---
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]].