Differentiability in the Complex Plane 2

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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.

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 Only at Specific Points

• Pathological Cases. Squared modulus. f(z) = |z|². Differentiable only at z = 0:

1. At a = 0, $\lim_{h \to 0} \frac{f(h)-f(0)}{h} = \lim_{h \to 0} \frac{|h|^2}{h} = \lim_{h \to 0} \frac{h\bar{h}}{h} = \lim_{h \to 0} \bar{h} = 0$.
2. At a ≠ 0, $\lim_{h \to 0} \frac{f(a+h)-f(a)}{h} = \lim_{h \to 0} \frac{(a+h)\overline{(a+h)} − a\bar a}{h} = \lim_{h \to 0} \frac{a\bar{a}+a\bar{h}+h\bar{a}+h\bar{h}− a\bar a}{h} = \lim_{h \to 0} \frac{ a\bar h + \bar a h + |h|²}{h} = \lim_{h \to 0} a\frac{\bar{h}}{h}+\bar{a}+\frac{|h|^2}{h}$

As h→0, the |h|²/h term tends to 0

Two directional limits

1. Along the real axis, h real (h = t ∈ ℝ), $\frac{\bar{h}}{h} = \frac{t}{t}=1, \lim_{h \to 0} \frac{f(a+h)-f(a)}{h} = a·1 + \bar{a}$ = 2ℜ(a).
2. Along the imaginary axis, h = it, t ∈ ℝ, $\frac{\bar{h}}{h} = \frac{-it}{it}=-1, \lim_{h \to 0} \frac{f(a+h)-f(a)}{h} = a·(-1) + \bar{a} = \bar{a} - a = −2iℑ(a).$ Unless both ℜ(a) = 0 and ℑ(a) = 0, these two limits disagree, therefore the full complex‐derivative does not exist at any a ≠ 0.

• Product with Real Part: f(z) = z⋅Re(z). Differentiable only at z = 0:

1. At z = 0, $\lim_{h \to 0} \frac{f(h)-f(0)}{h} = \lim_{h \to 0} \frac{h·Re(h)}{h} = \lim_{h \to 0} Re(h) = 0$. The limit exists, so f(z) is differentiable at z = 0.
2. At z ≠ 0. Let z = x + iy ≠ 0.

f(z) = z⋅Re(z) = (x + iy)·x = x² + ixy.

Compute the limit along two paths:

Real Axis (h = t ∈ ℝ), $\lim_{t \to 0} \frac{f(z+t)-f(z)}{t} = \lim_{t \to 0} \frac{(x+t)^2+i(x+t)y-x^2-ixy}{t} = \lim_{t \to 0} \frac{x^2 +2xt + t^2+ixy +ity -x^2-ixy}{t} \lim_{t \to 0} \frac{+2xt + t^2+ity }{t} = \lim_{t \to 0} 2x + t + iy = 2x + iy$

Imaginary Axis (h = it ∈ ℝ), $\lim_{t \to 0} \frac{f(z+it)-f(z)}{it} = \lim_{t \to 0} \frac{x^2 + ix(y+t)-x^2-ixy}{it} = \lim_{t \to 0} \frac{ixy + ixt -ixy}{it} \lim_{t \to 0} \frac{ixt}{it} = x$

The limits differ unless x = y = 0 (i.e., z = 0). Thus, the derivative does not exist for z ≠ 0.

Rational Functions (Differentiable Away from Poles)

• Reciprocal Function: f(z) = 1/z. Domain: ℂ∖{0} (pole at z = 0). Derivative: $f'(z) = \frac{-1}{z^2} \forall z \ne 0$

$\lim_{h \to 0} \frac{f(z+h)-f(z)}{h} = \lim_{h \to 0} \frac{\frac{1}{z+h}-\frac{1}{z}}{h} = \lim_{h \to 0} \frac{z - (z+h)}{hz(z+h)} = \lim_{h \to 0} \frac{-1}{z(z+h)} = \frac{-1}{z^2}$

f(z) is differentiable everywhere except at z = 0, where it has a pole (a type of isolated singularity).

• Rational Example: f(z) = $\frac{z}{z^2+1}$ Domain: ℂ∖{i, -i} (pole at z = ±i). Derivative: $f'(z) = \frac{1-z^2}{(z^2+1)^2} \forall z \ne ±i$

$\lim_{h \to 0} \frac{f(z+h)-f(z)}{h} = \lim_{h \to 0} \frac{\frac{z+h}{(z+h)^2+1}-\frac{z}{z^2+1}}{h} = \lim_{h \to 0} \frac{(z+h)(z^2+1)-z((z+h)^2+1)}{((z+h)^2+1)(z^2+1)h}$

Expand the numerator: $(z+h)(z^2+1)-z((z+h)^2+1) = z(z^2+1) + h(z^2+1) -z(z^2 +2zh + h^2 + 1) = z^3 + z + hz^2 + h -z^3 -2z^2h -zh^2 -z = hz^2 + h -2z^2h -zh^2 = h(1 -z^2 -zh)$

Substitute back into the limit: $\lim_{h \to 0} \frac{(z+h)(z^2+1)-z((z+h)^2+1)}{((z+h)^2+1)(z^2+1)h} = \lim_{h \to 0} \frac{h(1 -z^2 -zh)}{((z+h)^2+1)(z^2+1)h} = \lim_{h \to 0} \frac{1 -z^2 -zh}{((z+h)^2+1)(z^2+1)} = \frac{1 -z^2}{(z^2+1)^2}$

Final Result: $f'(z) = \frac{1 -z^2}{(z^2+1)^2}, \forall z \in ℂ$ {i, -i}. The derivative exists everywhere except at the poles z = ±i, where the denominator becomes zero.

It matches the derivative obtained via quotient rule: $\frac{d}{dz}\frac{u}{v} = \frac{u'v -uv'}{v^2} \leadsto \frac{z^2+1-z·2z}{(z^2+1)^2} = \frac{1 -z^2}{(z^2+1)^2}$

• Rational Example 2: Let D = ℂ - {i}, f: D → ℂ be defined as f(z) = $\frac{3z}{z-i}$. Then, f is differentiable at every point z₀ ∈ D (∀z₀ ∈ D).

Let z₀ ∈ D, $\lim_{h \to 0} \frac{f(z_0+h)-f(z_0)}{h} = \lim_{h \to 0} \frac{\frac{3(z_0+h)}{z_0+h-i}-\frac{3z_0}{z_0-i}}{h} =[\text{Combine over a common denominator}] \lim_{h \to 0} \frac{1}{h}(\frac{3(z_0+h)(z_0-i)-3z_0(z_0+h-i)}{(z_0+h-i)(z_0-i)})$ =[Expand the numerator:]

$\lim_{h \to 0} \frac{1}{h}(\frac{3z_0²+3z_0h-3iz_0-3hi-3z_0²-3z_0h+3z_0i)}{(z_0+h-i)(z_0-i)}) =[\text{Thus the quotient becomes}] \lim_{h \to 0} \frac{1}{h}(\frac{-3hi}{(z_0+h-i)(z_0-i)})$ =

$\lim_{h \to 0} (\frac{-3i}{(z_0+h-i)(z_0-i)}) = \frac{-3i}{(z_0-i)(z_0-i)} = \frac{-3i}{(z_0-i)²}$

$f'(z_0) = \frac{-3i}{(z_0-i)²}$. Since z₀ was taken arbitrary in the domain D, f is differentiable everywhere on its domain, with $f'(z) = \frac{-3i}{(z-i)²}, \forall z \ne i$.

Rules of differentiation for complex functions

1. Linearity Rule: If f(z) and g(z) are differentiable functions, and a and b are complex constants, then the function af(z)+bg(z) is also differentiable, and its derivative is given by: $\frac{d}{dz}$[af(z) + bg(z)] = af′(z) + bg′(z).
   Proof via Limit Definition:
   $\frac{d}{dz}[f(z) + g(z)] = \lim_{h \to 0} \frac{f(z+h)+g(z+h)-f(z)-g(z)}{h} = \lim_{h \to 0} (\frac{f(z+h)-f(z)}{h} + \frac{g(z+h)-g(z)}{h}) = f′(z) + g′(z).$
   $\frac{d}{dz}[cf(z)] = \lim_{h \to 0} \frac{cf(z+h)-cf(z)}{h} = c·\lim_{h \to 0} \frac{f(z+h)-f(z)}{h} = c·f′(z).$
2. Product Rule: If f(z) and g(z) are differentiable functions, then their product f(z)g(z) is also differentiable, and its derivative is given by: $\frac{d}{dz}$[f(z) · g(z)] = f′(z)g(z) + f(z)g′(z).
3. Quotient Rule: If f(z) and g(z) are differentiable functions, and g(z) ≠ 0, then their quotient $\frac{f(z)}{g(z)}$ is also differentiable, and its derivative is given by: $\frac{d}{dz} \frac{f(z)}{g(z)} = \frac{f′(z)g(z) - f(z)g′(z)}{(g(z)^2)}$
4. Chain Rule: If f(z) is a differentiable function and g(w) is a differentiable function of w, then the composition g(f(z)) is also differentiable, and its derivative is given by: $\frac{d}{dz}$[g(f(z))] = g′(f(z))·f′(z).
5. Power Rule: If n is an integer, then the function zⁿ is differentiable, and its derivative is given by: $\frac{d}{dz}[z^n] = nz^{n-1}$.
6. Exponential Rule: The exponential function e^z is differentiable, and its derivative is given by: $\frac{d}{dz}[e^z] = e^{z}$.
7. Logarithmic Rule: The principal branch of the logarithmic function log(z) (where z ≠ 0) is differentiable, and its derivative is given by: $\frac{d}{dz}[log(z)] = \frac{1}{z}$.
8. Trigonometric Rules: The trigonometric functions sin(z) and cos(z) are differentiable, and their derivatives are given by: $\frac{d}{dz}[sin(z)] = cos(z), \frac{d}{dz}[cos(z)] = -sin(z)$

These rules form the basis for differentiating complex functions.