MATH3401 — Complex Analysis

Lecture 1 — The Complex Field

Joseph Grotowski: [email protected]. 06-206 (Tues 10-12, Thurs 11-12, or by appointment).

Introduction

This course is compulsory for many students. Its prerequisites are MATH2000 and MATH2400 (or equivalents). This is the capstone course for math majors and math undergraduate degrees.

The techniques and problem solving strategies in this course will be beneficial in many ways.

Lecture recordings will be on the Blackboard. Tutorials start in week 1.

Assessment

The best 5 of 6 assignments together count for 20%. The midsemester counts for 20% (one page of handwritten notes is allowed, single-sided). The final exam counts for 60% (one page of handwritten notes, double-sided).

Complex analysis

Cool stuff

A ‘nice’ result which shows all different parts of maths coming together: $e^{i\pi} = -1.$

There are a number of fascinating things about this equation. Despite $i$ not appearing in this expression, it still delves into complex analysis. $\int_{0}^\infty \frac {\sin x} x \,dx = \frac \pi 2$ A reasonable question is does this integral even converge? If we replace $\sin x$ with $1$, the integral diverges by the $p$-test. Arguing the integral exists is a bad time without complex analysis, but is really really nice with complex. This will be done towards the end of the semester, making use of contour integrals around a path in the complex plane.

In the more applied realm, we can also do things with fluid flow. A very expensive method would be constructing a physical model then running experiments. With complex analysis, we can perform analysis on a straight pipe, then map to the pipe above without having to build the channel. We can just tweak the parameters in the map to test different scenarios. This is called a conformal transoformation.

Similarly, Joukowski transformations can be used to model air flow around a wing.

We can also get nice results about series like $\begin{aligned} \frac 1 {1^2} + \frac 1{2^2} + \frac1 {3^2} + \cdots &= \frac {\pi^2} 6 \\ \frac 1 {1^2} - \frac 1{2^2} + \frac1 {3^2} - \frac 1 {4^2} +\cdots &= \frac {\pi^2} {12} \\ \sum_{k=1}^\infty \frac 1 {1 + 4 k^2 \pi^2}& = \frac 1 2 \left(\frac 1 {e-1} - \frac 1 2\right) \end{aligned}$

Riemann Zeta

$\zeta (s) = \sum_{n=1}^\infty \frac 1 {n^s} = \prod_{p\ \text{prime}}(1-p^{-s})^{-1}$ (The product of primes result is from Euler. This is called the Riemann zeta function)

Riemann hypothesis: $\zeta$ has infinitely many non-trivial zeros and they all lie on the line $\text{Re}(s) = 1/2$.

Note that the expression for $\zeta$ only makes sense for $\text{Re} s) > 1$, so we need to extend it to $\mathbb C$ via analytic continuation. In doing this, the trivial zeros are $-2, -4, -6, \dots$

Lecture 2 — Complex Numbers

$\renewcommand\Re{\operatorname{Re}} \renewcommand\Im{\operatorname{Im}}$

Complex numbers have been around for a while.

• In 1545, Cardano looked at real roots of $x^3 + ax + b$. He found $5 + \sqrt{-15}$ and called it “mental torture”.
• In 1575, the algebraic rules for $i$ were first described.
• In 1629, $a+\sqrt{-b}$ was called “solutions impossibles” by Girard.
• In 1637, Descartes discovered imaginary numbers.
• From 1831 onwards, complex numbers.

$\mathbb C$ as a field

B.C section 1-3

$\begin{aligned} \mathbb N &= \{1, 2, 3, \ldots \} \\ \mathbb N_0 &= \{0, 1, 2, 3, \ldots \} \\ \mathbb Z &= \{0, \pm1, \pm2, \pm3, \ldots \} \\ \mathbb Q &= \{p/q : p, q \in \mathbb Z, q \ne 0 \} \\ \mathbb R &= \text{real numbers} \\ \mathbb C &= \text{complex numbers} \end{aligned}$

Note that $\mathbb Q$ is actually equivalence classes of “quotients” of integers because certain expressions are equivalent (see MATH2401). $\mathbb R$ can be defined in several technical ways, such as Dedekind cuts or limits of sequences.

$\mathbb C$ can be represented in various (equivalent) ways:

• $z = (x, y) \in \mathbb C$, for $x, y \in \mathbb R$. This is the “complex plane”. Alternatively, $z = x(1, 0) + y(0, 1)$.
• $z = x + ij$, where $x = \operatorname{Re}(z)$ and $y = \operatorname{Im}(z)$.

$i$ is the complex number represented by $(0,1)$. We say $\mathbb R \subset \mathbb C$ by identifying the complex number $x + 0i$ with the real number $x$.

Addition in $\mathbb C$

$\begin{aligned} (x_1, y_1) + (x_2, y_2) &= (x_1 + x_2, y_1 + y_2) \\ (x_1 + iy_1) + (x_2 + iy_2) &= (x_1 + x_2) + i(y_1 + y_2) \end{aligned}$

Multiplication in $\mathbb C$

Denoted by $\times$ or $\cdot$ or juxtaposition (that is, putting things next to each other). $\begin{aligned} (x_1, y_1)\cdot(x_2, y_2) &= (x_1x_2 - y_1y_2, y_1x_2 + x_1y_2) \\ (x_1 + iy_1) \cdot (x_2 + iy_2) &= (x_1 x_2 - y_1y_2) + i(y_1x_2 + x_1y_2) \end{aligned}$ The definition of multiplication formally applies if we use the usual rules for algebra in $\mathbb R$ and set $i^2 = -1$.

Note: Multiplication of two complex numbers sums their angles (where positive is CCW) and multiples their radius.

$\mathbb C$ is a field

With this addition and multiplication, $\mathbb C$ is a field. Check: $\mathbb C$ must be closed under the binary operations $+$ and $\cdot$.

F2: $+$ has identity $0+0i$ and inverse $(-x) + i(-y)$. F5: $\cdot$ has identity $1 + 0i$ and inverse $z^{-1} = 1/(x+iy) \cdot (x-iy)/(x-iy) = \frac x{x^2+y^2} - i\frac{y}{x^2+y^2}$

Since $\mathbb C$ is a field, it holds: $z_1, z_2 = 0 \implies z_1 = 0 \text{ or }z_2 = 0$. This is the null-factor law and holds because on all fields. Also, we have $(z_1z_2)^{-1} = z_1^{-1}z_2^{-1}$.

Note: $i^2 = -1$ and $(-i)^2 = -1$. These are the only two solutions of $z^2 = -1$ in the complex numbers (we cannot check this yet). This is due to the Fundamental Theorem of Algebra.

Remark: $\mathbb C$ is not ordered and, in fact, cannot be ordered. Thus, $i$ is no more special then $-i$.

B.C. 4, 5

Given $z = x + iy \in \mathbb C$, there are a few useful functions to have: - modulus: $|\cdot| : \mathbb C \to \mathbb [0, \infty)$, where $|z| = \sqrt{x^2 + y^2}$, - real part: $\operatorname{Re}(z) = x$, imaginary part: $\operatorname{Im}(z) = y$ (both $\mathbb C \to \mathbb R$),

Lecture 3 — Functions of Complex Numbers

Complex conjugate

The complex conjugate is defined as a function $\bar \cdot : \Complex \to \Complex$, where $(x + iy) \mapsto (x - iy)$. Geometrically, this reflects a complex number about the real axis.

Properties

$\begin{aligned} z = \bar z \iff \operatorname{Im}(z)&= 0 \text{ (i.e. z} \in \mathbb R \text{)} \\ \overline {(\bar z)} &= z \\ \overline {zw} &= \bar z \bar w \\ \overline{z+w} &= \bar z + \bar w \\ \overline {z^{-1}} &= (\bar z)^{-1}, z \ne 0 \\ |z|^2 &= z \bar z \\ \operatorname{Re}(z) &= \frac{z + \bar z} 2 \\ \operatorname{Im}(z) &= \frac{z - \bar z} 2 \end{aligned}$

A very useful property (from MATH1051) is the triangle inequality: $|z+w| \le |z| + |w|.$ Proof: More specifically using the cosine rule, $|z+w|^2 = |z|^2 + |w|^2 - 2|z||w|\cos A.$ This is a true and exact statement. However, in analysis, we often want to make these statements less precise but more useful. Because $-1 \le \cos \le 1$, $\begin{aligned} |z+w|^2 &\le |z|^2 + |w|^2 + 2|z||w| \\ &= (|z| + |w|)^2\\ \implies |z+w| &\le |z| + |w| \end{aligned}$

Polar coordinates

B.C. 6-9

Given a complex number $z = x+iy$, we can find $r$ and $\theta$ such that $x = r \cos \theta, \text{ and }\ y = r \sin \theta.$ Then, we can also write it using Euler’s formula (as a formal convention for the moment): $z = re^{i\theta} = r(\cos \theta + i \sin \theta).$ Remark: this formula follows formally from the Taylor series of $e^{i\theta}$.

Here, $\theta$ is an (as opposed to the) argument of the complex number $z$. We write $\theta = \arg z$. Here, $\arg$ is not a (single-valued) function. Given a $\theta$, we can always take $\theta + 2\pi$ which will satisfy the $x$ and $y$ equations. Also, for $z=0$, any $\theta$ will work.

To make $\arg$ a function, we need to restrict its range. There are two options: $0$ to $2 \pi$ and $-\pi$ to $\pi$. In complex analysis, we normally use the second. Specifically, $\operatorname{Arg} z$ is defined to be the unique values of $\theta$ such that $-\pi < \arg z \le \pi$.

Examples: - $\operatorname{Arg}(1+i) = \pi / 4$ but $\arg (1+i) = \ldots, -7\pi/4, \pi/4, 9\pi/4, \ldots$. - $\operatorname{Arg}(-1) = \pi$. - $\operatorname{Arg}(0)$ is undefined, but $\arg (0) = \mathbb R$.

In summary, $\operatorname{Arg}$ is a function $\mathbb C \setminus \{0\} \to (-\pi, \pi]$. Alternative notation for $\mathbb C \setminus \{0\}$ is $\mathbb C^*$ or $\mathbb C_*$.

Note: $\begin{aligned} |e^{i\theta}| &= 1 \text{ (easy to check)}\\ (e^{i\theta})^{-1} = e^{-i\theta} &= \overline {e^{i\theta}} \\ (re^{i\theta})(\rho e^{i\phi}) &= (r\rho) e^{i(\theta + \phi)} \\ \implies |zw| &= |z| |w|,\enspace \arg (zw) + \arg z \arg w \end{aligned}$ However, the last equality does not necessarily hold for $\operatorname{Arg}$. For example, with $z = w = {(-1 + i)/\sqrt 2}$

De Moivre’s formula

$z = re^{i\theta} \implies z^n = r^n e^{in\theta}, \quad n \in \mathbb Z.$ In particular, $e^{in\theta} = (\cos \theta + i \sin \theta)^n = \cos(n\theta) + i \sin (n\theta).$

Lecture 4 — Functions as Mappings

Roots of a complex number

What is the number $z$ such that $z^n$ gives us the original number? By the fundamental theorem of algebra, we know there ar exactly $n$ $n$-th roots in $\mathbb C$.

Consider the $n$-th roots of $z = re^{i\theta}$, for $z \in \mathbb C_*$. That is, we want all $w \in \mathbb C$ such that $w^n = z$. Notation: $\exp (\xi) = e^\xi$.

Then, we can use de Moivre’s theorem “in reverse” to see that $z$ has $n$ distinct roots: $\left\{ r^{1/n}\exp\left(\frac{i\theta}n\right), r^{1/n}\exp\left(\frac{i\theta}n + \frac{i2\pi}n\right), \ldots, r^{1/n}\exp\left(\frac{i\theta}n + \frac{i2\pi(n-1)}n\right) \right\}$

B.C. 13 (8th ed 12-13)

Functions & mappings

Suppose we have $\Omega \subseteq \mathbb C$ and a function $f : \Omega \to \mathbb C$ can be viewed as a mapping on $\Omega$, the domain of $f$. If $\Omega$ is not specified, then we take $\Omega$ to be as large as possible.

Example: For $f(z) = 1/z$ we can take $\Omega = \mathbb C \setminus\{0\}$, so $f : \mathbb C_* \to \mathbb C$. As notation, we can also write $f : z \mapsto 1/x$, or $w=1/z$, or just $1/z$ if the meaning is clear.

The usual notation is $w : (x,y) \mapsto (u,v)$, i.e. $w(x+iy)=u(x+iy)+iv(x+iy)$ or $w(x,y)=u(x,y)+iv(x,y)$.

This notation is not completely rigorous; $u$ is both a function from $\mathbb C$ and from $\mathbb R^2$. We could introduce a map $\varphi : (x,y)\mapsto (x+iy)$ but this is excessively verbose. There is no real problem with this, but be aware.

Definitions

• The domain of $f$ is $\Omega$, also denoted $\operatorname{dom} f$.
• The range of $f$ is $f(\Omega) = \operatorname{Range}f=\{w : w = f(z) \text{ for some }z \in \Omega\}$.
• The inverse of $f$ is $f^{-1}(\xi) = \{z \in \Omega : f(z) = \xi\}$. Note that this returns sets. If there is only one element, we can just call that element the “inverse”.

Examples: - Consider $f(z) = 1/z$. $\operatorname{dom}f = \mathbb C_\star$. $f^{-1}(\xi)=1/z$ is a function $\mathbb C_* \to \mathbb C_*$. - For $g(z) = 1/(1-|z|^2)$. $\operatorname{dom}g = \mathbb C \setminus \{z : |z| = 1\}$. The function is $g : \{z : z \ne 1\} \to \mathbb R$. The inverse is not a function. - For $h(z) = z^n$ where $h : \mathbb C \to \mathbb C$, the inverse is also not a function.

Geometric intuition

Let’s aim to get a geometric picture of what a given $f$ does.

Examples: - $w = 1+z$ moves each point one unit to the right (in the positive real direction). - $re^{i\theta} \mapsto re^{i(\theta+\pi/2)}$ rotates points through an angle of $\pi/2$ in the counter-clockwise direction about the origin.

For new and unfamiliar mappings, break them down into compositions of known or easy maps.

Examples: - $w = Az + b$ where $A, b \in \mathbb C$ and $A \ne 0$. We can think of $A$ as a dilation and rotation, then $+b$ as a translation. - For $z \mapsto Az$, write $A=a e^{i\alpha}$ for $\alpha, a \in \mathbb R$. This gives us $re^{i\theta}\mapsto ar e^{i(\theta+\alpha)}$. Specifically, it dilates the modulus by a factor of $a=|A|$ and rotates through $\alpha = \arg A$. - For $z \mapsto z+b$ where $b = b_1+b_2i$, $b_1, b_2 \in \mathbb R$. This translates $b_1$ to the right and $b_2$ up. If negative, goes in the opposite direction.

Note: The maps above have domain and image $\mathbb C$.

Lecture 5 — Mappings 2

Another very important map to look at is $z \mapsto 1/z$ on $\mathbb C_*$. We can write this as the composition of two slightly more complicated functions.

Define $\xi(z) = z/|z|^2$ on $\mathbb C_*$ and $\eta(z) = \bar z$. For $z \in \mathbb C_*$, we can compose these two as $\begin{aligned} \eta \circ \xi(z) = \eta(\xi(z)) = \overline {\left(\frac z {|z|^2}\right)} = \frac{\bar z}{|z|^2} = \frac {\bar z}{z \bar z} = \frac 1 z. \end{aligned}$

$\xi$ is called inversion, with respect to the unit circle. $\eta$ is just reflection about the real axis.

For $w = 1/z = \bar z / |z|^2$ we can write it as $x+iy \mapsto u+iv$, where $w=\frac{x-iy}{x^2 + y^2} \quad\implies\quad u = \frac{x}{x^2+y^2}, \quad v=\frac{-y}{x^2+y^2}.$ We can use this to show the following statement: $1/z$ maps circles and lines in the $z$=plane to circles and lines in the $w$-plane. Note that this does not require circles to map to circles, or lines to map to lines.

The key point is both circles and lines in the $z$-plane can be represented as $A(x^2+y^2)+Bx+Cy+D=0,\quad\text{ where } B^2+C^2 > 4AD$ for $A,B,C,D \in \mathbb R$. If $A=0$, then the equation is a circle. The inequality constraint tells us that $\begin{aligned} \left(x+\frac B {2A}\right)^2 + \left(y+\frac C {2A}\right)^2 = \left(\frac{\sqrt{B^2+C^2-4AD}}{2A}\right)^2. \end{aligned}$ Note, for $w = 1/z$, the $u$ and $v$ expressions earlier tell us that $w$ has the form $D(u^2+v^2)+Bu-Cv+A=0$ which is a circle or line.

Terminology

• $f : \Omega \to \mathbb C$ is 1-to-1 or injective if $f(z) = f(\xi)$ implies $z = \xi$.
• $f : \Omega \to \Lambda$, where $\Lambda \subseteq \mathbb C$, is onto or surjective if for all $\eta \in \Lambda$, there exists (one or more) $z \in \Omega$ such that $f(z) = \eta$.

Examples: Affine transformations are bijections $\mathbb C \to\mathbb C$ , and $1/z$ is a binection $\mathbb C_* \to \mathbb C_*$.

Möbius transformations

B.C. 99 (8th ed 93)

Let $a, b, c, d \in \mathbb C$ where $ad-bc \ne 0$. Then, $w = T(z) = \frac{az+b}{cz+d}$ is called a Möbius (or linear fractional) transformation. The natural domain of definition is - if $c = 0$, then $\operatorname{dom}w = \mathbb C$ (because $c = 0\implies d \ne 0$), or - if $c \ne 0$, then $\operatorname{dom}w = \mathbb C \setminus \{-d/c\}$.

Let’s try to understand $T$ geometrically.

Claim: $T$ is injective and surjective from $\mathbb C \to \mathbb C$. Proof. For $c = 0$, then to prove injectiveness suppose $T(z) = T(\xi)$. We want to show $z = \xi$. Substituting into the formula for $T$, $\frac a d z + \frac b d = \frac a d \xi + \frac b d \implies z = \xi.$ To prove it is surjective, given $w \in \mathbb C$, we need $z \in \mathbb C$ such that $T(z) = w$. The value $z = d/a(w-b/d)$ satisfies this.

For $c \ne 0$, consider $\begin{aligned} w &= \frac{az+b}{cz+d} = \frac{a(z+d/c) - ad/c + b}{c(z+d/c)}\\ &= \frac a c + \left(\frac{bc-ad}c\right)\frac 1 {cz-d} \end{aligned}$ This is a composition of a linear transformation, $1/z$ and another linear transformation.

Thus, $T$ is the composition of linear and $1/z$ maps. That is, $Z_1 = cz+d, \quad W = 1/Z_1, \quad w = \frac a z + \frac{bc-ad}cW.$ In both cases, Möbius transformations are compositions of maps previously studied. This means they are bijective.

Lecture 6 — Möbius Transformations 2 & The Extended Complex Plane

Recall that $T(z) = w = \frac{az+b}{cz+d}$ ($ad-bc \ne 0$) is a Möbius transformation.

It can be rewritten as $Azw + Bz + Cw + D = 0$ where $A = c$, $B = -a$, $C=d$, $D=b$. This is called the implicit form.

Recall that case 1 was $c = 0$, which reduces $T$ to a linear transformation which is a bijection $\mathbb C \to \mathbb C$. Case 2 was also a bijection from $\mathbb C \setminus \{-d/c\} \to \mathbb C \setminus \{a/c\}$, with inverse $T^{-1}(w) = \frac{-dw+b}{cw-a}.$

A question might be can we extend $T$ to a function $\mathbb C \to \mathbb C$ in case 2? In particular, such that the extension is injective and surjective. The answer is yes, by “plugging the hole”. We simply define $T(-d/c) = a/c.$ However, this is unsatisfying because the function becomes discontinuous.

An important concept: We are going to extend $\mathbb C$ to the extended complex plane, written $\bar{ \mathbb C}$. This is done by adding a point at infinity, which is called $\infty$. We can think of the complex plane as a sphere with the origin at one pole and this $\infty$ at the other, with distances expanding as you go further from $0$.

We then define $T(-d/c) = \infty$ and $T(\infty) = a/c$. This extends $T$ to a map $\bar {\mathbb C }\to \bar {\mathbb C}$ which is injective and surjective.

Remark: $\bar {\mathbb C}$ is a topological space and the above extension is continuous. A topology on a set is a space with so-called “open sets”. Intuitively, points can be ‘nearby’ to other points.

$\bar {\mathbb C}$ can be visualised as the Riemann sphere. The origin $0+0i$ is at the south pole. A point on the complex plane is mapped uniquely to a point on the sphere. This is done by picking the point on the sphere’s surface on the line between the point and the north pole. “Infinity” can be thought of as the north pole.

A few final remarks on Möbius transformations. Given 3 distinct points in $z_1, z_2, z_3\in\bar{ \mathbb C}$ and 3 different distinct points $w_1, w_2, w_3 \in \bar {\mathbb C}$, there exists a unique Möbius transformation $T$ such that $T(z_1) = w_1, \ T(z_2)=w_2, \text{ and }T(z_3)=w_3.$ In fact, $T$ is given by $\frac{(w-w_1)(w_2-w_3)}{(w-w_3)(w_2-w_1)} = \frac{(z-z_1)(z_2-z_3)}{(z-z_3)(z_2-z_1)}.$ In practice, it may be easier to directly solve for $a,b,c,d$ than using the above expression.

Note: How does this work with infinity? $\begin{aligned} T(\infty) = a/c &\iff \lim_{|z|\to\infty} T(z) = a/c\\ T(-d/c) = \infty &\iff \lim_{z\to -d/c} 1/T(z) = 0 \end{aligned}$

Lecture 7 — Exponential Maps

A note on coronavirus about the recent mail from Joanne Wright, the DVC(A).

Recall the Möbius transformation, and note that is is unique up to scaling for $\lambda > 0$. $w = \frac{az+b}{cz+d} = \frac{\lambda az+\lambda b}{\lambda cz+\lambda d}$

Remark: Any map from the inside of a (upper half) half-plane to the inside of a circle has the form $w = e^{-i\alpha} \frac{z-z_0}{z-z_0}\quad \text{ for some }\alpha \in \mathbb R, z_0 \in \mathbb C, \operatorname{Im} z_0 > 0.$

Exponential map

B.C. 103 (8Ed 104)

$z \mapsto e^z = \exp x = w, \quad \operatorname{dom} w = \mathbb C.$ Given a $z = x+iy$ for $x, y \in \mathbb R$, $w = e^z = e^{x+iy} = e^x e^{iy} = e^x (\cos y + i \sin y) = u+iv\\[0.7em] \begin{aligned} \text{ where }\quad u &= e^x \cos y\\ v &= e^x \sin y. \end{aligned}$ This is easier to see by writing $w = \rho e^{i\phi}$ where $\rho = e^x$, $\phi = y + 2k\pi$ for $k \in \mathbb Z$. This function is periodic in $\mathbb C$.

Images under exp

Properties

Many of the properties of the real $\exp$ extend to $\mathbb C$. Such as - $e^0 = 1$. - $e^{-z} = 1/e^z$. - $e^{z_1+z_2} = e^{z_1}e^{z_2}$. - $e^{z_1-z_2} = e^{z_1}/e^{z_2}$. - $(e^{z_1})^{z_2} = e^{z_1z_2}$.

However, some things do not extend: - $e^x > 0~\forall x \in \mathbb R$ but, for example, $e^{i\phi} = -1$. - $x \mapsto e^x$ is monotone increasing for $x \in \mathbb R$ but $z \mapsto e^z$ is periodic with period $2\pi i$.

Note: As in $\mathbb R$, $e^z = 0$ has no solution in $\mathbb C$. If there was some $z = x+iy$ such that $e^z = 0$, then $e^x e^{iy} = 0 \implies e^x = 0$ because $|e^{iy}| = 1$, contradiction.

Inverses

B.C. 31-33 (8Ed 30-32)

We have a function $f : \Omega \to \mathbb C$. Then, $g : \operatorname{Range}f \to \Omega$ is an inverse of $f$ if $g \circ f : \Omega \to \Omega$ is the identity. That is, $(g \circ f)(z) = z$ for all $z \in \Omega$.

Example: $z \mapsto z+1$ and $z \mapsto z-1$ are inverses for $\mathbb C \to \mathbb C$. $z \mapsto 1/z$ is its own inverse $\mathbb C_* \to \mathbb C_*$.

Lecture 8 — Logarithm

Inverse of the exponential

The inverse of the exponential! It’s probably too much to hope for $\log = \log_e$ to be the inverse, because $\exp$ is periodic (with period $2\pi i$) in $\mathbb C$.

Begin with $e^w = z$. Write $z = re^{i\Theta}, r > 0$, where $\Theta = \operatorname{Arg} z \in (-\pi, \pi]$.

We can make our calculations clearer by using polar coordinates in the domain and rectangular coordinates in the range. That is, $w = u+iv \implies z=e^w = e^{u+iv}=e^ue^{iv}\\ \implies e^u = r,\quad v=\Theta + 2k\pi, \quad k \in \mathbb Z.$ So $u = \ln r$, which (notation in this course) means logarithm with base $e$ of the positive real number $r$. Thus, $\begin{aligned} w &= u+iv \\ &= \ln r + i(\Theta + 2k\pi) \quad k \in \mathbb Z \\ &= \ln |z| + i \arg z \end{aligned}$ This defines the multi-valued function $\log : \mathbb C_* \to \mathbb C_*$. $\begin{aligned} \exp (\log z) &= z\\ \log(\exp z) &= z + 2k\pi i \end{aligned}$ We can check the the properties of $\log$ translate into $\mathbb C$. For example, (note that this is a statement of multi-valued functions) - $\log (z\xi) = \log z + \log \xi$. - $\log (z / \xi) = \log z - \log \xi$.

As with $\operatorname{Arg}$ and $\arg$, we can define the principal logarithm, denoted $\operatorname{Log} : \mathbb C_* \to \mathbb C_*$, as $\operatorname{Log} z = \ln |z| + i \operatorname{Arg} z$ This function is single-valued but has the disadvantage of being discontinuous on the negative real axis and $0$, since $\operatorname{Arg}$ is discontinuous there. Indeed, $\operatorname{Log}$ and $\operatorname{Arg}$ are not even defined at $0$.

As with $\operatorname{Arg}$, it may be the case that $\operatorname{Log}(z_1 z_2) \ne \operatorname{Log} z_1 + \operatorname{Log} z_2$.

Complex exponents

Remark: In reals, we could define dsomething like $2^{\sqrt 2}$ as $\lim_{n\to\infty}2^{a_n}$ where $\{a_n\}\to\sqrt 2$. This doesn’t quite work in complex.

Set $z^c = \exp(c \log z)$. Because $\log$ is multi-valued, this may result in a sequence of outputs. For $c \in \mathbb N$ and $1/c \in \mathbb Z$, we recover the formulas from the fourth lecture.

Remark: B.C. defines $z^{1/n}$ as a multi-valued function and defines the principal value as $\operatorname{PV}(z^{1/n}) = |z|^{1/n}\exp (i\operatorname{Arg} z / n).$ Similarly for $z \mapsto z^c$, $\operatorname{PV}(z^{c}) = \exp(c \operatorname{Log} z) = \exp (c \ln |z| + ic \operatorname{Arg}a).$

Example: As a concrete example, doable but easy to make mistakes, $\begin{aligned} \operatorname{PV}[(1-i)^{4i}] &= \exp(4i (\ln |1-i| + i\operatorname{Arg}(1-i)))\\ &= \exp (4i \ln \sqrt 2 -4(-\pi/4))\\ &= e^{\pi}\exp(4i\ln \sqrt 2) \\ &= e^\pi (\cos(2\ln 2) + i\sin (2\ln 2)) \end{aligned}$

Sometimes, we need to use a different single-valued $\operatorname{Log}$ or $\operatorname{Arg}$. For example, if we need to integrate around a contour excluding the $-i$ axis. In this case, we would define $\operatorname{\mathcal {Arg}} z$ such that $-\pi/2 < \arg z \le 3\pi/2$. This leads to an alternative single-valued $\mathcal {Log}$ and derived functions.

Next: square roots, branch cuts.

Lecture 9 — Branch Cuts & Trigonometric Functions

B-C 108.

A branch is a half-open interval of the form $\alpha \le \theta < \alpha + 2\pi$ or $\alpha < \tilde \theta \le \alpha + 2\pi$ of $\mathbb R$.

This is good because we can define a single-valued $\operatorname{Arg}$ with values in this interval, a single-valued $\operatorname{Log}$, as well as a single-valued branch of, for example, $z^{1/2}$.

A branch cut is a subset of $\mathbb C$, of the form $\{z : \arg z = \alpha\}\cup \{0\}$. This is where a particular branch is discontinuous.

For example, $\operatorname{PV}(z^{1/2})$ which maps $\begin{aligned} z &\mapsto |z|^{1/2} \exp \left(\frac{i\operatorname{Arg}z } 2\right) \\ re^{i\theta}&\mapsto \sqrt r \exp(i\theta/2) \end{aligned}$ The branch is $-\pi < \theta \le \pi$ and the branch cut is the negative real axis union with zero.

Consider the behaviour of $z \mapsto z^{1/2}$ under two different branches, $-\pi < \theta \le \pi$ and $0 \le \theta < 2\pi$. Exercise: Repeat for $(z-z_0)^{1/2}$.

Trigonometric functions

B-C 37-39 (8Ed 34-35)

For $x \in \mathbb R$, $\begin{aligned} e^{ix} &= \cos x + i\sin x \\ e^{-ix} &= \cos x - i \sin x\\ \implies \cos x &= \frac{e^{ix}+e^{-ix}}2\\ \implies \sin x &= \frac{e^{ix}-e^{-ix}}{2i} \end{aligned}$ We can use these expressions to define $\cos$ and $\sin$ on $\mathbb C$. Specifically, $\cos z = \frac{e^{iz}+e^{-iz}}2\quad \text{and}\quad \sin z = \frac{e^{iz}-e^{-iz}}{2i}.$ This gives us the following properties: - $\cos z = \cos (-z)$ - $\sin z = - \sin (-z)$ - $\cos(z+\xi) = \cos z \cos \xi - \sin z \sin \xi$ - $\sin (z+\xi) = \sin z \cos \xi + \cos z \sin \xi$ - $\sin^2 z + \cos^2 z = 1$ (this does not imply that they are bounded in $\mathbb C$) - $\sin (z+\pi/2) = \cos z$ - $\sin (z-\pi/2) = -\cos z$ (these two proven using properties of exp)

Hyperbolic functions

On $\mathbb R$, the hyperbolic functions were $\sinh x = \frac{e^x-e^{-x}}2\\ \cosh x = \frac{e^x + e^{-x}}2$ Recall that $\sinh$ is somewhat like a exaggerated cubic and $\cosh$ is not unlike a steeper periodic parabola. Also, $\cosh$ can be used to model a hanging cable with weight.

Similarly to the first trig functions, we can define the hyperbolic functions on $\mathbb C$ as $\cosh z = \frac{e^{z}+e^{-z}}2\quad \text{and}\quad \sinh z = \frac{e^{z}-e^{-z}}{2}.$ Interestingly, $\sin (iy) = i \sinh y \quad \text{and}\quad \cos(iy) = \cosh y.$ Tke $z = x$ and $\xi = iy$ in the sum formulas and we get $\begin{aligned} \sin(x+iy) &= \sin x \cos(iy) + \cos x + \sin (iy)\\ &= \sin x \cosh y + i \cos x \sinh y\\ \cos(x+iy) &= \cos x \cosh y - i\sin x \sinh y \end{aligned}$ Together, the two above equalities imply $\sin(z+2\pi) = \sin z$ and $\cos(z+2\pi) = \cos z$. Additionally, we have $\cosh^2 z = 1+\sinh^2 z$ and $\begin{aligned} |\sin z|^2 &= \sin^2 x \cosh^2 y + \cos^2 x \sinh^2 y \\ &= \sin^2 x(1+\sinh^2y) +(1-\sin^2x)\sinh^2 y\\ &= \sin^2 x + \sinh^2 y\\ |\cos z|^2 &= \cos^2x + \sinh^2 y \end{aligned}$