generated from smyalygames/quartz
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6.1 KiB
Markdown
121 lines
6.1 KiB
Markdown
---
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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. $\mid e^{i \phi} \mid = 1$,
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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 of [[Complex Conjugation]], as defined in the lecture.
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# Triangle Inequality
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Definition of [[Triangle Inequality]], as defined in the lecture.
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But we want to show a similar result for $\mathbb{C}$.
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## Lemma
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Let $z \in \mathbb{C}$, then we can take a look at the $\mathrm{Re}$ and $\mathrm{Im}$:
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$$- \mid z \mid \leq Re z \leq \mid z \mid$$
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$$- \mid z \mid \leq Im z \leq \mid z \mid.$$
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### Proof
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Recall that for $z = x + iy$ we have $\mid z \mid = \sqrt{ x^2 + y^2 }$. Then the results follow from the following inequalities ($a, b \in \mathbb{R}$):
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$$-\sqrt{ a^2 + b^2 } \leq - \sqrt{ a^2 } \leq \; \mid a \mid \; \leq \sqrt{ e^2 } \leq \sqrt{ a^2 + b^2 }.$$
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This gives the result.
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We use this to prove the triangle inequality for $\mathbb{C}$.
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## Proposition
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Let $z_{1}, z_{2} \in \mathbb{C}$ we have
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$$\mid z_{1} + z_{2} \mid \; \leq \; \mid z_{1} \mid + \mid z_{2} \mid.$$
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### Proof
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We compute:
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$$\mid z_{1} + z_{2} \mid^2 \; = (z_{1} + z_{2})(\overline{z_{1}} + \overline{z_{2}})$$
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$$= z_{1}\overline{z_{1}} + z_{1} \overline{z_{2}} + z_{2} \overline{z_{1}} + z_{2} \overline{z_{2}} = \; \mid z_{1} \mid^{2} + 2 \mathrm{Re} (z_{1} \overline{z_{2}}) \mid z_{2} \mid^2$$
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(Noting that $z_{2}\overline{z_{1}} = \overline{z_{1}\overline{z_{2}}}$).
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Next we use the property that $\mathrm{Re} z \leq \; \mid z \mid$. Then
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$$\mid z_{1} + z_{2} \mid^2 \; \leq \; \mid z_{1} \mid^2 + 2 \mid z_{1} \overline{z_{2}} \mid + \mid z_{2} \mid^2$$
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$$= \; \mid z_{1} \mid^2 + \; 2 \mid z_{1} \mid \times \mid \overline{z_{2}} \mid + \mid z_{2} \mid^2$$
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$$= \; \mid z_{1} \mid^2 + \; 2 \mid z_{1} \mid \mid z_{2} \mid + \mid z_{2} \mid^2$$
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$$= (\mid z_{1} \mid + \mid z_{2} \mid)^2.$$
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Taking square roots of both (non-negative) sides, we get the result.
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This means that $\mathbb{C}$ is a **[[Norm|normed]] space**.
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# Limits
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Before being able to define a limit, we need to define a [[Complex Functions|complex function]], from the lecture.
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## Definition
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[[Complex Limits]] as defined in the lecture
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## Proposition
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Let $f$ and $g$ be [[Complex Functions|complex functions]].
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Suppose that $\lim_{ z \to z_{0} } f(z)$ and $\lim_{ z \to z_{0} } g(z)$ exist. Then:
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1. For $a, b \in \mathbb{C}$ we have $\lim_{ z \to z_{0} } (af(z) + bg(z)) = a \lim_{ z \to z_{0} } f(z) + b \lim_{ z \to z_{0} } g(z)$,
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2. We have $\lim_{ z \to z_{0} } (f(z) \times g(z)) = \left(\lim_{ z \to z_{0} } f(z) \right) \times \left(\lim_{ z \to z_{0} } g(z) \right)$.
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### Example
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(Computing the [[Complex Limits|limit]], but will not happen again, just for understanding)
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Let $f(z) = z^2$. We want to show that $\lim_{ z \to 2 } f(z) = 4$. Fix $\varepsilon \gt 0$ and observe that
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$$f(z) - 4 = z^2 - 4 = (z - 2)^2 + 4(z - 2).$$
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Write $\mid z - 2 \mid \; \lt \delta$ for some $\delta$. Then using the [[Triangle Inequality|triangle inequality]], we get
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$$\mid f(z) - 4 \mid \; = \; \mid(z-2)^2 + 4 (z-2) \mid$$
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$$\leq \; \mid z - 2 \mid^2 + \; 4 \mid z - 2 \mid$$
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$$\lt \delta^2 + 4 \delta.$$
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It suffices to pick $\delta$ such that $\delta^2 + 4 \delta \leq \varepsilon$ (which is possible).
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A much easier way to compute the following:
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First $\lim_{ z \to z_{0} } z = z_{0}$, then
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$$\lim_{ z \to 2 } z^2 = \left( \lim_{ z \to 2 } z \right) \times \left( \lim_{ z \to 2 } z \right) = 2 \times 2 = 4$$ |