Quartz sync: Mar 20, 2025, 12:22 PM

content/Definitions/Complex Analysis/Complex Conjugation.md

@@ -3,7 +3,19 @@ Let $z = x + iy$. Then its **complex conjugate** is
 $$\bar{z} := x-iy.$$
 # Properties
 The following hold:
-1. $\mid z \mid^2 = z \bar{z}$,
+1. $\mid z \mid^2 = z \bar{z}$, ($=r^2$)
 2. $z + \bar{z} = 2 \mathrm{Re} z$,
 3. $z - \bar{z} = 2i \mathrm{Im} z$,
 4. $\overline{r e^{i \phi}} = r e^{-i \phi}$.
+
+> [!note]+ Note that (4) implies that
+>   
+> $\mid \bar{z} \mid = \mid z \mid$.
+> 
+> Also write that $z = r e^{i \phi}$ and $z' = r' e^{i \phi'}$.
+> 
+> Then $zz' = rr'e^{i(\phi + \phi')}$. Then (1) implies that
+> 
+> $\mid z z' \mid = rr' = \mid z \mid \mid z' \mid$.
+> 
+> (Nice interplay between complex multiplication with absolute values).

content/Definitions/Complex Analysis/Complex Functions.md

@@ -0,0 +1 @@
+A **complex function** is a rule $f : D \to \mathbb{C}$, that assigns [[Complex Numbers|complex numbers]] to a subset $D$ of, a complex plane, $\mathbb{C}$.

content/Definitions/Complex Analysis/Complex Limits.md

@@ -0,0 +1,7 @@
+# Definition
+We say that $L \in \mathbb{C}$ is the limit of a [[Complex Functions|function]] ($f : D \to \mathbb{C}$), $f$ at $z_{0} \in \mathbb{C}$ if the following holds:
+
+For every $\varepsilon \gt 0$ there exists $\delta \gt 0$ such that, if $\mid z - z_{0} \mid \; < \delta$ then $\mid f(z) - L \mid \; \lt \varepsilon$.
+(Then we write $L = \lim_{ z \to z_{0} } f(z)$)
+
+Observe that this uses the absolute value of complex numbers.

content/Definitions/Complex Analysis/Triangle Inequality.md

@@ -0,0 +1,4 @@
+# Definition
+We have an absolute value $\mid x \mid$ for real numbers, which satisfies
+$$\mid x + y \mid \leq \mid x \mid + \mid y \mid,$$
+called the **triangle inequality**.

content/Lectures/Lecture 18 - Complex Analysis.md

@@ -54,7 +54,7 @@ Let $\phi, \; \theta \in \mathbb{R}$. Then:
 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. $\mid e^{i \phi} = 1$,
+5. $\mid e^{i \phi} \mid = 1$,
 6. $\frac{d e^{i \phi}}{d \phi} = i e^{i \phi}$.
 ### Proof (only for 3.)
 According to a previous result, we have:
@@ -63,4 +63,59 @@ $$= \cos \phi - i \sin \phi$$
 $$= e^{- i \phi},$$
 using $\cos(-\phi) = \cos \phi$ and $\sin(-\phi) = -\sin \phi$.
 # Complex Conjugation
-Definition [[Complex Conjugation]] as defined in the lecture.
+Definition of [[Complex Conjugation]], as defined in the lecture.
+# Triangle Inequality
+Definition of [[Triangle Inequality]], as defined in the lecture.
+
+But we want to show a similar result for $\mathbb{C}$.
+## Lemma
+Let $z \in \mathbb{C}$, then we can take a look at the $\mathrm{Re}$ and $\mathrm{Im}$:
+$$- \mid z \mid \leq Re z \leq \mid z \mid$$
+$$- \mid z \mid \leq Im z \leq \mid z \mid.$$
+### Proof
+Recall that for $z = x + iy$ we have $\mid z \mid = \sqrt{ x^2 + y^2 }$. Then the results follow from the following inequalities ($a, b \in \mathbb{R}$):
+$$-\sqrt{ a^2 + b^2 } \leq - \sqrt{ a^2 } \leq \; \mid a \mid \; \leq \sqrt{  e^2 } \leq \sqrt{ a^2 + b^2 }.$$
+This gives the result.
+
+We use this to prove the triangle inequality for $\mathbb{C}$.
+## Proposition
+Let $z_{1}, z_{2} \in \mathbb{C}$ we have
+$$\mid z_{1} + z_{2} \mid \; \leq \; \mid z_{1} \mid + \mid z_{2} \mid.$$
+### Proof
+We compute:
+$$\mid z_{1} + z_{2} \mid^2 \; = (z_{1} + z_{2})(\overline{z_{1}} + \overline{z_{2}})$$
+$$= z_{1}\overline{z_{1}} + z_{1} \overline{z_{2}} + z_{2} \overline{z_{1}} + z_{2} \overline{z_{2}} = \; \mid z_{1} \mid^{2} + 2 \mathrm{Re} (z_{1} \overline{z_{2}}) \mid z_{2} \mid^2$$
+(Noting that $z_{2}\overline{z_{1}} = \overline{z_{1}\overline{z_{2}}}$).
+
+Next we use the property that $\mathrm{Re} z \leq \; \mid z \mid$. Then
+$$\mid z_{1} + z_{2} \mid^2 \; \leq \; \mid z_{1} \mid^2 + 2 \mid z_{1} \overline{z_{2}} \mid + \mid z_{2} \mid^2$$
+$$= \; \mid z_{1} \mid^2 + \; 2 \mid z_{1} \mid \times \mid \overline{z_{2}} \mid + \mid z_{2} \mid^2$$
+$$= \; \mid z_{1} \mid^2 + \; 2 \mid z_{1} \mid \mid z_{2} \mid + \mid z_{2} \mid^2$$
+$$= (\mid z_{1} \mid + \mid z_{2} \mid)^2.$$
+Taking square roots of both (non-negative) sides, we get the result.
+
+This means that $\mathbb{C}$ is a **[[Norm|normed]] space**.
+# Limits
+Before being able to define a limit, we need to define a [[Complex Functions|complex function]], from the lecture.
+## Definition
+[[Complex Limits]] as defined in the lecture
+## Proposition
+Let $f$ and $g$ be [[Complex Functions|complex functions]].
+
+Suppose that $\lim_{ z \to z_{0} } f(z)$ and $\lim_{ z \to z_{0} } g(z)$ exist. Then:
+1. For $a, b \in \mathbb{C}$ we have $\lim_{ z \to z_{0} } (af(z) + bg(z)) = a \lim_{ z \to z_{0} } f(z) + b \lim_{ z \to z_{0} } g(z)$,
+2. We have $\lim_{ z \to z_{0} } (f(z) \times g(z)) = \left(\lim_{ z \to z_{0} } f(z) \right) \times \left(\lim_{ z \to z_{0} } g(z) \right)$.
+### Example
+(Computing the [[Complex Limits|limit]], but will not happen again, just for understanding)
+
+Let $f(z) = z^2$. We want to show that $\lim_{ z \to 2 } f(z) = 4$. Fix $\varepsilon \gt 0$ and observe that
+$$f(z) - 4 = z^2 - 4 = (z - 2)^2 + 4(z - 2).$$
+Write $\mid z - 2 \mid \; \lt \delta$ for some $\delta$. Then using the [[Triangle Inequality|triangle inequality]], we get
+$$\mid f(z) - 4 \mid \; = \; \mid(z-2)^2 + 4 (z-2) \mid$$
+$$\leq \; \mid z - 2 \mid^2 + \; 4 \mid z - 2 \mid$$
+$$\lt \delta^2 + 4 \delta.$$
+It suffices to pick $\delta$ such that $\delta^2 + 4 \delta \leq \varepsilon$ (which is possible).
+
+A much easier way to compute the following:
+First $\lim_{ z \to z_{0} } z = z_{0}$, then
+$$\lim_{ z \to 2 } z^2 = \left( \lim_{ z \to 2 } z \right) \times \left( \lim_{ z \to 2 } z \right) = 2 \times 2 = 4$$