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

The truth will set you free, but first it will piss you off, Joe Klaas

Recall

Definition (Limit-based Continuity). A function $f: U \to \mathbb{C}$ is continuous at a point a ∈ U if $\lim_{z \to a} f(z) = f(a)$

This requires three conditions:

1. f(a) is defined (i.e., a is in the domain U and f maps a to $f(a) \in \mathbb{C}$).
2. The limit $\lim_{z \to a} f(z)$ exists.
3. The limit equals f(a).

   This aligns with the intuitive idea that “small changes in input near a produce small changes in output near f(a).”

Formal definition (Epsilon-Delta). f is continuous at a if and only if: $\forall \varepsilon > 0, \; \exists \delta > 0 : |z - a| < \delta \Rightarrow |f(z) - f(a)| < \varepsilon$.

Global Continuity

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

1. Every complex polynomial p(z) = $a_nz^n + ··· + a_1z + a_0$ is continuous everywhere in ℂ, e.g., f(z) = z, the identity function is trivially continuous; f(z) = z³ + 2z + 5.
   Constants are continuous. For f(z) = c (constant), |f(z) - f(z₀)| = |c - c| = 0 < ε for any ε, δ. ✓
   The identity function is continuous. For f(z) = z, |f(z) - f(z₀)| = |z - z₀|. Given ε > 0, choose δ = ε. ✓
   Powers are continuous. If f(z) = zⁿ is continuous and g(z) = z is continuous, then f·g = $z^{n+1}$ is continuous (by the product rule).
   Polynomials are sums and products. Apply sum and product rules (proven below). ∎

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, $f(z) = \frac{z^2 + 1}{z - 2}$ continuous on ℂ - except at poles, e.g., z = 2, where the denominator vanishes (f is undefined).

3. Algebra of continuous functions. If f and g are continuous at a point z₀, then sum/difference f ± g, scalar multiple c·f, product f·g, and quotient 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.
   Theorem. If f and g are continuous at z₀, then f + g is continuous at z₀. Let ε > 0. Since f and g are continuous at z₀:
   $\exists \delta_1 > 0 : |z - z_0| < \delta_1 \implies |f(z) - f(z_0)| < \frac{\varepsilon}{2}, \exists \delta_2 > 0 : |z - z_0| < \delta_2 \implies |g(z) - g(z_0)| < \frac{\varepsilon}{2}$
   Let δ = min(δ₁, δ₂). For |z - z₀| < δ:
   $|(f+g)(z) - (f+g)(z_0)| = |[f(z) - f(z_0)] + [g(z) - g(z_0)]| \leq |f(z) - f(z_0)| + |g(z) - g(z_0)| < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon$

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.
   sin(z) and cos(z) an be expressed in terms of the exponential: $\sin z = \frac{e^{iz} - e^{-iz}}{2i}, \quad \cos z = \frac{e^{iz} + e^{-iz}}{2}$
   Since eᶻ is continuous and sums/products/quotients of continuous functions are continuous, sin and cos are continuous.

5. Conjugation is continuous. The function f(z) = $\bar z$ is also continuous, since $|f(z) - f(z_0)| = |\overline{z} - \overline{z_0}| = |\overline{z - z_0}| = |z - z_0|$.
   Given ε > 0, choose δ = ε. Then, $|z - z_0| < \delta = \varepsilon \implies |f(z) - f(z_0)| = |z - z_0| < \varepsilon$
   Conjugation is an isometry —it preserves distances!

   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.

6. The Modulus Function f(z) = |z| is continuous on ℂ.
   By the reverse triangle inequality: $\big||z| - |z_0|\big| \leq |z - z_0|$
   Given ε > 0, choose δ = ε. Then |z - z₀| < δ implies $\big||z| - |z_0|\big| \lt \varepsilon$. ∎

7. The functions Re(z) and Im(z) are continuous on ℂ.
   For z = x + iy and z₀ = x₀ + iy₀:
   $|\text{Re}(z) - \text{Re}(z_0)| = |x - x_0| \leq |z - z_0|, |\text{Im}(z) - \text{Im}(z_0)| = |y - y_0| \leq |z - z_0|$
   Choose δ = ε in each case. ∎

8. Composition of Continuous Functions. If f: D → ℂ is continuous at z₀ and g: E → ℂ is continuous at f(z₀), where f(D) ⊆ E, then g ∘ f is continuous at z₀, e.g., $h(z) = e^{z^2 + 1}$ is continuous everywhere because it’s the composition of polynomial $z^2 + 1$ (continuous) and exponential $e^w$ (continuous).
   Let ε > 0. Since g is continuous at w₀ = f(z₀): $\exists \eta > 0 : |w - w_0| < \eta \implies |g(w) - g(w_0)| < \varepsilon$
   Since f is continuous at z₀: $\exists \delta > 0 : |z - z_0| < \delta \implies |f(z) - f(z_0)| < \eta$
   For |z - z₀| < δ: |f(z) - f(z₀)| < η. So |g(f(z)) - g(f(z₀))| < ε. Therefore, g ∘ f is continuous at z₀.

9. Principal Branches. The principal logarithm $\text{Log}(z)$ is continuous on ℂ \ (-∞, 0]. The principal square root $\sqrt{z}$ is continuous on ℂ \ (-∞, 0].

Types of Discontinuities

Definition. A function f: D → ℂ is discontinuous at z₀ if it fails to be continuous at z₀. This can happen if:

1. $z₀ \notin D$: f is not defined at z₀
2. $\lim_{z \to z_0} f(z)$ does not exist: The limit fails
3. $\lim_{z \to z_0} f(z) \neq f(z_0)$: The limit exists but doesn’t match the value.

Classification of Discontinuities

The classification of discontinuities in complex analysis hinges on how a function behaves near a point $z_0$, where it fails to be continuous.

• Removable. The limit $\lim_{z \to z_0} f(z)$ exists but either f(z₀) is undefined or $\lim_{z \to z_0} f(z) \ne f(z_0)$, e.g., $f(z) = \frac{z^2-1}{z-1}$ at z = 1.
  The limit $\lim_{z \to 1} \frac{z^2-1}{z-1} = \lim_{z \to 1}(z + 1) = 2$ exists. However, f(1) is undefined (division by zero).
  Redefining f(1) = 2 removes the Discontinuity. A “hole” in the graph that can be “plugged” by redefining the function at $z_0$ to match the limit.
• Jump Discontinuity. The limit $\lim_{z \to z_0} f(z)$ depends on the path of approach to $z_0$, leading to different limiting values, e.g., $f(z) = \arg(z)$ on the negative real axis.
  Approaching $z_0$ from above (Im(z) > 0) gives $arg(z) \to \pi$.
  Approaching $z_0$ from below (Im(z) < 0) gives $arg(z) \to -\pi$.
  The limit does not exist due to conflicting path-dependent values.
  Intuition: A “jump” between values when traversing different directions in the complex plane. Common in multi-valued functions (e.g., logarithms, roots) with branch cuts.
• Essential Discontinuity. The limit $\lim_{z \to z_0} f(z)$ doesn’t exist, not even in the extended complex plane (i.e., it is neither a finite complex number nor infinity), e.g., $f(z) = e^{1/z}$ at z = 0.
  Let $w = 1/z$. As $z \to 0$, $w$ goes to infinity, but in different directions in the complex plane.
  Approach along the Positive Real Axis ($z \to 0^+$). Let $z = x$ where $x > 0$. $w = \frac{1}{x} \to +\infty$. $f(z) = e^{1/x} \to +\infty$. The function blows up to infinity (like a pole).
  Approach along the Negative Real Axis ($z \to 0^-$). Let $z = x$ where $x < 0$. $w = \frac{1}{x} \to -\infty$. $f(z) = e^{1/x} = \frac{1}{e^{|1/x|}} \to 0$. The function decays rapidly to zero.
  Just these two paths are enough to prove the limit doesn’t exist (one path goes to $\infty$, the other to $0$). But it gets weirder.
  Approach along the Imaginary Axis ($z \to 0$ via imaginary $iy$).
  Let $z = iy$. Then, $1/z = 1/(iy) = -i/y$. $f(z) = e^{-i/y} = \cos(-1/y) + i\sin(-1/y)$ The magnitude is constant: $|e^{-i/y}| = 1$. The function stays on the unit circle forever. However, as $y \to 0$, the angle $-1/y$ goes to $-\infty$.
  The value spins around the unit circle infinitely many times with increasing frequency. It does not settle on any specific point on the circle.
  This is a chaotic behavior near $z_0$, with no predictable pattern. Essential singularities are “worse” than poles and cannot be “removed” or classified by polynomial growth.
• Pole. $|f(z)| \to \infty$ as $z \to z_0$. Poles are isolated singularities where the function blows up., e.g., f(z) = 1/z at z = 0.
  Poles are classified by order, e.g., simple for 1/z, pole of order 2 for $\frac{1}{z^2}$.
  This is a “blow-up” singularity where the function escapes to infinity.

Laurent‑Series Classification of Isolated Singularities

Let $f(z) = \sum_{n=-\infty}^{\infty} a_n (z - z_0)^n$ be the Laurent expansion of f in a punctured neighborhood of $z_0$. The nature of the singularity at $z_0$ is determined entirely by the principal part, i.e., the terms with negative powers.

• Removable Singularity. $a_{-1}=a_{-2}=a_{-3}=\cdots =0.$ The Laurent series reduces to an ordinary Taylor series. The function is bounded near $z_0$, and one can define $f(z_0)=a_0$ to make f analytic at $z_0$.

Intuition: No negative powers $\implies$ nothing “blows up” $\implies$ the singularity can be removed.

• Pole of Order m. If there are finitely many negative powers, $a_{-m}\neq 0,\quad a_{-(m+1)}=a_{-(m+2)}=\cdots =0.$ There are finitely many negative‑power terms, with the most negative being $(z-z_0)^{-m}$. The function behaves like $f(z)\sim \frac{a_{-m}}{(z-z_0)^m}$ near $z_0$, so it blows up in a controlled, algebraic way

Intuition: Finitely many negative powers $\implies$ controlled divergence $\implies$ a pole.

• An essential singularity occurs when there are infinitely many negative powers $a_{-1},a_{-2},a_{-3},\dots$. The principal part is infinite. This is the hallmark of an essential singularity.

Interpretation: Infinitely many negative powers $\implies$ unbounded, non‑algebraic, unpredictable behavior $\implies$ essential singularity.

Theorems describing the “Chaos”

The behavior of essential singularities is not just “undefined”; it is aggressively dense.

1. Casorati-Weierstrass Theorem. If $f$ has an essential singularity at $z_0$, then for any neighborhood $U$ of $z_0$, the image $f(U)$ is dense in $\mathbb{C}$. No matter how small a circle you draw around the singularity, the function values inside that circle come arbitrarily close to every complex number.
2. Picard’s Great Theorem. In any neighborhood of an essential singularity, the function takes on every single complex value infinitely often, with at most one exception.
   For $e^{1/z}$: The function takes every value $v \in \mathbb{C}$ infinitely often, except for $v = 0$.

Common Pitfalls & Counterexamples

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

$f(z) = \begin{cases} 1, &z \ne 0 \\\\ 0, &z = 0 \end{cases}$ fails continuity at the “hole” z = 0 because the function is defined as 1 everywhere except at $0$, where it jumps to $0$. The limit as $z\to 0$ is $1$, so redefining $f(0)=1$ makes it continuous.

$f(z) = \frac{z^2-1}{z-1}$ is undefined at $z=1$, but for $z\neq 1$ it simplifies to $z+1$. The limit as $z\to 1$ is $2$, so the singularity is removable by defining $\tilde{f}(1)=2$. The extended function $\tilde{f}(z)=z+1$ is entire (analytic everywhere).

In complex analysis, a point $z_0$ is a removable singularity if $f$ is analytic in a punctured neighborhood and the limit exists (and is finite). In such cases, we can always redefine $f$ at $z_0$ to make it analytic there.

• Piecewise-Defined Function. $f(z) = \begin{cases} z & |z| < 1 \\ \bar{z} & |z| \geq 1 \end{cases}$

At z₀ = eⁱᶿ (on the unit circle): (i) From inside (|z| < 1): f(z) = $z \to e^{i\theta}$; (ii) From outside (|z| > 1): f(z) = $\overline{z} \to e^{-i\theta}$

For θ ≠ 0, π (i.e., z₀ not on the real axis): eⁱᶿ ≠ e⁻ⁱᶿ, so discontinuous, e.g., discontinuity at i, (i) from inside $f(i) \to i$; (ii) from outside $f(i) \to -i$. For θ = 0 or π (i.e., z₀ = ±1): eⁱᶿ = e⁻ⁱᶿ, so continuous at ±1.

• Dirichlet-Type Function, $f(z) = \begin{cases} 1 & \text{Re}(z), \text{Im}(z) \in \mathbb{Q} \\ 0 & \text{otherwise} \end{cases}$

This function is discontinuous everywhere.

At any z₀ ∈ ℂ: (i) every neighborhood contains points where f = 1 (rationals are dense); (ii) every neighborhood contains points where f = 0 (irrationals are dense)

So f oscillates between 0 and 1 in every neighborhood —no limit exists anywhere.

Topological Characterization of Continuity

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).

The Open Preimage Theorem. Let D ⊆ ℂ and f a function, f: D → ℂ. Then, f is continuous on D if and only if the preimage of every open set in ℂ is open in D (as a subspace).

This theorem (Open Set Characterization) connects analysis (ε-δ) to topology (open sets), generalizing continuity to spaces beyond metric spaces.

For any set U ⊆ ℂ: $f^{-1}(U) = \{z \in D : f(z) \in U\}$, the theorem states that f continuous ⟺ ∀U open in ℂ, f⁻¹(U) is open in D.

Proof.

(⟹ Direction)

1. Assume that f is continuous on D. Let U ⊆ ℂ be open. We aim to show that f⁻¹(U) is open in D.
2. Case 1: If f⁻¹(U) = ∅, it’s trivially open. ✓
3. Case 2: Let f⁻¹(U) ≠ ∅. Pick any z₀ ∈ f⁻¹(U).
4. Then, by definition w₀ = f(z₀) ∈ U. Since U is open, there exists r > 0 such that B(w₀; r) ⊆ U.
5. Since f is continuous at z₀, there exists δ > 0 such that: $z \in D, \; |z - z_0| < \delta \implies |f(z) - f(z_0)| < r$
6. That is: $f(D \cap B(z_0; \delta)) \subseteq B(w_0; r) \subseteq U$
7. This means: $D \cap B(z_0; \delta) \subseteq f^{-1}(U)$. So z₀ has a neighborhood (in D) contained entirely in f⁻¹(U).
8. Since z₀ was taken arbitrarily, f⁻¹(U) is indeed open in D. $\blacksquare$

(⟸ Continuity ⇒ open–preimage)

1. Assume that the preimage of every open set is open in D. We aim to show that f is continuous at each z₀ ∈ D.
2. Let z₀ ∈ D and ε > 0. The open ball B(f(z₀); ε) is obviously open in ℂ.
3. By hypothesis, f⁻¹(B(f(z₀); ε)) is open in D.
4. Since z₀ ∈ f⁻¹(B(f(z₀); ε)) (because f(z₀) ∈ B(f(z₀); ε)), an open set, there exists δ > 0 such that: $D \cap B(z_0; \delta) \subseteq f^{-1}(B(f(z_0); \varepsilon))$
5. This means that for z ∈ D with |z - z₀| < δ, we have $z \in f^{-1}(B(f(z₀); ε)) \iff f(z) \in B(f(z₀); ε) \iff |f(z) - f(z₀)| \lt ε$. And this is just the $\varepsilon-\delta$ definition of continuity, i.e., f is continuous at z₀. ∎

Corollary. f: $D \to \mathbb{C}$ is continuous iff the preimage of every closed set is closed (in D).

Proof.

1. A set C is closed iff its complement ℂ \ C is open.
2. Relate preimages via complements. For any $C \subseteq \mathbb{C}$, $f^{-1}(C)=D\setminus f^{-1}(\mathbb{C}\setminus C).$
3. Translate “closed” into “open” in D. A subset $F\subseteq D$ is closed in D iff $D\setminus F$ is open in D. So f⁻¹(C) is closed in D ⟺ f⁻¹(ℂ \ C) is open in D.
4. Apply the open preimage theorem to ℂ \ C. f is continuous, ℂ \ C is open, so f⁻¹(ℂ \ C) is open in D.
5. Conversely, if the preimage of every closed set is closed, then in particular the preimage of every open set’s complement is closed, so its complement in D is open.

Important Results

Theorem (Image of Compact Sets). Let $f: D \to \mathbb{C}$ be continuous. If K ⊆ D is compact, then f(K) is compact.

Complex Extreme Value Theorem. Let $f: K \to \mathbb{C}$ be continuous where K is compact. If f is real-valued ($f: K \to \mathbb{R}$), then f attains its maximum and minimum on K. More generally: If $f: K \to \mathbb{C}$ is continuous and K is compact, then |f| attains its maximum on K.

Definition. A function f: D → ℂ is uniformly continuous on D if $\forall \varepsilon > 0, \; \exists \delta > 0 : \forall z, w \in D, \; |z - w| < \delta \implies |f(z) - f(w)| < \varepsilon$. Idea: The same δ works for ALL points simultaneously.

Theorem. If f: K → ℂ is continuous and K is compact, then f is uniformly continuous on K.

Proof.

1. Suppose for the sake of contradiction that f is not uniformly continuous on K. Then, $\exists \varepsilon_0 > 0 : \forall n \in \mathbb{N}, \; \exists z_n, w_n \in K : |z_n - w_n| < \frac{1}{n} \text{ but } |f(z_n) - f(w_n)| \geq \varepsilon_0$
2. Since K is compact, ${z_n}$ has a convergent subsequence $z_{n_k} \to z^* \in K$.
3. Since $|z_{n_k}-w_{n_k}| < 1/n_k \to 0$, we have $w_{n_k} \to z^*$ also.
4. By continuity: $z_{n_k} \to z^* \in K, w_{n_k} \to z^*$, then $f(z_{n_k}) \to f(z^*)$ and $f(w_{n_k}) \to f(z^*)$.
5. But $|f(z_{n_k}) - f(w_{n_k})| \geq \varepsilon_0$ for all k.
6. Taking the limit: $|f(z^*) - f(z^*)| \geq \varepsilon_0$, i.e., $0 \geq \varepsilon_0$. Contradiction! ∎

Component-wise Continuity. Let f(z) = u(x, y) + iv(x, y) where z = x + iy. Then, $f \text{ is continuous at } z_0 = x_0 + iy_0 \iff \begin{cases} u \text{ is continuous at } (x_0, y_0) \\ v \text{ is continuous at } (x_0, y_0) \end{cases}$

Theorem (Continuity vs. Differentiability). If f is differentiable at z₀, then f is continuous at z₀.

Proof.

1. Suppose f’(z₀) exists. Then, $f(z) - f(z_0) = \frac{f(z) - f(z_0)}{z - z_0} \cdot (z - z_0), \quad z \neq z_0.$
2. Now take the limit as $z \to z_0$: $\frac{f(z) - f(z_0)}{z - z_0} \to f'(z_0)$ (by definition of derivative), $(z - z_0) \to 0$
3. Using the product rule for limits (valid because both limits exist and are finite), we get $f(z) - f(z_0) \to f'(z_0) \cdot 0 = 0$
4. Therefore $f(z) \to f(z_0)$, i.e., f is continuous at z₀. ∎

The Converse is False! Continuity does not imply differentiability in the complex sense. Differentiability in $\mathbb{C}$ is a much stronger condition (it requires the limit of the difference quotient to exist and be independent of the direction of approach). Here are three classic counterexamples:

1. $f(z) = |z|^2$
   Continuity: Everywhere continuous, as it is a polynomial in x and y. Let z = x + iy, then we have $|z|^2 = x^2 + y^2$.
   Differentiability: Using the definition, one finds $\frac{f(z)-f(0)}{z-0} = \frac{|z|^2}{z} = \frac{z\overline{z}}{z} = \overline{z} \to 0 \quad \text{as } z\to 0,$
   so f’(0) = 0. But for $z_0 \neq 0$, the limit of the difference quotient depends on the direction of approach (e.g., along horizontal vs. vertical lines), so f is not differentiable at any other point.
2. $f(z) = \overline{z}$ (complex conjugation).
   Continuity: $\overline{z} = x - iy$ is continuous everywhere.
   Differentiability: The Cauchy–Riemann equations are nowhere satisfied: $u_x = 1,v_y = -1$, so $u_x \neq v_y$; and the limit $\lim_{h\to 0} \frac{\overline{z+h} - \overline{z}}{h} = \lim_{h\to 0} \frac{\overline{h}}{h}$ does not exist (it gives 1 for real , and -1 for purely imaginary h). Hence f is nowhere differentiable.
3. $f(z) = \operatorname{Re}(z) = x$.
   Continuity: Clearly continuous (it is the projection onto the real axis).
   Differentiability: Again the Cauchy–Riemann equations fail $u_x = 1, v_y = 0$, and the difference quotient $\frac{\operatorname{Re}(z+h) - \operatorname{Re}(z)}{h} = \frac{\operatorname{Re}(h)}{h}$ has no limit as $ h\to 0 $ (e.g., real h gives 1, imaginary h gives 0). So f is nowhere differentiable.