (the outside) is not bounded.

A sequence is a function s:s: \mathbb{N}\rightarrow \mathbb{C}. We use the notation sns_n to mean the same thing as s(n)s(n) for sequences. A sequence sns_n converges to LL \in \mathbb{C} if for every ϵ>0\epsilon > 0, there is an N>0N > 0 such that |snL|<ϵ|s_n - L | < \epsilon for every nNn \ge N. Intuitively, this means that for every open disk around LL, there can only be a finite number of nn such that sns_n is not in the disk. When a sequence converges, the number it converges to is called its limit.

If we don’t know the limit of a sequence, we can still use Cauchy’s criterion to show that it must converge. A Cauchy sequence is a sequence sns_n such that for every ϵ>0\epsilon > 0, there exists N>0N > 0 such that |snsm|<ϵ|s_n - s_m | < \epsilon whenever m,nNm, n \ge N.

Theorem (Cauchy’s Criterion). If sns_n is a Cauchy sequence, then it converges to some limit in \mathbb{C}.

We talked about how the Cauchy criterion applies to sequences of real numbers and the complex numbers, in fact the property that Cauchy sequences converge in a set is known as completeness and it is one of the defining properties of the real numbers.

We used completeness to prove the following theorem about real number sequence:

Theorem (Monontone convergence). If sns_n is a sequence of real numbers that is monotone (non-decreasing) and bounded, then it converges.

We ran out of time, but a related application of completeness is the following result.

Theorem. If AA \subset \mathbb{C} and AcA^c are both nonempty sets, then AA has at least one boundary point.


Week 3 Notes

Tentative Schedule

Day Section Topic
Mon, Sep 9 7.1 Sequences and completeness
Wed, Sep 11 7.2 Series
Fri, Sep 13

Mon, Sep 9

The complex numbers are a metric space because they have a distance function d(z,w)=|zw|d(z,w) = |z-w| that satisfies the following properties for all x,y,zCx, y, z \in C:

  1. Definiteness d(x,x)=0d(x,x) = 0
  2. Positivity d(x,y)>0d(x,y) > 0 if xyx \ne y
  3. Symmetry d(x,y)=d(y,x)d(x,y) = d(y,x)
  4. Triangle Inequality d(x,z)d(x,y)+d(y,z)d(x,z) \le d(x,y) + d(y,z)

It helps when working with absolute values in \mathbb{C} to remember that |zw||z-w| is the distance between zz and ww. We didn’t have time for a proof of the last theorem from Friday, so we worked through a proof using the fact that \mathbb{C} is a complete metric space today. Recall that complete means that all Cauchy sequences converge. We proved the theorem by proving the following claims:

  1. Any line segment that connects a point in AA to a point in AcA^c contains a line segment that is half as long and also connects a point in AA to a point in AcA^c.

  2. The midpoints of the line segments above form a Cauchy sequence.

  3. The limit of the midpoints is a boundary point of AA.

This proof definitely doesn’t work without assuming completeness. For example, if A={x:x<π}A = \{x \in \mathbb{Q} : x < \pi \}, then AA has no boundary points in \mathbb{Q}. After discussing that example, we talked about infinite series.

A series of complex numbers k=0ak\sum_{k = 0}^\infty a_k converges if its sequence of partial sums Sn=k=0nakS_n = \sum_{k = 0}^n a_k converges. It helps to know some example infinite series, so we talked about these three:

Then we reviewed geometric series which are series of the form k=0ark\sum_{k = 0}^\infty a r^k where aa is the first term and rr is the common ratio. We proved:

Theorem. A geometric series k=0ark\sum_{k = 0}^\infty a r^k converges if and only if |r|<1|r| < 1, and in that case k=0ark=a1r.\sum_{k = 0}^\infty a r^k = \frac{a}{1-r}.

Wed, Sep 11

Class was canceled since I was out with COVID.

Fri, Sep 13


Week 4 Notes

Tentative Schedule

Day Section Topic
Mon, Sep 16 2.1 Limits and continuity
Wed, Sep 18 2.2 Differentiability and holomorphicity
Fri, Sep 20 2.2 The Cauchy-Riemann equations

Mon, Sep 16

Today we reviewed the questions from the infinite series workshop. Then we talked about functions f:f: \mathbb{C}\rightarrow \mathbb{C}. We defined limits of functions several ways, and we did the following examples.

  1. limz0zz\lim_{z \rightarrow 0} \frac{\bar{z}}{z}.

  2. Exercise 2.6

  3. Use the ϵδ\epsilon-\delta definition of limits to prove that if limzz0f(z)=L0\lim_{z \rightarrow z_0} f(z) = L \ne 0 then limzz01f(z)=1L\lim_{z \rightarrow z_0} \frac{1}{f(z)} = \frac{1}{L}.

    Consider |1f(z)1L|=|Lf(z)||Lf(z)|\left| \frac{1}{f(z)} - \frac{1}{L} \right| = \frac{|L - f(z)|} {|L f(z)|} We can use the fact that f(z)Lf(z) \rightarrow L to make the top as small as we want. We just need to make sure that the bottom doesn’t get close to zero at the same time. The trick is to use the triangle inequality to show that |f(z)|>|L|ϵ|f(z)| > |L| - \epsilon when |Lf(z)|<ϵ|L - f(z)| < \epsilon,

  4. Use the sequential definition of limits to prove the product rule for limits, i.e., limzz0f(z)g(z)=(limzz0f(z))(limzz0g(z))\lim_{z \rightarrow z_0} f(z) g(z) = (\lim_{z \rightarrow z_0} f(z)) (\lim_{z \rightarrow z_0} g(z) ).

Wed, Sep 18

We introduced the complex derivative for functions f:Df: D \rightarrow \mathbb{C}:

f(z)=limh0f(z+h)f(z)h.f'(z) = \lim_{h \rightarrow 0} \frac{f(z+h) - f(z)}{h}.

  1. Find the derivative of f(z)=z2f(z) = z^2.

What does it mean for a function to be differentiable? For functions f:nmf: \mathbb{R}^n \rightarrow \mathbb{R}^m, we say that ff is differentiable at a vector xnx \in \mathbb{R}^n if there is a matrix JJ (called the Jacobian) such that f(x+Δx)f(x)+JΔxf(x + \Delta x) \approx f(x) + J \Delta x more specifically, limΔx0f(x+Δx)(f(x)+JΔx)Δ=0.\lim_{\Delta x \rightarrow 0} \frac{f(x + \Delta x) - (f(x) + J \Delta x)}{\|\Delta \|} = 0. The expression f(x)+JΔxf(x) + J \Delta x is an affine linear approximation of ff near xx.

  1. Show that multiplication by a+iba + ib is a linear transformation on \mathbb{C} that corresponds to multiplying vectors in 2\mathbb{R}^2 by the matrix (abba).\begin{pmatrix} a & -b \\ b & a \end{pmatrix}. Note that the multiplication by a complex number can rotate and/or scale, but it cannot reflect or skew.

A complex function f:f: \mathbb{C}\rightarrow \mathbb{C} can be thought of as a real function f:22f: \mathbb{R}^2 \rightarrow \mathbb{R}^2, and f(z)f'(z) corresponds to the Jacobian matrix J=(Ref/xRef/yImf/xImf/y)J = \begin{pmatrix} \partial \operatorname{Re}f / \partial x & \partial \operatorname{Re}f/ \partial y \\ \partial \operatorname{Im}f / \partial x & \partial \operatorname{Im}f/ \partial y \end{pmatrix}

  1. Observe that if z=x+iyz = x+iy, then z2=x2y2+2xyiz^2 = x^2 - y^2 + 2 x y i. Find the Jacobian of the function f(x,y)=(x2y22xy)f(x,y) = \begin{pmatrix} x^2 - y^2 \\ 2xy \end{pmatrix} and show that it corresponds to the complex number 2z2z.

  2. Show that the function f(z)=zf(z) = \bar{z} is not complex differentiable. Explain why that makes sense, even though the corresponding map on 2\mathbb{R}^2, f(x,y)=(xy)f(x,y) = \begin{pmatrix} x \\ -y \end{pmatrix}, does have a Jacobian matrix.

Fri, Sep 20

We introduced the Cauchy-Riemann equations.

Theorem. Suppose f:Df: D \rightarrow \mathbb{C} is complex differentiable at z0z_0, and f(x+iy)=u(x,y)+iv(x,y)f(x+iy) = u(x,y) + i v(x,y). Then ff satisfies the Cauchy-Riemann equations ux=vy and vx=uy.\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y} \text{ and } \frac{\partial v}{\partial x} = -\frac{\partial u}{\partial y}. Conversely, if f(x+iy)=u(x,y)+iv(x,y)f(x+iy) = u(x,y) + i v(x,y) and the partial derivatives of uu and vv are continuous in a disk around z0=x0+iy0z_0 = x_0 + i y_0 and satisfy the Cauchy-Riemann equations, then ff is complex differentiable at z0z_0.

  1. f(z)=ezf(z) = e^z

  2. This is Exercise 2.19 in the book: Define f(z)=0f(z) = 0 if zz is on either purely real or purely imaginary, and f(z)=1f(z) = 1 otherwise. Show that ff satisfies the Cauchy–Riemann equations at z=0z = 0, yet ff is not differentiable at z=0z = 0. Why doesn’t this contradict the theorem about a complex function being complex differentiable if and only if it satisfies the Cauchy-Riemann equations?

We say that a function ff that is complex differentiable in an open disk around z0z_0 \in \mathbb{C} is holomorphic at z0z_0.

Then we talked about smooth paths which are paths that are differentiable. If γ\gamma is a smooth path in \mathbb{C} and ff is holomorphic at every point on the curve, then the chain rule holds and says that ddtf(γ(t))=f(γ(t))γ(t).\frac{d}{dt} f(\gamma(t)) = f'(\gamma(t)) \cdot \gamma'(t).

Idea. Notice that the angle of the tangent vector gets rotated by the argument of f(γ(t))f'(\gamma(t)). If two different smooth curves both intersect at a point z0z_0, and ff is holomorphic at z0z_0, then since their tangent vectors are both rotated by the same angle, the angle between the two tangent vectors stays the same. This implies that holomorphic maps are conformal, that is they preserve angles between curves.

  1. f(z)=z2/5f(z) = z^2 / 5 (see Complex Grapher)

We also looked at the function eze^z on the complex grapher and saw that it was also conformal.


Week 5 Notes

Tentative Schedule

Day Section Topic
Mon, Sep 23 3.4 Exponential and trigonometric functions
Wed, Sep 25 3.5 Logarithms and complex exponentials
Fri, Sep 27 3.1 Möbius transforms

Mon, Sep 23

We started by reviewing the properties of the exponential function including its algebraic properties, domain, range and that it is periodic with period 2πi2\pi i. Then we introduced the complex natural logarithm which is the inverse of eze^z. There is one problem with defining the inverse: eze^z is an \infty-to-1 function, every w{0}w \in \mathbb{C}\backslash \{0\} has infinitely many pre-images. So we have two options:

  1. logz\log z denotes the multivalued inverse of eze^z. It has the form logz=ln|z|+iargz\log z = \ln |z| + i \operatorname{arg}z where argz\operatorname{arg}z is the multivalued argument function.

  2. Logz\operatorname{Log}z is the principal branch of logz\log z. It is a single valued function of the form Logz=ln|z|+iArgz\operatorname{Log}z = \ln |z| + i \operatorname{Arg}z where Argz\operatorname{Arg}z is the single valued principal branch of the argument function that takes values in (π,π](-\pi, \pi].

We did the following exercises.

  1. Calculate log(1+i)\log(1+i).

  2. If z=x+iyz = x+iy where x>0x > 0, then Argz=arctan(y/x)\operatorname{Arg}z = \arctan(y/x). Calculate the Cauchy-Riemann equations for Logz\operatorname{Log}z to verify that Logz\operatorname{Log}z is complex differentiable when x>0x > 0 (in fact it is complex differentiable everywhere except at its branch cut (,0](-\infty, 0]). What is the derivative of Logz\operatorname{Log}z?

Wed, Sep 25

We looked at the Cauchy-Riemann equations for Logz=12ln(x2+y2)+iarctan(yx)\operatorname{Log}z = \tfrac{1}{2} \ln(x^2 + y^2) + i \arctan \left( \tfrac{y}{x} \right) again. We also did the following.

  1. Give an example where Log(z+w)Logz+Logw\operatorname{Log}(z+w) \ne \operatorname{Log}z + \operatorname{Log}w.

We introduced principle values of a complex power zα=eαLogzz^\alpha = e^{\alpha \operatorname{Log}z} for any α,z\alpha, z \in \mathbb{C}, with z0z \ne 0. We calculated the following examples.

  1. i1/3i^{1/3}

  2. (1+i)2i(1+i)^{2-i}

  3. Find all other values of (1+i)2i(1+i)^{2-i} in addition to the principal value.

  4. Find all solutions of the equation e1/z=ie^{1/z} = i.

We finished by talking about the reciprocal map f(z)=1zf(z) = \tfrac{1}{z}. We looked at how it appears to transform lines & circles into lines and circles. We’ll look at why next time.

Fri, Sep 27

We started by proving this theorem about the reciprocal map.

Theorem. Let f(z)=1zf(z) = \tfrac{1}{z}. Then ff transforms lines and circles to lines in circles.

The key to the proof is the fact that the solution of the algebraic equation (with real coefficients) a(x2+y2)+b1x+b2y+c=0a(x^2 + y^2) + b_1 x + b_2 y + c = 0 is a circle (possibly degenerate to a point or \varnothing) if a0a \ne 0, and it is a line if a=0a = 0.

It helps to think of the reciprocal map f(z)=1zf(z) = \tfrac{1}{z} as a bijection (1-to-1 and onto map) from {}{}\mathbb{C}\cup \{ \infty\} \rightarrow \mathbb{C}\cup \{ \infty \}. We define 1=0 and 10=.\frac{1}{\infty} = 0 \text{ and } \frac{1}{0} = \infty. We call {}\mathbb{C}\cup \{ \infty \} the extended complex plane.

The reciprocal map is a special case of an important family of bijections on {}\mathbb{C}\cup \{\infty\} called Möbius transforms. A Möbius transform (also known as a Linear Fractional Transform) is a map f(z)=az+bcz+df(z) = \frac{a z + b}{c z + d} where a,b,c,da, b, c, d \in \mathbb{C} satisfy adbc0ad - bc \ne 0. We proved the following facts.

  1. A Möbius transform always has two fixed points in {}\mathbb{C}\cup \{ \infty \}.

  2. For any invertible matrix A=(a11a12a21a22)A = \begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix}, we can define an associated Möbius transform TA(z)=a11z+a12a21z+a22.T_A(z) = \frac{a_{11} z + a_{12}}{a_{21} z + a_{22}}. Then if A,B2×2A, B \in \mathbb{C}^{2 \times 2} are any two intertible matrices, TATB=TAB.T_A \circ T_B = T_{AB}.

In particular the inverse of a Möbius transform can be found by inverting its matrix: TA1=TA1.T_A^{-1} = T_{A^{-1}}. Notice also that if you multiply a matrix by a constant, the Möbius transform doesn’t change, so TcA=TA.T_{cA} = T_A. That is convenient because the inverse of a 2-by-2 matrix is (abcd)1=1adbc(dbca).\begin{pmatrix} a & b \\ c & d \end{pmatrix}^{-1} = \frac{1}{ad - bc} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}. Therefore you can invert a Möbius transform f(z)=az+bcz+df(z) = \frac{a z + b}{c z + d} by swapping aa and dd and making bb and cc negative: f1(z)=dzbcz+a.f^{-1}(z) = \frac{d z - b}{-cz + a}.


Week 6 Notes

Tentative Schedule

Day Section Topic
Mon, Sep 30 3.2 Möbius transforms - con’d
Wed, Oct 2 Review
Fri, Oct 4 Midterm 1

We started by proving that

Theorem. Any Möbius transformation maps lines & circles to lines & circles.

To prove this, we showed that if c0c \ne 0, then T(z)=az+bcz+dT(z) = \frac{az + b}{cz + d} can be expressed as a composition of three maps T3(z)=(bcdac)z+ac,T2(z)=1z, and T1(z)=cz+d.T_3(z) = \left(\frac{bc-da}{c} \right) z + \frac{a}{c}, ~~~~~ T_2(z) = \frac{1}{z}, \text{ and } T_1(z) = cz + d.

Theorem 2. A Möbius transform is uniquely determined by where it maps any three points in {}.\mathbb{C}\cup \{ \infty \}.

Proof. If z1,z2,z3z_1, z_2, z_3 are any three distinct points in {}\mathbb{C}\cup \{\infty\}, suppose that there are two Möbius transforms TT and SS such that S(zi)=T(zi)S(z_i) = T(z_i) for each ii. Then S1TS^{-1} T has three distinct fixed points which is impossible unless S1=T1S^{-1} = T^{-1} which means that SS and TT are the same. \square

  1. Find a Möbius transform that sends \infty to 2, ii to \infty, and 11 to 11.

  2. Find a general formula for a Möbius transform that sends z1,z2,z3z_1, z_2, z_3 to w1w_1, w2w_2, w3w_3. Hint: It might help to start with a formula to send z1z_1 to 00, z2z_2 to 11 and z3z_3 to \infty.

  3. Find a Möbius transform that leaves +1+1 and 1-1 fixed but maps 0 to \infty. Actually, this wasn’t a great example because it turned out to be the reciprocal map!

  4. How does the Möbius transform from the previous example transform the circles in this figure where the blue circle is centered at ii?

Video: Möbius Transformations Revealed

Wed, Oct 2

We reviewed for the exam by talking about these two problems.

  1. How does the Möbius transform T(z)=ziz1T(z) = \dfrac{z-i}{z-1} transform the three shapes shown below?
1 i
  1. Suppose sns_n is a sequence in \mathbb{C} such that |sn+1sm+1|0.9|snsm||s_{n+1} - s_{m + 1}| \le 0.9 |s_n - s_m| for all m,nm, n \in \mathbb{N}.

    1. Show that sns_n is bounded by finding an upper bound for |sn||s_n|.

    2. Show that sns_n is a Cauchy sequence by finding an upper bound for |snsm||s_n - s_m| when n>mn > m.


Week 7 Notes

Tentative Schedule

Day Section Topic
Mon, Oct 7 4.1 Complex integrals
Wed, Oct 9 4.3 Cauchy’s theorem
Fri, Oct 11 7.3 & 7.4 Power series

Mon, Oct 7

We introduced complex integrals, which are defined for any piecewise smooth path γ:[a,b]\gamma: [a, b] \rightarrow \mathbb{C} by γf(z)dz=abf(γ(t))γ(t)dt.\int_\gamma f(z) \, dz = \int_a^b f(\gamma(t)) \, \gamma'(t) \, dt. We talked about why this definition makes sense and we did these examples.

  1. γz2dz\int_\gamma z^2 \, dz on the upper half of the unit circle from 11 to 1-1. We started by using Python to numerically approximate the integral with a Riemann sum:
from cmath import *

n = 1000
total = 0
for k in range(1000):
    delta_z = exp(1j * pi * (k + 1) / n) - exp(1j * pi * k / n)
    z = exp(1j * pi * k / n)
    total += z ** 2 * delta_z
    
print(total)
  1. γ(z¯)2dz\int_\gamma (\overline{z})^2 \, dz on the path γ(t)=t(1+i)\gamma(t) = t(1+i) with t[0,1]t \in [0,1].

  2. δ(z¯)2dz\int_\delta (\overline{z})^2 \, dz on the path δ(t)=t+it2\delta(t) = t+it^2 with t[0,1]t \in [0,1]. (Link: Sympy code)

from sympy import *
x, y, z = symbols('x y z')
t = symbols('t',real=True)

f = conjugate(z)**2
g = t+t**2*I

print(integrate(f.subs(z,g)*diff(g,t),(t,0,1)))
  1. Use the formula length(γ)=ab|γ(t)|dt\operatorname{length}(\gamma) = \int_a^b |\gamma'(t)| \, dt to find the length of the unit circle.

Wed, Oct 7

Today we introduced

Cauchy’s Theorem. If ff is holomorphic on a simply connected open set DD and CC is a simple, closed, piecewise, smooth curve in DD, then Cf(z)dz=0.\oint_C f(z) \, dz = 0.

We defined what simply connected means (intuitively it means that DD has no holes), and we reviewed simple closed curves. We also talked about why the specific parametrization of a curve does not matter for integrals (see Proposition 4.2), but the orientation does, and the symbol \oint indicates that the orientation of a simple closed curve is positive (counterclockwise). We calculated

  1. |z|=11zdz\oint_{|z| = 1} \frac{1}{z} \, dz

and saw that Cauchy’s theorem does not apply because the reciprocal function is not holomorphic at 0.

Then we talked about how to prove Cauchy’s theorem using

Green’s Theorem. If P(x,y)P(x,y) and Q(x,y)Q(x,y) are real-valued functions with continuous partial derivatives in a simply connected domain DD with a piecewise smooth boundary curve CC, then CPdx+Qdy=D(QxPy).\oint_C P \, dx + Q \, dy = \iint_D \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right).

Here is a nice video explanation of the intuition behind Green’s theorem. One problem with using Green’s theorem to prove Cauchy’s theorem is that it requires us to assume that f(z)f'(z) is continuous (so that the partial derivatives are all continuous. It turns out that is true for all holomorphic functions, as we will see later. The Complex Variables book has a different proof of Cauchy’s theorem that doesn’t require this assumption.

We finished by starting to talk about how useful Cauchy’s theorem is. One application of Cauchy’s theorem is that it implies that every holomorphic function has a holomorphic antiderivative. You can define the antidervative F(z)F(z) of a holomorphic function ff on a simply connected open domain this way: F(z)=γf(z)dzF(z) = \int_\gamma f(z) \, dz where γ\gamma is any piecewise smooth path from a fixed z0Dz_0 \in D to zz. We talked about why F(z)F(z) is well-defined.

Friday, Oct 11

Last time we showed that when a function f(z)f(z) is holomorphic on a simply connected open set DD, then the integral on any two paths γ1\gamma_1 and γ2\gamma_2 that both start at a point z0z_0 and end at z1z_1 and stay inside DD must be the same:

γ1f(z)dz=γ2f(z)dz.\int_{\gamma_1} f(z) \, dz = \int_{\gamma_2} f(z) \, dz.

In other words, the value of the integral is independent of the path. Today we proved some important theorems to make working with complex integeral easier.

Evaluation Theorem. If ff has a holomorphic antiderivative FF in a path connected open domain DD, then for any piecewise smooth path γ:[a,b]D\gamma: [a,b] \rightarrow D, γf(z)dz=F(γ(b))F(γ(a)).\int_\gamma f(z) \, dz = F(\gamma(b)) - F(\gamma(a)).

This theorem follows immediately from the definition of a complex path integral and the chain rule, combined with the evaluation theorem from real variable calculus. An application of this theorem is the following.

  1. Calculate γzndz\int_\gamma z^n \, dz for any piecewise smooth path γ\gamma that begins at zz \in \mathbb{C} and ends at ww \in \mathbb{C}. Note: this solution works as long as n1n \ne -1 with the caveat that the path shouldn’t pass through 0 if nn is negative. Why won’t this work if n=1n = -1?

ML-Inequality. If ff is a continuous complex-valued function on a path connected open domain DD and γ\gamma is a piecewise smooth path in DD, then |γf(z)dz|maxzrange(γ)|f(z)|length(γ).\left| \int_\gamma f(z) \, dz \right| \le \max_{z \in \operatorname{range}(\gamma)} |f(z)| \cdot \operatorname{length}(\gamma).

This is like a triangle inequality for complex integrals. In fact, you can use the triangle inequality to prove it.

Antiderivative Theorem. If ff is holomorphic in a simply connected open domain DD, then ff has a holomorphic antiderivative FF in DD.

Last time we used independence of path to define F(z)=γf(z)dzF(z) = \int_\gamma f(z) \, dz where γ\gamma is any path in DD from z0z_0 to zz. We just have to show that F(w)=f(w)F'(w) = f(w) for every wDw \in D to prove that this function FF is the antiderivative needed for the antiderivative theorem. Two key ideas in the proof are the following:

  1. Let δ\delta be a parametrization of the line segment from ww to w+hw+h in DD. Show that δf(w)dz=hf(w)\int_\delta f(w) \, dz = h f(w).

  2. Use the ML-inequality to estimate |δf(z)f(w)dzh|.\left| \frac{\int_\delta f(z) - f(w) \, dz}{h} \right|.


Week 8 Notes

Tentative Schedule

Day Section Topic
Mon, Oct 14 Fall break, no class
Wed, Oct 16 4.4 Cauchy’s integral formula
Fri, Oct 18 5.1 Holomorphic implies infinitely differentiable

Wed, Oct 16

Cauchy Integral Formula. Let ff be holomorphic in a simply connected open domain DD \subseteq \mathbb{C} that contains a simple, closed, piecewise smooth curve CC around a point ww. Then f(w)=12πiCf(z)zwdz.f(w) = \frac{1}{2 \pi i} \, \oint_C \frac{f(z)}{z-w} \, dz.

That’s kind of weird if you think about it. This theorem says that the value of ff at a point inside a curve is completely determined by the values of ff on the curve. The proof required two key insights:

  1. If ff is complex differentiable at ww, then f(z)=f(w)+f(w)(zw)+ϵ(z),f(z) = f(w) + f'(w) (z-w) + \epsilon(z), where the error term ϵ(z)\epsilon(z) satisfies limzwϵ(z)zw=0\lim_{z \rightarrow w} \frac{\epsilon(z)}{z-w} = 0.

  2. If C1C_1 and C2C_2 are two different positively oriented piecewise smooth closed paths in a region where ff is holomorphic, then C1f(z)zwdz=C2f(z)zwdz.\int_{C_1} \frac{f(z)}{z-w} \, dz = \int_{C_2} \frac{f(z)}{z-w} \, dz.

Use the Cauchy integral formula to evaluate the following integrals.

  1. Cezz1dz\int_C \frac{e^z}{z-1} \, dz where CC is the square with vertices at 10,10i,10,10i10, 10i, -10, -10i. (https://youtu.be/NJap6Vm5mEk)

  2. |zi|=11z2+1dz\oint_{|z-i|=1} \frac{1}{z^2+1} \, dz

  3. |z|=3ezz22zdz\oint_{|z|=3} \frac{e^z}{z^2 - 2z} \, dz


Week 9 Notes

Tentative Schedule

Day Section Topic
Mon, Oct 21 8.1 Power series and holomorphic functions
Wed, Oct 23 5.3 Entire functions & Liouville’s theorem
Fri, Oct 25 9.1 Classification of singularities

Week 10 Notes

Tentative Schedule

Day Section Topic
Mon, Oct 28 8.3 Laurent series
Wed, Oct 30 9.2 Residues
Fri, Nov 1 9.3 The argument principle and Rouche’s theorem

Week 11 Notes

Tentative Schedule

Day Section Topic
Mon, Nov 4 8.2 Zeroes of analytic functions
Wed, Nov 6 8.2 The maximum modulus principle
Fri, Nov 8 Conformal mappings

Week 12 Notes

Tentative Schedule

Day Section Topic
Mon, Nov 11 Riemann mapping theorem
Wed, Nov 13 Review
Fri, Nov 15 Midterm 2

Week 13 Notes

Tentative Schedule

Day Section Topic
Mon, Nov 18 10.1 Applications of the residue theorem
Wed, Nov 20 10.2 Binomial coefficients
Fri, Nov 22 6.1 Harmonic functions

Week 14 Notes

Tentative Schedule

Day Section Topic
Mon, Nov 25 Conformal mappings & heat flows
Wed, Nov 27 Thanksgiving break, no class
Fri, Nov 29 Thanksgiving break, no class

Week 15 Notes

Tentative Schedule

Day Section Topic
Mon, Dec 2 Fourier transform
Wed, Dec 4 The heat equation
Fri, Dec 6 The Dirichlet problem
Mon, Dec 9 Recap & review