ACIT4330-Page/content/Lectures/Lecture 19 - Derivatives.md

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19	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 function (follows from the proposition).

Example

Consider f(z) = \; \mid z \mid^{2} = z \bar{z}. This is not going to be 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.md, and the Drawing 2025-03-24 11.00.27.excalidraw.light.svg%%

= \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 at z \neq 0.

Cauchy-Riemann Equations

A Complex Functions 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, 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 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.md, and the Drawing 2025-03-24 11.13.06.excalidraw.light.svg%%

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 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 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 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.md, and the Drawing 2025-03-24 11.32.41.excalidraw.light.svg%%

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 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, 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 (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, we get that f = u + iv is Holomorphic.