2.1 Limits, Derivative, Continuity of Complex Functions

Introduction

Only analytic or holomorphic functions can be freely differentiated and integrated.

There are: Real functions of a real variable, real functions of a complex variable, complex functions of a real variables, and complex function of a real variable.

Generally $z, w \in C,$ $t \in R$

Limits and Continuity

Function Limit

lim_{x \to a} f (x) = A

$⟺ \forall ε > 0, \exists δ > 0, 0 < | x - a | < δ ⟹ | f (x) - A | < ε$

Same as real analysis

Properties

noting $| a b | = | a | \cdot | b |$ and $| a + b | \leq | a | + | b |$

\begin{aligned} lim_{x \to a} \overset{―}{f (x)} = \bar{A} \\ lim_{x \to a} Im f (x) = Im A \\ lim_{x \to a} Re f (x) = Re A \end{aligned}

Continuity

\begin{array}{r} lim_{x \to a} = f (a) \end{array}

Same as real analysis

\forall ε > 0, \exists δ > 0, \forall z \in C \cup {\infty}, d (z, a) < δ ⟹ d (f (z), f (z)) < ε

The sum $f (x) + g (x)$ and product $f (x) g (x)$ are continuous if $f$ and $g$ are continuous.
So will the quotient $\frac{f (x)}{g (x)}$ if $g (a) \neq 0$ .

$Re f (x), Im f (x),$ and $| f (x) |$ will all be continuous

Derivative

Same as real analysis

f^{'} (a) = lim_{x \to a} \frac{f (x) - f (a)}{x - a}

Real vs Complex

Let $f (z) \in R$ whose derivative exists at $z = a$

⟹ \frac{f (a + h) - f (a)}{h}

Exists and is real, like

f^{'} (a) = lim_{Δ x \to 0} \frac{f (a + Δ x) - f (a)}{Δ x}

Now taking the imaginary path:

⟹ \frac{f (a + i h) - f (a)}{i h}

f^{'} (a) = lim_{Δ y \to 0} \frac{f (a + i Δ y) - f (a)}{i Δ y}

Exists and is purely imaginary. Therefore $f^{'} (a)$ must equal zero or it doesn't exist.

Converting to Real Case

For $f : R \to C$

z (t) = x (t) + i y (t) ⟹ z^{'} (t) = x^{'} (t) + i y^{'} (t)

Analytic Functions

A class of analytic functions made of complex functions of complex variables are those that possess a derivative wherever the function is defined. These are also called holomorphic functions.

\begin{aligned} f^{'} (z) = lim_{h \to 0} \frac{f (z + h) - f (z)}{h} \\ ⟹ f (z + h) - f (z) = \frac{h \cdot (f (z + h))}{h} \\ ⟹ lim_{h \to 0} (f (z + h) - f (z)) = 0 \cdot f^{'} (z) = 0 \\ ⟹ continuous \end{aligned}

If $f (z) = u (z) + i v (z)$ then $u (z)$ and $v (z)$ are continuous.