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content/Definitions/Complex Analysis/Derivatives/Cauchy-Riemann Equations.md

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+# Definition
+A [[Complex Functions|complex function]] $f(z)$ can be seen as a function of two real variables. Hence
+$$f(z) = f(x + iy) = f(x, y).$$
+When interpreted appropriately.
+
+> [!example] For instance
+> $$f(z) = z^2 = (x + iy)^2 = x^2 - y^2 + 2ixy$$
+
+Suppose $f$ is [[Differentiable|differentiable]], we should expect **relations** between $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$.

content/Definitions/Complex Analysis/Derivatives/Differentiable.md

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+# Definition
+A function $f : D \to \mathbb{C}$ is **differentiable** at $z \in \mathbb{C}$ if the following limit exists
+$$\lim_{ \Delta z \to \infty } \frac{f(z + \Delta z) - f(z)}{\Delta z}.$$
+We denote by $f'(z)$.

content/Definitions/Complex Analysis/Derivatives/Entire.md

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+# Definition
+Let $f: D \to \mathbb{C}$ be a function with domain $D$.
+
+We say that $f$ is **entire** if $f$ is **entire** if $f$ is [[Differentiable|differentiable]] everywhere (so $D = \mathbb{C}$).

content/Definitions/Complex Analysis/Derivatives/Holomorphic.md

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+# Definition
+Let $f: D \to \mathbb{C}$ be a function with domain $D$.
+
+We say that $f$ is **holomorphic** at $z_{0}$ if $f$ is [[Differentiable|differentiable]] in an [[Open Sets|open]] disk (entered at $z_{0}$).
+
+> [!info] Remark on naming of "holomorphic"
+> Many sources use the term **analytic** instead of holomorphic.
+

content/Lectures/Lecture 19 - Derivatives.md

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+---
+lecture: 19
+date: 2025-03-24
+---
+# Differentiable
+## Definition
+A function $f : D \to \mathbb{C}$ is **differentiable** at $z \in \mathbb{C}$ if the following limit exists
+$$\lim_{ \Delta z \to \infty } \frac{f(z + \Delta z) - f(z)}{\Delta z}.$$
+We denote by $f'(z)$.
+
+The same thing is also defined here: [[Differentiable]]
+## Example
+Consider $f(z) = z$. We have:
+$$\frac{f(z + \Delta z) - f(z)}{\Delta z} = \frac{z + \Delta z}{\Delta z} = 1 .$$
+Hence $f'(z) = 1$.
+
+
+We have general results to compute derivatives, as in the real case.
+# Proposition
+Suppose $f$ and $g$ are differentiable at $z \in \mathbb{C}$. Then:
+1. $(af(z) + bg(z))' = af'(z) + bg'(z)$ for any $a, b \in \mathbb{C}$,
+2. $(f(z)g(z))' = f'(z) g(z) + f(z) g'(z)$,
+3. Also a quotient rule.
+
+Some terminologies which will be important to understand the rest:
+- [[Holomorphic]]
+- [[Entire]]
+
+> [!example]
+> Any polynomial $P(z) = \sum_{k = 0}^{n} a k z^{k}$ is an [[Entire|entire]] function (follows from the proposition).
+
+## Example
+Consider $f(z) = \; \mid z \mid^{2} = z \bar{z}$. This is not going to be [[Holomorphic|holomorphic]].
+
+We compute:
+$$\frac{f(z+ \Delta z) - f(z)}{\Delta z} = \frac{\mid z + \Delta z \mid^{2} = \mid z \mid^{2}}{\Delta z}$$
+> [!note]
+> We say that $z = x + iy$, hence $\Delta z = a + ib$.
+
+![[Drawing 2025-03-24 11.00.27.excalidraw.dark.svg]]
+%%[[Drawing 2025-03-24 11.00.27.excalidraw.md|🖋 Edit in Excalidraw]], and the [[Drawing 2025-03-24 11.00.27.excalidraw.light.svg|light exported image]]%%
+ $$= \frac{(x + a)^2 + (y + b)^2 - x^2 - y^2}{a + ib} = \frac{a^2 + b^2 + 2(ax + by)}{a + ib}.$$
+Consider the real direction.
+
+> [!note] Write
+> $$\Delta z = \Delta x$$
+
+$$\lim_{ \underbrace{\Delta x \to 0}_{a \to 0} } \frac{f(z + \Delta x) - f(z)}{\Delta x} = \lim_{ a \to 0 } \frac{a^2 + 2ax}{a} = 2x$$
+
+Similarly for the imaginary axis
+> [!note] Write
+> $$\Delta z = i \Delta y$$
+
+$$\lim_{ \underbrace{i \Delta y \to 0}_{b \to 0} } \frac{f(z + i \Delta y) - f(z)}{i \Delta y} = \lim_{ b \to 0 } \frac{b^2 + 2by}{ib} = -2iy.$$
+This means that $f$ is **not** [[Differentiable|differentiable]] at $z \neq 0$.
+# Cauchy-Riemann Equations
+A [[Complex Functions|complex function]] $f(z)$ can be seen as a function of two real variables. Hence
+$$f(z) = f(x + iy) = f(x, y).$$
+When interpreted appropriately.
+
+> [!example] For instance
+> $$f(z) = z^2 = (x + iy)^2 = x^2 - y^2 + 2ixy$$
+
+Suppose $f$ is [[Differentiable|differentiable]], we should expect **relations** between $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$.
+
+The same thing is also defined here: [[Cauchy-Riemann Equations]]
+# Theorem
+Let $z_{0} = x_{0} + iy_{0}$.
+1. Suppose $f$ is [[Differentiable|differentiable]] at $z_{0}$. Then $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$ exists and have $$\frac{\partial f}{\partial x}(x_{0}, y_{0}) = -i \frac{\partial f}{\partial y} (x_{0}, y_{0}).$$
+> [!info]- Visual Representation
+> 
+> ![[Drawing 2025-03-24 11.13.06.excalidraw.dark.svg]]
+%%[[Drawing 2025-03-24 11.13.06.excalidraw.md|🖋 Edit in Excalidraw]], and the [[Drawing 2025-03-24 11.13.06.excalidraw.light.svg|light exported image]]%%
+
+2. Suppose $f$ is such that:
+	- $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$ exists at $(x_{0}, y_{0})$,
+	- they are [[Continuous|continuous]] in a small disk centred at $(x_{0}, y_{0})$.
+	Then $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$ satisfy the condition above and $f$ is [[Differentiable|differentiable]] at $z_{0}$.
+3. In both cases, we have $f'(z_{0}) = \frac{\partial f}{\partial x}(x_{0}, y_{0})$.
+## Proof
+### For 1.
+Since $f$ is [[Differentiable|differentiable]] at $z_{0}$, the following limits exists
+$$f'(z_{0}) = \lim_{ \Delta z \to 0 } \frac{f(z_{0} + \Delta z) - f(z_{0})}{\Delta z}.$$
+> [!info]- Visual representation
+> ![[Drawing 2025-03-24 11.32.41.excalidraw.dark.svg]]
+%%[[Drawing 2025-03-24 11.32.41.excalidraw.md|🖋 Edit in Excalidraw]], and the [[Drawing 2025-03-24 11.32.41.excalidraw.light.svg|light exported image]]%%
+
+$$f(z) = f(x, y)$$
+
+First we take $\Delta z$ **real**, so $\Delta z = \Delta x$. Then
+$$f'(z_{0}) = \lim_{ \Delta x \to 0 } \frac{f(x_{0} + \Delta x) - f(x_{0}, y_{0})}{\Delta x} = \frac{\partial f}{\partial x}(x_{0}, y_{0}).$$
+
+
+Next take $\Delta z$ to be purely imaginary ($\Delta x, \Delta y \in \mathbb{R}$), so $\Delta z = i \Delta y$
+($z_{0} + i \Delta y = x_{0} + i(y_{0} + \Delta y)$)
+
+Then:
+$$f'(z_{0}) = \lim_{ \Delta y \to 0 } \frac{f(x_{0}, y_{0} + \Delta y) = f(x_{0}, y_{0})}{i \Delta y} = -i \frac{\partial f}{\partial y}(x_{0}, y_{0}).$$
+The two expressions must coincide, since $f$ is [[Differentiable|differentiable]] at $z_{0}$.
+### For 2.
+It is more complicated to prove. You can find the proof in the references on Canvas for the course.
+### For 3.
+Shown in [[#For 1.]]
+# Decomposition
+We can decompose any $f : D \to \mathbb{C}$ into its **real** and **imaginary** part. We write
+$$f(z) = f(x, y) = u(x, y) + iv(x, y).$$
+or $\mathrm{Re}f = u$ and $\mathrm{Im} f = v$
+## Corollary
+With notation as before, we have
+$$\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y},\ \frac{\partial v}{\partial x} = -\frac{\partial u}{\partial y}.$$
+### Proof
+By the [[#Theorem|previous theorem]], we have
+$$\frac{\partial f}{\partial x} = -i \frac{\partial f}{\partial y}.$$
+Write $f = u + iv$. Then
+$$\frac{\partial u}{\partial x} + i \frac{\partial v}{\partial x} = -i \frac {\partial u}{\partial y} + \frac{\partial v}{\partial y}.$$
+Comparing real and imaginary parts gives the result.
+## Example
+Consider $f(z) = z^2$.
+This is seen to be [[Holomorphic|holomorphic]] (in at least two ways).
+
+One way is
+$$f'(z) = (zz)' = z'z + zz' = zz.$$
+
+Another way is to use the [[Cauchy-Riemann Equations]]. We have
+$$z^2 = (x + iy)^2 = (x^2 - y^2) + 2ixy.$$
+So here
+$$\mathrm{Re}f = u = x^2 - y^2,\ \mathrm{Im}f = v = 2xy.$$
+So one checks that $u$ and $v$ satisfy the [[Cauchy-Riemann Equations]].
+Since the derivatives are [[Continuous|continuous]], we get that $f = u + iv$ is [[Holomorphic|holomorphic]].

content/index.md

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 - [[Lecture 17 - Lp Spaces]]
 # Complex Analysis
 - [[Lecture 18 - Complex Analysis]]
+- [[Lecture 19 - Derivatives]]