Complex Differentiation

For any $z \in \mathbb C$ , there exist unique real numbers $\mathrm{Re}(z) := x, \mathrm{Im}(z):= y \in \mathbb R$ such that $z = x + iy$ . Recall that $\mathbb C$ forms a normed space under the usual Euclidean norm

$\| z \| := \sqrt{\mathrm{Re}(z)^2 + \mathrm{Im}(z)^2}, \quad z \in \mathbb C.$

Therefore, we can discuss limits and continuity without worry in the sense of $\epsilon$ – $\delta$ limits: $f(z) \to f(z_0)$ if and only if for any $\epsilon > 0$ , there exists $\delta > 0$ such that

$0 < \| w \| < \delta \quad \Rightarrow \quad \| f(z_0+w) - f(z_0) \| < \epsilon.$

Differentiation, however, requires more special features of $\mathbb C$ , and it turns out to have even more special features than even $\mathbb R^2$ .

Definition 1. A function $f : \mathbb C \to \mathbb C$ is said to be complex-differentiable at $z_0$ if there exists a (unique) complex number $f'(z_0)$ such that

$\displaystyle \lim_{w \to 0} \frac{f(z_0 + w) - f(z_0)}{w} = f'(z_0).$

We say that $f$ is complex-differentiable on $K \subseteq \mathbb C$ if $f$ is complex-differentiable at every $z_0 \in K$ .

Example 0. For any fixed $w_0 \in \mathbb C$ , the constant function $f := w_0$ is complex-differentiable on $\mathbb C$ with $f' = 0$ : for any $z_0 \in \mathbb C$ ,

$\displaystyle \lim_{ w \to 0 } \frac{ f(z_0 + w) - f(z_0) }{ w } = \lim_{ w \to 0 } \frac{ w_0 - w_0 }{ w } = 0.$

Example 1. The function $f := \mathrm{id}_{\mathbb C} : \mathbb C \to \mathbb C$ (i.e. $f(z) = z$ ) is complex-differentiable on $\mathbb C$ , and $f' = 1$ : for any $z_0 \in \mathbb C$ ,

$\displaystyle \lim_{ w \to 0 } \frac{ f(z_0 + w) - f(z_0) }{ w } = \lim_{ w \to 0 } \frac{ (z_0 + w) - z_0 }{ w } = 1.$

Theorem 1. Let $f, g : \mathbb C \to \mathbb C$ be complex-differentiable at $z_0 \in \mathbb C$ .

$f \pm g$ is complex-differentiable at $z_0$ ,
for any $\alpha \in \mathbb C$ , $\alpha f$ is complex-differentiable at $z_0$ ,
$f \cdot g$ are complex-differentiable at $z_0$ ,
if $g(z_0) \neq 0$ , then $1/g$ and $f/g$ are complex-differentiable at $z_0$ .

Proof. Same computations as in the case of real analysis, assuming that the map $f : \mathbb C \backslash \{0\} \to \mathbb C$ defined by $f(z) = 1/z$ is continuous, which we prove below in the special case $z_0 = 1$ to reiterate:

Fix $\epsilon > 0$ . By algebra,

$\displaystyle \left\| \frac{1}{1 + w} - 1 \right\| = \frac{1}{ \| 1 + w \| } \cdot \| w \|.$

Therefore, we set $\delta := 1/2$ so that $\| w \| < \delta < 1$ :

$\displaystyle \|1 + w\| \geq | \|1\| - \| w \| | = \|1\| - \| w \| > \|1\| - \delta = \frac{1}{2}.$

Plugging in to our original estimate,

$\displaystyle \frac 1{\|1 + w\|} \leq 2\quad \Rightarrow \quad \left\| \frac{1}{1 + w} - 1 \right\| = \frac{1}{ \| 1 + w \| } \cdot \| w \| \leq 2 \cdot \|w \|.$

Therefore, further stipulate that $\delta < \epsilon / 2$ (so that consolidating we set $\delta := \min\{1, \epsilon\}/4$ ) and we are done.

Corollary 1. Polynomials and rational functions are all complex-differentiable whenever they are well-defined.

Theorem 2. If $f$ is complex differentiable at $z_0$ , then it is continuous at $z_0$ .

Proof. Take the limit $w \to 0$ of the decomposition

$\displaystyle f(z_0 + w) = f(z_0) + \frac{ f(z_0 + w) - f(z_0) }{w} \cdot w.$

Example 2. The map $f(z) = z^2$ is differentiable by Corollary 1. Writing $z = x + i y$ ,

$f(x + i y) = (x + i y)^2 = (x^2 - y^2) + i \cdot (2xy).$

Now we note that $f(z_0) \in \mathbb C$ , which means we can also write $f(z_0) = u_0 + i v_0$ uniquely. Due to $\mathbb C$ being algebraically rather similar to $\mathbb R^2$ , they are fundamentally “portable” from one to another: letting $\iota$ denote the usual vector space isomorphism from $\mathbb R^2$ to $\mathbb C$ given by $\iota(x,y) = x+iy$ , for any $f : \mathbb C \to \mathbb C$ , define

$\mathbf f := \iota^{-1} \circ f \circ \iota : \mathbb R^2 \to \mathbb R^2.$

Denoting $\mathbf f_1 \equiv f_1 := u : \mathbb R^2 \to \mathbb R$ and $\mathbf f_2 \equiv f_2 := v : \mathbb R^2 \to \mathbb R$ , it is not hard to verify that

$f(x+iy) = u(x,y) + i \cdot v(x,y),\quad (x,y) \in \mathbb R^2,$

so that we can interpret $u = \mathrm{Re} \circ f$ and $v = \mathrm{Im} \circ f$ , further emphasising the two-dimensional nature of $\mathbb C$ , which probes a brief study of multivariable calculus.

Definition 2. Let $u : \mathbb R^2 \to \mathbb R$ be a function, and denote the function by $u(x, y)$ . For any $x_0, y_0 \in \mathbb R$ , define the maps $\tilde u (x_0), \tilde u (y_0)$ by $\tilde u (x_0) := u(x_0,\cdot)$ and $\tilde u (y_0) := u(\cdot, y_0)$ . Define the partial derivatives at $(x_0, y_0)$ by

$\begin{aligned} \frac{\partial u}{\partial x}(x_0, y_0) &\equiv u_x(x_0,y_0) := (\tilde u (y_0))'(x_0) ,\\ \frac{\partial u}{\partial y}(x_0, y_0) &\equiv u_y(x_0,y_0) := (\tilde u (x_0))'(y_0) \end{aligned}$

whenever they exist, and we abbreviate

$\displaystyle \frac{\partial u}{\partial x} = u_x \quad \frac{\partial u}{\partial y} = u_y.$

Example 3. Using Example 2, define $u(x, y) = x^2 - y^2$ and $v(x, y) = 2xy$ . To evaluate $u_x$ , fix $y_0$ , so that $(\tilde u(y_0))(x) = x^2 - y_0^2$ . Using usual differentiation,

$\displaystyle (\tilde u(y_0))'(x) = \frac{\mathrm d}{\mathrm dx}(x^2 - y_0^2) = 2x.$

Using partial-derivative notation,

$\displaystyle u_x = \frac{\partial u}{\partial x} = \frac{\partial}{\partial x} (x^2 - y^2) = 2x.$

Therefore, when taking $\frac{\partial}{\partial x}$ , we can regard $y$ as a constant. Similarly,

$\displaystyle u_x = 2x = v_y,\quad u_y = \frac{\partial}{\partial y}(x^2 - y^2) = -2y= -v_x.$

The connections $u_x = v_y$ and $u_v = -v_x$ are not coincidental, but an immediate consequence of $f$ being differentiable.

Theorem 3 (Cauchy-Riemann Equations). Given $f : \mathbb C \to \mathbb C$ , write $f_{\#} = u + i v : \mathbb R^2 \to \mathbb C$ , where $u, v : \mathbb R^2 \to \mathbb R^2$ . If $f$ is complex-differentiable at $z_0 = x_0 + i y_0$ , then $u_x, u_y, v_x, v_y$ satisfy the Cauchy-Riemann equations at $(x_0, y_0)$ :

$\displaystyle u_x(x_0, y_0) = v_y(x_0, y_0),\quad u_y(x_0, y_0) = -v_x(x_0, y_0).$

Proof. Fix $z_0 = x_0 + i y_0$ . Since $f$ is complex-differentiable at $z_0$ , there exists $f'(z_0) := r_0 + i s_0 \in \mathbb C$ such that

$\displaystyle f'(z_0) = \lim_{w \to 0} \frac{ f(z_0 + w) - f(z_0) }{ w } = f'(z_0).$

Writing $w = h + i k \in \mathbb R$ ,

$\begin{aligned} f'(z_0)&= \lim_{w \to 0} \frac{ f ( (x_0 + h) + i \cdot (y_0 + k)) - f (x_0 + i \cdot y_0) }{ h } \\ &= \lim_{w \to 0} \frac{ u(x_0 + h, y_0 + k) + i v(x_0 + h, y_0 + k) - (u(x_0, y_0) + i v(x_0, y_0)) }{ w } \\ &= \lim_{w \to 0} \frac{ u(x_0 + h, y_0 + k) - u(x_0, y_0) }{ w } + i\cdot \lim_{w \to 0} \frac{ v(x_0 + h, y_0 + k) - v(x_0, y_0) }{ w }. \end{aligned}$

In the case $k = 0$ (corresponding to $w = h \in \mathbb R$ ),

$\begin{aligned} r_0 + i s_0 &= \lim_{h \to 0} \frac{ u(x_0 + h, y_0) - u(x_0, y_0) }{ h } + i\cdot \lim_{h \to 0} \frac{ v(x_0 + h, y_0) - v(x_0, y_0) }{ h } \\ &= (\tilde u(y_0))'(x_0) + i \cdot (\tilde v(y_0))'(x_0) \\ &= u_x(x_0, y_0) + i \cdot v_x(x_0, y_0). \end{aligned}$

Similarly, in the case $h = 0$ (corresponding to $w = ik \in \mathbb R[i]$ ),

$\begin{aligned} r_0 + i s_0 &= \lim_{ik \to 0} \frac{ u(x_0, y_0+k) - u(x_0, y_0) }{ ik } + i\cdot \lim_{ik \to 0} \frac{ v(x_0, y_0 + k) - v(x_0, y_0) }{ ik } \\ &= -i \cdot (\tilde u(x_0))'(y_0) + (\tilde v(x_0))'(y_0) \\ &= v_y(x_0, y_0) - i \cdot u_y(x_0, y_0) . \end{aligned}$

Therefore,

$u_x(x_0, y_0) = r_0 = v_y(x_0, y_0),\quad u_y(x_0, y_0) = -s_0 = -v_x(x_0, y_0).$

Does this result hold in the reverse direction? To answer this question, we will need to think about multivariable differentiability, and we will explore this topic next time.

—Joel Kindiak, 9 Aug 25, 1820H

KindiakMath

Complex Differentiation

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Complex Differentiation

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