Complex Analysis

Definitions
- Functions
Theorems
Holomorphic functions
Möbius transformations
Complex integration
Series Expansions for Holomorphic Functions
- Stuff
- Analytic continuation
Cauchy Residue

Definitions

A multivalued function is a function which assigns to each point a set of values rather than a single value to each point, as a regular function would.

We define the open ball (or open disk since the complex plane is 2-dimensional) in the same way we do for the reals:

$\begin{equation*} D_\varepsilon(z_0) = \{ z \in \mathbb{C} : |z - z_0| < \varepsilon \} \end{equation*}$

and the closed disk

$\begin{equation*} \overline{D}_\varepsilon(z_0) = \{ z \in \mathbb{c} : |z - z_0| \le \varepsilon \} \end{equation*}$

Open and closed sets in the complex plane then follow the same definition as for open and closed sets in the 2-dimension real case.

$\begin{equation*} \big( \cos \theta + i \sin \theta \big)^n = \cos (n \theta) + i \sin (n \theta) \end{equation*}$

Functions

The *complex exponential function is the function $\exp: \mathbb{C} \to \mathbb{C}$ defined by

$\begin{equation*} \exp(z) = \exp(x + iy) = e^x e^{iy} = e^x \big( \cos y + i \sin y \big) \end{equation*}$

$\begin{equation*} \cos(z) := \frac{\exp(iz) + \exp(-iz)}{2} \end{equation*}$

and

$\begin{equation*} \sin(z) := \frac{\exp(iz) - \exp(-iz)}{2i} \end{equation*}$

Let $z \in \mathbb{C}$ be non-zero. We define the multivalued function $\log(z)$ by

$\begin{equation*} \log(z) := \{ w \in \mathbb{C} \mid \exp(w) = z \} \end{equation*}$

We call any element $w$ of $\log(z)$ a logarithm of $z$ .

The principal branch of the logarithm function is the function $\text{Log}: \mathbb{C} \setminus \{ 0 \} \to \mathbb{C}$ defined by

$\begin{equation*} \text{Log}(z) := \ln |z| + i \text{Arg}(z) \end{equation*}$

for non-zero $z \in \mathbb{C}$ , where $\text{Arg}$ is the principal value of the argument function.

Or more general

The branch cut of a complex function is defined to be the point in the principal branch where the function is no longer holomorphic.

Theorems

Holomorphic functions

Notation

$z \in \mathbb{C}$ is almost always denoted $z = x + iy$
$\arg(z) = \{ \theta : z = |z|e^{i \theta} \} = \text{preim}_{z(\theta)} \big( \{ z \} \big)$
$\text{Arg}(z) = \{ \theta : z = |z| e^{i \theta}, \quad \theta \in (- \pi, \pi] \}$ , i.e. the unique $\theta$ in the range $(- \pi, \pi]$
$\mathbb{C}^\times = \mathbb{C} \ \backslash \ \{ 0 \}$ is the complex plane without origin => can be equipped with multiplication operation to form an (Abelian) group
complex-valued function of a real variable means $f(x) = u(x) + i v(x)$ , i.e. input is real and we can decompose the output into a real part and an imaginary part
$\partial = \frac{1}{2} \bigg( \frac{\partial}{\partial x} - i \frac{\partial}{\partial y} \bigg)$
$\bar{\partial} = \frac{1}{2} \bigg( \frac{\partial}{\partial x} + i \frac{\partial }{\partial y} \bigg)$

Differential Geometric view on `arg` function

Let

$\Sigma = \{ (z, \theta) \mid z \in \mathbb{C}^\times, \theta \in \arg(z) \}$
$\pi: \Sigma \to \mathbb{C}^\times$ defined by $(z, \theta) \mapsto z$ , which is then surjective and continuous
$\mathbb{C}^\times$

Apparently, we call a complex manifold such as the one above a Riemann surface, and I honestly believe this is something I ought to return to once I have a better understanding of what complex manifolds are.

Observe then that this defines a fibre bundle, since for any two points $p, p' \in \mathbb{C}^\times$ we have the fibres be isomorphic, i.e. $F_p \cong F_{p'}$ .

$\Sigma$ is the plane defined by Cartesian product $\mathbb{C}^\times \times \mathbb{R}$ since $z \in \mathbb{C}^\times$ and $\theta \in \mathbb{R}$ . Thus, $\Sigma$ is in fact a topological manifold (probably equipped with the standard topology, but I haven't checked). Further we observe that $\mathbb{C}^\times$ is also a topological manifold. Thus, we our fibre bundle as claimed earlier.

Complex-valued functions

Suppose we have a complex-valued function of a real value

$\begin{equation*} f(x) = u(x) + i v(x) \end{equation*}$

We then say that $f(x)$ is continuous at $x_0$ if and only if both $u(x)$ and $v(x)$ are continuous at $x_0$ .

If $f(x)$ is instead a complex-valued function of a complex variable, let $z = x + i y$ , then

$\begin{equation*} f(z) = f(x + iy) = u(x, y) + i v(x, y) \end{equation*}$

We then say that $f(z$ is continuous at $z_0$ if and only if for any $\varepsilon > 0$ there exists $\delta > 0$ s.t.

$\begin{equation*} |z - z_0| < \delta \quad \implies \quad |f(z) - f(z_0)| < \varepsilon \end{equation*}$

where $u(x, y)$ and $v(x, y)$ are real valued functions, and this is thus equivalent of $u$ and $v$ being continuous at the point $(x_0, y_0)$ .

Complex differentiability and holomorphicity

A function $f: U \to \mathbb{C}$ in a neighborhood $U$ of $z_0$ is differentiable (everywhere in $U$ ) if and only if its real and imaginary parts $u$ and $v$ are continuously differentiable and obey the Cauchy-Riemann equations (everywhere in $U$ ):

$\begin{equation*} \frac{\partial u}{\partial x} = \frac{\partial v}{\partial y} \qquad \text{and} \qquad \frac{\partial v}{\partial x} = - \frac{\partial u}{\partial y} \end{equation*}$

This can be seen by considering the "normal" definition of differentiability at a point, and observing the definition changes depending on which direction we approach the point from (vertical or horizontal direction).

Important: it's crucial that the function be differentiable in a neighborhood $U$ of $z_0$ not just at the point $z_0$ !

We say that the complex-valued function $f$ is holomorphic at $z_0$ if it is differentiable everywhere in some neighbourhood $U$ of $z_0$ .

We say that a function is entire if it is holomorphic in the whole complex plane.

Let $h: \mathbb{R}^2 \to \mathbb{R}$ .

Then $h$ is harmonic if it satisfies the Laplace equation:

$\begin{equation*} \frac{\partial^2 h}{\partial x^2}(x, y) + \frac{\partial^2 h}{\partial y^2}(x, y) = 0 \end{equation*}$

Let $U \subseteq \mathbb{R}^2$ be open, and let $u : U \to \mathbb{R}$ be harmonic.

We say a harmonic function $v: U \to \mathbb{R}$ is the harmonic conjugate of $u$ if the complex-valued function

$\begin{equation*} f = u + iv \end{equation*}$

is holomorpic on $U$ .

Solving differential equations

This techinque can be used to solve differential equations on "ugly" domains.

Consider the following:

Polynomials and rational functions

Complex powers

Let $\alpha, z \in \mathbb{C}$ with $z \ne 0$ . Then we define $\alpha$ th power of $z$ by

$\begin{equation*} z^{\alpha} = \{ \exp(\alpha w) : w \in \log(z) \} \end{equation*}$

Unless stated otherwise, a complex power is defined in the principal branch.

Let $\alpha, z \in \mathbb{C}$ with $z \ne 0$ . Then

if $\alpha \in \mathbb{Z}$ there is exactly one value of $z^\alpha$
if $\alpha = \frac{p}{q}$ where $p, q$ are coprime, with $q \ne 0$ there are exactly $q$ values of $z^{\alpha}$
if $\alpha \notin \mathbb{Q}$ or $\alpha \notin \mathbb{R}$ , there are infinitely manu values of $z^\alpha$

Let $q \in \mathbb{N}$ . Then the $q$ values

$\begin{equation*} 1^{\frac{1}{q}} = 1, w, \dots, w^{q - 1} \end{equation*}$

where

$\begin{equation*} w := e^{\frac{2 \pi i}{q}} \end{equation*}$

are the $q$ roots of unity.

Graphing complex functions

Hard to visualize as 2D + 2D becomes 4D
Good idea to treat each of the different 3D plots separately

Let $U \subseteq \mathbb{C}$ be open, and let $f: U \to \mathbb{C}$ .

We say $f$ is conformal if $f$ preserves angles: i.e. if the angle between the images under $f$ of two straight lines in $U$ are equal to the angle between the two straight lines themselves.

By applying this definition to tangents of differentiable curves, more generally we can say the same about the angles between curves at certain points.

We're saying the push-forward of $f_{p_*}: T_p U \to T_{f(p)} \mathbb{C}$ , is angle-preserving.

Therefore, any diffeomorphism between $U$ and $\mathbb{C}$ is angle-preserving.

Consider $f(z) = \sqrt{z}$ . For $z \ne 0$ , $\sqrt{z} = \{ \pm w \}$ . Then we consider the surface

$\begin{equation*} \Sigma = \{ (z, w) \mid w^2 = z \} \subseteq \mathbb{C}^2 \cong \mathbb{R}^4 \end{equation*}$

and the function

$f: \Sigma \to \mathbb{C}$ defined by $f(z, w) = w$ .

This is the projection of a bundle; further, $(\Sigma, f, \mathbb{C})$ forms a fibre bundle (right?)

Let $U \subseteq \mathbb{C}$ be open, and let $f: U \to \mathbb{C}$ be holomorphic.

Then $f$ preserves angles at every $z_0 \in U$ where $f'(z_0) \ne 0$ .

This theorem is useful for establishing the image of a function (if it's a polytope (?)) since we can simply compute the mapped values at the edges (i.e. curves from vertex to vertex) and, knowing that the angles are preserved, immediately know how the edges between the mapped vertices look.

Möbius transformations

A Möbius transformation is a function of the form

$\begin{equation*} f(z) = \frac{az + b}{cz + d} \end{equation*}$

where $a, b, c, d \in \mathbb{C}$ are such that $ad \ne bc$ .

Observe that $f(z)$ is not defined on the entirety of $\mathbb{C}$ , which leads us to defining the extended complex plane.

We'll often consider the case $ad - bc = 1$ , if nothing else is specified.

That is, a Mobius transformation is defined on $\text{SL}(2, \mathbb{C})$ .

The extended complex plane is the set $\tilde{\mathbb{C}} = \mathbb{C} \cup \{ \infty \}$ , where $\infty$ is just some object $\infty \notin \mathbb{C}$ .

We extend the usual arithmetic operations in the following way: for $a \in \mathbb{C}$ and non-zero $b \in \mathbb{C}$ ,

$\begin{equation*} a + \infty = \infty, \quad b \cdot \infty = \infty, \quad \frac{b}{0} = \infty, \quad \frac{b}{\infty} = 0 \end{equation*}$

Consider the coordinates $(X, Y, Z)$ describing $\mathbb{R}^3$ . We identify the complex plane with the plane defined by $Z = 0$ , and a complex number $X + i Y$ with the point $(X, Y, 0)$ .

The Riemann sphere is the unit sphere $S^2$ in $\mathbb{R}^3$ defined by

$\begin{equation*} S^2 := \{ \big( X, Y, Z \big) \in \mathbb{R}^3 \mid X^2 + Y^2 + Z^2 = 1 \} \end{equation*}$

and we consider the "north pole" to be the point $N := (0, 0, 1)$ .

The Riemann sphere therefore has two charts:

For all points in the complex plane, the chart is the identity map from the sphere (with $\infty$ removed) to the complex plane.
For $\infty$ , the chart neighborhood is the sphere (with the origin removed), and the chart is given by sending $\infty$ to $0$ and all other points $z$ to $\frac{1}{z}$ .

Let $\phi: \mathbb{C} \to S^2$ such that the three points $N$ , $z$ , and $\phi(z)$ are colinear.

It is clear that

$\begin{equation*} \lim_{|z| \to \infty} \phi(z) = (0, 0, 1) \end{equation*}$

thus we define also $\phi(\infty) = N$ and thereby consider $\phi$ as being defined on the extended complex plane, i.e. $\phi: \tilde{\mathbb{C}} \to S^2$ .

The map $\phi: \tilde{\mathbb{C}} \to S^2$ is evidently bijective, so it has an inverse $\psi: S^2 \to \tilde{\mathbb{C}}$ . This function $\psi$ is the sterographic projection.

Better description:

The unit sphere in $\mathbb{R}^3$ is the set of pints $(x, y, z)$ such that $x^2 + y^2 + z^2 = 1$ .

Let $N = (0, 0, 1)$ be the "north pole", and let $S^2$ be the rest of the sphere.

The plane $z = 0$ (xy-plane) runs through the center of the sphere; the "equator" is the intersection of the sphere with this plane.

For any point $P$ on $S^2$ , there is a unique line through $N$ and $P$ , and this line intersects the plane $z = 0$ in exactly one point $P'$ . We define the stereographic projection of $P$ to be this point $P'$ in the plane.

$\begin{equation*} \phi(z) = \phi(x + iy) = \bigg( \frac{2x}{|z|^2 + 1}, \frac{2y}{|z|^2 + 1}, \frac{|z|^2 - 1}{|z|^2 + 1} \bigg) \end{equation*}$

and

$\begin{equation*} \psi(X, Y, Z) = \begin{cases} \frac{X + iY}{1 - Z} & (X, Y, Z) \ne N, \\ \infty & (X, Y, Z) = N \end{cases} \end{equation*}$

Stereographic projection maps a circle to either a circle or a straight line (a "circline").

Makes some people say that straight lines are circles of infinite radius.

A translation is a Möbius transformation of the form

$\begin{equation*} f(z) = z + b, \quad b \in \mathbb{C} \end{equation*}$

which corresponds to the matrix

$\begin{equation*} \begin{pmatrix} 1 & b \\ 0 & 1 \end{pmatrix} \end{equation*}$
A rotation is a Möbius transformation of the form

$\begin{equation*} f(z) = az, \quad \text{where} \quad |a| = 1 \end{equation*}$

so that $a = e^{i \theta}$ for some $\theta$ , which corresponds to the matrix

$\begin{equation*} \begin{pmatrix} e^{i \theta / 2} & 0 \\ 0 & e^{- i \theta / 2} \end{pmatrix} \end{equation*}$
A dilation is a Möbius transformation of the form

$\begin{equation*} f(z) = rz, \quad \text{where} \quad r > 0 \end{equation*}$

which corresponds to the matrix

$\begin{equation*} \begin{pmatrix} \sqrt{r} & 0 \\ 0 & 1 / \sqrt{r} \end{pmatrix} \end{equation*}$
An inversion is a Möbius transformation of the form

$\begin{equation*} f(z) = \frac{1}{z} \end{equation*}$

which corresponds to the matrix

$\begin{equation*} \begin{pmatrix} 0 & i \\ i & 0 \end{pmatrix} \end{equation*}$

We say that a Möbius transformation $f$ fixes the point of infinity if $f(\infty) = \infty$ .

Translations, rotatons, and dilations fix the point at infinity, while inversions do not.

Let $f$ be a Möbius transformation. Then $f$ is a composition of a finite number of translations, rotations, diluations, and, if and only if $f$ does not fix the point at infinity, one inversion.

Möbius transformations map circlines to circlines.

Let $z_2, z_3, z_4 \in \tilde{\mathbb{C}}$ be three distinct points. Then there exists a unqiue Möbius transformation $f$ such that

$\begin{equation*} f(z_2) = 1, \quad f(z_3) = 0, \quad f(z_4) = \infty \end{equation*}$

This is useful because we by simply knowing how the Möbius transformation maps the three different points, we can tell what it does to circles or lines.

Let $z_1, z_2, z_3, z_4 \in \tilde{\mathbb{C}}$ be distinct points.

The cross-ratio $[z_1, z_2, z_3, z_4]$ of the four points is the image of $z_1$ under the Möbius transformation which sends $(z_2, z_3, z_4)$ to $(1, 0, \infty)$ .

Complex integration

Complex integrals

Let $[a, b] \subseteq \mathbb{R}$ be a interval, and $f : [a, b] \to \mathbb{C}$ of the form

$\begin{equation*} f = u + iv \end{equation*}$

Then $f$ is integrable if its real and imaginary parts $u, v: [a, b] \to \mathbb{R}$ are integrable in the usual (real) sense, and we define the integral of $f$ by

$\begin{equation*} \int_{a}^{b} f(t) \ dt := \int_{a}^{b} u(t) \ dt + i \int_{a}^{b} v(t) \ dt \end{equation*}$

It will usually suffice to observe that continuous functions are integrable.

Example

$\begin{equation*} \begin{split} | \int_{\Gamma} f(z) \ dz | &= | \int_{t_0}^{t_1} f \big( \gamma(t) \big) \gamma'(t) \ dt \\ &\le \int_{t_0}^{t_1} | f \big( \gamma(t) \big) \gamma'(t)| \ dt \\ &\le \int_{t_0}^{t_1} \max_{z \in \Gamma} |f(z)| |\gamma'(t)| \ dt \\ &= \max_{z \in \Gamma} |f(z)| \int_{t_0}^{t_1} |\gamma'(t)| \ dt \\ &= \max_{z \in \Gamma} | f(z)| \ \ell(\Gamma) \end{split} \end{equation*}$

where $\ell(\Gamma)$ denotes the arc-length of $\Gamma$ .

$\Gamma$ is by def. closed and bounded, i.e. a compact set
$f$ is continuous

Hence $\max_{z \in \Gamma} |f(z)|$ is in fact bounded i.e. finite.

Contour integrals

Let $z_0, z_1 \in \mathbb{C}$ be distinct.

Then a (parametrized) curve $\Gamma$ connecting $z_0$ and $z_1$ is a continuous function

$\begin{equation*} \begin{split} \gamma : [t_0, t_1] \to \mathbb{C}, & \quad t_0, t_1 \in \mathbb{R}, \quad t_0 < t_1 \\ \text{such that} & \quad \gamma(t_0) = z_0, \quad \gamma(t_1) = z_1 \end{split} \end{equation*}$

Writing $z_0 = x_0 + i y_0$ and $z_1 = x_1 + i y_1$ , we decompose $z$ into a real and imaginary parts for continuous real functions $x, y: [t_0, t_1] \to \mathbb{R}$ , so

$\begin{equation*} \gamma(t) = x(t) + i y(t), \quad \forall t \in [t_0, t_1] \end{equation*}$

We say the curve $\Gamma$ is regular if $\gamma$ is continuously differentiable and $\gamma'(0) \ne 0$ forall $t \in [t_0, t_1]$ .

A curve $\Gamma$ from $z_0$ to $z_1$ in $\mathbb{C}$ is a contour if it is a finite union of regualr curves, which together joint $z_0$ with $z_1$ , i.e. there exists

$\begin{equation*} \gamma_i : [t_0^i, t_1^i] \to \mathbb{C}, \quad i = 1, \dots, n \end{equation*}$

such that

$\begin{equation*} \gamma(t) = \begin{cases} \gamma_1(t) & \quad \text{if } t \in [t_0, t_1] \\ \gamma_2(t) & \quad \text{if } t \in [t_1, t_2] \\ & \vdots \\ \gamma_n(t) & \quad \text{if } t \in [t_{n - 1}, t_n] \end{cases} \end{equation*}$

such that $\gamma(t)$ is continuous.

For a continuous function $f: \Gamma \to \mathbb{C}$ , we define the contour integral of $f$ along $\Gamma$ by

$\begin{equation*} \int_{\Gamma} f(z) \ dx = \sum_{i=1}^{n} \int_{\Gamma_i} f(z) \ dz \end{equation*}$

Let $\Gamma$ be a curve in $\mathbb{C}$ , and let $f: \Gamma \to \mathbb{C}$ be continuous. Then

$\begin{equation*} \abs{\int_{\Gamma} f(z) \ dz} \le \max_{z \in \Gamma} |f(z)| \ \ell(\Gamma) \end{equation*}$

Let $D \subseteq \mathbb{R}^2$ . We will say that $D$ is a domain if $D$ is open and every two points in $D$ can be connected by a contour which lies wholly on $D$ .

Let $D \subseteq \mathbb{C}$ be a domain, and $f : D \to \mathbb{C}$ be continous. Then the following are equivalent:

$f$ has an antiderviative $F$ on $D$
$\int_{\Gamma} f(z) \ dz = 0$ for all closed contours $\Gamma$ in $D$
all contour integrals $\int_{\Gamma} f(z) \ dz$ are independent of path $\Gamma$ , and depend only on the endpoints.

Cauchy's Integral Theorem

Let $\Gamma$ be a contour in $\mathbb{C}$ .

Then $\Gamma$ is simple if it has no self-intersections, except possible at the endpoints, i.e. $\gamma(t) \ne \gamma(s)$ for all distinct $s, t \in [t_0, t_1]$ , unless $s = t_0$ and $t = t_0$ and $\Gamma$ is a closed contour.

A loop is a simple, closed contour.

Let $\Gamma$ be a loop in $\mathbb{C}$ . Then $\Gamma$ defines two regions in the $\mathbb{C}$ , with $\Gamma$ as their common boundary:

a boundary domain, the interior of $\Gamma$
an unbounded domain, the exterior of $\Gamma$

Let $\Gamma$ be a loop in $\mathbb{C}$ . We say $\Gamma$ is positively-oriented if as we move along the curve in the direction of parametrization, the interior is on the LHS.

Otherwise state, all loops should be assumed to be positively-oriented.

$D$ is said to be simply connected if the interior of every loop on $D$ lies wholly in $D$ .

Let $z_0 \in \mathbb{C}$ and $\Gamma$ be a loop in $\mathbb{C}$ which does not pass through $z_0$ . Then

$\begin{equation*} \int_{\Gamma} \frac{1}{z - z_0} \ dz = \begin{cases} 2\pi i & \text{if } z_0 \text{ lies in the interior of $\Gamma$} \\ 0 & \text{otherwise} \end{cases} \end{equation*}$

Since $z_0 \ne \Gamma$ , we clearly have two cases:

$z_0 \in \text{Int}(\Gamma)$ : integral is zero due to Cauchy Integral theorem
$z_0 \notin \text{Int}(\Gamma)$ : Consider the following figure: Since $z_0 \notin \text{Int}(\Gamma)$ this figure basically "encapsulates" all possible loops which have $z_0$ outside!

Let $\Gamma$ be a loop, and $f$ be holomorphic inside and on $\Gamma$ .

Then,

$\begin{equation*} \int_{\Gamma} f(z) \ dz = 0 \end{equation*}$

We break the integrand $f$ , aswell as the differential $dz$ into their real and imaginary components:

$\begin{equation*} \begin{split} f & = u + iv \\ dz &= dx + i dy \end{split} \end{equation*}$

In this case we have

$\begin{equation*} \oint_{\gamma} f(z) \ dz = \oint_{\gamma} (u + iv) (dx + i dy) = \oint_{\gamma} (u \ dx - v \ dy) + i \oint_{\gamma} (v \ dx + u \ dy) \end{equation*}$

By Green's Theorem, we may then replace the integrals around the closed contour $\gamma$ with an area integral throughout the domain $D$ that is enclosed by $\gamma$ as follows:

$\begin{equation*} \oint_{\gamma} (u dx - v dy) = \iint_D \bigg( - \frac{\partial v}{\partial x} - \frac{\partial u}{\partial y} \bigg) \ dx dy \end{equation*}$

and for imaginary part,

$\begin{equation*} \oint_{\gamma} (v dx + u dy) = \iint_D \bigg( \frac{\partial u}{\partial x} - \frac{\partial v}{\partial y} \bigg) dx dy \end{equation*}$

However, being the real and imaginary parts of a function holomorphic in the domain $D$ , $u$ and $v$ must satisfy the Cauchy-Riemann equations there:

$\begin{equation*} \begin{split} \frac{\partial u}{\partial x} &= \frac{\partial v}{\partial y} \\ \frac{\partial u}{\partial y} &= \frac{\partial v}{\partial x} \end{split} \end{equation*}$

We therefore find that both integrads (and hence their integrals) are zero:

$\begin{equation*} \begin{split} \iint_D \bigg( - \frac{\partial v}{\partial x} - \frac{\partial u}{\partial y} \bigg) \ dx dy &= \iint_D \bigg( \frac{\partial u}{\partial y} - \frac{\partial u}{\partial y} \bigg) \ dx dy = 0 \\ \iint_D \bigg( \frac{\partial u}{\partial x} - \frac{\partial v}{\partial y} \bigg) \ dx dy &= \iint_D \bigg( \frac{\partial u}{\partial x} - \frac{\partial u}{\partial x} \bigg) \ dx dy = 0 \end{split} \end{equation*}$

Which gives us the result

$\begin{equation*} \oint_{\gamma} f(z) \ dz = 0 \end{equation*}$

as wanted.

Let $\Gamma$ be a loop, $z_0 \in \text{Int}(\Gamma)$ , and $f$ be holomorphic inside and on $\Gamma$ . Then

$\begin{equation*} f(z_0) = \frac{1}{2 \pi i} \int_{\Gamma} \frac{f(z)}{z - z_0} \ dz \end{equation*}$

Let $\Gamma$ be a loop, $f$ be holomorphic inside and on $\Gamma$ , and $z$ lie inside $\Gamma$ .

Then $f$ is infinitively differentiable at $z$ and, for all positive integers $n$ ,

$\begin{equation*} f^{(n)}(z) = \frac{n!}{2\pi i} \int_{\Gamma} \frac{f(w)}{(w - z)^{n + 1}} \ dw \end{equation*}$

Let $D \subseteq \mathbb{C}$ be a domain, $f: D \to \mathbb{C}$ be continuous, and suppose that

$\begin{equation*} \int_{\Gamma} f(z) \ dz = 0 \end{equation*}$

for all loops $\Gamma$ inside $D$ .

Then $f$ is holomorphic.

Liouville's Theorem and its applications

Let $D \subseteq \mathbb{C}$ be a domain, $z_0 \in D$ and $R > 0$ be such that $D_R(z_0) \subseteq D$ , $f$ be holomorphic on $D$ , and $M > 0$ be such that $|f(z) \le M$ for all $z \in D$ .

Then for all $n \in \mathbb{N}$ , we have that

$\begin{equation*} |f^{(n)}(z_0)| \le \frac{n! M}{R^n} \end{equation*}$

Let $f$ be holomorphic on $\mathbb{C}$ and be bounded, i.e. satisfying

$\begin{equation*} | f(z) | \le M, \quad M > 0, \forall z \in \mathbb{C} \end{equation*}$

then $f$ is constant.

Maximum Modulus Principle

$z_0 \in D \subseteq \mathbb{C}$ and $R > 0$ be such that the closed disc $\overline{D}_R(z_0) \subseteq D$ , and $f$ be holomorphic on $D$ .

Then

$\begin{equation*} f(z_0) = \frac{1}{2 \pi} \int_{0}^{2 \pi} f(z_0 + R e^{it}) \ dt \end{equation*}$

Let:

$f$ be holomorphic on a domain $D$
$z_0 \in D$
$R > 0$ s.t. $\bar{D}_R(z_0) \subseteq D$

Fix $0 < r \le R$ , then

$\begin{equation*} f(z_0) = \frac{1}{2 \pi i} \int_{C_r} \frac{f(z)}{z - z_0} \ dz \end{equation*}$

where $C_r$ is the circle of radius $r$ centred at $z_0$ (Cauchy Integral formula).

$C_r$ is parametrized by $\gamma: [0, 1] \to \mathbb{C}$ , given by

$\begin{equation*} \gamma(t) = z_0 + r \exp \big( i t \big) \end{equation*}$

So, using the definition of the contour integral

$\begin{equation*} \begin{split} f(z_0) &= \frac{1}{2 \pi i} \int_{0}^{2 \pi} \frac{f \big( z_0 + r \exp(it) \big)}{z_0 + r \exp(it) - z_0} ir \exp(it) \ dt \\ &= \frac{1}{2 \pi} \int_{0 }^{2 \pi} f \big( z_0 + r \exp(it) \big) \ dt \end{split} \end{equation*}$

Suppose

$\begin{equation*} \max_{z \in \overline{D}_R(z_0)} | f(z)| = |f(z_0)| \end{equation*}$

i.e. maximum value is attained at the center of the circle. Then,

$\begin{equation*} \begin{split} f(z_0) &= \frac{1}{2 \pi} \Big| \int_{0}^{2 \pi} f \big( z_0 + r \exp(it) \big) \ dt \Big| \\ & \le \frac{1}{2 \pi} \int_{0}^{ 2 \pi} | f(z_0 + r \exp(it) ) | \ dt \\ & \le \frac{1}{2 \pi} \int_{0}^{2 \pi} |f(z_0)| \ dz = |f(z_0)| \end{split} \end{equation*}$

hence

$\begin{equation*} |f(z_0)| = \frac{1}{2 \pi} \int_{0}^{2 \pi} \Big| f\Big(z_0 + r \exp(it) \Big) \Big| \ dt \end{equation*}$

which, since $f(z_0)$ is clearly constant, we can use the linearity of the integral to get

$\begin{equation*} \int_{0}^{2 \pi} \Big( |f(z_0) | - |f(z_0 + r \exp(it))| \Big) \ dt = 0 \end{equation*}$

which is a non-negative continuous function, which implies that the expression inside the integral is zero for all $t \in [0, 2\pi]$ , i.e.

$\begin{equation*} |f(z_0) | - \Big| f \big( z_0 + r \exp(it) \big) \Big| = 0, \quad \forall t \in [0, 2 \pi] \end{equation*}$

which is true for every point in $\overline{D}_R(z_0)$ is of this form, for some $0 < r < R$ , and some $t \in [0, 2 \pi]$ .

Hence $|f|$ is constant on $\overline{D}_R(z_0)$ .

Let $D \subseteq \mathbb{C}$ , and $f$ be holomorphic and bounded on $D$ , i.e.

$\begin{equation*} |f(z)| \le M, \quad \forall z \in D \end{equation*}$

for some $M > 0$ .

If $|f(z)|$ achives its maximum at $z_0 \in D$ , then $f$ is constant on $D$ .

Series Expansions for Holomorphic Functions

Stuff

Let $\sum_{j=0}^{\infty} a_j (z - z_0)^j$ be a power series.

Then $\exists R \in [0, \infty) \cup \left\{ \infty \right\}$ s.t.

$\sum_{j=0}^{\infty} a_j (z - z_0)^j$ converges if $|z - z_0| < R$
$\sum_{j=0}^{\infty} a_j (z - z_0)^j$ converges uniformly on $\bar{D}_r(z_0)$ for all $0 \le r < R$
$\sum_{j=0}^{\infty} a_j (z - z_0)^j$ diverges if $|z - z_0| > R$

$\sum_{j=0}^{\infty} a_j (z - z_0)^j$ is holomorphic on its disc of convergence $D_R(z_0)$ , where $R$ is the radius of convergence.

To proof holomorphicity of convergent power series, one could do as follows:

Fix $z \in D_R(z_0)$ . Then $z \in \bar{D}_r(z_0)$ for some $0 < r < R$ . Series converges uniformly on $\bar{D}_r(z_0)$ , thus partial sums are holomorphic, which implies series is holomorphic at $z$ .

Let $U \subseteq \mathbb{C}$ be open, and $f: U \to \mathbb{C}$ .

Then we say that $f$ is analytic if at everypoint $z \in U$ , $f$ can be expressed as a convergent power series.

Suppose $f$ is holomorphic on $D_R(z_0)$ . Then the Taylor series

$\begin{equation*} \sum_{j=0}^{\infty} \frac{f^{(j)}(z_0)}{j!} (z - z_0)^j, \quad \forall z \in D_R(z_0) \end{equation*}$

converges to $f(z)$ on $D_R(z_0)$ , and converges uniformly on $\overline{D}_R(z_0)$ for all $r < R$

where

$\begin{equation*} \frac{f^{(j)}(z_0)}{j!} = \frac{1}{2 \pi i} \int_{\Gamma} \frac{f(w)}{(w - z_0)^{j + 1}} \ dw \end{equation*}$

for any loop $\Gamma \subseteq D_R(z_0)$ with $z_0 \in \text{Int}(\Gamma)$ .

It's REALLY important to realize that this Taylor series is only on some disk $D_R(z_0)$ !!!

Therefore, if we know $f$ is holomorphic on some open subset $U \subseteq \mathbb{C}$ , then if we want to talk about the Taylor series on any point $z_0 \in U$ , still can only say something about this on some disk $D_R(z_0) \subseteq U$ !!

The Taylor series at some point $z_1 \in U$ is NOT necessarily the same as the Taylor series at some other point $z_2 \in U$ .

Let $\Gamma$ be the circle of radius $\rho$ centered at $z_0$ where $r < \rho < R$ . By CIF we know that for $z \in D_r(z_0)$ we have

$\begin{equation*} f(z) = \frac{1}{ 2 \pi i} \int_{\Gamma} \frac{f(w)}{w - z} \ dw \end{equation*}$

$\begin{equation*} \begin{split} f(z) &= \frac{1}{2 \pi i} \int_{\Gamma} \frac{f(w)}{(w - z_0) - (z - z_0)} \ dw \\ &= \frac{1}{2 \pi i} \int_{\Gamma} \frac{f(w)}{w - z_0} \frac{1}{1 - \frac{z- z_0}{w - z_0}} \ dw \\ &= \frac{1}{2 \pi i} \int_{\Gamma} \frac{f(w)}{w - z_0} \sum_{j=0}^{\infty} \Bigg( \frac{z - z_0}{w - z_0} \Bigg)^j \ dw \end{split} \end{equation*}$

where we've used the fact that

$\begin{equation*} |z - z_0| < r < \rho = | w - z_0 |, \quad \text{for } w \in \Gamma \end{equation*}$

thus

$\begin{equation*} \left| \frac{z - z_0}{w - z_0} < \frac{r}{\rho} < 1 \right| \end{equation*}$

hence we could recognize the fraction above as the convergent series. Further, due to the series being convergent, we can interchange the summation and integration

$\begin{equation*} \begin{split} f(z) &= \frac{1}{2 \pi i} \sum_{j=0}^{\infty} \int_{\Gamma} \frac{f(w)}{w - z_0} \Bigg( \frac{z - z_0}{w - z_0} \Bigg)^j \\ &= \sum_{j=0}^{\infty} \Bigg( \frac{1}{2 \pi i} \int_{\Gamma} \frac{f(w)}{(w - z_0)^{j + 1}} \ dw \Bigg) (z - z_0)^j \\ &= \sum_{j = 0}^{\infty} \frac{f^{(j)}(z_0)}{j!} (z - z_0)^j \end{split} \end{equation*}$

which is just the Taylor series!

Taylor series of a $f$ (if it exists) is unique, i.e. if $f$ is holomorphic on $D_R(z_0)$ and

$\begin{equation*} f(z) = \sum_{j=0}^{\infty} a_j (z - z_0)^j, \quad \forall z \in D_R(z_0) \quad \implies a_j = \frac{f^{(j)}(z_0)}{j!} \end{equation*}$

A anulus is

$\begin{equation*} A_{r, R}(z_0) = \left\{ z \in \mathbb{C} : r < |z - z_0| < R \right\} \end{equation*}$

$\begin{equation*} A_{0, R}(z_0) = D_R(z_0) \setminus \left\{ z_0 \right\} \end{equation*}$

Suppose $f$ is holomorphic on the anulus $A_{r, R}(z_0)$ .

Then

$\begin{equation*} f(z) = \sum_{j= - \infty}^{\infty} a_j (z - z_0)^j, \quad \text{where} \quad a_j = \frac{1}{2 \pi i} \int_{\Gamma} \frac{f(w)}{(w - z_0)^{j + 1}} \ dw \end{equation*}$

for any loop $\Gamma \subseteq A_{r_1, R}(z_0)$ with $z_0 \in \text{Int} (\Gamma)$ for all $z \in A_{r_1, R}(z_0)$ , and the series converges uniformly on $\overline{A}_{r, R}(z_0)$ for all $r < r_1 \le R_1 < R$ .

We say $z_0$ is a singularity if $f$ is not holomorphic at $z_0$ .

Suppose $f$ has a singularity at $z_0$ .

If there exists $R > 0$ s.t. $f$ is holomorphic on the punctured disk $D_R(z_0) \setminus \left\{ z_0 \right\}$ , $f$ has a Laurent series centred at $z_0$ , valid on this disk

$\begin{equation*} f(z) = \sum_{j = -\infty}^{\infty} a_j \big( z - z_0 \big)^j \end{equation*}$

Then ONE of the following is true:

$a_j = 0$ for all $j < 0$ ( $z_0$ is a removable singularity)
$\exists m \ge 1$ such that $a_{-m} \ne 0$ but $a_j 0$ for all $j < -m$ ( $z_0$ is a pole of order $m$ )
$\exist$ infinitively many negative $j$ s.t. $a_j \ne 0$ ( $z_0$ is an essential singularity)

If you have a removable singularity and then consider the integral along some loop around $z_0$ .

Then the integral is the integral along some loop on which $f$ is holomorphic, hence it's zero.

We can therefore redefine $f$ to take on the value $a_0$ at the singularity point $z_0$ , and we got ourselves a holomorphic function on the non-punctured disk centered at $z_0$ !

$z_0$ is a zero of holomorphic function $f$ if $f(z_0)$ .

Further, we say it's a zero of order $m$ if

$\begin{equation*} f(z_0) = f'(z_0) = \dots = f^{(m - 1)}(z_0) = 0 \end{equation*}$

but

$\begin{equation*} f^{(m)}(z_0) \ne 0 \end{equation*}$

Then

$\begin{equation*} f(z) = \sum_{j=m}^{\infty} \frac{f^{(j)}(z_0)}{j!} (z - z_0)^j \end{equation*}$

Let $z_0 \in \mathbb{C}$ , $U \subseteq \mathbb{C}$ be a neighbourhood of $z_0$ , $f$ be holomorphic on $U$ , and such that $f(z_n) = 0$ for a sequence of distinct points $z_n \in U$ which converge to $z_0$ .

Then $f$ is identically zero on some disc centered at $z_0$ .

Let $f, g$ be holomorphic at $z_0$ be a zero of $g$ of order $m$ .

Then $f / g$ has

has a pole of order $m$ at $z_0$ if $f(z_0) \ne 0$
has a pole of order $m - k$ (if $k < m$ ) at $z_0$ , if $z_0$ is a zero of order $k$ of $f$ , OR $z_0$ is a removable singularity if $k \ge m$

Analytic continuation

Suppose $f$ is holomorphic on a domain $D_1$ with $z_0 \in D_1$ and $D_r(z_0) \subseteq D$ .

Then

$\begin{equation*} f_1(z) = \sum_{j=0}^{\infty} \frac{f^{(j)}(z_0)}{j!} (z - z_0)^j, \quad \forall z \in D_r(z_0) \end{equation*}$

Power series above might have radius of convergence $R > r$

$\begin{equation*} f_2(z) = \sum_{j=0}^{\infty} \frac{f^{(j)}(z_0)}{j!} (z - z_0)^j \end{equation*}$

is a holomorphic on $D_R(z_0)$ with $D_R(z_0) \nsubseteq D$ , i.e.

$\begin{equation*} f_1(z) = f_2(z), \quad \forall z \in D_r(z_0) \end{equation*}$

One might then as does $f_1 = f_2$ on $D \cap D_R(z_0)$ ?

Yes, $f$ is an analytic continuation of $f_1$ to $D \cup D_R(z_0)$ where

$\begin{equation*} f(z) = \begin{cases} f_1(z) & z \in D \\ f_2(z) & z \in D_R(z_0) \end{cases} \end{equation*}$

and $f$ is well-defined.

Let $D \subseteq \tilde{D} \subseteq \mathbb{C}$ be domain, and $f: D \to \mathbb{C}$ be holomorphic.

We say that a holomorphic function $F: \tilde{D} \to \mathbb{C}$ is an analytic continuation of $f$ if

$\begin{equation*} F(z) = f(z), \quad \forall z \in D \end{equation*}$

Let $D \subseteq \mathbb{C}$ be a domain, $z_0 \in D$ , $f$ be holomorphic on $D$ and such that

$\begin{equation*} f(z) = 0, \quad \forall z \in D_R(z_0) \end{equation*}$

for some $R > 0$ . Then

$\begin{equation*} f(z) = 0, \quad \forall z \in D \end{equation*}$

Let $D \subseteq \mathbb{C}$ be a domain, $z_0 \in D$ , $f, g: D \to \mathbb{C}$ be holomorphic and such that

$\begin{equation*} f(z) = g(z), \quad \forall z \in D_R(z_0) \end{equation*}$

for some $R > 0$ .

Then $f(z) = g(z)$ for all $z \in D$ .

Let $D \subseteq \mathbb{C}$ be a domain, $z_0 \in \mathbb{C}$ , and $f$ be holomorpic on $D$ and such that

$\begin{equation*} f(z_n) = 0 \end{equation*}$

for a sequence of distinct points $z_n \in \mathbb{C}$ which converge to $z_0$ . Then

$\begin{equation*} f(z) = 0, \quad \forall z \in D \end{equation*}$

Let $D \subseteq \mathbb{C}$ be a domain, $z_0 \in \mathbb{C}$ , and $f, g$ be holomorphic on $D$ and such that

$\begin{equation*} f(z_n) = g(z_n) \end{equation*}$

for a sequence of distinct points $z_n \in \mathbb{C}$ which converge to $z_0$ . Then

$\begin{equation*} f(z) = g(z), \quad \forall z \in D \end{equation*}$

Let $f, g$ be holomorphic at $z_0$ , where $z_0$ is zero of order $m$ . Then

if $z_0$ is not zero of $f$ , then $f / g$ has a pole of order $m$ at $z_0$
if $z_0$ is a zero of order $k$ of $f$ , then $f / g$ has a pole of order $m - k$ at $z_0$ if $m > k$ , and has a removable singularity at $z_0$ otherwise.

Cauchy Residue

Theorem

Let $z_0 \in \mathbb{C}$ , $f$ be holomorphic on the punctured disc $D_R'(z_0)$ for some $R > 0$ , with an isolated singularity at $z_0$ , and $\Gamma$ be a loop inside $D_R'(z_0)$ , with $z_0 \in \text{Int}(\Gamma)$ . Then

$\begin{equation*} \int_{\Gamma} f(z) \ dz = 2\pi i a_{-1} \end{equation*}$

where $a_{-1}$ is the coefficient of the $(z - z_0)^{-1}$ term in the Laurent expansion of $f$ centred at $z_0$ ,

$\begin{equation*} f(z) = \sum_{j = - \infty}^{\infty} a_j (z - z_0)^j \end{equation*}$

Let $z_0 \in \mathbb{C}$ and $f$ be holomorphic on the punctured disc $D_R'(z_0)$ , for some $R > 0$ , with an isolated singularity at $z_0$ .

Then the residue of $f$ at $z_0$ , is

$\begin{equation*} \text{Res}(f, z_0) = a_{-1} \end{equation*}$

where $a_{- 1}$ is the term in the Laurent series of $f$ centered at $z_0$ .

Let $z_0 \in \mathbb{C}$ , and $f$ be holomorphic on the punctured dist $D_R'(z_0)$ for some $R > 0$ , with removable singularity at $z_0$ . Then

$\begin{equation*} \text{Res}(f, z_0) = 0 \end{equation*}$

Let $z_0 \in \mathbb{C}$ and $f$ be holomorphic on the punctured disc $D_R'(z_0)$ , for some $R > 0$ , with a pole of order $m$ at $z_0$ . Then

$\begin{equation*} \text{Res}(f, z_0) = \lim_{z \to z_0} \frac{1}{(m - 1)!} \frac{d^{m - 1}}{dz^{m - 1}} \Big( (z - z_0)^m f(z) \Big) \end{equation*}$

Let $z_0 \in \mathbb{C}$ and $g$ , $h$ be holomorphic on $D_R'(z_0)$ for some $R > 0$ , such that $h$ has a simple zero ( $m = 1$ ) at $z_0$ , while $g(z_0) \ne 0$ . Then, defining $f = g / h$ we have

$\begin{equation*} \text{Res}(f, z_0) = \frac{g(z_0)}{h'(z_0)} \end{equation*}$

Let $\Gamma$ be a loop, and $f$ be holomorphic inside and on $\Gamma$ except for finitely many isolated singularities $z_1, \dots, z_k \in \text{int}(\Gamma)$ . Then

$\begin{equation*} \int_{\Gamma} f(z) \ dz = 2 \pi i \sum_{j=1}^{k} \text{Res}(f, z_j) \end{equation*}$

Let $D \subseteq \mathbb{C}$ be a domain.

A function $f$ is meromorphic on $D$ if for all $z \in D$ , either $f$ has a pole at $z$ or $f$ is holomorphic at $z$ .

Application: trigonometric integrals

Integrals of the form

$\begin{equation*} \int_{0}^{2 \pi} R(\cos \theta, \sin \theta) \ d \theta \end{equation*}$

for ration function $R$ , can often be evaluated by considering a contour integral of appropriate function around the unit circle centered at $0$ .

On $C_1(0)$ with $z = e^{i \theta}$ we have

$\begin{equation*} \cos \theta = \text{Re}(z) = \frac{z + \overline{z}}{2} = \frac{1}{2} \Bigg( z + \frac{1}{z} \Bigg) \end{equation*}$

and

$\begin{equation*} \sin \theta = \text{Im}(z) = \frac{z - \overline{z}}{2i} = \frac{1}{2i} \Bigg( z - \frac{1}{z} \Bigg) \end{equation*}$

Defining $f(z) = \frac{1}{iz} R \Big( \frac{z + \frac{1}{z}}{2}, \frac{z - \frac{1}{z}}{2} \Big)$ , we therefore have that

$\begin{equation*} f \big( \exp(i \theta) \big) = \frac{1}{i \exp(i \theta)} R (\cos \theta, \sin \theta) \end{equation*}$

If $\Gamma = C_1(0)$ parametrized by $\gamma: [0, 2 \pi] \to \mathbb{C}$ defined by $\gamma(\theta) = \exp(i \theta)$ , then

$\begin{equation*} \int_{\Gamma} f(z) \ dz = \int_{0}^{2 \pi} R(\cos \theta, \sin \theta) \ d \theta \end{equation*}$

It basically comes down to rewriting $\cos \theta$ and $\sin \theta$ with $z = e^{i \theta}$ , which then often provides us with a rational function of which it's substantially esaier to obtain the singularities, thus the residue, hence the integral around $\Gamma$ .

Improper Integrals

We define the Cauchy principal value of the integral

$\begin{equation*} \int_{-\infty}^{\infty} f(x) \ dx \end{equation*}$

$\begin{equation*} \text{p.v.} \Bigg( \int_{-\infty}^{\infty} f(x) \ dx \Bigg) := \lim_{\rho \to \infty} \int_{-\rho}^{\rho} f(x) \ dx \end{equation*}$

Let $P = R / Q$ be a rational function, where

$\begin{equation*} \deg(Q) \ge \deg(P) + 1 \end{equation*}$

ad $a \in \mathbb{R}$ such that $a \ne 0$ . Then

$\begin{equation*} \begin{split} \lim_{\rho \to \infty} \int_{C_p^+} \exp(iaz) \frac{P(z)}{Q(z)} \ dz = 0 \quad & \text{if } a > 0 \\ \lim_{\rho \to \infty} \int_{C_p^-} \exp(iaz) \frac{P(z)}{Q(z)} \ dz = 0 \quad & \text{if } a < 0 \end{split} \end{equation*}$

where $C_p^+$ and $C_p^-$ are the semicircular contours from $- \rho$ to $\rho$ in the upper and lower half-plane, respectively.

Let

$D \subseteq \mathbb{C}$ be a domain
$z_0 \in D$ , $f$ be meromorphic on $D$ with a simple pole at $z = z_0$
$A_r$ be the circular arc parametrized by $\gamma(\theta) = z_0 + r \exp(i \theta)$ for $\theta \in [\theta_0, \theta_1]$ for some $0 \le \theta_0 < \theta_1 \le 2 \pi$ .