Limits of Complex Functions. Definition, Examples, and Rules

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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.

Calc I/II (Real ε–δ limits): $\lim_{z \to a} g(z)=L$ means |g(x) - L| < ε whenever 0 < ∣x − a∣ < δ.
Calc III (Multivariable): Limits in ℝ² use $0 < \sqrt{(x-a)^2+(y-b)^2} < δ$, and the same ideas carry over to sequences of functions, continuity definitions, and on to power‐series convergence.

Examples

• A piecewise polynomial

Let $f: \overline{B(0; 1)} \rarr \Complex$

f(z) = $\begin{cases} 3z², |z| < 1 \\\\ 3, |z| = 1 \end{cases}$

At z₀ = 1. If z → 1 from inside the unit circle ∣z∣ < 1, then f(z) = 3z² → 3·1² = 3. If z actually equals 1, f(1)=3. Hence $\lim_{z \to 1} f(z)=3$.

Furthermore, at other boundary points |z₀| = 1, z₀ ≠ ±1, z₀ = e^iθ, θ ≠ 0, π, as z → e^iθ along points |z| < 1, f(z) = 3z² ➞3e^2iθ ≠ 3. However, at the boundary itself f(e^iθ) = 3. Thus, $\lim_{z \to e^{i\theta}} f(z)$ does not exist for θ ≠ 0, π.

• A rational‐type singularity

Let g(z) = $\frac{1}{z-2}, D = \Complex -${2}.

As z → 2, |z - 2| → 0, hence g(z) = $\frac{1}{z-2} \leadsto \infin$ or simply “no finite limit.”

As z → ∞, g(z)→0.

• A continuous function

If g(z) = e^z ∀z ∈ ℂ, then for every z₀ ∈ ℂ, $\lim_{z \to z_0} e^z=e^{z_0}$

Proof:

0 < |z - z₀| < δ, claim: $|e^z-e^{z_0}|=|e^{z_0}||e^{z-z_0}-1|< \varepsilon$

Set: M = $|e^{z_0}|$ (just a constant depending on z₀), we want $M |e^{z - z_0} - 1| < \varepsilon ↭|e^{z - z_0} - 1| < \frac{\varepsilon}{M}$

Let h = z - z₀, so h → 0 as z → z₀. Since the function e^h is continuous at 0 -Calc I/II (Real ε–δ limits)-, we can choose δ small enough such that: $|h| = |z - z_0| < \delta \Rightarrow |e^h - 1| < \frac{\varepsilon}{M}$

Hence, $|e^z - e^{z_0}| = |e^{z_0}||e^{z-z_0}-1| = M|e^h - 1| < \varepsilon$. Therefore: $\lim_{z \to z_0} e^z = e^{z_0}$

• No limit at the branch cut (-∞, 0]

Let g(z) = $\sqrt{z}$ defined on the punctured complex plane D = ℂ - {0}. That is, we’re excluding the origin but not yet specifying a branch cut.

However, to make $\sqrt{z}$ single-valued, we must choose a branch — and a conventional one is to remove the negative real axis (i.e. use the branch cut along (-∞, 0]). The domain of definition for this branch is ℂ ∖ (−∞,0], meaning the negative real axis is excluded. define: $\sqrt{z} = \exp\left( \frac{1}{2} \log z \right)$ where $\log z = \ln |z| + i \arg(z)$, and $\arg(z) \in (-\pi, \pi)$. This forces the square root to be single-valued on the domain minus the branch cut.

Why the Limit Doesn’t Exist on (-∞, 0):

However, we can still consider the behavior as points on this axis (x < 0) are approached from within the domain (i.e., from the upper or lower half-plane).

• From above (upper half-plane): $z = x + i\varepsilon$
• From below (lower half-plane): z = $x - i\varepsilon$

Then, the argument of z changes drastically: $\arg(z) \to \pi \text{ as } \varepsilon \to 0^+$ because z is in the second quadrant

$\arg(z) \to -\pi \text{ as } \varepsilon \to 0^-$ because z is in the third quadrant

So we get two different square roots: $\lim_{\varepsilon \to 0^+} \sqrt{x \pm i\varepsilon} = \sqrt{|x|} e^{i\pi/2} = i \sqrt{|x|} = i\sqrt{-x}, \lim_{\varepsilon \to 0^-} \sqrt{x \pm i\varepsilon} = \sqrt{|x|} e^{-i\pi/2} = -i \sqrt{|x|} = -i\sqrt{-x}$.

Since $i\sqrt{-x} \ne -i\sqrt{-x}$ for any x < 0 (as they are complex conjugates and not equal), the two-sided limit does not exist at any point x on (-∞, 0). This discontinuity is often described as a “jump” across the branch cut. At z = 0, the function is undefined, so the limit does not exist trivially.

Limit rules

Suppose D ⊆ ℂ, f, g: D → ℂ, and z₀ is a limit point of D. If $\lim_{z \to z_0} f(x) = L_1$ and $\lim_{z \to z_0} g(x) = L_2$, then the following limit laws hold:

1. Uniqueness of the Limit. If $\lim_{z \to z_0} f(x)$ exists, then it is unique.

   If $\lim_{z \to z_0} f(x) = L$ and also $\lim_{z \to z_0} f(x) = M$, then for any ε>0 eventually ∣f(z) − L∣ < ε and ∣f(z) − M∣<ε. By the triangle inequality, ∣L − M∣ < 2ε. Letting ε→0 forces L = M.

2. Sum and Difference. $\lim_{z \to z_0} (f(x) \plusmn g(x)) = L_1 \plusmn L_2$

3. Product. $\lim_{z \to z_0} (f(x) · g(x)) = L_1 · L_2$

4. Quotient. $\lim_{z \to z_0} (\frac{f(x)}{g(x)}) = \frac{L_1}{L_2}$ provided L₂ ≠ 0.

5. Since $\lim_{z \to z_0} z = z_0, ∀z_0\in\Complex$, then $\lim_{z \to z_0} f(z) = f(z_0), ∀z_0\in dom(f)$, any polynomial or more generally any rational function f(z) is continuous on its domain, e.g., $f(z)=\frac{(3+i)z²+2z-i}{(1+i)z⁴-z³+1}, \lim_{z \to z_0}f(z) = \lim_{z \to z_0}\frac{(3+i)z²+2z-i}{(1+i)z⁴-z³+1} = \frac{(3+i)z_0²+2z_0-i}{(1+i)z_0⁴-z_0³+1} =f(z_0)$

6. You can combine limits with constant factors, e.g. $\lim_{z \to z_0} c·f(z) = c·\lim_{z \to z_0} f(z)$.

7. The sandwich (squeeze) theorem also holds if h(z) ≤ f(z) ≤ k(z) and $\lim_{z \to z_0} h(z) = \lim_{z \to z_0} k(z) = L \leadsto \lim_{z \to z_0} f(z) = L$.

Limits That “Go to Infinity”

A function blows up if its values grow without bound as z → z₀. Definition. Let D ⊆ ℂ, let f: D ⟶ ℂ, and z₀ be a limit point of D. We say that $\lim_{z \to z_0} f(z) = \infty$ if for every M > 0, there exists δ > 0 such that ∀z ∈ D and 0 < |z -z₀| < δ ⇒ |f(z)| > M.

Equivalently, $\lim_{z \to z_0} f(z) = \infty$ if for every M > 0, there exists δ > 0 such that ∀z ∈ D ∩ B’(z₀; δ), then f(z) ∈ {w: |w| > M} -neighborhood of infinity-.

Example. $\lim_{z \to 0} \frac{1}{z} = \infty$

Given M > 0, there is a δ = ¹⁄_M > 0 such that ∀z ∈ D and 0 < |z| < δ = ¹⁄_M, |¹⁄_z| > ¹⁄_δ = M↭ z∈D ∩ B’(0; ¹⁄_M), then ¹⁄_z ∈ {w: |w| > M} ∎

Limits as z → ∞

Definition. Let D ⊆ ℂ, let f: D ⟶ ℂ. We say that $\lim_{z \to \infty} f(z) = L$ if for every ε > 0, there exists M > 0 such that ∀z ∈ D, |f(z)-L| < ε wherever z ∈ {w: |w| > M} ↭ |f(z)-L| < ε, ∀z ∈ D, ∣z∣ > M.

Examples

• $\lim_{z \to \infty} \frac{1}{z} = 0$

Given ε > 0, take M = 1/ε. Then, if ∣z∣ > M, $|\frac{1}{z}-0| = \frac{1}{|z|} \le \frac{1}{M} = \epsilon$

• $\lim_{z \to \infty} \frac{3z²}{(1+i)z²-z+2} = \frac{3}{1+i}$

$|\frac{3z²}{(1+i)z²-z+2} - \frac{3}{1+i}| = |\frac{3z²(1+i)-3z²(1+i)+3z-6}{(1+i)((1+i)z²-z+2)}| = |\frac{3z-6}{(1+i)((1+i)z²-z+2)}| =[\text{Then, dividing both numerator and denominator by z² within the absolute value}]$

$|\frac{\frac{3}{z}-\frac{6}{z²}}{(1+i)((1+i)-\frac{1}{z}+\frac{2}{z²})}|$ [*]

$|(1+i)-\frac{1}{z}+\frac{2}{z²}| \ge[\text{Recall the reverse triangle inequality}|a-b| \ge |a| - |b|]\sqrt{2}-|\frac{1}{z}-\frac{2}{z²}|$

$|z| \ge M \leadsto \frac{1}{|z|} \le \frac{1}{M} \leadsto [\text{Triangular inequality}] |\frac{1}{z}-\frac{2}{z²}| \le |\frac{1}{z}| + |\frac{-2}{z^2}| = |\frac{1}{z}| + |\frac{2}{z^2}| \le \frac{1}{M}+\frac{2}{M²}$

$|(1+i)-\frac{1}{z}+\frac{2}{z²}| \ge \sqrt{2}-|\frac{1}{z}-\frac{2}{z²}| \ge \sqrt{2}+\frac{1}{M}+\frac{2}{M²}$

[*] $\le \frac{\frac{3}{M}+\frac{6}{M²}}{\sqrt{2}(\sqrt{2}+\frac{1}{M}+\frac{2}{M²})} = \frac{3}{\sqrt{2}}\frac{\frac{1}{M}+\frac{2}{M²}}{\sqrt{2}+\frac{1}{M}+\frac{2}{M²}} = \frac{3}{\sqrt{2}}\frac{M+2}{\sqrt{2}M²+M+2M} \le[\text{Given M large enough,} M+2 \le 2M, \sqrt{2}M²+M+2M \le 2\sqrt{2}M²] \frac{3}{\sqrt{2}}\frac{2M}{2\sqrt{2}M²} = \frac{3}{2M}$

In other words, for large M, [*] is arbitrary small, hence $\lim_{z \to \infty} \frac{3z²}{(1+i)z²-z+2} = \frac{3}{1+i}$.