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+# Definition
+Let $z = x + iy$. Then its **complex conjugate** is
+$$\bar{z} := x-iy.$$
+# Properties
+The following hold:
+1. $\mid z \mid^2 = z \bar{z}$,
+2. $z + \bar{z} = 2 \mathrm{Re} z$,
+3. $z - \bar{z} = 2i \mathrm{Im} z$,
+4. $\overline{r e^{i \phi}} = r e^{-i \phi}$.
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+To be defined later...
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+# Definition
+$\mathbb{C}$ is obtained from $\mathbb{R}$ by adjoining the imaginary unit $i$ such that $i^2 = -1$.
+
+In particular, a complex number is of the form
+$$z = x+iy$$
+with $x$, $y$ being real, we write
+$$\mathrm{Re} z = x, \; \mathrm{Im} z = y.$$
+# Absolute Value
+Let $z = x+iy$. Its **absolute value** (or [[Norm|norm]]) is defined by
+$$\mid z \mid \sqrt{ x^2 + y^2 }.$$
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+---
+lecture: 18
+date: 2025-03-20
+---
+# Overview of Complex Analysis of the Course
+- Will do **analysis** using complex numbers.
+- **Concrete Goal:** compute integrals such as $\int_{-\infty}^{+\infty} \frac{e^{i \omega t}}{t^2 + a^2} \, dt = \frac{\pi}{a}e^{-a\mid \omega \mid}$ ($a \gt 0$) using complex techniques.
+# Complex Numbers
+The definition of a [[Complex Numbers]] as defined in the lecture.
+
+All algebraic properties follow from the definition. For instance
+$$(x_{1}+iy_{1})(x_{2}+iy_{2}) = (x_{1}x_{2} -y_{1}y_{2}) + i(x_{1}y_{2} + y_{1}x_{2}).$$
+This operation is **commutative**, that is $z_{1}z_{2} = z_{2}z_{1}$.
+## Proposition
+Let $z = x+iy$, and it should be non-zero. Then there exists another complex number $z^{-1} \in \mathbb{C}$ such that $zz^{-1} = 1$. It is given by
+$$z^{-1} = \frac{x-iy}{x^2 + y^2}$$
+### Proof
+We compute
+$$(x + iy)(x-iy) = x^2 + y^2.$$
+This is non-zero. Then
+$$zz^{-1} = (x+iy) \frac{x-iy}{x^2 + y^2} = 1$$
+> [!info] Remark
+> This means that $\mathbb{C}$ is a **field** like $\mathbb{R}$.
+
+## Definition - Absolute Value
+Let $z = x+iy$. Its **absolute value** (or [[Norm|norm]]) is defined by
+$$\mid z \mid \sqrt{ x^2 + y^2 }.$$
+An **argument** for $z$ is a real number $\phi \in \mathbb{R}$ such that
+$$x = \mid z \mid \cos \phi, \; y = \mid z \mid \sin \phi.$$
+This allows us to identify complex numbers with points in the plane $\mathbb{R}^2$:
+
+![[Drawing 2025-03-20 10.50.15.excalidraw.dark.svg]]
+%%[[Drawing 2025-03-20 10.50.15.excalidraw.md|🖋 Edit in Excalidraw]], and the [[Drawing 2025-03-20 10.50.15.excalidraw.light.svg|light exported image]]%%
+(note that angles are only defined up to multiples of $2\pi$).
+
+You could multiply pairs in $\mathbb{R}^2$ by
+$$(x_{1},y_{1}) \times (x_{2}, y_{2}) = (x_{1}x_{2}, y_{1}y_{2}).$$
+But this **does not** correspond to the multiplication of complex numbers.
+# Polar Form
+- Using the previous definitions, we can write any $z \in \mathbb{C}$ as:
+$$z = x + iy = \mid z \mid (\cos \phi + i \sin \phi).$$
+## Definition
+We introduce the following notations:
+$$r := \mid z \mid,$$
+$$e^{i\phi} := \cos \phi + i \sin \phi.$$
+Then we have that $z = r e^{i\phi}$, which we call the **polar form** of $z$.
+> [!info] Remark
+> For now, $e^{i\phi}$ is just a symbol. Later we will identify it with the [[Complex Exponential Function]]
+
+We have that $e^{i\phi}$ shares many properties with $e^{x}$.
+## Proposition
+Let $\phi, \; \theta \in \mathbb{R}$. Then:
+1. $e^{i \phi} e^{i \theta} = e^{i (\phi + \theta)}$,
+2. $e^{i 0} = 1$,
+3. $\frac{1}{e^{i \phi}} = e^{-i \phi}$,
+4. $e^{i(\phi + 2 \pi k)} = e^{i \phi}$, for $k \in \mathbb{Z}$,
+5. $\mid e^{i \phi} = 1$,
+6. $\frac{d e^{i \phi}}{d \phi} = i e^{i \phi}$.
+### Proof (only for 3.)
+According to a previous result, we have:
+$$(\cos \phi + i \sin \phi)^{-1} = \frac{\cos \phi - i \sin \phi}{\overbrace{(\cos \phi)^2 + (\sin \phi)^2}^{=1}}$$
+$$= \cos \phi - i \sin \phi$$
+$$= e^{- i \phi},$$
+using $\cos(-\phi) = \cos \phi$ and $\sin(-\phi) = -\sin \phi$.
+# Complex Conjugation
+Definition [[Complex Conjugation]] as defined in the lecture.
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# The Inverse Image
The Inverse Image uses a [[Inverse Function]] $f^{-1} (z)$ of $Z \subset Y$ written $f: x \to y$ is $f^{-1}(z) \equiv \{ x \in X \mid f(x) \in Z \}$.
# Complex Numbers
-In [[Complex Numbers]], $\mathbb{C} = \mathbb{R} \times \mathbb{R}$ with usual addition of vectors
+In [[Complex Numbers in Sets]], $\mathbb{C} = \mathbb{R} \times \mathbb{R}$ with usual addition of vectors
## Multiplying Vectors
Multiply vectors by adding their angles multiplying their lengths.
diff --git a/content/index.md b/content/index.md
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- [[Lecture 15]]
- [[Lecture 16]]
- [[Lecture 17 - Lp Spaces]]
+# Complex Analysis
+- [[Lecture 18 - Complex Analysis]]
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