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# Definition
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Let $z = x + iy$. Then its **complex conjugate** is
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$$\bar{z} := x-iy.$$
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# Properties
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The following hold:
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1. $\mid z \mid^2 = z \bar{z}$,
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2. $z + \bar{z} = 2 \mathrm{Re} z$,
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3. $z - \bar{z} = 2i \mathrm{Im} z$,
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4. $\overline{r e^{i \phi}} = r e^{-i \phi}$.
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To be defined later...
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content/Definitions/Complex Analysis/Complex Numbers.md
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content/Definitions/Complex Analysis/Complex Numbers.md
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# Definition
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$\mathbb{C}$ is obtained from $\mathbb{R}$ by adjoining the imaginary unit $i$ such that $i^2 = -1$.
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In particular, a complex number is of the form
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$$z = x+iy$$
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with $x$, $y$ being real, we write
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$$\mathrm{Re} z = x, \; \mathrm{Im} z = y.$$
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# Absolute Value
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Let $z = x+iy$. Its **absolute value** (or [[Norm|norm]]) is defined by
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$$\mid z \mid \sqrt{ x^2 + y^2 }.$$
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content/Lectures/Lecture 18 - Complex Analysis.md
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content/Lectures/Lecture 18 - Complex Analysis.md
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---
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lecture: 18
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date: 2025-03-20
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---
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# Overview of Complex Analysis of the Course
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- Will do **analysis** using complex numbers.
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- **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.
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# Complex Numbers
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The definition of a [[Complex Numbers]] as defined in the lecture.
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All algebraic properties follow from the definition. For instance
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$$(x_{1}+iy_{1})(x_{2}+iy_{2}) = (x_{1}x_{2} -y_{1}y_{2}) + i(x_{1}y_{2} + y_{1}x_{2}).$$
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This operation is **commutative**, that is $z_{1}z_{2} = z_{2}z_{1}$.
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## Proposition
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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
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$$z^{-1} = \frac{x-iy}{x^2 + y^2}$$
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### Proof
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We compute
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$$(x + iy)(x-iy) = x^2 + y^2.$$
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This is non-zero. Then
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$$zz^{-1} = (x+iy) \frac{x-iy}{x^2 + y^2} = 1$$
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> [!info] Remark
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> This means that $\mathbb{C}$ is a **field** like $\mathbb{R}$.
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## Definition - Absolute Value
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Let $z = x+iy$. Its **absolute value** (or [[Norm|norm]]) is defined by
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$$\mid z \mid \sqrt{ x^2 + y^2 }.$$
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An **argument** for $z$ is a real number $\phi \in \mathbb{R}$ such that
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$$x = \mid z \mid \cos \phi, \; y = \mid z \mid \sin \phi.$$
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This allows us to identify complex numbers with points in the plane $\mathbb{R}^2$:
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![[Drawing 2025-03-20 10.50.15.excalidraw.dark.svg]]
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%%[[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]]%%
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(note that angles are only defined up to multiples of $2\pi$).
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You could multiply pairs in $\mathbb{R}^2$ by
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$$(x_{1},y_{1}) \times (x_{2}, y_{2}) = (x_{1}x_{2}, y_{1}y_{2}).$$
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But this **does not** correspond to the multiplication of complex numbers.
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# Polar Form
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- Using the previous definitions, we can write any $z \in \mathbb{C}$ as:
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$$z = x + iy = \mid z \mid (\cos \phi + i \sin \phi).$$
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## Definition
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We introduce the following notations:
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$$r := \mid z \mid,$$
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$$e^{i\phi} := \cos \phi + i \sin \phi.$$
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Then we have that $z = r e^{i\phi}$, which we call the **polar form** of $z$.
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> [!info] Remark
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> For now, $e^{i\phi}$ is just a symbol. Later we will identify it with the [[Complex Exponential Function]]
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We have that $e^{i\phi}$ shares many properties with $e^{x}$.
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## Proposition
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Let $\phi, \; \theta \in \mathbb{R}$. Then:
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1. $e^{i \phi} e^{i \theta} = e^{i (\phi + \theta)}$,
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2. $e^{i 0} = 1$,
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3. $\frac{1}{e^{i \phi}} = e^{-i \phi}$,
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4. $e^{i(\phi + 2 \pi k)} = e^{i \phi}$, for $k \in \mathbb{Z}$,
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5. $\mid e^{i \phi} = 1$,
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6. $\frac{d e^{i \phi}}{d \phi} = i e^{i \phi}$.
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### Proof (only for 3.)
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According to a previous result, we have:
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$$(\cos \phi + i \sin \phi)^{-1} = \frac{\cos \phi - i \sin \phi}{\overbrace{(\cos \phi)^2 + (\sin \phi)^2}^{=1}}$$
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$$= \cos \phi - i \sin \phi$$
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$$= e^{- i \phi},$$
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using $\cos(-\phi) = \cos \phi$ and $\sin(-\phi) = -\sin \phi$.
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# Complex Conjugation
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Definition [[Complex Conjugation]] as defined in the lecture.
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# The Inverse Image
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# The Inverse Image
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The Inverse Image uses a [[Inverse Function]] $f^{-1} (z)$ of $Z \subset Y$ written $f: x \to y$ is $f^{-1}(z) \equiv \{ x \in X \mid f(x) \in Z \}$.
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The Inverse Image uses a [[Inverse Function]] $f^{-1} (z)$ of $Z \subset Y$ written $f: x \to y$ is $f^{-1}(z) \equiv \{ x \in X \mid f(x) \in Z \}$.
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# Complex Numbers
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# Complex Numbers
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In [[Complex Numbers]], $\mathbb{C} = \mathbb{R} \times \mathbb{R}$ with usual addition of vectors
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In [[Complex Numbers in Sets]], $\mathbb{C} = \mathbb{R} \times \mathbb{R}$ with usual addition of vectors
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## Multiplying Vectors
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## Multiplying Vectors
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Multiply vectors by adding their angles multiplying their lengths.
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Multiply vectors by adding their angles multiplying their lengths.
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@ -25,3 +25,5 @@ title: ACIT4330 Lecture Notes
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- [[Lecture 15]]
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- [[Lecture 15]]
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- [[Lecture 16]]
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- [[Lecture 16]]
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- [[Lecture 17 - Lp Spaces]]
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- [[Lecture 17 - Lp Spaces]]
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# Complex Analysis
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- [[Lecture 18 - Complex Analysis]]
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