Limits in Complex Analysis

Do, or do not. There is no try, Yoda.

Limits in Complex Analysis

In complex analysis, we encounter two related but distinct concepts:

1. Limit of a sequence. A sequence of complex numbers $\{z_n\}$ has a limit $L$ ($\lim_{n \to \infty} z_n = L$) if the terms eventually get “stuck” arbitrarily close to $L$ (terms approach L as index grows).
2. Limit of a function For a function $f: U \to \mathbb{C}$ where $U \subseteq \mathbb{C}$, we want to describe the behavior of f(z) as z gets close to a point a. $\lim_{z \to a} f(z) = L$ means that the function’s output approaches L as its input approaches a.

Notice that the variable z can approach a from any direction in the complex plane —not just along the real line! For the limit to exist, f(z) must approach the same value L regardless of the approach path.

Limits of Sequences

Definition. Let $\{z_n\}$ be a sequence of complex numbers and $z \in \mathbb{C}$. We say that $\{z_n\}$ converges to $z$, written as $\lim_{n \to \infty} z_n = z$ or $z_n \to z$, if: $\forall \varepsilon > 0, \; \exists N \in \mathbb{N} : n \geq N \Rightarrow |z_n - L| < \varepsilon$

In simple terms, the limit of a sequence {zₙ} is the value “L” that the terms of the sequence get arbitrarily close to as n (the index of the terms) becomes larger and larger (approaches infinity).

Geometric Interpretation:

For any open disk $\mathbb{B}(L; \varepsilon)$ centered at L with radius $\varepsilon$, there is a cutoff index $N$ such that every term $z_N, z_{N+1}, z_{N+2}, \dots$ lies inside $\mathbb{B}(L; \varepsilon)$. The sequence eventually enters and never leaves any neighborhood of L.

Imagine drawing a target around L. If you make the bullseye (radius $\varepsilon$) extremely tiny, eventually all the shots (terms $z_n$) must land inside it and stay there (you can always find a point in the sequence N such that all terms after that point lie inside this circle or are within a distance of ε from the limit L).

Example. Consider the sequence $z_n = \frac{1+i}{n}$, $\lim_{n \to \infty} \frac{1+i}{n} = 0$

Proof: Let $\varepsilon > 0$ be given. We want to find $N$ such that for all $n \ge N$, $|z_n - 0| < \varepsilon$.

$$|z_n - 0| = \left| \frac{1+i}{n} \right| = \frac{|1+i|}{n} = \frac{\sqrt{1^2+1^2}}{n} = \frac{\sqrt{2}}{n}$$

We need $\frac{\sqrt{2}}{n} < \varepsilon$, which is equivalent to $n > \frac{\sqrt{2}}{\varepsilon}$. Choose $N = \lceil \frac{\sqrt{2}}{\varepsilon} \rceil + 1$.

Then, for any $n \ge N$: $|z_n - 0| = \frac{\sqrt{2}}{n} \le \frac{\sqrt{2}}{N} < \varepsilon$. Thus, the limit is 0. ∎

Component-wise Convergence Theorem. Convergence of Complex Sequences and Real/Imaginary Parts A sequence of complex numbers {zₙ} converges to a complex number z if and only if the sequences of their real parts {xₙ} and imaginary parts {yₙ} converge to the real and imaginary parts of z, respectively. $zₙ \to z$ where zₙ = xₙ + iyₙ ∀n, z = x + iy, if and only if $x_n \to x$ and $y_n \to y$

Limits of Functions

For a function $f: U \to \mathbb{C}$ where $U \subseteq \mathbb{C}$, we want to describe the behavior of f(z) as z gets close to a point a. Notice that the variable z can approach a from any direction in the complex plane —not just along the real line!

For the limit to exist, f(z) must approach the same value L regardless of the approach path.

Definition. Let $f: U \to \mathbb{C}$ be a function defined on a set $U \subseteq \mathbb{C}$, and let a be a limit point of U (meaning every neighborhood of a contains points of U other than a itself). We say: $\lim_{z \to a} f(z) = L$ if for every ε > 0, there exists δ > 0 such that: $0 < |z - a| < \delta \text{ and } z \in U \Rightarrow |f(z) - L| < \varepsilon$ where:

• ε > 0 indicates how close we want f(z) to be to L (tolerance in output).
• δ > 0 tells us how close z must be to a to achieve that tolerance (input restriction).
• 0 < |z - a| (z is close to a but not equal to a) and |z - a| < δ, in words z is within a “punctured” δ-disk around a. We exclude z = a because (i) f might not be defined at a (e.g., f(z) = 1/z at z = 0); (ii) the limit describes behavior near a, not at a itself; (iii) f(a) might exist but differ from the limit (removable discontinuity).
• Then, |f(z) - L| < ε, that is, f(z) is within an ε-disk around L.

All z in the punctured δ-disk around a land in the corresponding ε-disk around L.

Theorem (Sequential Characterization). $\lim_{z \to a} f(z) = L \iff \text{for every sequence } \{z_n\} \text{ with } z_n \to a \text{ and } z_n \neq a, \text{ we have } f(z_n) \to L$

Utility. 🛠️ To prove a limit does NOT exist, find two sequences approaching a that give different limits under f.

Example: Show $\lim_{z \to 0} \frac{\bar{z}}{z}$ does not exist.

Approach along real axis: Let $z_n = 1/n$ (real). Then $\frac{\bar{z_n}}{z_n} = \frac{1/n}{1/n} = 1 \to 1$.

Approach along imaginary axis: Let $z_n = i/n$. Then $\frac{\bar{z_n}}{z_n} = \frac{-i/n}{i/n} = -1 \to -1$.

Different limits, so $\lim_{z \to 0} \frac{\bar{z}}{z}$ does not exist. ∎

Properties of Limits

Let $\lim_{z \to a} f(z) = L$ and $\lim_{z \to a} g(z) = M$. Let $c \in \mathbb{C}$.

1. Uniqueness: If a limit exists, it is unique. ($L_1 = L_2$).
2. Basic Limits: Identity Function. $\lim_{z \to a} z = a$. Let ε > 0. Choose δ = ε. Then, $0 < |z - a| < \delta = \varepsilon \Rightarrow |z - a| < \varepsilon$
   Constant Function. For any constant c ∈ ℂ: $\lim_{z \to a} c = c$. Let ε > 0. Choose δ = 1 (or any positive number). Then, $0 < |z - a| < \delta \Rightarrow |c - c| = 0 < \varepsilon$
3. Algebraic Rules:
   Sum/Difference: $\lim_{z \to a} [f(z) \pm g(z)] = L \pm M$
   Scalar Multiplication: $\lim_{z \to a} [c \cdot f(z)] = c L$ for any $c \in \mathbb{C}$
   Product: $\lim_{z \to a} [f(z) g(z)] = L \cdot M$
   Quotient: $\lim_{z \to a} \frac{f(z)}{g(z)} = \frac{L}{M}$ (provided $M \ne 0$)
   Conjugate: $\lim_{z \to a} \overline{f(z)} = \overline{L}$
   Modulus: $\lim_{z \to a} |f(z)| = |L|$
4. Polynomials: For any polynomial $P(z)$, $\lim_{z \to a} P(z) = P(a)$.
5. Rational Functions: For rational $R(z) = P(z)/Q(z)$, $\lim_{z \to a} R(z) = R(a)$ if $Q(a) \ne 0$.

Uniqueness of Limits Theorem. If $\lim_{z \to a} f(z)$ exists, it is unique.

Proof. Suppose $\lim_{z \to a} f(z) = L_1$ and $\lim_{z \to a} f(z) = L_2$.

Let ε > 0. There exist δ₁, δ₂ such that: (i) 0 < |z - a| < δ₁ ⟹ |f(z) - L₁| < ε/2 and (ii) 0 < |z - a| < δ₂ ⟹ |f(z) - L₂| < ε/2

Choose δ = min(δ₁, δ₂). For any z with 0 < |z - a| < δ: $|L_1 - L_2| = |L_1 - f(z) + f(z) - L_2| \leq |L_1 - f(z)| + |f(z) - L_2| < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon$

Since ε was arbitrary, |L₁ - L₂| = 0, so L₁ = L₂. ∎

Sum Rule. $\lim_{z \to a} [f(z) + g(z)] = L_f + L_g$

Proof.

1. Let ε > 0.
2. Since $f(z) \to L_f$, there exists δ₁ > 0 such that: $0 < |z - a| < \delta_1 \Rightarrow |f(z) - L_f| < \frac{\varepsilon}{2}$
3. Since $g(z) \to L_g$, there exists δ₂ > 0 such that: $0 < |z - a| < \delta_2 \Rightarrow |g(z) - L_g| < \frac{\varepsilon}{2}$
4. Let δ = min(δ₁, δ₂). For 0 < |z - a| < δ: $|[f(z) + g(z)] - [L_f + L_g]| = |[f(z) - L_f] + [g(z) - L_g]|\leq |f(z) - L_f| + |g(z) - L_g| < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon \blacksquare$

Scalar Multiple Rule. $\lim_{z \to a} [c \cdot f(z)] = c \cdot L_f$

Proof.

1. c = 0. Then, c·f(z) = 0 for all z, and c·L_f = 0. By the constant function limit, $\lim_{z \to a} 0 = 0$. ✓
2. c ≠ 0. Let ε > 0. Since $f(z) \to L_f$, there exists δ > 0 such that: $0 < |z - a| < \delta \Rightarrow |f(z) - L_f| < \frac{\varepsilon}{|c|}$

For 0 < |z - a| < δ: $|c \cdot f(z) - c \cdot L_f| = |c| \cdot |f(z) - L_f| < |c| \cdot \frac{\varepsilon}{|c|} = \varepsilon \blacksquare$

Limit of products is product of limits. $\lim_{z \to a} [f(z) \cdot g(z)] = L_f \cdot L_g$

Proof.

1. We use the identity: $f(z)g(z) - L_f L_g = f(z)g(z) - L_f g(z) + L_f g(z) - L_f L_g = [f(z) - L_f]g(z) + L_f[g(z) - L_g]$
   Therefore, $|f(z)g(z) - L_f L_g| \leq |f(z) - L_f| \cdot |g(z)| + |L_f| \cdot |g(z) - L_g|$
2. Bound |g(z)|. Since $g(z) \to L_g$, there exists δ₃ > 0 such that: $0 < |z - a| < \delta_3 \Rightarrow |g(z) - L_g| < 1$
   Then, by triangle inequality: $|g(z)| = |g(z) - L_g + L_g| \leq |g(z) - L_g| + |L_g| < 1 + |L_g|$
3. Let ε > 0. Choose: (i) δ₁ such that 0 < |z - a| < δ₁ ⟹ $|f(z) - L_f| < \frac{\varepsilon}{2(1 + |L_g|)}$; (ii) δ₂ such that 0 < |z - a| < δ₂ ⟹ $|g(z) - L_g| < \frac{\varepsilon}{2(1 + |L_f|)}$
4. Let δ = min(δ₁, δ₂, δ₃). For 0 < |z - a| < δ: $|f(z)g(z) - L_f L_g| < \frac{\varepsilon}{2(1 + |L_g|)} \cdot (1 + |L_g|) + |L_f| \cdot \frac{\varepsilon}{2(1 + |L_f|)} < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon$

Lemma (Reciprocal). If $\lim_{z \to a} g(z) = L_g \neq 0$, then $\lim_{z \to a} \frac{1}{g(z)} = \frac{1}{L_g}$

Proof.

1. Keep g(z) away from zero. Since $|L_g| > 0$ and $g(z) \to L_g$, there exists δ₁ > 0 such that: $0 < |z - a| < \delta_1 \Rightarrow |g(z) - L_g| < \frac{|L_g|}{2}$. By the reverse triangle inequality: $|g(z)| \geq |L_g| - |g(z) - L_g| > |L_g| - \frac{|L_g|}{2} = \frac{|L_g|}{2}$. So $\frac{1}{|g(z)|} < \frac{2}{|L_g|} (\star)$.
2. $\left|\frac{1}{g(z)} - \frac{1}{L_g}\right| = \left|\frac{L_g - g(z)}{g(z) \cdot L_g}\right| = \frac{|g(z) - L_g|}{|g(z)| \cdot |L_g|} (\dagger)$.
3. For 0 < |z - a| < δ₁: $\left|\frac{1}{g(z)} - \frac{1}{L_g}\right| =[(\dagger)] \frac{|g(z) - L_g|}{|g(z)| \cdot |L_g|} <[(\star)] \frac{2|g(z) - L_g|}{|L_g| \cdot |L_g|} = \frac{2|g(z) - L_g|}{|L_g|^2}$
4. Choose δ. Let ε > 0. Choose δ₂ such that 0 < |z - a| < δ₂ ⟹ $|g(z) - L_g| < \frac{\varepsilon |L_g|^2}{2}$.
5. Let δ = min(δ₁, δ₂). For 0 < |z - a| < δ: $\left|\frac{1}{g(z)} - \frac{1}{L_g}\right| < \frac{2}{|L_g|^2} \cdot \frac{\varepsilon |L_g|^2}{2} = \varepsilon \blacksquare$

Theorem (Quotient Rule). $\lim_{z \to a} \frac{f(z)}{g(z)} = \frac{L_f}{L_g}$ (when $L_g \neq 0$)

Proof. Combining the product rule and reciprocal lemma we get: $\frac{f(z)}{g(z)} = f(z) \cdot \frac{1}{g(z)} \to L_f \cdot \frac{1}{L_g} = \frac{L_f}{L_g} \blacksquare$

Theorem (Power Rule). For any $n \in \mathbb{N}$: $\lim_{z \to a} z^n = a^n$

Proof by induction.

1. Base case (n = 0): $\lim_{z \to a} 1 = 1 = a^0$. ✓
2. Inductive step: Assume $\lim_{z \to a} z^n = a^n$. Then, $\lim_{z \to a} z^{n+1} = \lim_{z \to a} (z^n \cdot z) =[\text{Product Limit Rule}] \left(\lim_{z \to a} z^n\right) \cdot \left(\lim_{z \to a} z\right) =[\text{Inductive hypothesis}] a^n \cdot a = a^{n+1}$ ∎

Corollary (Polynomial Limit). If $p(z) = c_0 + c_1 z + c_2 z^2 + \cdots + c_n z^n$, then, $\lim_{z \to a} p(z) = p(a)$

Proof. Apply sum and scalar multiple rules: $\lim_{z \to a} p(z) = \lim_{z \to a} \sum_{k=0}^{n}c_k z^k = \sum_{k=0}^{n} c_k \lim_{z \to a} z^k = \sum_{k=0}^{n} c_k a^k = p(a)$

Linear Combination of Limits Theorem. Suppose $\lim_{z \to a} f_j(z) = L_j$ for j = 1, 2, …, n. Then, for any constants $c_1, ..., c_n \in \mathbb{C}$: $\boxed{\lim_{z \to a} \sum_{j=1}^{n} c_j f_j(z) = \sum_{j=1}^{n} c_j L_j}$

Proof by induction.

1. Base case (n = 1): This is the scalar multiple rule. ✓
2. Inductive step: Assume true for n. Then, $\lim_{z \to a} \sum_{j=1}^{n+1} c_j f_j(z) = \lim_{z \to a} \left[\sum_{j=1}^{n} c_j f_j(z) + c_{n+1} f_{n+1}(z)\right] = \lim_{z \to a} \sum_{j=1}^{n} c_j f_j(z) + \lim_{z \to a} c_{n+1} f_{n+1}(z) = \sum_{j=1}^{n} c_j L_j + c_{n+1} L_{n+1} = \sum_{j=1}^{n+1} c_j L_j $∎

Examples

• Basic Polynomial $\lim_{z \to 2+i} (z^2 + 3z - 1) =[\text{Polynomial limit theorem}] (2+i)^2 + 3(2+i) - 1 = (4 + 4i - 1) + (6 + 3i) - 1 = 8 + 7i$
• Rational Function. $\lim_{z \to i} \frac{z^2 + 1}{z - i} = \lim_{z \to i} \frac{(z+i)(z-i)}{z-i} = \lim_{z \to i}(z + i) = i + i = 2$
• Limit Does Not Exist. $\lim_{z \to 0} \frac{z}{\bar{z}} =[\text{Write } z = re^{i\theta}] \lim_{z \to 0} \frac{re^{i\theta}}{re^{-i\theta}} = \lim_{z \to 0} e^{2i\theta}$.
  As $z \to 0, r \to 0$ but θ can be any value depending on the approach direction.
  Along positive real axis (θ = 0): $e^{0} = 1$, along positive imaginary axis (θ = π/2): $e^{i\pi} = -1$, and along ray at angle θ: $e^{2i\theta}$.
  Different directions give different limits, so the limit does not exist.
• $\lim_{z \to 0} \frac{\sin z}{z}$. ‘From the power series, we get $\sin z = z - \frac{z^3}{3!} + \frac{z^5}{5!} - \cdots$.
  For $z \neq 0$: $\frac{\sin z}{z} = 1 - \frac{z^2}{3!} + \frac{z^4}{5!} - \cdots$
  As $z \to 0$, each term with positive power of z vanishes: $\lim_{z \to 0} \frac{\sin z}{z} = 1$

Cauchy Sequences and Completeness

Definition. A sequence $\{z_n\}$ is a Cauchy sequence if the terms become arbitrarily close to each other as the sequence progresses. Formally, $\forall \varepsilon > 0, \exists N \in \mathbb{N} \text{ such that } \forall n, m \ge N, |z_n - z_m| < \varepsilon$ where:

• ε indicates how close the terms should be.
• N tells us the point in the sequence after which the terms are guaranteed to be as close as required.
• Then, for all natural numbers n and m greater than or equal to N, the distance between zₙ and zₘ is less than ε.