Differentiability in the Complex Plane

Simplicity is the ultimate sophistication, Anonymous

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.

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.

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

Differentiability in the Complex Plane

Definition. Let D ⊆ ℂ, a ∈ D, and f: D → ℂ be a function. The derivative of f at the point a is defined to be $\lim_{h \to 0} \frac{f(a+h)-f(a)}{h}$ if this limits exists. If this limit exists and we express it as f'(a) or $\frac{df}{dz}|_{z=a}$, we say that f is differentiable at the point a. Otherwise, if this limit does not exist, then we say that f is not differentiable at a.

Definition. A function f that is differentiable at every point in its domain is said to be differentiable.

Functions Differentiable Everywhere

• Constant Functions: f(z)=c (where c ∈ ℂ is constant). Derivative: f′(z) = 0 for all z.

$f′(z)= \lim_{h \to 0} \frac{f(z+h)-f(z)}{h} = \lim_{h \to 0} \frac{c-c}{h} = lim_{h \to 0} \frac{0}{h} = lim_{h \to 0} 0 = 0.$ ∎ This shows that the derivative exists and is zero for all z∈ℂ.

• Identity function: f(z) = z is differentiable everywhere with f’(z) = 1

$f′(z)= \lim_{h \to 0} \frac{f(z+h)-f(z)}{h} = \lim_{h \to 0} \frac{z+h-z}{h} = lim_{h \to 0} \frac{h}{h} = lim_{h \to 0} 1 = 1.$ ∎

• Power functions: f(z) = zⁿ are differentiable everywhere, and its derivative is: f’(z) = nzⁿ⁻¹ ∀z∈ℂ

$f′(z)= \lim_{h \to 0} \frac{f(z+h)-f(z)}{h} = \lim_{h \to 0} \frac{{z+h}^n-z^n}{h} =[\text{Using the binomial theorem}] \lim_{h \to 0} \frac{\sum_{k=0}^n {n \choose k}z^{n-k}h^k-z^n}{h} = \lim_{h \to 0} \sum_{k=1}^n {n \choose k}z^{n-k}h^{k-1}$

As h → 0, all terms with h^k-1, k ≥ 2 vanish, leaving $f′(z)= {n \choose 1}z^{n-1} = nz^{n-1}$ ∎

• Polynomials and Rational functions

A polynomial function in the complex variable z is of the form: f(z) = aₙzⁿ + aₙ₋₁zⁿ⁻¹ + … + a₁z + a₀ where aᵢ ∈ ℂ for i = 0, 1, …, n, and aₙ ≠ 0. The differentiability of polynomials follows from several key facts:

1. Constant functions: f(z) = c are differentiable everywhere with f’(z) = 0
2. Identity function: f(z) = z is differentiable everywhere with f’(z) = 1
3. Power functions: f(z) = zⁿ are differentiable everywhere with f’(z) = nzⁿ⁻¹
4. Complex derivatives follow the same algebraic rules as real derivatives: Sum rule: (f + g)’ = f’ + g’; Product rule: (fg)’ = f’g + fg’; and constant multiple rule: (cf)’ = cf'.

Since polynomials are finite sums of products of differentiable functions (constants and powers of z), they are differentiable everywhere.

Examples

1. Linear Polynomial, f(z) = az + b, f’(z) = a. Differentiable at every point in ℂ.
2. Quadratic Polynomial, f(z) = az² + bz + c, P’(z) = 2az + b. Differentiable at every point in ℂ, e.g, f(z) = z² + 3z + 1. Derivative: f′(z) = 2z + 3.
3. General Case. For f(z) = $\sum_{i=0}^n a_kz^k, f'(z) = \sum_{i=1}^n ka_kz^{-1},$ differentiable everywhere in ℂ.
4. Let D = ℂ - {i} and let f(z): D → C be defined by $f(z) = \frac{3z}{z-i}$. Then, f is differentiable at every point $z_0 \in D, lim_{h \to 0} \frac{f(z_0+ h) - f(z_0)}{h} = lim_{h \to 0} \frac{\frac{3(z_0+ h)}{(z_0+ h)-i}-\frac{3z_0}{z_0-i}}{h} = lim_{h \to 0} \frac{1}{h}(\frac{3(z_0+ h)(z_0-i)-3z_0(z_0+ h-i)}{(z_0+ h-i)(z_0-i)}) = lim_{h \to 0} \frac{1}{h}(\frac{3z_0^2+3hz_0-3iz_0-3hi-3z_0^2-3z_0h+3iz_0}{(z_0+ h-i)(z_0-i)}) = lim_{h \to 0} \frac{1}{h}(\frac{-3hi}{(z_0+ h-i)(z_0-i)}) = lim_{h \to 0} \frac{-3i}{(z_0+ h-i)(z_0-i)} = \frac{-3i}{(z_0-i)^2}$

This limit exists and we also know that $f'(z_0) = \frac{-3i}{(z_0-i)^2}$. Since z₀ is an arbitrary point in dom(f), f is differentiable.

• Exponential Function. f(z) = e^z. Derivative: f’(z) = e^z

$f′(z)= \lim_{h \to 0} \frac{f(z+h)-f(z)}{h} = \lim_{h \to 0} \frac{e^{z+h}-e^z}{h} = e^z·\lim_{h \to 0} \frac{e^{h}-1}{h}$

Expand e^h as a Taylor series: $e^h = 1 + h + \frac{h^2}{2!} + \frac{h^3}{3!} + \cdots, \frac{e^h-1}{h} = 1 + \frac{h}{2!} + \frac{h^2}{3!} + \cdots$

As h→0, higher-order terms vanish, leaving: $\lim_{h \to 0}\frac{e^h-1}{h} = 1, f′(z)= e^z·\lim_{h \to 0} \frac{e^{h}-1}{h} = e^z·1 = e^z$ ∎

• Trigonometric Functions: f(z) = sin(z). Derivative f’(z) = cos(z) (using $sin(z) = \frac{e^{iz}-e^{-iz}}{2i}$)

f(z) = sin(z), $f′(z)= \lim_{h \to 0} \frac{f(z+h)-f(z)}{h} = \lim_{h \to 0} \frac{sin(z+h)-sin(z)}{h} =$

Substitute the exponential form of sine, $sin(z) = \frac{e^{iz}-e^{-iz}}{2i}$:

= $\lim_{h \to 0} \frac{1}{h} (\frac{e^{i(z+h)}-e^{-i(z+h)}}{2i}-\frac{e^{iz}-e^{-iz}}{2i}) = \lim_{h \to 0} \frac{e^{i(z+h)}-e^{-i(z+h)}-e^{iz}+e^{-iz}}{2ih} = \lim_{h \to 0} \frac{e^{iz}e^{ih}-e^{-iz}e^{-ih}-e^{iz}+e^{-iz}}{2ih} = \lim_{h \to 0} \frac{e^{iz}(e^{ih}-1)-e^{-iz}(e^{-ih}-1)}{2ih} $

$= \frac{e^{iz}}{2i}\lim_{h \to 0} \frac{(e^{ih}-1)}{h} + \frac{e^{-iz}}{2i}\lim_{h \to 0} \frac{-(e^{-ih}-1)}{h}$

$\lim_{h \to 0} \frac{(e^{ih}-1)}{h} = \lim_{h \to 0}\frac{cos(h)+isin(h)-1}{h} = \lim_{h \to 0} \frac{cos(h)-1}{h} + i \lim_{h \to 0}\frac{sin(h)}{h} = \text{L'Hôpital's rule, limits are of the indeterminante form 0/0} = \lim_{h \to 0} \frac{-sin(h)}{1} + i\lim_{h \to 0}\frac{cos(h)}{1} = -sin(0) + icos(0) = 0 + i = i$. Using a completely similar reasoning, $\lim_{h \to 0} \frac{-(e^{-ih}-1)}{h} = i$

$= \frac{e^{iz}}{2i}\lim_{h \to 0} \frac{(e^{ih}-1)}{h} + \frac{e^{-iz}}{2i}\lim_{h \to 0} \frac{-(e^{-ih}-1)}{h} =[\text{Substitute Limits:}] = \frac{e^{iz}·i + e^{-iz}·i}{2i} = \frac{e^{iz} + e^{-iz}}{2} = cos(z)$ ∎

f(z) = cos(z), $f′(z)= \lim_{h \to 0} \frac{f(z+h)-f(z)}{h} = \lim_{h \to 0} \frac{cos(z+h)-cos(z)}{h} =$

Substitute the exponential form of cosine, $cos(z) = \frac{e^{iz}+e^{-iz}}{2}$

$\lim_{h \to 0} \frac{\frac{e^{i(z+h)}+e^{-i(z+h)}}{2}-\frac{e^{iz}+e^{-iz}}{2}}{h} = \lim_{h \to 0}\frac{e^{i(z+h)}+e^{-i(z+h)}-e^{iz}-e^{-iz}}{2h}$

Expand numerator: $e^{i(z+h)}+e^{-i(z+h)}-e^{iz}-e^{-iz} = e^{iz}e^{ih}+e^{-iz}e^{-ih}-e^{iz}-e^{-iz} = e^{iz}(e^{ih}-1)+e^{-iz}(e^{-ih}-1)$

$\lim_{h \to 0}\frac{e^{i(z+h)}+e^{-i(z+h)}-e^{iz}-e^{-iz}}{2h} = \lim_{h \to 0}\frac{e^{iz}(e^{ih}-1)+e^{-iz}(e^{-ih}-1)}{2h} = \frac{e^{iz}}{2}\lim_{h \to 0}\frac{e^{ih}-1}{h} + \frac{e^{-iz}}{2}\lim_{h \to 0}\frac{e^{-ih}-1}{h}$

Apply limits (Euler’s formula, then L’Hôpital’s rule): $\lim_{h \to 0}\frac{e^{ih}-1}{h} = i, \lim_{h \to 0}\frac{e^{-ih}-1}{h} = -i$

$\frac{e^{iz}}{2}\lim_{h \to 0}\frac{e^{ih}-1}{h} + \frac{e^{-iz}}{2}\lim_{h \to 0}\frac{e^{-ih}-1}{h} = \frac{i}{2}(e^{iz}-e^{-iz}) = \frac{i}{2}(cos(z)+isin(z)-cos(-z)-isin(-z)) =[\text{Use even/odd properties of sine and cosine}] \frac{i}{2} (cos(z)+isin(z)-cos(z)+isin(z)) = \frac{2i^2sin(z)}{2} = -sin(z)$. Therefore, f(z) = cos(z), f’(z) = -sin(z) ∎

f(z) = tan(z), $f′(z)= \lim_{h \to 0} \frac{f(z+h)-f(z)}{h} = \lim_{h \to 0} \frac{tan(z+h)-tan(z)}{h} = lim_{h \to 0} \frac{\frac{sin(z+h)}{cos(z+h)}-\frac{sin(z)}{cos(z)}}{h} = lim_{h \to 0} \frac{sin(z+h)cos(z)-cos(z+h)sin(z)}{hcos(z+h)cos(z)}$

Since sin (a - b) = sin a cos b - cos a sin b,

$lim_{h \to 0} \frac{sin(z+h)cos(z)-cos(z+h)sin(z)}{hcos(z+h)cos(z)} = lim_{h \to 0} \frac{sin(z+h-z)}{hcos(z+h)cos(z)} = lim_{h \to 0} \frac{sin(h)}{hcos(z+h)cos(z)} = lim_{h \to 0} \frac{sin(h)}{h}·lim_{h \to 0} \frac{1}{cos(z+h)cos(z)} = 1·\frac{1}{cos^2(z)} = sec^2(z)$.

1. cos(z) and sin(z) are entire differentiable everywhere in ℂ.
2. tan(z) is differentiable except where cos(z) = 0, i.e., z = π/2 + kπ, k ∈ ℤ.

These simple examples serve as building blocks for more complex analytic functions and illustrate the consistency between real and complex differentiation in these cases.

Functions Differentiable (or almost) Nowhere

• Complex conjugate. f(z) = $\bar{z}$. Fails at all a, $\lim_{h \to 0} \frac{f(a+h)-f(a)}{h} = \lim_{h \to 0} \frac{\overline{a+h}-\overline{a}}{h}=[\text{Conjugation distributes over addition}] \lim_{h \to 0} \frac{\overline{h}}{h}$ does not exist (e.g., h ∈ ℝ gives 1, h imaginary gives −1).

• Real Part Function. f(z) = Re(z). Fails at all a, $\lim_{h \to 0} \frac{f(a+h)-f(a)}{h} = \lim_{h \to 0} \frac{Re(a+h)-Re(a)}{h} = \lim_{h \to 0} \frac{Re(h)}{h}$ does not exist (e.g., h ∈ ℝ gives 1, h pure imaginary gives 0).

• Let f: ℂ → ℂ be a complex function defined by f(z) = z·Re(z). This function is differentiable only at 0.

Let z ∈ ℂ, $\lim_{h \to 0} \frac{f(z+h)-f(z)}{h} = \lim_{h \to 0} \frac{(z+h)\mathbb{Re}(z+h)-z\mathbb{Re}(z)}{h} = \lim_{h \to 0} \frac{z\mathbb{Re}(z)+z\mathbb{Re}(h)+h\mathbb{Re}(z)+h\mathbb{Re}(h)-z\mathbb{Re}(z)}{h} = \lim_{h \to 0} \frac{+z\mathbb{Re}(h)+h\mathbb{Re}(z)+h\mathbb{Re}(h)}{h} = \lim_{h \to 0} \frac{z\mathbb{Re}(h)}{h} + \mathbb{Re}(z) + \mathbb{Re}(h) = \lim_{h \to 0} (\frac{z\mathbb{Re}(h)}{h}) + \mathbb{Re}(z)$

If z ≠ 0, then $\lim_{h \to 0} (\frac{z\mathbb{Re}(h)}{h})$ does not exist.

1. If h → 0 through purely imaginary numbers ⇒ Re(h) = 0, Re(h)/h = 0 for any arbitrary h approaching zero.
2. Whereas if h → 0 through real numbers ⇒ Re(h) = h, Re(h)/h = 1 for any arbitrary h approaching zero. Therefore, $\lim_{h \to 0} (\frac{z\mathbb{Re}(h)}{h})$ does not exist.
3. If z = 0, $\frac{z\mathbb{Re}(h)}{h} = 0$ for any arbitrary h approaching zero, so the limit does exist and is zero.

Conclusion: f(z) = z·Re(z) is differentiable only at z = 0 and f’(0) = 0.

Definition of Complex Differentiability

The derivative of a complex function f(z) at a point z₀ is defined by the limit f’(z₀) = $\lim_{h \to 0}\frac{f(z₀+h)-f(z₀)}{h}$, provided this limit exists and is independent of the path h takes to approach zero (in the two-dimensional complex plane). This path independence is a much stronger condition than for real differentiability.

This definition can also be expressed as follows: Δz = z − z₀, f’(z₀) = $\lim_{Δz \to 0}\frac{f(z₀+Δz)-f(z₀)}{Δz}$. Typically, the subscript in z₀ is dropped for simplicity and the change in the function value w = f(z) corresponding to a change Δz in the input is denoted by Δw = f(z + Δz) − f(z). Hence, the derivative could be written as $\frac{dw}{dz} = \lim_{Δz \to 0}\frac{Δw}{Δz}$. The key aspect to consider is that the limit must exist and be the same regardless of the path h (or Δz) takes as it approaches zero in the complex plane.

If this limit exists, f(z) is said to be complex differentiable at z₀ and the value of the limit is denoted by f’(z₀). If a function is differentiable at every point in an open set U ⊆ ℂ is said to be analytic or holomorphic on U.