02. Paths and Integration

2.1 Paths

Path

Path in the Complex Plane is continuous function

$$\gamma :[a,b]\to \mathbb{C}$$

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Closed Path

Let $\gamma :[a,b]\to \mathbb{C}$ be a path then

$\gamma$ is a closed path if

$$\gamma (a)=\gamma (b)$$

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Image of a Path

Let $\gamma :[a,b]\to \mathbb{C}$ be a path then

Image of path ${\gamma }^{\ast }$ is defined as

$$\{z\in C:z=\gamma (t)\text{ for some }t\in [a,b]\}$$

Note that by abuse of notation the image of $\gamma$ is also denoted by $\gamma$

Alternative Definition (Personal)

$$\{\gamma (t):t\in [a,b]\}$$

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Differentiable Path

Let $\gamma :[a,b]\to \mathbb{C}$ be a path then

$\gamma$ is differentiable if it’s real and imaginery parts are differentiable as real-valued functions

Equivalently, $\gamma$ is differentiable at $t_0\in [a,b]$ if

$$\underset{t\to t_0}{\lim }\frac{\gamma (t)-\gamma (t_0)}{t-t_0}\text{ exists}$$

in which cases the limit is denoted as ${\gamma }^{\prime }(t_0)$

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Path is $C^1$

Let $\gamma :[a,b]\to \mathbb{C}$ be a path then

$\gamma$ is $C^1$ if it is differentiable and it’s derivative ${\gamma }^{\prime }(t)$ is continuous

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Piecewise $C^1$ Path

Let $\gamma :[a,b]\to \mathbb{C}$ be a path

$\gamma$ is piecewise $C^1$ if

1. Continuous on $[a,b]$
2. Interval $[a,b]$ can be divided into subintervals on each of which $\gamma$ is $C^1$
   So there is a finite sequence $a=a_0<a_1<\cdots a_m=b$ such that

$$\gamma {\mid }_{[a_i,a_{i+1}]}\text{ is }{C}^1$$

Note that in particular it is not necessary that the left-hand and right-hand derivatives of $\gamma$ are equal at $a_i$

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Opposite Path

Let $\gamma :[a,b]\to \mathbb{C}$ then

Opposite Path ${\gamma }^{-1}$ is defined as

$${\gamma }^{-}:[0,b-a]\to \mathbb{C}\text{ where }{\gamma }^{-1}(t)=\gamma (b-t)$$

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Concatenated Paths

Let ${\gamma }_1[a,b]\to \mathbb{C}$ and ${\gamma }_2[c,d]\to \mathbb{C}$ be two paths such that

$${\gamma }_1(b)={\gamma }_2(c)$$

Concatenated Paths of ${\gamma }_1$ and ${\gamma }_2$ is defined as
${\gamma }_1\star {\gamma }_2:[a,b+d-c]$ where

$${\gamma }_1\star {\gamma }_2(t)=\{\begin{array}{ll}{\gamma }_1(t)& t\le b\\ {\gamma }_2(t-b+c)& t\ge b\end{array}$$

Note that a piecewise $C^1$ path is a finite concatenation of $C^1$ paths

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Composite Functions on Paths lemma

Let $\gamma :[c,d]\to \mathbb{C}$
Let $s:[a,b]\to [c,d]$

Suppose that $s$ is differentiable at $t_0$ and $\gamma$ is differentiable at $s_0=s(t_0)$ then

$\gamma \circ s$ is differentiable at $t_0$ with derivative

$$(\gamma \circ s{)}^{\prime }(t_0)=s^{\prime }(t_0)\cdot {\gamma }^{\prime }(s(t_0))$$

Proof

Let $ϵ:[c,d]\to \mathbb{C}$ be given by $ϵ(s_0)=0$ and

$$\gamma (x)=\gamma (s_0)+{\gamma }^{\prime }(s_0)(x-s_0)+(x-s_0)ϵ(x)$$

so that this equation holds for all $x\in [c,d]$

By the definition of ${\gamma }^{\prime }(s_0)$ then $ϵ(x)\to 0$ as $x\to s_0$ so $ϵ$ is continuous at $t_0$

Substituting $x=s(t)$ then for all $t\ne t_0$

$$\frac{\gamma (s(t))-\gamma (s_0)}{t-t_0}=\frac{s(t)-s(t_0)}{t-t_0}\cdot ({\gamma }^{\prime }(s(t))+ϵ(s(t)))$$

As $s(t)$ is differentiable at $t_0$ then $s(t)$ is continuous at $t_0$ hence

$$ϵ(s(t))\to 0\text{ as }t\to t_0$$

Hence taking the limit as $t\to t_0$ then

$$(\gamma \circ s{)}^{\prime }(t_0)=s^{\prime }(t_0)({\gamma }^{\prime }(s_0)+0)=s^{\prime }(t_0){\gamma }^{\prime }(s(t_0))$$

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Reparameterization of Paths

Let $\varphi :[a,b]\to [c,d]$ be continuously differentiable with

$$\varphi (a)=c\text{ and }\varphi (b)=d$$

Let $\gamma :[c,d]\to \mathbb{C}$ be a $C^1$-path

Setting $\tilde{\gamma }=\gamma \circ \varphi$ by Composite functions on paths then

$$\tilde{\gamma }:[a,b]\to \mathbb{C}\text{ is also a }{C}^1\text{-path}$$

which has the same image as $\gamma$

Hence $\tilde{\gamma }$ is a reparameterization of $\gamma$

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Equivalent Paths

Let ${\gamma }_1:[a,b]\to \mathbb{C}$ and ${\gamma }_2:[c,d]\to \mathbb{C}$ be two parametrised paths

${\gamma }_1$ and ${\gamma }_2$ are equivalent if there exists continuously differentiable bijective function

$$s:[a,b]\to [c,d]$$

such that

1. $s^{\prime }(t)>0$ for all $t\in [a,b]$
2. ${\gamma }_1={\gamma }_2\circ s$

Note that ${\gamma }_1$ is piecewise $C^1$ if and only if ${\gamma }_2$ is piecewise $C^1$

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Oriented Curves

As equivalence between paths is a equivalence relation then

The equivalence class of $\gamma$ is

$$[\gamma ]$$

where condition $s^{\prime }(t)>0$ ensures the paths are traversed in the same direction

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Constant Path

Let $z_0\in \mathbb{C}$ then

Define $\gamma :[a,b]\to \mathbb{C}$ by

$$\gamma (t)=z_0\text{ for all }t\in [a,b]$$

Hence ${\gamma }^{\prime }(t)=0$ for all $t\in [a,b]$

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Line Segment path

Let $a,b\in \mathbb{C}$ then

Line Segment Path ${\gamma }_{[a,b]}:[0,1]\to \mathbb{C}$ between $a$ and $b$ is defined by

$${\gamma }_{[a,b]}(t)=a+t(b-a)$$

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2.2 Integration Along a Path

Riemann Integrability lemma

Let $[a,b]$ be a closed interval
Let $S\subset [a,b]$ be a finite set

If $f$ is a bounded continuous function on $[a,b]\setminus S$ then

$$f\text{ is Riemann Integrable on }[a,b]$$

Proof

Case for complex-values functions follows from real case by real and imaginary parts

So suppose $f:[a,b]\setminus S\to \mathbb{R}$
Let $a=x_0<x_1<\cdots <x_k=b$ be any partition of $[a,b]$ which includes the elements of $S$

Then on each open interval $(x_i,x_{i+1})$

$$f\text{ is bounded and continuous hence integrable}$$

Thus

$${\int }_a^{b}f(t)dt={\int }_{x_0}^{x_1}f(t)dt+{\int }_{x_1}^{x_2}f(t)dt+\cdots +{\int }_{x_{k-1}}^{x_k}f(t)dt$$

By standard additivity properties of the integral then
${\int }_a^{b}f(t)dt$ is independent of any choices

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Triangle Inequality for Integrals lemma

Suppose that $f:[a,b]\to \mathbb{C}$ is a complex-valued function then

$$\left|{\int }_a^{b}f(t)dt\right|\le {\int }_a^b|f(t)|dt$$

Proof

If $f(t)=u(t)+iv(t)$ then $|f(t)|=\sqrt{u(t{)}^2+v(t{)}^2}$
Hence if $f$ is integrable then $|f(t)|$ is also integrable

Suppose

$${\int }_a^{b}f(t)dt=r{e}^{i\theta }\text{ where }r\in [0,\infty )\text{ and }\theta \in [0,2\pi )$$

Taking the components of $f$ in the direction of $e^{i\theta }$ and $e^{i(\theta +\pi /2)}=i{e}^{i\theta }$ so

$$f(t)=\tilde{u}(t)e^{i\theta }+i\tilde{v}(t)e^{i\theta }$$

for some real-valued functions $\tilde{u}(t)$ and $\tilde{v}(t)$

By the choice of $\theta$ then

$$r{e}^{i\theta }={\int }_a^{b}f(t)dt=e^{i\theta }{\int }_a^b\tilde{u}(t)dt$$

Note that $\tilde{v}$ disappears as $r$ is real (after factorising $e^{i\theta }$ on the right)

Hence

$$\left|{\int }_a^{b}f(t)dt\right|=\left|{\int }_a^b\tilde{u}(t)dt\right|\le {\int }_a^b|\tilde{u}(t)|dt\le {\int }_a^b|f(t)|dt$$

where we use the triangle inequality for Riemann Integrals of real-valued functions

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Integral of a Function along a Path

Let $f:\mathbb{C}\to \mathbb{C}$

If $\gamma :[a,b]\to \mathbb{C}$ is a piecewise-$C^1$ path then
Integral of $f$ along $\gamma$ is defined as

$${\int }_{\gamma }f(z)dz={\int }_a^{b}f(\gamma (t)){\gamma }^{\prime }(t)dt$$

Note that for the integral to exist then $f(\gamma (t))$ and ${\gamma }^{\prime }(t)$ to be bounded and continuous at all but finitely many $t$

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Integrals of a Function along Equivalent Paths lemma

Let $\gamma :[a,b]\to \mathbb{C}$ and $\tilde{\gamma }:[c,d]\to \mathbb{C}$ be piecewise $C^1$ paths

If $\gamma$ and $\tilde{\gamma }$ are Equivalent paths then
For any continuous function $f:\mathbb{C}\to \mathbb{C}$ then

$${\int }_{\gamma }f(z)dz={\int }_{\tilde{\gamma }}f(z)dz$$

Note that the integral only depends on the oriented curve $[\gamma ]$

Proof

Since $\tilde{\gamma }$ is equivalent to $\gamma$ there is continuously differentiable function

$$s:[c,d]\to [a,b]\text{ with }s(c)=a,s(d)=b\text{ and }{s}^{\prime }(t)>0\text{ for all }t\in [c,d]$$

Suppose that $\gamma$ is $C^1$ then by chain rule

$$\begin{array}{rl}{\int }_{\tilde{\gamma }}f(z)dz& ={\int }_c^{d}f(\gamma (s(t)))\cdot (\gamma \circ s{)}^{\prime }(t)dx\\ & ={\int }_c^{d}f(\gamma (s(t)))\cdot {\gamma }^{\prime }(s(t))\cdot s^{\prime }(t)dt\\ & ={\int }_a^{b}f(\gamma (s))\cdot {\gamma }^{\prime }(s)ds\\ & ={\int }_{\gamma }f(z)dz\end{array}$$

Otherwise if $a=x_0<x_1<\cdots <x_n=b$ is a decomposition of $[a,b]$ into subintervals
Such that $\gamma$ is $C^1$ on $[x_i,x_{i+1}]$ for $1\le i\le n-1$ then

Since $s$ is a continuous increasing bijection then
Points $s^{-1}(x_0)<\cdots <s^{-1}(x_n)$ is a decomposition of $[c,d]$
Hence

$$\begin{array}{rl}{\int }_{\tilde{\gamma }}f(z)dz& ={\int }_c^{d}f(\gamma (s(t)))\cdot {\gamma }^{\prime }(s(t))\cdot s^{\prime }(t)dt\\ & =\sum _{i=0}^{n-1}{\int }_{s^{-1}(x_i)}^{s^{-1}(x_{i+1})}f(\gamma (s(t)))\cdot {\gamma }^{\prime }(s(t))\cdot s^{\prime }(t)dt\\ & =\sum _{i=0}^{n-1}{\int }_{x_i}^{x_{i+1}}f(\gamma (x))\cdot {\gamma }^{\prime }(x)dx\\ & ={\int }_a^{b}f(\gamma (x))\cdot {\gamma }^{\prime }(x)dx\\ & ={\int }_{\gamma }f(z)dz\end{array}$$

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Length of a Path

Let $\gamma :[a,b]\to \mathbb{C}$ be a $C^1$ path then

Length of $\gamma$ is defined as

$$\ell (\gamma )={\int }_a^b|{\gamma }^{\prime }(t)|dt$$

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Integral of a Path with respect to Arc-Length

Let $f:U\to \mathbb{C}$
Let $\gamma :[a,b]\to \mathbb{C}$ be a $C^1$ path with ${\gamma }^{\ast }\subseteq U$ then

Integral of $f$ along $\gamma$ with respect to arc-length is defined as

$${\int }_{\gamma }f(z)|dz|={\int }_a^{b}f(\gamma (t))|{\gamma }^{\prime }(t)|dt$$

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Properties of Integrals along Paths

Let $f,g:U\to \mathbb{C}$ be continuous functions on an open subset $U\subseteq \mathbb{C}$
Let $\gamma ,\eta :[a,b]\to \mathbb{C}$ be piecewise-$C^1$ paths whose images lie in $U$ then

1. Linearity
   For $\alpha ,\beta \in \mathbb{C}$

$${\int }_{\gamma }(\alpha f(z)+\beta g(z))dz=\alpha {\int }_{\gamma }f(z)dz+\beta {\int }_{\gamma }g(z)dz$$

2. Orientation
   If ${\gamma }^{-}$ denotes the opposite path to $\gamma$

$${\int }_{\gamma }f(z)dz=-{\int }_{{\gamma }^{-}}f(z)dz$$

3. Additivity
   If $\gamma \star \eta$ is the concatenation of paths $\gamma ,\eta$ in $U$

$${\int }_{\gamma \star \eta }f(z)dz={\int }_{\gamma }f(z)dz+{\int }_{\eta }f(z)dz$$

4. Estimation Lemma

$$\left|{\int }_{\gamma }f(z)dz\right|\le \underset{z\in {\gamma }^{\ast }}{\sup }|f(z)|\cdot \ell (\gamma )$$

Proof $(1),(2),(3)$ As $f,g$ are continuous and $\gamma ,\eta$ are piecewise $C^1$ then the integrals are all well-defined

For

Functions

\begin{align*} &f(\gamma(t))\gamma'(t) \\ &f(\eta(t))\eta'(t) \\ &g(\gamma(t))\gamma'(t)$ \\ &g(\eta(t))\eta'(t) \end{align*}

are all Riemann Integrable

$(1)$ is immediate from linearity of the Riemann Integral
$(2)$ follows from definitions and using $({\gamma }^{-}{)}^{\prime }(t)=-{\gamma }^{\prime }(a+b-t)$
$(3)$ follows from corresponding additivity property of Riemann Integrals

For $(4)$
$\gamma ([a,b])$ is compact in $\mathbb{C}$ as it is the image of a compact set $[a,b]$ under a continuous map
So $|f|$ is bounded on $\gamma ([a,b])$ so ${\sup }_{z\in \gamma ([a,b])}|f(z)|$ exists
Hence

$$\begin{array}{rl}\left|{\int }_{\gamma }f(z)dz\right|& =\left|{\int }_a^{b}f(\gamma (t)){\gamma }^{\prime }(t)dt\right|\\ & \le {\int }_a^b|f(\gamma (t))|\cdot |{\gamma }^{\prime }(t)|dt\\ & \le \underset{z\in {\gamma }^{\ast }}{\sup }|f(z)|\cdot {\int }_a^b|{\gamma }^{\prime }(t)|dt\\ & =\underset{z\in {\gamma }^{\ast }}{\sup }|f(z)|\cdot \ell (\gamma )\end{array}$$

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Integrals of Uniform Converging Functions along a Path

Let $f_n:U\to \mathbb{C}$ be a sequence of continuous functions on an open subset $U\subset \mathbb{C}$
Suppose $\gamma :[a,b]\to \mathbb{C}$ is a path such that ${\gamma }^{\ast }\subseteq U$

If $(f_n)$ converges uniformly to function $f$ on image $\gamma$ then

$${\int }_{\gamma }{f}_n(z)dz\to {\int }_{\gamma }f(z)dz$$

Proof

$$\begin{array}{rl}\left|{\int }_{\gamma }f(z)dz-{\int }_{\gamma }{f}_n(z)dz\right|& =\left|{\int }_{\gamma }(f(z)-f_n(z))dz\right|\\ & \le \underset{z\in {\gamma }^{\ast }}{\sup }\{|f(z)-f_n(z)|\}\ell (\gamma )\end{array}$$

Since $f_n\to f$ uniformly on ${\gamma }^{\ast }$ then

$$\sup \{|f(z)-f_n(z)|:z\in {\gamma }^{\ast }\}\to 0\text{ as }n\to \infty$$

Hence this implies the result

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Primitives

Let $U\subseteq \mathbb{C}$ be an open set
Let $f:U\to \mathbb{C}$ be a continuous function

If there exists differentiable function $F:U\to \mathbb{C}$ with

$$F^{\prime }(z)=f(z)$$

then

$$F\text{ is a primitive for }f\text{ on }U$$

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Derivative of a Composition of a Function with a Path lemma

Let $U$ be an open subset of $\mathbb{C}$
Let $f:U\to \mathbb{C}$ be a holomorphic function

If $\gamma :[a,b]\to U$ is a piecewise $C^1$-path then

$f(\gamma (t))$ is differentiable at any $t$ where $\gamma$ is differentiable with

$$\frac{d}{dt}(f(\gamma (t)))=f^{\prime }(\gamma (t))\cdot {\gamma }^{\prime }(t)$$

Proof

Assume that $\gamma$ is differentiable at $t_0\in [a,b]$
Let $z_0=\gamma (t_0)\in U$

By definition of $f^{\prime }$ there exists function $ϵ(z)$ such that

$$f(z)=f(z_0)+f^{\prime }(z_0)(z-z_0)+ϵ(z)(z-z_0)$$

where $ϵ(z)\to 0=ϵ(z_0)$ as $z\to z_0$ then

$$\frac{f(\gamma (t))-f(\gamma (t_0))}{t-t_0}=f^{\prime }(z_0)\cdot \frac{\gamma (t)-\gamma (t_0)}{t-t_0}+ϵ(\gamma (t))\cdot \frac{\gamma (t)-\gamma (t_0)}{t-t_0}$$

As $t\to t_0$ then

$$\begin{array}{rl}{f}^{\prime }(z_0)\cdot \frac{\gamma (t)-\gamma (t_0)}{t-t_0}& \to f^{\prime }(z_0){\gamma }^{\prime }(t_0)\\ \frac{\gamma (t)-\gamma (t_0)}{t-t_0}& \to {\gamma }^{\prime }(t_0)\end{array}$$

As $\frac{\gamma (t)-\gamma (t_0)}{t-t_0}$ is bounded as $t\to t_0$ and $\gamma (t)$ is continuous at $t_0$ by differentiability there so

$$ϵ(\gamma (t))\to ϵ(\gamma (t_0))=ϵ(z_0)=0$$

Hence as $t\to t_0$

$$\frac{f(\gamma (t))-f(\gamma (t_0))}{t-t_0}\to f^{\prime }(z_0){\gamma }^{\prime }(t_0)$$

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04 - Fundamental Theorem of Calculus

Fundamental Theorem of Calculus

Let $U\subseteq \mathbb{C}$ be open
Let $f:U\to \mathbb{C}$ be an continuous function
Let $\gamma :[a,b]\to U$ be a piecewise $C^1$ path in $U$

If $F:U\to \mathbb{C}$ is a primitive for $f$ then

$${\int }_{\gamma }f(z)dz=F(\gamma (b))-F(\gamma (a))$$

If $\gamma$ is a closed path then the integral is then $0$

Proof

Suppose that $\gamma$ is $C^1$ then

$$\begin{array}{rl}{\int }_{\gamma }f(z)dz& ={\int }_{\gamma }{F}^{\prime }(z)dz\\ & ={\int }_a^b{F}^{\prime }(\gamma (t)){\gamma }^{\prime }(t)dt(\text{by chain rule})\\ & ={\int }_a^b\frac{d}{dt}(F\circ \gamma )(t)dt\\ & =F(\gamma (b))-F(\gamma (a))\end{array}$$

However if $\gamma$ is piecewise $C^1$ then there exists
Partition $a=a_0<a_1<\cdots <a_k=b$ such that $\gamma$ is $C^1$ on $[a_i,a_{i+1}]$ for $i=0,1,\cdots ,k-1$
Hence obtain telescoping sum

$$\begin{array}{rl}{\int }_{\gamma }f(z)dz& ={\int }_a^{b}f(\gamma (t)){\gamma }^{\prime }(t)dt\\ & =\sum _{i=0}^{k-1}{\int }_{a_i}^{a_{i+1}}f(\gamma (t)){\gamma }^{\prime }(t)dt\\ & =\sum _{i=0}^{k-1}(F(\gamma (a_{i+1}))-F(\gamma (a_i)))\\ & =F(\gamma (b))-F(\gamma (a))\end{array}$$

If $\gamma$ is closed $\gamma (a)=\gamma (b)$ hence the integral of $f$ along the closed path is $0$

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Zero Derivative Function is Constant corollary

Let $U$ be a domain
Let $f:U\to \mathbb{C}$ be a function

If $f^{\prime }(z)=0$ for all $z\in U$ then

$$f\text{ is constant}$$

Proof

Let $z_0\in U$
As open connected set of a normed space is path-connected then

Any point $w$ of $U$ can be joined to $z_0$ by a piecewise $C^1$-path

$$\gamma :[0,1]\to U$$

so that $\gamma (0)=z_0$ and $\gamma (1)=w$

So by Fundamental Theorem of Calculus then

$$f(w)-f(z_0)={\int }_{\gamma }{f}^{\prime }(z)dz=0$$

Hence $f$ is constant

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05 - Existence of Primitives Theorem

Existence of Primitives Theorem

Let $U$ be a domain
Let $f:U\to \mathbb{C}$ be a continuous function

If for any closed path $\gamma$ in $U$ has ${\int }_{\gamma }f(z)dz=0$ then

$$f\text{ has a primitive}$$

Proof

Fix $z_0\in U$

For any $z\in U$ define

$$F(z)={\int }_{\gamma }f(z)dz$$

where $\gamma :[a,b]\to U$ with $\gamma (a)=z_0$ and $\gamma (b)=z$

Need to show that $F(z)$ is independent of the choice of $\gamma$
Suppose ${\gamma }_1,{\gamma }_2$ are two paths with ${\gamma }_1(a)={\gamma }_2(a)=z_0$ and ${\gamma }_1(b)={\gamma }_2(b)=z$ then
Let $\gamma ={\gamma }_1\star {\gamma }_2^{-}$ so $\gamma$ is closed so using Properties of integrals along paths

$$0={\int }_{\gamma }f(z)dz={\int }_{{\gamma }_1}f(z)dz+{\int }_{{\gamma }_2^{-}}f(z)dz={\int }_{{\gamma }_1}f(z)dz-{\int }_{{\gamma }_2}f(z)dz$$

Hence

$${\int }_{{\gamma }_1}f(z)dz={\int }_{{\gamma }_2}f(z)dz$$

Need to show that $F$ is differentiable with $F^{\prime }(z)=f(z)$
Fix $w\in U$ and $ϵ>0$ such that $B(w,ϵ)\in U$

Let any path $\gamma :[a,b]\to U$ with $\gamma (a)=z_0$ and $\gamma (b)=w$
Let straight line path $s:[0,1]\to U$ from $w$ to $z$ be given by

$$s(t)=w+t(z-w)$$

If $z\in B(w,ϵ)\subseteq U$ then concatenation of $\gamma$ with $s$ is a path ${\gamma }_1$ from $z_0$ to $z$ so

$$\begin{array}{rl}F(z)-F(w)& ={\int }_{{\gamma }_1}f(z)dz-{\int }_{\gamma }f(z)dz\\ & =\left({\int }_{\gamma }f(z)dz+{\int }_{s}f(z)dz\right)-{\int }_{\gamma }f(z)dz\\ & ={\int }_{s}f(z)dz\end{array}$$

However for $z\ne w$ then

$$\begin{array}{rl}\left|\frac{F(z)-F(w)}{z-w}-f(w)\right|& =\left|\frac{1}{z-w}\left({\int }_0^{1}f(w+t(z-w))(z-w)dt\right)-f(w)\right|\\ & =\left|{\int }_0^1(f(w+t(z-w))-f(w))dx\right|\\ & \le \underset{t\in [0,1]}{\sup }|f(w+t(z-w))-f(w)|\to 0\text{ as }z\to w\end{array}$$

as $f$ is continuous at $w$

Hence $F$ is differentiable at $w$ with derivative

$$F^{\prime }(w)=f(w)$$

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2.3 Cauchy’s Theorem

06 - Cauchy or Cauchy-Goursat Theorem

Cauchy or Cauchy-Goursat Theorem

Let $U\subset \mathbb{C}$ be a domain
Let $\gamma$ be a closed curve such that $\gamma$ and all bounded components of $\mathbb{C}\setminus {\gamma }^{\ast }$ are inside $U$

Let $f$ be a function holomorphic in $U$ then

$${\int }_{\gamma }f(z)dz=0$$

Proof - Lecture Notes

Refer to lecture notes (page 48)

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Convex

Let $X$ be a subset in $\mathbb{C}$ then

$X$ is convex if
For all $z,w\in \mathbf{x}$ then

$$\text{line segment between }z\text{ and }w\text{ is contained in }X$$

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Star-Like

Let $X$ be a subset in $\mathbb{C}$ then

$X$ is star-like if
There exists point $z_0$ in $X$ such that for all $w\in X$ then

$$\text{line segment between }{z}_0\text{ and }w\text{ is contained in }X$$

so that $X$ is star-like with respect to $z_0$

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2.4 Deformation Theorem and Homotopy

Homotopic

Let $U$ be an open set in $\mathbb{C}$

Let $a,b\in U$
Let ${\gamma }_0:[0,1]\to U,{\gamma }_1:[0,1]\to U$ be two paths in $U$ such that

$${\gamma }_0(0)={\gamma }_1(0)=a\text{ and }{\gamma }_0(1)={\gamma }_1(1)=b$$

${\gamma }_1$ and ${\gamma }_2$ are homotopic in $U$ if there exists continuous function $h:[0,1]\times [0,1]\to U$ with

$$\begin{array}{rlrlrl}h(0,s)& =a& & h(1,s)& & =b\\ h(t,0)& ={\gamma }_0(t)& & h(t,1)& & ={\gamma }_1(t)\end{array}$$

Note that $h$ acts like a interpolation between the two curves

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Null Homotopic

Let $U$ be an open set in $\mathbb{C}$

Closed curve $\gamma$ is null homotopic in $U$ if

$$\gamma \text{ is homotopic to a constant path}$$

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Simply Connected

Let $U$ be a domain in $C$

$U$ is simply connected if for every $a,b\in U$ any two paths ${\gamma }_1,{\gamma }_2$ from $a$ to $b$ then

$${\gamma }_1\text{ and }{\gamma }_2\text{ are homotopic}$$

Equivalently, any closed curve is null homotopic

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Annulus

$$A(x_0,r,R)=\{z\in \mathbb{C}:0\le r<|z-z_0|<R\le \infty \}$$

Note that annulus is not simply connected

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07 - Homotopy Cauchy's Theorem

Homotopy Cauchy's Theorem / Deformation Theorem

Let $U$ be a domain in $\mathbb{C}$
Suppose that ${\gamma }_1$ and ${\gamma }_2$ are two paths in $U$ and are homotopic in $U$

Let $f:U\to \mathbb{C}$ be a holomorphic function then

$${\int }_{{\gamma }_1}f(z)dz={\int }_{{\gamma }_2}f(z)dz$$

Proof - Non-Examinable

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08 - Cauchy's Theorem for Simply Connected Domains

Cauchy's Theorem for Simply Connected Domains

Suppose $U$ is a simply connected domain
Let $a,b\in U$

Let $f:U\to C$ be a holomorphic function on $U$ then

If ${\gamma }_1,{\gamma }_2$ are paths from $a$ to $b$ then

$${\int }_{{\gamma }_1}f(z)dz={\int }_{{\gamma }_2}f(z)dz$$

In particular if $\gamma$ is a closed oriented curve then ${\int }_{\gamma }f(z)dz=0$
so any holomorphic function on $U$ has a primitive

Proof

Since $U$ is simply connected then
Any two paths from $a$ to $b$ are homotopic so first part is by Homotopy Cauchy’s Theorem

In a simply connected domain, any closed path

$$\gamma :[0,1]\to U\text{ with }\gamma (0)=\gamma (1)=a$$

is homotopic to some constant path $c(t)=z_0$ hence

$${\int }_{\gamma }f(z)dz={\int }_{c}f(z)dz=0$$

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2.5 Winding Numbers

Existence of a Continuous Argument

Let $\gamma :[0,1]⟹\mathbb{C}\setminus \{0\}$ be a path then

There exists continuous function $a:[0,1]\to \mathbb{R}$ such that

$$\gamma (t)=|\gamma (t)|e^{2\pi ia(t)}$$

If $a,b$ are two such functions that satisfy the above then there exists $n\in \mathbb{Z}$ such that

$$a(t)=b(t)+n\text{ for all }t\in [0,1]$$

So $a(t_0)$ at any $t_0$ uniquely determines $a(t)$ for all $t$

Proof

Replacing $\gamma (t)$ with $\frac{\gamma (t)}{|\gamma (t)|}$, assume that $|\gamma (t)|=1$

Since $\gamma$ is continuous on a compact set, then $\gamma$ is uniformly continuous
So there is $\delta >0$ such that

$$|\gamma (s)-\gamma (t)|<\sqrt{3}\text{ for all }s,t\text{ with }|s-t|<\delta$$

Choose integer $n>0$ such that $n>\frac{1}{\delta }$ so on subinterval $\left[\frac{j}{n},\frac{j+1}{n}\right]$ then

$$|\gamma (s)-\gamma (t)|<\frac{\sqrt{3}}{2}$$

So for any half-plane in $\mathbb{C}$ then continuous argument function can be defined

If $|z_1|=|z_2|=1$ and $|z_1-z_3|<\sqrt{3}$ then angle between $z_1$ and $z_2$ is at most $\frac{2\pi }{3}$

So there exists continuous function $a_i:\left[\frac{j}{n},\frac{j+1}{n}\right]\to \mathbb{R}$ such that

$$\gamma (t)=e^{2\pi i{a}_j(t)}\text{ for }t\in \left[\frac{j}{n},\frac{j+1}{n}\right]$$

as $\gamma \left(\left[\frac{j}{n},\frac{j+1}{n}\right]\right)$ must lie in an arc of length at most $\frac{2\pi }{3}$

As $e^{2\pi i{a}_j(j/n)}=e^{2\pi i{a}_{j-1}(j/n)}$ and $a_{j-1}\left(\frac{j}{n}\right)$ and $a_i\left(\frac{j}{n}\right)$ differ by an integer then
Successively adjust $a_j$ for $j>1$ by an integer to get continuous function $a:[0,1]\to \mathbb{C}$
so that

$$\gamma (t)=e^{2\pi ia(t)}$$

Uniqueness statement follows from as

$$e^{2\pi i(a(t)-b(t))}=1$$

with $a(t)-b(t)\in \mathbb{Z}$ so since $[0,1]$ is connected then $a(t)-b(t)$ is constant

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Winding Number

Let $\gamma :[0,1]\to \mathbb{C}\setminus \{0\}$ be a closed path and

$$\gamma (t)=|\gamma (t)|e^{2\pi ia(t)}$$

As $\gamma (0)=\gamma (1)$ then

$$I(\gamma ,0):=a(1)-a(0)\in \mathbb{Z}\text{ is known as the winding number}$$

where $I(\gamma ,0)$ is the winding number of $\gamma$ around $0$

If $z_0$ is not in the image of $\gamma$, it is still defined in the same fashion where
Let $t:\mathbb{C}\to \mathbb{C}$ be defined by $t(z)=z-z_0$ so

$$I(\gamma ,z_0)=I(t\circ \gamma ,0)$$

where the quantity is called the index of $z_0$ with respect to $\gamma$

Uniqueness

Uniquely determined by path $\gamma$ as function $a$ is unique up to an integer

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Cauchy Integral Formula for the Winding Number lemma

Let $\gamma$ be a piecewise $C^1$ closed path
Let $z_0\in \mathbb{C}$ be a point not in the image of $\gamma$ then

Winding number $I(\gamma ,z_0)$ of $\gamma$ around $z_0$ is given by

$$I(\gamma ,z_0)=\frac{1}{2\pi i}{\int }_{\gamma }\frac{1}{z-z_0}dz$$

Proof

If $\gamma :[0,1]\to \mathbb{C}$ then write

$$\gamma (t)=z_0+r(t)e^{2\pi ia(t)}$$

where $r(t)=|\gamma (t)-z_0|>0$ is continuous and existence of $a(t)$ is guaranteed by Existence of a Continuous Argument

So

$$\begin{array}{rl}{\int }_{\gamma }\frac{1}{z-z_0}dz& ={\int }_0^1\frac{1}{r(t)e^{2\pi ia(t)}}(r^{\prime }(t)+r(t)\cdot 2\pi i{a}^{\prime }(t))e^{2\pi ia(t)}dt\\ & ={\int }_0^1\frac{r^{\prime }(t)}{r(t)}+2\pi i{a}^{\prime }(t)dt\\ & =[\log (r(t))+2\pi ia(t){]}_0^1\\ & =2\pi i(a(1)-a(0))\end{array}$$

as $r(1)=r(0)=|\gamma (0)-z_0|$

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Entire

Function $f:\mathbb{C}\to \mathbb{C}$ is entire if it is complex diffferentiable on the whole complex plane

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General Cauchy Integral Formula for the Winding Number

Let $U$ be an open set in $\mathbb{C}$
Let $\gamma :[0,1]\to U$ be a closed path

If $f(z)$ is a continuous function on ${\gamma }^{\ast }$ then

$$I_f(\gamma ,w)=\frac{1}{2\pi i}{\int }_{\gamma }\frac{f(z)}{z-w}dz$$

is analytic in $w$

Particularly if $f(z)=1$ then $w\mapsto I(\gamma ,w)$ is a continuous function on $\mathbb{C}\setminus {\gamma }^{\ast }$
Since it is integer-valued then it is constant on connected components of $\mathbb{C}\setminus {\gamma }^{\ast }$

Proof

By translation, assume $z_0=0$

Since $\mathbb{C}\setminus {\gamma }^{\ast }$ is open, there exists some $r>0$ such that

$$B(0,2r)\cap {\gamma }^{\ast }=\varnothing$$

For $w\in B(0,r)$ and $z\in {\gamma }^{\ast }$ then

$$\left|\frac{w}{z}\right|<\frac{1}{2}$$

Since ${\gamma }^{\ast }$ is compact then

$$M=\sup \{|f(z)|:z\in {\gamma }^{\ast }\}\text{ is finite}$$

Hence

$$\left|f(z)\frac{w^n}{z^{n+1}}\right|=|f(z)||z{|}^{-1}{\left|\frac{w}{z}\right|}^n<\frac{M}{2r}{\left(\frac{1}{2}\right)}^n\forall z\in {\gamma }^{\ast }$$

By Weierstrass M-test then series

$$\sum _{n=0}^{\infty }\frac{f(z)w^n}{z^{n+1}}=\sum _{n=0}^{\infty }\frac{f(z)}{z}{\left(\frac{w}{z}\right)}^n=\frac{f(z)}{z}{\left(1-\frac{w}{z}\right)}^{-1}=\frac{f(z)}{z-w}$$

is viewed as a function of $z$ converges uniformly on ${\gamma }^{\ast }$ to $\frac{f(z)}{z-w}$

Hence for all $w\in B(0,r)$ then

$$I_f(\gamma ,w)=\frac{1}{2\pi i}{\int }_{\gamma }\frac{f(z)}{z-w}dz=\sum _{n=0}^{\infty }\left(\frac{1}{2\pi i}{\int }_{\gamma }\frac{f(z)}{z^{n+1}}dz\right)w^n$$

Thus $I_f\gamma ,w$ is given by a power series in $B(0,r)$ and hence analytic and holomorphic

Finally, if $f=1$ then as $I_1(\gamma ,z)=I(\gamma ,z)$ is integer-valued then
It must be constant on any connected component of $\mathbb{C}\setminus {\gamma }^{\ast }$ as required

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Derivatives of the General Cauchy Integral Formula

Let

$$g(w)=I_f(\gamma ,w)$$

at $z_0$ for some continuous function $f$ then

$$g^{(n)}(z_0)=\frac{n!}{2\pi i}{\int }_{\gamma }\frac{f(z)}{(z-z_0{)}^{n+1}}dz$$

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09 - Jordan Curve Theorem

Jordan Curve Theorem

Let $\gamma$ be a simple closed curve then

$\mathbb{C}\setminus {\gamma }^{\ast }$ has precisely one bounded and one unbounded component

1. Interior of $\gamma$: the bounded component (inside region of $\gamma$)
2. Exterior of $\gamma$: the unbounded component (outside region of $\gamma$)

Note that the components refer to the regions where $\gamma$ cuts $\mathbb{C}$

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Positively Oriented

Let $U$ be an open set in $\mathbb{C}$
Let $\gamma :[0,1]\to U$ be a closed curve

$\gamma$ is positively oriented then if
Winding number around any point from bounded component is $1$

Otherwise $\gamma$ is negatively oriented

Equivalent Definition $w$ be a point from the bounded component then

Let

$\gamma$ is positively oriented if

$$\frac{1}{2\pi i}{\int }_{\gamma }\frac{1}{z-w}dz=1$$

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Homotopy between Interior Closed Paths and Interior Circle

Let $U$ be a domain
Let $\gamma$ be a simply positively oriented curve such that $\gamma$ and its interior are inside $U$

Let $r>0$ such that $B(w,r)$ is inside $\gamma$
Denote the positively oriented circle of radius $r$ around $w$ by ${\gamma }_r$ then

$$\gamma \text{ is homotopic to }{\gamma }_r\text{ inside }U\setminus \{w\}$$

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