Continuity of Complex Functions. Definitions, Criteria, and Examples

You have enemies? Good. That means you’ve stood up for something, sometime in your life, Winston Churchill

Recall

A complex number is specified by an ordered pair of real numbers (a, b) ∈ ℝ² and expressed or written in the form z = a + bi, where a and b are real numbers, and i is the imaginary unit, defined by the property i² = −1 ⇔ i = $\sqrt{-1}$, e.g., 2 + 5i, $7\pi + i\sqrt{2}.$ ℂ= { a + bi ∣a, b ∈ ℝ}.

Definition. Let D ⊆ ℂ be a set of complex numbers. A complex-valued function f of a complex variable, defined on D, is a rule that assigns to each complex number z belonging to the set D a unique complex number w, f: D ➞ ℂ.

• The set D is called the domain of the complex function f, D = Dom(f).
• The set of all actual outputs {f(z): z ∈ D} is called the range (or image) of the complex function f, also denoted f(D) = {f(z): z ∈ D}.

We often call the elements of D as points. If z = x+ iy ∈ D, then f(z) is called the image of the point z under f. f: D ➞ ℂ means that f is a complex function with domain D. We often write f(z) = u(x ,y) + iv(x, y), where u, v: ℝ² → ℝ are the real and imaginary parts.

Definition. Let D ⊆ ℂ, $f: D \rarr \Complex$ be a function and z₀ be a limit point of D (so arbitrarily close points of D lie around z₀, though possibly z₀ ∉ D). A complex number L is said to be a limit of the function f as z approaches z₀, written or expressed as $\lim_{z \to z_0} f(z)=L$, if for every epsilon ε > 0, there exist a corresponding delta δ > 0 such that |f(z) -L| < ε whenever z ∈ D and 0 < |z - z₀| < δ.

Why 0 < |z - z₀|? We exclude z = z₀ itself because the limit cares about values near z₀, not at z₀ itself. When z₀ ∉ D, you cannot evaluate f(z₀), so you only care about z approaching z₀. When z₀ ∈ D, you still want the function’s nearby behavior; this separates “limit” from “value.”

Equivalently, if ∀ε >0, ∃ δ > 0: (for every ε > 0, there exist a corresponding δ > 0) such that whenever z ∈ D ∩ B'(z₀; δ), f(z) ∈ B(L; ε) ↭ f(D ∩ B'(z₀; δ)) ⊂ B(L; ε).

If no such L exists, then we say that f(z) does not have a limit as z approaches z₀. This is exactly the same ε–δ formulation we know from real calculus, but now z and L live in the complex plane ℂ, and neighborhoods are round disks rather than intervals.

Continuity in the complex plane

Definition. Let D ⊆ ℂ. A function f: D → ℂ is said to be continuous at a point z₀ ∈ D if given any arbitrarily small ε > 0, there is a corresponding δ > 0 such that |f(z) - f(z₀)| < ε whenever z ∈ D and |z - z₀| < δ.

In words, arbitrarily small output‐changes ε can be guaranteed by restricting z to lie in a sufficiently small disk of radius δ around z₀.

Alternative (Sequential) Definition. Let D ⊆ ℂ. A function f: D → ℂ is said to be continuous at a point z₀ ∈ D if for every sequence {z_n}^∞_n=1 such that z_n ∈ D ∀n∈ℕ & z_n → z₀, we have $\lim_{z_n \to z_0} f(z_n) = f(z_0)$ .

Global Continuity

Definition. A function f: D → ℂ is said to be continuous if it is continuos at every point in its domain (∀z₀ ∈ D), e.g.,

1. Every complex polynomial p(z) = $a_nz^n + ··· + a_1z + a_0$ is continuos everywhere in ℂ, e.g., f(z) = z, the identity function is trivially continuous; f(z) = z³ + 2z + 5.
2. Rational functions. Any rational function $r(z) = \frac{p(z)}{q(z)}$ where p,q are polynomials, is continuous on its domain D = {z: q(z) ≠ 0}, e.g, $f(z) = \frac{z^2 + 1}{z - 2} $ continuos on ℂ - except at poles, e.g., z = 2, where the denominator vanishes.
3. Algebra of continuous functions. If f and g are continuous at a point z₀, then f ± g, f·g, and f/g (provided g(z₀) ≠ 0) are also continuous at z₀. These follow directly from the ε–δ definitions and the corresponding real‐analysis proofs.
4. Elementary transcendental functions. The exponential e^z, the trigonometric functions sin(z), cos(z), and their inverses (on appropriate domains) are all continuous on ℂ.
5. Conjugation is continuous. The function f(z) = $\bar z$ is also continuous, since $|f(z) - f(z_0)| = |\overline{z} - \overline{z_0}| = |z - z_0|$

   Since the modulus is preserved under conjugation, the “distance” between image points matches the distance in the domain. This means small changes in z lead to small changes in f(z), satisfying the definition of continuity. It ties small input changes to small output changes.

Common Pitfalls & Counterexamples

• A function with a jump or point‐removal, e.g.,

$f(x) = \begin{cases} 1, &z \ne 0 \\\\ 0, &z = 0 \end{cases}$ fails continuity at the “hole” z = 0.

• A function defined differently on rationals vs. irrationals (the complex analog of the Dirichlet function) is nowhere continuous.

Discontinuities arise only where a function is “patched” or forced to misbehave: division by zero (f(z) = $\frac{1}{z}$. It is not continuos at z = 0 where it is undefined), piecewise jumps ($f(z) = \begin{cases} z, &|z| < 1 \\\\ \overline{z}, &|z| \ge 1 \end{cases}$ It is discontinuous at the boundary |z| = 1 due to a sudden change in definition) or should I even say, pathological constructions.

Continuity in the complex plane is exactly the same ε–δ notion you have already studied (supposedly, hopefully,...) in single‐variable real calculus —only now disks in ℂ replace intervals in ℝ. If you draw an arbitrarily small circle around z₀, continuity demands that f(z) stays within an arbitrarily small circle around f(z₀).

Theorem. Let D ⊆ ℂ and f a function, f: D → ℂ. f is continuous on D if and only if the inverse image of every open set in ℂ is open in D.

Proof.

→) Continuity ⇒ open–preimage

Let U ⊆ ℂ be an open set in the complex numbers. We must show f^-1(U) = {z ∈ D : f(z) ∈ U} is open as a subset of D.

If f^-1(U) = ∅, then since ∅ is vacuously open (in D), the statement holds true.

Otherwise, f^-1(U) ≠ ∅, let us pick an arbitrary element z₀ ∈ f^-1(U), w₀ = f(z₀) ∈ U.

Since, by assumption, U is an open set in ℂ, there exists ∃r > 0 s.t. B(w₀; r) ⊆ U. Furthermore, since f is continuous at z₀, there is δ > 0, s.t. |f(z)-f(z₀)| < r ∀z ∈ D and |z - z₀| < δ.

Hence z ∈ D ∩ B(z₀, δ) → f(z) ∈ B(w₀, r) ↭ f(B(z₀, δ) ∩ D) ⊆ B(w₀, r) ⊆ U.

In other words, ∀U open set in the complex numbers, ∀z₀ ∈ f^-1(U), ∃δ, f(B(z₀, δ) ∩ D) ⊆ U ↭[🚀] f^-1(U) is open in D∎

🚀 Since B(z₀, δ) ∩ D is an open subset of D containing z₀, we conclude every point of f^-1(U) has an open neighborhood inside it.

←) open–preimage condition ⇒ Continuity

Suppose the inverse image of every open set in ℂ is open in D.

Let z₀ ∈ D and let w₀ = f(z₀). We will show f is continuous at z₀.

Let ε > 0. Since B(w₀; ε) is an open set in ℂ, by hypothesis its preimage f^-1(B(w₀; ε)) is open in D and it contains z₀ (z₀ ∈ f^-1(B(w₀; ε))).

Therefore, there is a δ > 0 s.t. D ∩ B(z₀, δ) ⊆ f^-1(B(w₀; ε)) ⇒ f(D ∩ B(z₀, δ)) ⊆ B(w₀; ε).

Equivalently, ∀z∈ D, ∃δ > 0 s.t. |z - z₀| < δ ⇒ |f(z) - f(z₀)| < ε, f is continuous at z₀ ∎