2.2 Cauchy Riemann and Results

Analytic then Cauchy Riemann

The limit must be the same regardless of the direction that $h \to 0$

Say $f (z) = u (x, y) + i v (x, y)$ and $f^{'} (z)$ exists at $z_{0} = x_{0} + i y_{0}$

\begin{array}{r} \frac{\partial f}{\partial x} = - i \frac{\partial f}{\partial y} ⟹ {\begin{cases} \frac{\partial u}{\partial x} (x_{0}, y_{0}) = \frac{\partial v}{\partial y} (x_{0}, y_{0}) \\ \frac{\partial u}{\partial y} (x_{0}, y_{0}) = - \frac{\partial v}{\partial x} (x_{0}, y_{0}) \end{cases} \end{array}

Proof:

$Δ z = Δ x + i Δ y$ , $Δ z \to 0 ⟺ (Δ x, Δ y) \to 0$

\begin{aligned} lim_{Δ z \to 0} \frac{f (z_{0} + Δ z) - f (z_{0})}{Δ z} & = lim_{Δ z \to 0} [\frac{u (x_{0} + Δ x, y + Δ y) - u (x_{0}, y_{0}) + i v (x + Δ x, y + Δ y) - i v (x_{0}, y_{0})}{Δ z}] \\ = lim_{Δ z \to 0} [\frac{u (x_{0} + Δ x, y_{0} + Δ y) - u (x_{0}, y_{0})}{Δ z} + \\ i \frac{v (x + Δ x, y + Δ y) - v (x_{0}, y_{0})}{Δ z}] \end{aligned}

Along real line:
$Δ z = Δ x + i 0$

\begin{aligned} = lim_{Δ x \to 0} [\frac{u (x_{0} + Δ x, y_{0}) - u (x_{0}, y_{0})}{Δ x} + \\ i \frac{v (x + Δ x, y) - v (x_{0}, y_{0})}{Δ x}] \\ = \frac{\partial u}{\partial x} (x_{0}, y_{0}) + i \frac{\partial v}{\partial x} (x_{0}, y_{0}) \end{aligned}

Along complex line:
$Δ z = 0 + i Δ y$ note $\frac{1}{i} = - i$

\begin{aligned} = lim_{Δ y \to 0} [\frac{u (x_{0}, y_{0} + i Δ y) - u (x_{0}, y_{0})}{i Δ y} + \\ i \frac{v (x_{0}, y_{0} + i Δ y) - v (x_{0}, y_{0})}{i Δ y}] \\ = lim_{Δ y \to 0} [- i \frac{u (x_{0}, y_{0} + i Δ y) - u (x_{0}, y_{0})}{Δ y} + \\ \frac{v (x_{0}, y_{0} + i Δ y) - v (x_{0}, y_{0})}{Δ y}] \\ = - i \frac{\partial u}{\partial y} (x_{0}, y_{0}) + \frac{\partial v}{\partial y} (x_{0}, y_{0}) \end{aligned}

Corollaries

1. The $\bar{z}$ -Derivative (The Test for Analyticity)

Let $x = \frac{z + \bar{z}}{2}$ and $y = \frac{z - \bar{z}}{2 i}$ . The partials w.r.t. $\bar{z}$ are:

\frac{\partial x}{\partial \bar{z}} = \frac{1}{2}, \frac{\partial y}{\partial \bar{z}} = - \frac{1}{2 i} = \frac{i}{2}

By the chain rule, the derivative of $f (z) = u + i v$ w.r.t. $\bar{z}$ is:

\begin{aligned} \frac{\partial f}{\partial \bar{z}} & = \frac{\partial f}{\partial x} \frac{\partial x}{\partial \bar{z}} + \frac{\partial f}{\partial y} \frac{\partial y}{\partial \bar{z}} \\ = (u_{x} + i v_{x}) (\frac{1}{2}) + (u_{y} + i v_{y}) (\frac{i}{2}) \\ = \frac{1}{2} [(u_{x} + i v_{x}) + (i u_{y} - v_{y})] \\ = \frac{1}{2} [(u_{x} - v_{y}) + i (v_{x} + u_{y})] \end{aligned}

This expression equals $0$ if and only if $u_{x} = v_{y}$ and $v_{x} = - u_{y}$ .

Point 1: The Cauchy-Riemann equations are equivalent to the single compact equation:

\frac{\partial f}{\partial \bar{z}} = 0

(Analyticity means $f$ is a function of $z$ only, not $\bar{z}$ .)

2. The Jacobian (The Geometry of $f^{'} (z)$ )

The complex derivative is $f^{'} (z) = u_{x} + i v_{x}$ . Its squared magnitude is:

| f^{'} (z) |^{2} = (u_{x})^{2} + (v_{x})^{2}

The Jacobian determinant of the transformation $(x, y) \to (u, v)$ is:

J = det (\begin{matrix} u_{x} & u_{y} \\ v_{x} & v_{y} \end{matrix}) = u_{x} v_{y} - u_{y} v_{x}

If $f$ is analytic, we use the CR equations ( $u_{x} = v_{y}, u_{y} = - v_{x}$ ) on $J$ :

J = (u_{x}) (u_{x}) - (- v_{x}) (v_{x}) = (u_{x})^{2} + (v_{x})^{2}

Point 2: The magnitude of the complex derivative equals the scaling factor of the real map.

| f^{'} (z) |^{2} = J (u, v)

( $| f^{'} (z) |$ is the length-scaling factor, $J$ is the area-scaling factor.)

Cauchy-Riemann CTS then Analytic

Let $f (z) = u (x, y) + i v (x, y)$

If $u, v$ have continuous derivatives that satisfy the Cauchy-Riemann equations

\begin{array}{r} {\begin{cases} \frac{\partial u}{\partial x} (x_{0}, y_{0}) = \frac{\partial v}{\partial y} (x_{0}, y_{0}) \\ \frac{\partial u}{\partial y} (x_{0}, y_{0}) = - \frac{\partial v}{\partial x} (x_{0}, y_{0}) \end{cases} \end{array}

then $f$ is an analytic function

Lemma: Linear Approx. Error

If $u : R^{2} \to R$ is a $C^{1}$ function:

u (x + Δ x, y + Δ y) = u_{x} (x, y) Δ x + u_{y} (x, y) Δ y + E

Where $lim_{Δ z \to 0} \frac{E}{Δ z} = 0$

Proof:
Let $P (t) = u (x + t Δ x, y + t Δ y)$ $t \in [0, 1]$
$P (1) = u (x + Δ x, y + Δ y), P (0) = u (x, y)$
By MVT $\exists ξ \in (0, 1)$ such that $P (1) - P (0) = P^{'} (ξ)$

P^{'} (ξ) = u_{x} (x + ξ Δ x, y + ξ Δ y) Δ x + u_{y} (x + ξ Δ x, y + ξ Δ y) Δ y

\begin{aligned} ⟹ u (x + Δ x, y + Δ y) - u (x, y) & = [u_{x} (x + ξ Δ x, y + ξ Δ y) Δ x - u_{x} (x, y) Δ x \\ \underset{E}{\underset{⏟}{+ u_{y} (x + ξ Δ x, y + ξ Δ y) Δ y - u_{y} (x, y) Δ y]}} \\ + u_{x} (x, y) Δ y + u_{y} (x, y) Δ y \end{aligned}

Since $u_{x}, u_{y}$ are continuous If $ε > 0$ there exists a $δ > 0$ so that:

\begin{aligned} | u_{x} (x + ξ Δ x, y + ξ Δ y) - u_{x} (x, y) | & < ε Δ x \\ | u_{y} (x + ξ Δ x, y + ξ Δ y) - u_{y} (x, y) | & < ε Δ x \\ ⟹ | E | < ε Δ x + ε Δ y & \leq 2 ε | Δ z | \\ ⟹ \frac{E}{| Δ z |} & < 2 ε \\ ⟹ lim_{Δ z \to 0} \frac{E}{| Δ z |} = 0 \end{aligned}

Proof:

by Lemma since $u, v$ have continuous partials

\begin{aligned} u (x + Δ x, y + Δ y) - u (x, y) = \frac{\partial u}{\partial x} Δ x + \frac{\partial u}{\partial y} Δ y + E_{1} \\ v (x + Δ x, y + Δ y) - v (x, y) = \frac{\partial v}{\partial x} Δ x + \frac{\partial v}{\partial y} Δ y + E_{2} \end{aligned}

Where $\frac{E_{1}}{| Δ z |}, \frac{E_{2}}{| Δ z |} \to 0$ as $Δ z \to 0$

\begin{aligned} f (z + Δ z) - f (z) & = u (x + Δ x, y + Δ y) - u (x, y) \\ + i [v (x + Δ x, y + Δ y) - v (x, y)] \\ = \frac{\partial u}{\partial y} Δ x + \frac{\partial u}{\partial y} Δ y + E_{1} \\ + i [\frac{\partial v}{\partial x} Δ x + \frac{\partial v}{\partial y} Δ y + E_{2}] \\ = (\frac{\partial u}{\partial x} + i \frac{\partial v}{\partial x}) (Δ x + i Δ y) + E_{1} + i E_{2} \\ ⟹ lim_{Δ z \to 0} \frac{f (z + Δ z) - f (z)}{Δ z} & = \frac{\partial u}{\partial x} + i \frac{\partial v}{\partial x} \\ = \frac{\partial v}{\partial y} - i \frac{\partial u}{\partial y} \end{aligned}

Laplace Equations

A function $u$ that satisfies

\underset{Δ u}{\underset{⏟}{\frac{\partial^{2} u}{\partial x^{2}} + \frac{\partial^{2} u}{\partial y^{2}}}} = 0

is said to be harmonic

If two harmonic functions satisfy the Cauchy-Riemann Equations then $v$ (and any additional constant) is said to be the conjugate harmonic function of $u$ . $u$ is the conjugate harmonic function of $- v$ .

If $f = u + i v$ is an analytic function then $u, v$ are harmonic functions

$u$ is harmonic and there exists a harmonic conjugate $v$ such that CR holds.
Then $f = u + i v$ is analytic.

Not Laplace $⟹$ Not Analytic

So overall if $f (z) = u + i v$ is analytic:

CR equations hold: $u_{x} = v_{y}, u_{y} = - v_{x}$ and the are continuous
Harmonicity holds: $u_{x x} + u_{y y} = 0, v_{x x} + v_{y y} = 0$

Analytic then Cauchy Riemann

Proof:

Corollaries

1. The z¯-Derivative (The Test for Analyticity)

2. The Jacobian (The Geometry of f′(z))

Cauchy-Riemann CTS then Analytic

Lemma: Linear Approx. Error

Proof:

Laplace Equations

1. The $\bar{z}$ -Derivative (The Test for Analyticity)

2. The Jacobian (The Geometry of $f^{'} (z)$ )