Quartz sync: May 25, 2025, 4:30 PM

content/Definitions/Complex Analysis/Contractible Paths.md

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+A path $\gamma$ is called **contractible** if it is [[Homotopy|homotopic]] to a point (where a point is considered to be a closed constant path).

content/Definitions/Complex Analysis/Exponentials/Complex Exponential.md

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+The **complex exponential** $\exp : \mathbb{C} \to \mathbb{C}$ is defined by $\exp(z) := e^{x} (\cos(y) + i \sin(y))$, that is
+$$\exp(z) = e^{x} \times e^{iy}$$
+> [!note]
+> For $y=0$, it gives back $e^{x}$
+

content/Definitions/Complex Analysis/Exponentials/Complex Power.md

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+# Definition
+Let $\alpha \in \mathbb{C}$. We define
+$$z^{\alpha} = e^{\alpha\operatorname{Log} z}$$
+We observe that $z \mapsto z^{\alpha}$ is a single-valued function, because we are using the [[Principal Logarithm]].

content/Definitions/Complex Analysis/Exponentials/Principal Argument.md

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+# Definition
+Given a non-zero complex number $z \in \mathbb{C}$, its **principal argument** $\operatorname{Arg} z$ is the unique argument for $z$ such that
+$$-\pi \lt \operatorname{Arg} z \leq \pi.$$

content/Definitions/Complex Analysis/Exponentials/Principal Logarithm.md

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+# Definition
+The **principal logarithm** is the function $\operatorname{Log} : \mathbb{C} \setminus \{ 0 \} \to \mathbb{C}$ is defined by
+$$\operatorname{Log} z = \ln|z| + i \operatorname{Arg} z.$$
+>[!note] For $z = x \gt 0$
+>We get $\operatorname{Log} x = \ln x$.

content/Definitions/Complex Analysis/Integrals/Antiderivative.md

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+# Definition
+Let $F : D \to \mathbb{C}$ be a [[Holomorphic|holomorphic function]]. Suppose $F'(z) = f(z)$ for all $z \in D$. Then $F$ is an **antiderivative** of $f$ on $D$.

content/Definitions/Complex Analysis/Integrals/Cauchy's Theorem.md

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+# Definition
+Let $f : D \to \mathbb{C}$ be [[Holomorphic|holomorphic]]. Suppose $\gamma_{1}$ and $\gamma_{2}$ are two ([[Smooth Parametrization|parametrized]]) closed curves in $D$ such that $\gamma_{1} \sim \gamma_{2}$. Then we have
+$$\int_{\gamma_{1}} f = \int_{\gamma_{2}} f.$$
+# Proof
+The proof can be found [[Lecture 22 - Cauchy's theorem and Cauchy's Integral Formula#Proof of Cauchy's Theorem|here]].

content/Definitions/Complex Analysis/Integrals/Complex Integral.md

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+# Definition
+We start by considering the natural analogue of the real integral $\int_{a}^{b} f(t) \, dt$ along the segment $[a, b]$, defined as follows:
+
+Let $a, b \in \mathbb{R}$ and $f : [a,b] \to \mathbb{C}$ be continuous. Then we set
+$$\int_{a}^{b} f(t) \, dt := \int_{a}^{b} \mathrm{Re}f(t) \, dt + i \int_{a}^{b} \mathrm{Im}f(t) \, dt  $$
+Here $\mathrm{Re} f$ and $\mathrm{Im} f$ denote the real and imaginary parts of the complex-valued function $f$. Hence the integrals on the right-hand side are familiar integrals of real functions.
+
+# Example
+An example can be found in [[Lecture 21 - Integration, Antiderivatives, Homotopies#Example of a Complex Integral]].

content/Definitions/Complex Analysis/Integrals/Homotopy.md

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+# Definition
+Let $\gamma_{1}(t)$ and $\gamma_{2}(t)$ be two closed ([[Smooth Parametrization|parametrized]]) curves in a region $D$, with parameter $0 \leq t \leq 1$.
+They are called **homotopic** if there exists a continuous function $h : [0,1] \times [0, 1] \to D$ such that, for all $s, t \in [0,1]$, we have
+$$h(t,0) = \gamma_{1}(t),\ h(t,1) = \gamma_{2}(t),$$
+$$h(0, s) = h(1, s).$$
+In this case we use the notation $\gamma_{1} \sim \gamma_{2}$.
+# Example
+An example can be found [[Lecture 21 - Integration, Antiderivatives, Homotopies#Example of Homotopy|here]].

content/Definitions/Complex Analysis/Integrals/Integral along a Curve.md

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+# Definition
+We will define the integral along a curve $C$ in the complex plane. We will consider curves $C$ which admit a [[Smooth Parametrization]].
+
+Let $C$ be a smooth curve with parametrization $\gamma : [a,b] \to \mathbb{C}$. Let $f$ be a function which is continuous on $C$. Then we set
+$$\int_{C} f(z) \, dz := \int_{a}^{b} f(\gamma(t)) \gamma'(t) \, dt.$$
+# Example
+An example can be found in [[Lecture 21 - Integration, Antiderivatives, Homotopies#Example of Integrating a Complex Curve]].

content/Definitions/Complex Analysis/Integrals/Reparametrizations/Opposite Orientation.md

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+# Definition
+Let $\gamma : [a,b] \to \mathbb{C}$ be a [[Smooth Parametrization]] of a curve $C$.
+
+We say that $\gamma$ and $\sigma$ have the **opposite orientation** if $\alpha'(t) \leq 0$ for all $t \in [c,d]$.

content/Definitions/Complex Analysis/Integrals/Reparametrizations/Reparametrization.md

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+# Definition
+Let $\gamma : [a,b] \to \mathbb{C}$ be a [[Smooth Parametrization]] of a curve $C$.
+
+We say that $\sigma : [c,d] \to \mathbb{C}$ is a **reparametrization** of $\gamma$ if there exists a function $\alpha : [c, d] \to [a,b]$ such that $\sigma(t) = \gamma(\alpha(t))$ for all $t \in [c,d]$.

content/Definitions/Complex Analysis/Integrals/Reparametrizations/Same Orientation.md

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+# Definition
+Let $\gamma : [a,b] \to \mathbb{C}$ be a [[Smooth Parametrization]] of a curve $C$.
+
+We say that $\gamma$ and $\sigma$ have the **same orientation** if $\alpha'(t) \geq 0$ for all $t \in [c,d]$.

content/Definitions/Complex Analysis/Integrals/Smooth Parametrization.md

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+# Definition
+Define a curve $C$ in the complex plane. The curve $C$ admits a **smooth parametrization** when $C$ is described by a function $\gamma : [a,b] \to \mathbb{C}$ which is infinitely-differentiable.

content/Exam Preparation.md

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 Taken directly from Canvas:
 # Exam topic
 The following list is meant to provide a starting point for the type of questions (relating to the complex function theory part of the course) you will get at the exam (but this is not an exhaustive list). You are not expected to know the details of the proofs, but you need to show that you understand the concepts and know how to apply them.
-
-- Similarities and differences between $\mathbb{C}$ and $\mathbb{R}^2$.
-- Holomorphic functions and their properties.
-- Complex exponential and logarithm.
-- Integration in the complex plane.
-- Cauchy's theorem and integral formula.
-- Taylor and Laurent series.
-- Singularities and their classification.
-- Residues and the residue theorem.
-- Applications to the computation of integrals.
 # Relevant Questions
+## Topology and Measure Theory
 Here are 20 relevant exam questions in the topology and measure theory part of the course:
 
 1. What distinguishes the real numbers from the rational ones?
@@ -33,4 +24,14 @@ Here are 20 relevant exam questions in the topology and measure theory part of t
 17. (Not relevant: State the Riesz representation theorem.
 18. What is the Lebesgue measure on $\mathbb{R}^n$ ?
 19. What is a complex measure?
-20. State the Lebesgue-Radon-Nikodym theorem.)
+20. State the Lebesgue-Radon-Nikodym theorem.)
+## Complex Analysis
+- Similarities and differences between $\mathbb{C}$ and $\mathbb{R}^2$.
+- Holomorphic functions and their properties.
+- Complex exponential and logarithm.
+- Integration in the complex plane.
+- Cauchy's theorem and integral formula.
+- Taylor and Laurent series.
+- Singularities and their classification.
+- Residues and the residue theorem.
+- Applications to the computation of integrals.

content/Lectures/Lecture 20 - Complex Exponential, Logarithm, and Powers.md

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+---
+lecture: 20
+date: 2025-03-27
+---
+# Complex Exponential
+Definition of a complex exponential is given [[Complex Exponential|here]].
+## Proposition
+The function $\exp$ satisfies the following:
+1. $\exp(z) = e^{z}$ is [[Entire|entire]] and $\frac{de^{z}}{dz} = z$,
+2. $e^{z_{1}} e^{z_{2}} = e^{z_{1} + z_{2}}$
+3. $e^{z + 2\pi i n} = e^{z}$ for $n \in \mathbb{Z}$
+### Proof
+#### Proposition 1
+Writing $z = x + iy$ we get
+$$\frac{\partial e^{z}}{\partial x} = e^{x} (\cos(y) + i \sin(y)),$$
+$$\frac{\partial e^{z}}{\partial y} = e^{x} (-\sin(y) + i\cos(y)).$$
+These are [[Continuous|continuous]] and
+$$\frac{\partial e^{z}}{\partial x} = -i \frac{\partial e^{z}}{\partial y}.$$
+Hence $e^{z}$ is [[Holomorphic|holomorphic]] for any $z$ by the [[Cauchy-Riemann Equations]].
+#### Proposition 2
+We have
+$$e^{z_{1}} e^{z_{2}} = e^{x_{1}} e^{x_{2}} e^{iy_{1}} e^{iy_{2}}$$
+$$= e^{x_{1} + x_{2}} e^{i(y_{1} + y_{2})}$$
+$$= e^{z_{1} + z_{2}}$$
+Where we used the previous result for $e^{i \phi}$
+#### Proposition 3
+This follows from the periodicity of $\cos(y)$ and $\sin(y)$.
+
+>[!info]+ Another possible definition of $e^{z}$
+>$e^{z} =  \sum_{n=0}^{\infty} \frac{z^{n}}{n!}$
+>This leads to [[Euler's Formula]]:
+>$e^{ix} = \cos(x) + i \sin(x)$
+
+# The Complex Logarithm
+In the real case, $\log$ is the **inverse** function of the $\exp$. It is mapped as a one-to-one function $\exp : \mathbb{R} \to \mathbb{R}$. This means that:
+$$\ln(\exp(x)) = x,\ \forall x \in \mathbb{R},$$
+$$\exp(\ln(y)) = y,\ \forall y \gt 0.$$
+However, this is **not** the case in the complex case, $e^{z}$, since $e^{z + 2\pi i n} = e^{z}$, for $n \in \mathbb{Z}$.
+
+We [[Characteristic Function|characterize]] all solutions to the equation $e^{w} = z$.
+## Lemma
+Fix $z \in \mathbb{C}$. Then all solutions $w$ to $e^{w} = z$ are of the form
+$$w = \ln|z| + i \arg(z)$$
+for **any** argument of $z$.
+### Proof
+Write $w = x + iy$. Then we have
+$$z = |z| \times e^{i \arg(z)}.$$
+Comparing with the polar form of $e^{w}$, we get that:
+$$e^{x} = |z|,\ e^{i \arg(z)} = e^{iy}.$$
+The first gives $x = \ln|z|$, while the second gives $\arg(z) = \arg(y) + \text{multiples of}\ 2\pi$.
+
+If we want to define a complex log we have two possibilities:
+1. We can consider **multi-valued** functions (or [[Riemann Surfaces]]),
+2. We can make it into a **single-valued** function by making a imposing some restrictions on the argument.
+## Principal Argument Definition
+Given a non-zero complex number $z \in \mathbb{C}$, its [[Principal Argument]] $\operatorname{Arg} z$ is the unique argument for $z$ such that
+$$-\pi \lt \operatorname{Arg} z \leq \pi.$$
+Then we define the log as follows
+## Principal Logarithm Definition
+The **[[Principal Logarithm|principal logarithm]]** is the function $\operatorname{Log} : \mathbb{C} \setminus \{ 0 \} \to \mathbb{C}$ is defined by
+$$\operatorname{Log} z = \ln|z| + i \operatorname{Arg} z.$$
+>[!note]- For $z = x \gt 0$
+>We get $\operatorname{Log} x = \ln x$. (The right hand side is the usual log for real numbers)
+
+### Remark
+Here we use the following notations:
+- $\ln x$ is the usual log for real numbers
+- $\operatorname{Log} z$ is the [[Principal Logarithm]], as introduced
+- $\log z$ refers to other possible choices for the logarithm
+
+It tuns out that $\operatorname{Log} z$ is also [[Holomorphic]].
+# Complex Powers
+We use the [[Complex Exponential|complex exponential]] and the complex logarithm to define the [[Complex Power|complex powers]] of complex numbers.
+## Example
+Let us show that the [[Complex Power|complex power]]
+$$z^{\frac{1}{2}} = e^{\frac{1}{2} \operatorname{Log} z}$$
+gives the usual square-root for positive real numbers (so $z = x \gt 0$).
+
+We have
+$$x^{\frac{1}{2}} = e^{\frac{1}{2} \operatorname{Log} x} = e^{\frac{1}{2}(\ln|x| + i \operatorname{Arg} x)}.$$
+We have $|x| = x$, since $x \gt 0$, and $\operatorname{Arg} x = 0$. Then
+$$x^{\frac{1}{2}} = e^{\frac{1}{2} \ln x} = e^{\ln x^{\frac{1}{2}}} = \sqrt{ x }$$
+gives back the expected result.
+
+What happens if we make another choice for the logarithm? Equivalently, another choice for the argument, such as $\arg x = 2\pi$.
+Then, writing $\log z$ for the corresponding logarithm, we have
+$$x^{\frac{1}{2}} = e^{\frac{1}{2} \log x} = e^{\frac{1}{2}(\ln x + i \arg x)}$$
+$$= e^{\frac{1}{2} \ln x + i \pi}$$
+$$= e^{i \pi} \times e^{\frac{1}{2} \ln x}$$
+$$= -\sqrt{ x }$$
+This is related to $x^{2}$ as it is **not** one-to-one.
+All different choices fr the $\log$/$\arg$ lead to such different choices for the inner functions.
+## Remark
+Note that for $n \in \mathbb{Z}$ there is **no ambiguity** in defining $z^{n}$. Indeed, we have
+$$z^{n} = e^{n \operatorname{Log} z} = e^{n(\ln|z| + i(\theta + 2 \pi k))},\ k \in \mathbb{Z}$$
+$$= e^{n(\ln|z| + i \theta)}$$
+because $n \times k$ is again an integer (and $e^{i 2 \pi n} = 1$).
+## Example - Roots of Unity
+Let's look at an example of solving polynomial equations
+
+We want to find all solutions to the equation $z^{3} = 1$
+
+Writing $z = n e^{i \theta}$ we must have
+$$n^{3} e^{3 i \theta} = 1.$$
+Then $n^{3} = 1$, which gives $n = 1$. (Since $n \geq 0$)
+For angles we must have
+$$3 i \theta = 2 \pi i n,\ n \in \mathbb{Z}.$$
+This is rewritten as $\theta = \frac{2\pi}{3}n$.
+This leads to
+$$z_{0} = e^{i \frac{2 \pi}{3} \times 0} = 1,\ z_{1}=e^{i \frac{2 \pi}{3} \times 1} = - \frac{1}{2} + i \frac{\sqrt{ 3 }}{2},\ z_{3} = e^{i \frac{2 \pi}{3} \times 2} = - \frac{1}{2} - i \frac{\sqrt{ 3 }}{2}.$$

content/Lectures/Lecture 21 - Integration, Antiderivatives, Homotopies.md

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+---
+lecture: 21
+date: 2025-03-31
+---
+This lecture will cover how to integrate complex functions in the complex plane. It will be similar to the notion of line integral in $R^{2}$. In the next [[Lecture 22 - Cauchy's theorem and Cauchy's Integral Formula|lecture]], [[Cauchy's Theorem]] will be proved, which is one of the cornerstones of complex analysis.
+# Integration in the Complex Plane
+## Definition of the Integral
+$$\int_{a}^{b} f(t) \, dt := \int_{a}^{b} \mathrm{Re}f(t) \, dt + i \int_{a}^{b} \mathrm{Im}f(t) \, dt  $$
+A more detailed definition of an integral is provided [[Complex Integral|here]].
+### Example of a Complex Integral
+Consider the complex-valued function $f(t) = t + it^2$. Then
+$$\int_{0}^{1} f(t) \, dt = \int_{0}^{1} t \, dt + i \int_{0}^{1} t^{2} \, dt = \frac{1}{2} + \frac{i}{3}$$
+## Integral along a Curve
+$$\int_{C} f(z) \, dz := \int_{a}^{b} f(\gamma(t)) \gamma'(t) \, dt.$$
+The full definition can be found [[Integral along a Curve|here]].
+
+This notion is actually analogous to that of line integral of a integral of a *vector* field in $\mathbb{R}^{2}$. The reason is that we can identify complex numbers with vectors in $R^{2}$ by $z = x + iy \mapsto (x,y)$.
+
+>[!info] The meaning of $\int_{C}$
+>This is used to denote the integral along the curve. which has the parameters $[a,b]$, hence it gets simplified that way. It is the same as $\int_{C} \equiv \int_{a}^{b}$.
+
+> [!note] Note on future notation
+> The integral $\int_{C} f(z) \, dz$ will also be denoted simply by $\int_{C} f$.
+
+### Example of Integrating a Complex Curve
+Consider the complex curve $C$ [[Smooth Parametrization|parametrized]] by $\gamma(t) = 1 + it$ with $0 \leq t \leq 1$. Given the complex function $f(z) = z$, we compute
+$$\int_{C} f = \int_{0}^{1} (1+ it)i  \, dt = -\int_{0}^{1} t \, dt + i \int_{0}^{1}  \, dt  = -\frac{1}{2} + i.$$
+> [!note]
+> The integral $\int_{C} f$ depends on the curve $C$, in general.
+
+# Reparametrizations
+The complex integral $\int_{C} f$ was defined by choosing a [[Smooth Parametrization|smooth parametrization]] $\gamma : [a,b] \to \mathbb{C}$ of $C$. But the question is how does its value depend on such a choice? To figure this out, we need to introduce some terminology regarding curves.
+## Definitions
+- [[Reparametrization]]
+- [[Same Orientation]]
+- [[Opposite Orientation]]
+## Example of Reparametrization
+Let $C$ be the line segment with endpoints $0$ and $4i$. An obvious parametrization for this curve is given by $\gamma_{1} : [0,4] \to \mathbb{C}$ defined by $\gamma_{1}(t) = it$.
+
+Now consider $\gamma_{2} : [0,2] \to \mathbb{C}$ defined by $\gamma_{2}(t) = it^{2}$.
+We claim that $\gamma_{2}$ is a [[Reparametrization|reparamatrization]] of $\gamma_{1}$ and with the [[Same Orientation|same orientation]]. To show this, we observe that we have $\gamma_{2}(t) = \gamma_{1}(\alpha(t))$ with $\alpha(t) = t^{2}$ and moreover $\alpha'(t) = 2t \geq 0$ for all $t \in [0,2]$.
+
+On the other hand consider the parametrization $\gamma_{3} : [0, 4] \to \mathbb{C}$ defined by $\gamma_{3}(t) = i(4 - t)$.
+We have $\gamma_{3}(t) = \gamma_{1}(\beta(t))$ with $\beta(t) = 4 - t$ and $\beta'(t) = -1$.
+We conclude that $\gamma_{3}$ is a [[Reparametrization|reparamatrization]] of $\gamma_{1}$ with [[Opposite Orientation|opposite orientation]].
+## Proposition
+Let $C$ be a curve with a [[Smooth Parametrization|smooth parametrization]] $\gamma : [a,b] \to \mathbb{C}$. If $\sigma : [c,d] \to \mathbb{C}$ is a [[Reparametrization|reparametrization]] of $\gamma$ with the [[Same Orientation|same orientation]] we have
+$$\int_{a}^{b} f(\gamma(t))\gamma'(t) \, dt = \int_{c}^{d} f(\sigma(t))\sigma'(t) \, dt.$$
+If instead $\sigma$ has [[Opposite Orientation|opposite orientation]], the two integrals are equal up to minus sign..
+
+Let $\int_{C} f$ be defined with respect to a parametrization $\gamma$. If we write $\int_{-C} f$ for the integral computed with, to a reparametrization $\sigma$ with [[Opposite Orientation|opposite orientation]] as above, we have
+$$\int_{-C} f = - \int_{C} f.$$
+Notice that this is analogous to what we have for the line integral of a vector field in $R^{2}$.
+### Example
+Let $\gamma : [a,b] \to \mathbb{C}$ be a smooth parametrization of $C$. We define the opposite parametrization $\gamma_{op} : [-b,-a] \to \mathbb{C}$ by setting $\gamma_{op}(t) = \gamma(-t)$.
+
+It starts at $\gamma_{op}(-b) = \gamma(b)$ and ends at $\gamma_{op}(-a) = \gamma(a)$, hence it goes along the curve £C£ in the opposite direction of $\gamma$. We write the integral with respect to this parametrization as
+$$\int_{C} f = \int_{-b}^{-a} f(\gamma_{op}(t))\gamma'_{op}(t) \, dt.$$
+Now we introduce the variable $u = -t$ and observe that
+$$\gamma'_{op} = -\gamma'(u).$$
+Then change the of variable gives
+$$\int_{-C} f = \int_{b}^{a} f(\gamma_{op}(-u))(-\gamma'(u)) \; (-du)$$
+$$= -\int_{a}^{b} f(\gamma(u))\gamma'(u) \, du = -\int_{C} f.$$
+Hence integrals with respect to $\gamma$ and $\gamma_{op}$ differ by a minus sign.
+# Antiderivatives
+> [!info] Regarding the rest of this page
+> All the following curves will be considered as [[Smooth Parametrization|smooth]], and that they will be *simple*, meaning that the curve does not cross itself.
+> 
+> tl;dr: "smooth simple curve" will be referred as "curve" (or "path").
+
+Recall that in real analysis, antiderivatives can be used to conveniently compute integrals. This notion also makes sense for complex functions.
+## Definition
+Let $F : D \to \mathbb{C}$ be a [[Holomorphic|holomorphic function]]. Suppose $F'(z) = f(z)$ for all $z \in D$. Then $F$ is an **antiderivative** of $f$ on $D$.
+
+We have the following analogue of the fundamental theorem of calculus.
+## Theorem
+Let $f : D \to \mathbb{C}$ be continuous. Suppose that $F$ is an [[Antiderivative|antiderivative]] of $f$ on $D$. Let $\gamma : [a,b] \to \mathbb{C}$ be a [[Smooth Parametrization|parametrization]] of a curve $C$ in the region $D$. Then
+$$\int_{C} f = F(\gamma(b)) - F(\gamma(a)).$$
+### Proof
+Using the chain rule we compute
+$$\frac{dF(\gamma(t))}{dt} = F'(\gamma(t))\gamma'(t) = f(\gamma(t))\gamma'(t),$$
+where we used that $F$ is an [[Antiderivative|antiderivative]] of $f$, that is $F'(z) = f(z)$. Hence
+$$\int_{C} f = \int_{a}^{b} f(\gamma(t))\gamma'(t) \, dt  = \int_{a}^{b} \frac{dF(\gamma(t))}{dt} \, dt.$$Now we can use the fundamental theorem of calculus (strictly speaking after splitting $F$ into its real and complex parts), the chain rule, to conclude that
+$$\int_{C} f = F(\gamma(b) - F(\gamma(a))).$$
+> [!info] Remark
+> Continuing our comparison between complex numbers and $R^{2}$, we see that functions admitting an [[Antiderivative|antiderivative]] are the analogue of conservative vector fields.
+## Example
+Consider $f(z) = z$. It is easy to check that $F(z) = \frac{z^{2}}{2}$ is an [[Antiderivative|antiderivative]] of $f$ (for any $z$). Let $C$ be any curve starting at $1$ and ending at $1 + i$. Then we get
+$$\int_{C} f = \frac{(1 + i)^{2}}{2} - \frac{1^{2}}{2} = -\frac{1}{2} + i.$$
+In particular, this gives another way to do the computation in [[#Example of a Complex Integral]].
+# Homotopies
+This will be used for proving [[Cauchy's Theorem]].
+
+## Definition of a Homotopy
+Let $\gamma_{1}(t)$ and $\gamma_{2}(t)$ be two closed ([[Smooth Parametrization|parametrized]]) curves in a region $D$, with parameter $0 \leq t \leq 1$.
+They are called **homotopic** if there exists a continuous function $h : [0,1] \times [0, 1] \to D$ such that, for all $s, t \in [0,1]$, we have
+$$h(t,0) = \gamma_{1}(t),\ h(t,1) = \gamma_{2}(t),$$
+$$h(0, s) = h(1, s).$$
+In this case we use the notation $\gamma_{1} \sim \gamma_{2}$.
+
+> [!info] Conditions
+> The first two conditions mean that the curve $h$ interpolates between the two curves $\gamma_{1}$ and $\gamma_{2}$.
+> The third condition means that for fixed $s$ in the curve $h(t,s)$ is closed.
+
+Let us also note that the notion of homotopy makes sense for any paths, not necessarily closed. Here we insist that the interpolating curves are also closed.
+## Example of Homotopy
+a circle in the complex plane of radius $R$ (and centred at zero) is given by those $z \in \mathbb{C}$ such that $|z| = R$.
+We can [[Smooth Parametrization|parametrize]] this by
+$$\gamma(t) = Re^{2\pi it},\ 0 \leq t \leq 1.$$
+consider the curves $\gamma_{1}(t) = e^{2\pi it}$ and $\gamma_{2}(t) = 2e^{2\pi it}$, corresponding to radii $1$ and $2$ respectively.
+A [[Homotopy|homotopy]] between these two closed curves is given by
+$$h(t,s) = (1 + s) e^{2\pi it}.$$
+Indeed it is continuous, we have $h(t,0) = \gamma_{1}(t)$ and $h(t,1) = \gamma_{2}(t)$, and for any fixed $s$ we have that $h(t,s)$ [[Smooth Parametrization|parametrizes]] a circle (of radius $1 + s$).

content/Lectures/Lecture 22 - Cauchy's theorem and Cauchy's Integral Formula.md

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+---
+lecture: 22
+date: 2025-04-03
+---
+The content of [[Lecture 21 - Integration, Antiderivatives, Homotopies#Homotopies|Homotopies from the last lecture]] will be used in this lecture.
+# Cauchy's Theorem
+Let $f : D \to \mathbb{C}$ be [[Holomorphic|holomorphic]]. Suppose $\gamma_{1}$ and $\gamma_{2}$ are two ([[Smooth Parametrization|parametrized]]) closed curves in $D$ such that $\gamma_{1} \sim \gamma_{2}$. Then we have
+$$\int_{\gamma_{1}} f = \int_{\gamma_{2}} f.$$
+## Proof of Cauchy's Theorem
+> [!info]
+> This is also called the **Cauchy-Goursat theorem**.
+> Cauchy proved the theorem with some extra assumptions, which was later shown to be unnecessary by Goursat.
+
+We also assume the following:
+- The derivative of $f'$ is continuous
+- The [[Homotopy|homotopy]] $h(t,s)$ has continuous second partial derivatives
+(The assumptions are not necessary, but makes the proof a lot simpler).
+
+Fix $s \in [0,1]$ and let $\gamma_{s}$ be the closed curve $h_{s}(t) = h(t, s)$ with $0 \leq t \leq 1$, where $h$ is the [[Homotopy|homotopy]] between $\gamma_{1}$ and $\gamma_{2}$.
+
+We introduce a function $I : [0,1] \to \mathbb{C}$ by
+$$I(s) = \int_{\gamma_{s}} f.$$
+> [!note]
+> Note that $I(0) = \int_{\gamma_{1}} f$ and $I(1) = \int_{\gamma_{2}} f$.
+> Our main aim is to prove that $I(s)$ is constant, which in particular implies that $I(0) = I(1)$ and hence the result.
+
+Since $\gamma_{s}$ is [[Smooth Parametrization|parametrized]] by $h_{s}(t) = h(s,t)$, we have by definition of [[Complex Integral|complex integral]]
+$$I(s) = \int_{0}^{1} f(h(t,s)) \frac{\partial h}{\partial t} \, dt.$$
+We compute the derivative with respect to $s$, giving
+$$\frac{dI}{ds} = \frac{d}{ds} \int_{0}^{1} f(h(t,s)) \frac{\partial h}{\partial t} \, dt = \int_{0}^{1} \frac{\partial}{\partial s}\left( f(h(t,s)) \frac{\partial h}{\partial t} \right) \, dt$$
+$$= \int_{0}^{1} \left( f'(h(t,s)) \frac{\partial h}{\partial s} \frac{\partial h}{\partial t} + f(h(t,s)) \frac{\partial^{2} h}{\partial s \partial t} \right) \, dt.$$
+Since the second derivatives of $h$ are continuous, we can switch the order of the partial derivatives by [[Schwartz's Theorem|Schwartz's theorem]]. Then we can rewrite it as
+$$\frac{dI}{ds} = \int_{0}^{1} \left( f'(h(t,s)) \frac{\partial h}{\partial t} \frac{\partial h}{\partial s} + f(h(t,s)) \frac{\partial^{2} h}{\partial t \partial s} \right) \, dt = \int_{0}^{1} \frac{\partial}{\partial t}\left( f(h(t,s)) \frac{\partial h}{\partial s} \right) \, dt.$$
+Finally we use the fundamental theorem of calculus to get
+$$\frac{dI}{ds} = f(h(1,s)) \frac{\partial h}{\partial s}(1,s) - f(h(0,s)) \frac{\partial h}{\partial s}(0,s).$$
+Now since $\gamma_{1}$ and $\gamma_{2}$ are two [[Homotopy|homotopic]] closed curves, we have that $h(0,s) = h(1,s)$ for all $s \in [0,1]$.
+Clearly this also implies that their derivatives coincide, that is $\frac{\partial h}{\partial s}(0, s) = \frac{\partial h}{\partial s}(1,s)$.
+From this we conclude that $\frac{dI}{ds} = 0$, that is $I(s)$ is constant.
+# Contractible Paths
+A path $\gamma$ is called [[Contractible Paths|contractible]] if it is [[Homotopy|homotopic]] to a point (where a point is considered to be a closed constant path).
+
+In this case we get an important corollary, which is very useful in practice.
+## Corollary
+Let $f : D \to \mathbb{C}$ be [[Holomorphic|holomorphic]] and $\gamma \subset D$ be [[Contractible Paths|contractible]]. Then
+$$\int_{\gamma} f = 0.$$
+### Proof
+We have that $\gamma$ can be modified into a point by some [[Homotopy|homotopy]].
+Then by [[Cauchy's Theorem|Cauchy's theorem]]:
+$$\int_{\gamma} f = \int_{pt} f.$$
+But the integral over a point is always zero.
+### Another Proof
+We give another proof of [[Cauchy's Theorem|Cauchy's theorem]] (in the [[Contractible Paths|contractible]] setting) by using [[Green's Theorem]] from vector calculus.
+
+Let $C$ be a (simple, positively-oriented) closed curve in the plane and denoted by $R$ the region enclosed by $C$. Let $P(x,y)$ and $Q(x,y)$ be (real-valued) functions and assume that the first-derivatives are continuous.
+Then we have the identity
+$$\int_{C}(Pdx + Qdy) = \int\int_{R} \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \, dA.$$
+In this formula we use standard notations from vector calculus.

content/Lectures/Revision Lecture.md

@@ -0,0 +1,14 @@
+---
+date: 2025-05-19
+---
+# Exam Info
+- Not too many details are expected
+- Proofs are not expected - but they are appreciated if you can remember them
+- 1 hour max per exam (realistically more like 40 minutes)
+- Topology starts first, then it's complex analysis (the questions are basically in order)
+# Good to know
+- Lebesgue's Theorem
+# Doing Revision
+- Abstract some of the things from the more complex topics
+	- If you know some parts of that complex topic, then you can think of the parts that you know about, and think about how it relates
+	- You can also think of it informally

content/index.md

@@ -28,4 +28,9 @@ title: ACIT4330 Lecture Notes
 # Complex Analysis
 - [[Lecture 18 - Complex Analysis]]
 - [[Lecture 19 - Derivatives]]
-- [[Lecture 30 - Residue Theorem|Last Lecture - Residue Theorem]]
+- [[Lecture 20 - Complex Exponential, Logarithm, and Powers]]
+- [[Lecture 21 - Integration, Antiderivatives, Homotopies]]
+- [[Lecture 22 - Cauchy's theorem and Cauchy's Integral Formula]]
+- [[Lecture 30 - Residue Theorem|Last Lecture - Residue Theorem]]
+# End of Course
+- [[Revision Lecture]]