Complex Differentiability: Examples and Counterexamples

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Differentiable function

Definition. Let $f: \text{dom}(f) \to \mathbb{R}^m$ be a function, where its domain $\text{dom}(f) \subseteq \mathbb{R}^n$ is an open set. The function $f$ is said to be differentiable if it is differentiable at every point $\mathbf{x} \in \text{dom}(f)$.

Formally, $f$ is differentiable at a point $\mathbf{x}$ if there exists a linear map $L: \mathbb{R}^n \to \mathbb{R}^m$ (represented by an $m \times n$ matrix, the Jacobian matrix $Df(\mathbf{x})$), such that $\lim_{\mathbf{h} \to \mathbf{0}, \mathbf{h} \in \mathbb{R}^n} \frac{\| f(\mathbf{x} + \mathbf{h}) - f(\mathbf{x}) - Df(\mathbf{x})\mathbf{h} \|}{\|\mathbf{h}\|} = 0$

Here, $\| \cdot \|$ denotes a norm (e.g., the Euclidean norm) on $\mathbb{R}^n$ and $\mathbb{R}^m$. This limit must hold for sequences $\mathbf{h} \to 0$ from any direction within $\mathbb{R}^n$.

Examples of Differentiable Functions

• Trigonometric Functions: f(z) = sin(z).

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

Using a completely similar reasoning, $\lim_{h \to 0} \frac{-(e^{-ih}-1)}{h} = i$

$$ \begin{aligned} f'(z)= \lim_{h \to 0} \frac{f(z+h)-f(z)}{h} = \lim_{h \to 0} \frac{sin(z+h)-sin(z)}{h} = &[\text{Substitute the exponential form of sine } sin(z) = \frac{e^{iz}-e^{-iz}}{2i}]\\[2pt] &= \lim_{h \to 0} \frac{1}{h} (\frac{e^{i(z+h)}-e^{-i(z+h)}}{2i}-\frac{e^{iz}-e^{-iz}}{2i}) \\[2pt] &= \lim_{h \to 0} \frac{e^{i(z+h)}-e^{-i(z+h)}-e^{iz}+e^{-iz}}{2ih} = \lim_{h \to 0} \frac{e^{iz}e^{ih}-e^{-iz}e^{-ih}-e^{iz}+e^{-iz}}{2ih} \\[2pt] &=\lim_{h \to 0} \frac{e^{iz}(e^{ih}-1)-e^{-iz}(e^{-ih}-1)}{2ih}= \frac{e^{iz}}{2i}\lim_{h \to 0} \frac{(e^{ih}-1)}{h} + \frac{e^{-iz}}{2i}\lim_{h \to 0} \frac{-(e^{-ih}-1)}{h} \\[2pt] &[\text{Substitute Limits:}] = \frac{e^{iz}·i + e^{-iz}·i}{2i} = \frac{e^{iz} + e^{-iz}}{2} = cos(z). \end{aligned} $$$$ \begin{aligned} f′(z)= \lim_{h \to 0} \frac{f(z+h)-f(z)}{h} = \lim_{h \to 0} \frac{cos(z+h)-cos(z)}{h} = &[\text{Substitute the exponential form of cosine, } cos(z) = \frac{e^{iz}+e^{-iz}}{2}]\\[2pt] &=\lim_{h \to 0} \frac{\frac{e^{i(z+h)}+e^{-i(z+h)}}{2}-\frac{e^{iz}+e^{-iz}}{2}}{h} = \lim_{h \to 0}\frac{e^{i(z+h)}+e^{-i(z+h)}-e^{iz}-e^{-iz}}{2h} \\[2pt] &[\text{Expand numerator: }e^{i(z+h)}+e^{-i(z+h)}-e^{iz}-e^{-iz} = e^{iz}e^{ih}+e^{-iz}e^{-ih}-e^{iz}-e^{-iz} = e^{iz}(e^{ih}-1)+e^{-iz}(e^{-ih}-1)]\\[2pt] &=0+i(-2\pi)=-2\pi i. \\[2pt] &=\lim_{h \to 0}\frac{e^{iz}(e^{ih}-1)+e^{-iz}(e^{-ih}-1)}{2h} = \frac{e^{iz}}{2}\lim_{h \to 0}\frac{e^{ih}-1}{h} + \frac{e^{-iz}}{2}\lim_{h \to 0}\frac{e^{-ih}-1}{h} \\[2pt] &[\text{Apply limits (Euler's formula, then L'Hôpital's rule): } \lim_{h \to 0}\frac{e^{ih}-1}{h} = i, \lim_{h \to 0}\frac{e^{-ih}-1}{h} = -i] \\[2pt] &= \frac{i}{2}(e^{iz}-e^{-iz}) = \frac{i}{2}(cos(z)+isin(z)-cos(-z)-isin(-z)) =[\text{Use even/odd properties of sine and cosine}] \frac{i}{2} (cos(z)+isin(z)-cos(z)+isin(z)) = \frac{2i^2sin(z)}{2} = -sin(z) \end{aligned} $$

Therefore: (i) f(z) = sin(z), f’(z) = cos(z) and f(z) = cos(z), f’(z) = -sin(z) ∎

Another approach. $\sin z = \frac{e^{iz} - e^{-iz}}{2i}, \quad \cos z = \frac{e^{iz} + e^{-iz}}{2}$

Derivatives (Chain Rule applies): $\frac{d}{dz}(\sin z) = \frac{1}{2i} \frac{d}{dz}(e^{iz} - e^{-iz}) = \frac{1}{2i} (ie^{iz} - (-i)e^{-iz}) = \frac{i(e^{iz} + e^{-iz})}{2i} = \frac{e^{iz} + e^{-iz}}{2} = \cos z$

Similarly, $\frac{d}{dz}(\cos z) = \frac{1}{2} \frac{d}{dz}(e^{iz} + e^{-iz}) = \frac{1}{2} (ie^{iz} - ie^{-iz}) = \frac{i(e^{iz} - e^{-iz})}{2} = i^2 \frac{e^{iz} - e^{-iz}}{2i} = -\sin z$

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

Since sin (a - b) = sin a cos b - cos a sin b where a = z + h and b = z.

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

Proof using quotient rule: f(z) = tan(z) = sin(z)/cos(z)

$$ \begin{aligned} (\tan z)' = \frac{(\sin z)' \cos z - \sin z (\cos z)'}{\cos^2 z} &=\int_{2\pi}^{4\pi}t\cos t\,dt + i\int_{2\pi}^{4\pi} t\sin t\,dt \\[2pt] &= \frac{\cos z \cdot \cos z - \sin z \cdot (-\sin z)}{\cos^2 z}\\[2pt] &= \frac{\cos^2 z + \sin^2 z}{\cos^2 z} = \frac{1}{\cos^2 z} = \sec^2 z\\[2pt] &=0+i(-2\pi)=-2\pi i. \end{aligned} $$

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.

Algebraic Rules, Derivatives of Common Functions

Function f(z)	Derivative f’(z)	Domain of Differentiability
c (constant)	0	ℂ
z	1	ℂ
zⁿ	nzⁿ⁻¹	ℂ
Polynomial p(z)	p’(z)	ℂ
p(z)/q(z)	Quotient rule	ℂ \ {zeros of q}
eᶻ	eᶻ	ℂ
sin z	cos z	ℂ
cos z	-sin z	ℂ
tan z	sec² z	ℂ \ {π/2 + kπ}
sinh z	cosh z	ℂ
cosh z	sinh z	ℂ
log z	1/z	ℂ \ (-∞, 0] (principal branch)
zᵅ	αzᵅ⁻¹	ℂ \ (-∞, 0] (principal branch)

If f and g are complex differentiable at z₀, then:

Rule	Formula
Constant	(c)’ = 0
Sum	(f + g)’ = f’ + g'
Difference	(f - g)’ = f’ - g'
Constant Multiple	(cf)’ = cf'
Product	(fg)’ = f’g + fg'
Quotient	(f/g)’ = (f’g - fg’)/g² (where g ≠ 0)
Chain Rule	(f ∘ g)’(z₀) = f’(g(z₀)) · g’(z₀)
Power	(zⁿ)’ = nzⁿ⁻¹

Functions That Fail to Be Differentiable

Many functions that are perfectly smooth from a real analysis perspective fail to be complex differentiable. The reason: the limit must be path-independent.

• Complex conjugate. f(z) = $\bar{z}$ is nowhere complex differentiable.

$\lim_{h \to 0} \frac{f(a+h)-f(a)}{h} = \lim_{h \to 0} \frac{\overline{a+h}-\overline{a}}{h}=[\text{Conjugation distributes over addition}] \lim_{h \to 0} \frac{\overline{h}}{h}$ does not exist.
Path 1. h = t (real, t → 0), $\frac{\bar{h}}{h} = \frac{t}{t} = 1$
Path 2. h = it (imaginary, t → 0), $\frac{\bar{h}}{h} = \frac{-it}{it} = -1$ Different limits for different paths, hence the limit does not exist.
Consider that f(z) = z̄ is continuous everywhere, real-differentiable as a map ℝ² → ℝ²), but complex-differentiable nowhere.

• Real Part Function. f(z) = Re(z). is nowhere complex differentiable. $\lim_{h \to 0} \frac{f(a+h)-f(a)}{h} = \lim_{h \to 0} \frac{Re(a+h)-Re(a)}{h} = \lim_{h \to 0} \frac{Re(h)}{h}$.
  Path 1: h = t (real, t → 0), $\frac{\text{Re}(h)}{h} = \frac{t}{t} = 1$
  Path 2: h = it (imaginary, t → 0), $\frac{\text{Re}(h)}{h} = \frac{0}{it} = 0$
  Different limits for different paths, hence the limit does not exist.

• Imaginary Part Function f(z) = Im(z) is nowhere complex differentiable.. $\lim_{h \to 0} \frac{\text{Im}(h)}{h}$. Path 1: h = t (real): Im(t)/t = 0/t = 0. Path 2: h = it (imaginary): Im(it)/it = t/(it) = -i. Different limits → not differentiable.

• Modulus Function f(z) = |z|. f(z) = |z| is nowhere complex differentiable.

Proof at z₀ ≠ 0: Write $z₀ = r₀e^{iθ₀}$ with r₀ > 0. Consider: $\lim_{h \to 0} \frac{|z_0 + h| - |z_0|}{h}$

Path 1: $h = te^{iθ₀}$ (radial direction, t → 0⁺): $|z_0 + h| = |r_0 e^{i\theta_0} + te^{i\theta_0}| = (r_0 + t)$
$\frac{|z_0 + h| - |z_0|}{h} = \frac{r_0 + t - r_0}{te^{i\theta_0}} = \frac{1}{e^{i\theta_0}} = e^{-i\theta_0}$ Path 2: $h = ite^{iθ₀}$ (tangential direction, t → 0 real)
Then, $z_0+h=r_0e^{i\theta _0}+ite^{i\theta _0}=(r_0+it)e^{i\theta _0},$
so $|z_0+h|=|r_0+it|=\sqrt{r_0^2+t^2}.$
Hence, $\frac{|z_0+h|-|z_0|}{h}=\frac{\sqrt{r_0^2+t^2}-r_0}{ite^{i\theta _0}}$.
For small t, use the standard expansion $\sqrt{r_0^2+t^2}=r_0\sqrt{1+\frac{t^2}{r_0^2}}\approx r_0\left( 1+\frac{1}{2}\frac{t^2}{r_0^2}\right) =r_0+\frac{t^2}{2r_0}.$
Note: For small x, we have the classical expansion: $ \sqrt{1+x} = 1 + \frac{x}{2} - \frac{x^2}{8} + \cdots$. When x is tiny, the higher-order terms ($x^2, x^3,\dots$) are negligible, so: $\sqrt{1+x} \approx 1 + \frac{x}{2}$.
So $\sqrt{r_0^2+t^2}-r_0\approx \frac{t^2}{2r_0}$ and then
$\frac{|z_0+h|-|z_0|}{h}\approx \frac{\frac{t^2}{2r_0}}{ite^{i\theta _0}}=\frac{t}{2r_0}\cdot \frac{1}{ie^{i\theta _0}}\rightarrow 0\quad \mathrm{as\ }t\rightarrow 0.$
So along this path, the limit is 0. Since along one path the limit is $e^{-i\theta _0}$ and along another it is 0, the limit does not exist at z_0. Thus, f is not complex differentiable at any $z_0\neq 0$.

Not differentiable at $z_0=0$. $\lim _{h\rightarrow 0}\frac{|h|-|0|}{h}=\lim _{h\rightarrow 0}\frac{|h|}{h}.$

Path 1: real axis. Choice: h=t, $t\rightarrow 0^+$ real. Then, $\frac{|h|}{h}=\frac{|t|}{t}=1.$
Path 2: imaginary axis. Choice: h=it, $t\rightarrow 0^+ $real. Then, $\frac{|h|}{h}=\frac{|it|}{it}=\frac{t}{it}=-i.$

The two limits 1 and -i differ, so the limit does not exist at 0 either.

Conclusion: The complex derivative of f(z)=|z| does not exist at any point z$_0\in \mathbb{C}$. So f is nowhere complex differentiable.

• 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

$$ \begin{aligned} \lim_{h \to 0} \frac{f(z+h)-f(z)}{h} = \lim_{h \to 0} \frac{(z+h)\mathbb{Re}(z+h)-z\mathbb{Re}(z)}{h} &=\lim_{h \to 0} \frac{z\mathbb{Re}(z)+z\mathbb{Re}(h)+h\mathbb{Re}(z)+h\mathbb{Re}(h)-z\mathbb{Re}(z)}{h} \\[2pt] &= \lim_{h \to 0} \frac{+z\mathbb{Re}(h)+h\mathbb{Re}(z)+h\mathbb{Re}(h)}{h} = \lim_{h \to 0} \frac{z\mathbb{Re}(h)}{h} + \mathbb{Re}(z) + \mathbb{Re}(h) \\[2pt] &= \lim_{h \to 0} (\frac{z\mathbb{Re}(h)}{h}) + \mathbb{Re}(z). \end{aligned} $$

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.

If z = 0, h = h₁ + ih₂, $\frac{z\mathbb{Re}(h)}{h} = 0$ for any arbitrary h approaching zero.

$\frac{f(h) - f(0)}{h} =[\frac{z\mathbb{Re}(h)}{h} + \mathbb{Re}(z) + \mathbb{Re}(h)] 0 \cdot \frac{h_1}{h} + 0 + h_1 = h_1 = \text{Re}(h) \to 0$

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

• The Squared modulus function. f(z) = |z|² = $z\bar{z}$. We investigate its differentiability.

1. At a = 0, $\lim_{h \to 0} \frac{f(h)-f(0)}{h} = \lim_{h \to 0} \frac{|h|^2}{h} = \lim_{h \to 0} \frac{h\bar{h}}{h} = \lim_{h \to 0} \bar{h} = 0$. Hence, f′(0)=0; the function is differentiable at the origin.
2. At a ≠ 0, $\lim_{h \to 0} \frac{f(a+h)-f(a)}{h} = \lim_{h \to 0} \frac{(a+h)\overline{(a+h)} − a\bar a}{h} = \lim_{h \to 0} \frac{a\bar{a}+a\bar{h}+h\bar{a}+h\bar{h}− a\bar a}{h}$
   $= \lim_{h \to 0} \frac{ a\bar h + \bar a h + |h|²}{h} = \lim_{h \to 0} a\frac{\bar{h}}{h}+\bar{a}+\frac{|h|^2}{h} =[\star] \lim_{h \to 0} a\frac{\bar{h}}{h}+\bar{a}$

[$\star$] As $h \to 0$, the $\frac{|h|^2}{h} = \frac{h\bar{h}}{h} = \bar{h} \to 0$. Then, $\lim_{h \to 0} a\frac{\bar{h}}{h}+\bar{a}+\frac{|h|^2}{h} = a\frac{\bar{h}}{h}+\bar{a}$

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, $\frac{\bar{h}}{h} = \frac{-it}{it}=-1, \lim_{h \to 0} \frac{f(a+h)-f(a)}{h} = a·(-1) + \bar{a} = \bar{a} - a = −2iℑ(a).$

For the full limit to exist, these two directional limits must coincide. This requires $2\Re(a) = -2i\Im(a).$ The left side is real, the right side is purely imaginary (unless zero). 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.

Remark: This function is real‑differentiable everywhere (as a map $\mathbb{R}^2\to\mathbb{R}^2$), but fails the Cauchy–Riemann equations except at the origin. Indeed, with $u(x, y) = x^2+ y^2, v(x,y) = 0$, the Cauchy–Riemann equations $u_x = v_y, u_y = -v_x$ give 2x = 0 and 2y = 0, so only (0, 0) satisfies them.

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

For any 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 on its entire domain ℂ∖{0}, where it has a pole (an isolated singularity of order 1).

• 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: $f'(z) = \frac{1 -z^2}{(z^2+1)^2}, \forall z \in \mathbb{C} \setminus \{i, -i \}$. The derivative exists everywhere except at the poles z = ±i, where the denominator vanishes. The function is analytic on $ \mathbb{C} \setminus \{i, -i \}$.

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

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

$f'(z_0) = \frac{-3i}{(z_0-i)²}, \forall z \in \mathbb{C} \setminus \{ i \}$. Since z₀ was taken arbitrary in the domain D, f is differentiable everywhere on its domain, with $f'(z) = \frac{-3i}{(z-i)²}, \forall z \ne i$. The derivative inherits the pole at z = i (order 2), consistent with the original function's pole (order 1).