Complex Functions - Introduction

Complex Functions
Introduction

Complex Functions - Introduction

Topological introduction

Complex functions

Limits

Continuity

Topological introduction

We begin by establishing the topological structure of the complex plane, \mathbb{C}. The concepts of proximity and locality are defined through the metric d(z_1, z_2) = |z_1 - z_2|.

An open disk or neighborhood of a point z_0 \in \mathbb{C} with radius \varepsilon > 0 is the set N_\varepsilon(z_0) = \{z \in \mathbb{C} : |z - z_0| < \varepsilon\}.

Let \mathcal{S} be a subset of \mathbb{C}. We define:

  • A point z_0 \in \mathcal{S} is an interior point of \mathcal{S} if there exists a neighborhood N_\varepsilon(z_0) such that N_\varepsilon(z_0) \subseteq \mathcal{S}. The set of all interior points is the interior of \mathcal{S}, denoted \text{int}(\mathcal{S}).
  • A point z_0 \in \mathbb{C} is a boundary point of \mathcal{S} if every neighborhood N_\varepsilon(z_0) contains at least one point in \mathcal{S} and at least one point not in \mathcal{S}. The set of all boundary points is the boundary of \mathcal{S}, denoted \partial\mathcal{S}.
  • An open set is a set \mathcal{S} that consists entirely of interior points, i.e., \mathcal{S} = \text{int}(\mathcal{S}).
  • A closed set is a set that contains all of its boundary points, i.e., \partial\mathcal{S} \subseteq \mathcal{S}. The closure of a set \mathcal{S} is \overline{\mathcal{S}} = \mathcal{S} \cup \partial\mathcal{S}.

A set \mathcal{S} is connected if it cannot be expressed as the disjoint union of two non-empty open subsets. For open sets in \mathbb{C}, this is equivalent to the property of being path-connected: any two points z_1, z_2 \in \mathcal{S} can be joined by a continuous path \gamma: \to \mathcal{S} with \gamma(0) = z_1 and \gamma(1) = z_2.

With this framework, we formally define a domain as a non-empty, open, connected subset of \mathbb{C}. A closed domain is the closure of a domain. A domain \mathcal{D} is bounded if it is contained within some disk of finite radius; that is, there exists an R > 0 such that |z| < R for all z \in \mathcal{D}. Otherwise, \mathcal{D} is unbounded.

Complex functions

A function of a complex variable z is a map f from a domain \mathcal{D} \subseteq \mathbb{C} to the complex numbers \mathbb{C}. We write f: \mathcal{D} \to \mathbb{C}. The domain \mathcal{D} is the set of values for the independent variable. The set f(\mathcal{D}) = \{f(z) : z \in \mathcal{D}\} \subseteq \mathbb{C} is the image or range of the function.

Function domain \mathcal{D}

If for each w in the image, there is a unique z \in \mathcal{D} such that f(z) = w, the function is injective (or one-to-one). Such a function establishes a bijective correspondence between \mathcal{D} and f(\mathcal{D}). In this case, an inverse function \varphi: f(\mathcal{D}) \to \mathcal{D} exists, defined by \varphi(w) = z if and only if f(z) = w.

Any complex function f(z) can be decomposed by considering its effect on the real and imaginary parts of its argument z = x + iy. The value of the function is a complex number w = u + iv. Both u and v are real-valued functions of the real variables x and y. We express this as:

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

Here, u(x,y) = \text{Re}(f(z)) is the real part and v(x,y) = \text{Im}(f(z)) is the imaginary part of the function f(z). This representation establishes a direct correspondence between a complex function f: \mathbb{C} \to \mathbb{C} and a vector-valued function from \mathbb{R}^2 to \mathbb{R}^2, given by (x,y) \mapsto (u(x,y), v(x,y)).

Limits

The concept of a limit in complex analysis is analogous to that in real analysis, leveraging the metric on \mathbb{C}.

Let f be a function defined on a domain \mathcal{D}, and let z_0 be a point in \overline{\mathcal{D}}. We say that the limit of f(z) as z approaches z_0 is w_0, written as:

\lim_{z \to z_0} f(z) = w_0 if for every real \varepsilon > 0, there exists a real \delta > 0 such that for all z \in \mathcal{D} satisfying 0 < |z - z_0| < \delta, we have |f(z) - w_0| < \varepsilon.

An equivalent characterization can be formulated using sequences. The limit \lim_{z \to z_0} f(z) = w_0 if and only if for every sequence of points \{z_n\}_{n=1}^\infty in \mathcal{D} \setminus \{z_0\} that converges to z_0, the corresponding sequence of values \{f(z_n)\}_{n=1}^\infty converges to w_0.

The equivalence of these definitions is a standard result in metric spaces.

To demonstrate, assume the \varepsilon-\delta definition holds. Let \{z_n\} \to z_0. For any \varepsilon > 0, there is a corresponding \delta > 0. Since \{z_n\} \to z_0, there exists an integer N such that for all n > N, |z_n - z_0| < \delta. Consequently, |f(z_n) - w_0| < \varepsilon, which proves \lim_{n \to \infty} f(z_n) = w_0.

Conversely, we prove the contrapositive and we assume the \varepsilon-\delta definition fails. Then there exists an \varepsilon_0 > 0 such that for every \delta > 0, there is a point z_\delta \in \mathcal{D} with 0 < |z_\delta - z_0| < \delta but |f(z_\delta) - w_0| \ge \varepsilon_0. By choosing a sequence of deltas \{\delta_n = 1/n\}_{n=1}^\infty, we can construct a sequence of points \{z_n\} such that 0 < |z_n - z_0| < 1/n. This sequence \{z_n\} clearly converges to z_0. However, the sequence \{f(z_n)\} cannot converge to w_0, as |f(z_n) - w_0| \ge \varepsilon_0 for all n. This violates the sequential definition.

Continuity

A function f(z) defined on a domain \mathcal{D} is continuous at a point z_0 \in \mathcal{D} if the limit of the function at z_0 exists, is finite, and coincides with the function’s value at that point:

\lim_{z \to z_0} f(z) = f(z_0)

This definition implicitly requires that f is defined at z_0. A function is said to be continuous on a set if it is continuous at every point in the set.

A fundamental property is that the continuity of a complex function f(z) = u(x,y) + i v(x,y) is entirely equivalent to the continuity of its real and imaginary parts, u(x,y) and v(x,y), as functions from \mathbb{R}^2 to \mathbb{R}.

Theorem: The function f(z) = u(x,y) + i v(x,y) is continuous at z_0 = x_0 + i y_0 if and only if u(x,y) and v(x,y) are continuous at (x_0, y_0).

Proof: (\Rightarrow) Assume f is continuous at z_0. Let \varepsilon > 0 be given. By the definition of continuity for f, there exists a \delta > 0 such that if |z - z_0| < \delta, then |f(z) - f(z_0)| < \varepsilon. Let z = x + iy. The condition |z - z_0| < \delta is equivalent to \sqrt{(x-x_0)^2 + (y-y_0)^2} < \delta. We have the inequalities:

\begin{aligned} |u(x,y) - u(x_0, y_0)| & = |\text{Re}(f(z) - f(z_0))| \le |f(z) - f(z_0)| < \varepsilon \\ |v(x,y) - v(x_0, y_0)| & = |\text{Im}(f(z) - f(z_0))| \le |f(z) - f(z_0)| < \varepsilon \end{aligned}

Therefore, for any (x,y) such that \sqrt{(x-x_0)^2 + (y-y_0)^2} < \delta, we have |u(x,y) - u(x_0, y_0)| < \varepsilon and |v(x,y) - v(x_0, y_0)| < \varepsilon. This is precisely the definition of continuity for u and v at (x_0, y_0).

Let’s assume u(x,y) and v(x,y) are continuous at (x_0, y_0). Let \varepsilon > 0 be given. By the continuity of u and v, for \varepsilon/2 > 0, there exist \delta_1 > 0 and \delta_2 > 0 such that:

  • If \sqrt{(x-x_0)^2 + (y-y_0)^2} < \delta_1, then |u(x,y) - u(x_0, y_0)| < \varepsilon/2.
  • If \sqrt{(x-x_0)^2 + (y-y_0)^2} < \delta_2, then |v(x,y) - v(x_0, y_0)| < \varepsilon/2.

Let \delta = \min(\delta_1, \delta_2). If |z - z_0| = \sqrt{(x-x_0)^2 + (y-y_0)^2} < \delta, then both conditions are met.

Using the triangle inequality, we have:

\begin{aligned} |f(z) - f(z_0)| & = |(u(x,y) - u(x_0, y_0)) + i(v(x,y) - v(x_0, y_0))| \\ & \le |u(x,y) - u(x_0, y_0)| + |v(x,y) - v(x_0, y_0)| < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon \end{aligned}

This demonstrates that f is continuous at z_0.

This equivalence allows us to transfer many properties of continuous real-valued functions of two variables to complex functions.

For example, if f(z) and g(z) are continuous on a domain \mathcal{D}, then their sum f(z)+g(z) and product f(z)g(z) are also continuous on \mathcal{D}. The quotient f(z)/g(z) is continuous at all points in \mathcal{D} where g(z) \ne 0.

Polynomial functions P(z) = a_n z^n + \dots + a_1 z + a_0 are continuous on the entire complex plane. Rational functions P(z)/Q(z), where P and Q are polynomials, are continuous everywhere except at the roots of Q(z).

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