Relationship Between Complex Differentiability and the Cauchy–Riemann Equations

For every problem there is always, at least, a solution which seems quite plausible. It is simple and clean, direct, neat and nice, and yet very wrong, #Anawim, justtothepoint.com

Relationship Between Complex Differentiability and the Cauchy-Riemann Equations

Necessary Condition: Complex Differentiability Implies the Cauchy-Riemann Equations

If a function $f(z) = u(x, y) + iv(x, y)$ is complex-differentiable at a point $z_0 = x_0 + iy_0$, then its real and imaginary parts, $u(x,y)$ and $v(x,y)$, must satisfy the Cauchy–Riemann equations at that point: $\frac{\partial u}{\partial x}(x_0, y_0) = \frac{\partial v}{\partial y}(x_0, y_0), \quad \frac{\partial u}{\partial y}(x_0, y_0) = -\frac{\partial v}{\partial x}(x_0, y_0)$.

Sufficient Condition: Cauchy-Riemann Equations (with Continuous Partial Derivatives) Imply Complex Differentiability

Proposition. Let $f(z) = u(x, y) + iv(x, y)$ for $z = x + iy \in D$, where $D \subseteq \mathbb{C}$ is an open set. Assume that $u$ and $v$ have continuous first partial derivatives throughout $D$ and they satisfy the Cauchy–Riemann equations at a point $z \in D$. Then $f'(z)$ exists, i.e., $f$ is differentiable at $z$.

Consequences of the Cauchy-Riemann Equations

Theorem. Let $f = u + iv$ be a complex function $f(z) = u(x, y) + iv(x, y)$, where $z = x + iy$, which is differentiable (analytic) at every point in a domain $D \subseteq \mathbb{C}$. Suppose that $|f(z)| = k$ for all $z \in D$, where $k$ is some constant. Then $f$ must be constant on $D$.

This is a fundamental result that reveals a deep property of analytic functions: they cannot map an open region $D$ to a circle ($|f(z)| = k$ for all $z \in D$) without collapsing to a single point ($f \equiv c$).

Proof:

Recall that the magnitude of a complex number is: $f(z) = \sqrt{u(x, y)² + v(x, y)²}$

So the condition $|f(z)| = k$ for all $z \in D$ implies that $u(x, y)^2 + v(x, y)^2 = k^2$ for all $(x, y)$ corresponding to $z \in D$. This means that as $z$ varies over the domain $D$, the point $(u(x, y), v(x, y))$ always lies on a circle of radius $k$ centered at the origin in the $w$-plane.

🍀 Case 1: k = 0 (The Trivial Case)

$u(x, y)^2 + v(x, y)^2 = 0$. Since both $u(x, y)^2$ and $v(x, y)^2$ are non-negative, this implies $u(x, y) = v(x, y) = 0$ for all $z \in D$. Therefore $f(z) = 0$ for all $z \in D$, and $f$ is the zero function — clearly constant.

🍀 Case 2: k ≠ 0 (The Non-Trivial Case)

Now assume $k \neq 0$. We differentiate the identity $u^2 + v^2 = k^2$ with respect to each variable. The right-hand side is zero because $k^2$ is constant; on the left-hand side, applying the chain rule:

$\text{Differentiate with respect to } x: 2u\frac{\partial u}{\partial x} + 2v\frac{\partial v}{\partial x} = 0. \text{Differentiate with respect to } y: 2u\frac{\partial u}{\partial y} + 2v\frac{\partial v}{\partial y} = 0.$

Since $f$ is analytic on $D$, it satisfies the Cauchy–Riemann equations everywhere in $D$: $\frac{\partial v}{\partial x} = -\frac{\partial u}{\partial y}, \qquad \frac{\partial v}{\partial y} = \frac{\partial u}{\partial x}.$

Substituting these into the differentiated equations and dividing by two:

$$\begin{cases} u\dfrac{\partial u}{\partial x} - v\dfrac{\partial u}{\partial y} = 0, \\[8pt] u\dfrac{\partial u}{\partial y} + v\dfrac{\partial u}{\partial x} = 0. \end{cases}$$

Let us denote $A = \frac{\partial u}{\partial x}$ and $B = \frac{\partial u}{\partial y}$. The system becomes:

$$\begin{cases} uA - vB = 0, \\ uB + vA = 0. \end{cases}$$

This is a homogeneous linear system in the unknowns $A$ and $B$. We solve it by elimination.

Finding A: Multiply the first equation by $u$ and the second by $v$:

$$\begin{cases} u^2 A - uvB = 0, \\ uvB + v^2 A = 0. \end{cases}$$

Adding: $(u^2 + v^2)A = 0$. Since $u^2 + v^2 = k^2 \neq 0$, we conclude $A = \frac{\partial u}{\partial x} = 0$.

Finding B: Multiply the first equation by $v$ and the second by $u$:

$$\begin{cases} uvA - v^2 B = 0, \\ u^2 B + vuA = 0. \end{cases}$$

Subtracting the first from the second: $(u^2 + v^2)B = 0$. Again $u^2 + v^2 = k^2 \neq 0$, so $B = \frac{\partial u}{\partial y} = 0$.

Therefore, $\frac{\partial u}{\partial x} = 0 \quad \text{and} \quad \frac{\partial u}{\partial y} = 0$ everywhere in $D$. This means that $u$ has zero gradient throughout $D$, and hence zero directional derivatives in all directions.

---

Recall (Zero Gradient Implies Constancy on Connected Sets).

For a function u(x, y) that is differentiable (which is guaranteed in our case since f = u + iv is analytic, making u and v continuously differentiable), the directional derivative in the direction of a unit vector $\vec{v}$ = (a, b) is given by: $D_{\vec{v}}u = \nabla u · \vec{v} = \frac{\partial u}{\partial x}·a + \frac{\partial u}{\partial y}·b$

where $\nabla u = (\frac{\partial u}{\partial x}, \frac{\partial u}{\partial y})$ is the gradient of u, $\vec{v}$ is a unit vector (a² + b² = 1). The dot product captures how u changes in the direction of $\vec{v}$. If $\frac{\partial u}{\partial x} = \frac{\partial u}{\partial y} = 0 \leadsto D_{\vec{v}}u = \nabla u · \vec{v} = \frac{\partial u}{\partial x}·a + \frac{\partial u}{\partial y}·b = 0·a + 0·b = 0$, and this holds for any direction $\vec{v}$.

Both partial derivatives tell us how steep the surface is when moving in the x and y -direction respectively. If both of these are zero, there is neither slope in the x- direction, nor slope in the y -direction. Therefore, there can’t be a slope in any direction that’s a combination of x and y. Visually, the surface is perfectly flat at that point.

The directional derivative of a differentiable function u at a point x in the direction of a unit vector $\vec{v}$ is $D_{\vec{v}}u(x)=\nabla u(x)\cdot \vec{v}$. If $D_{\vec{v}}u(x)=0$ for every direction $\vec{v}$, then the inner product with any $\vec{v}$ vanishes. That can only happen when $\nabla u(x) = (0,0,\dots,0) \forall x$ (gradient is identically zero).

On a connected (in fact path-connected) region Ω, pick any two points P₁ and P₂. Choose a smooth curve C from P₁ to P₂. Then, the change in u along C is given by the line integral of its gradient: $u(P₂)-u(P₁)=\int_{C}\nabla u\cdot d\mathbf s=0.$ Since this holds for every pair of points, u takes the same value everywhere on Ω.

---

Since $D$ is a domain (by definition, an open connected set), u has zero directional derivatives in all directions, $u$ must be constant throughout $D$.

Now consider $v$. From the Cauchy–Riemann equations:

$$\frac{\partial v}{\partial x} = -\frac{\partial u}{\partial y} = 0, \qquad \frac{\partial v}{\partial y} = \frac{\partial u}{\partial x} = 0.$$

So $v$ also has zero partial derivatives everywhere in $D$, which means $v$ is constant on $D$ as well.

Since both $u$ and $v$ are constant on $D$, the function $f(z) = u(x, y) + iv(x, y)$ must also be constant on $D$. $\blacksquare$

---

Since D is a domain (by definition, an open connected set), u must be constant throughout D .

Now let’s look at v. From the Cauchy-Riemann equations: $\frac{\partial v}{\partial x} = -\frac{\partial u}{\partial y} = 0, \frac{\partial v}{\partial y} = \frac{\partial u}{\partial x} = 0$

So v also has zero partial derivatives everywhere in D, which means v is constant on D as well. Since both u and v are constant functions on D, their combination: f(z) = u(x, y) + iv(x, y) must also be constant on D.

Note. The proof relies on D being connected. If $D$ were disconnected (e.g., two disjoint, nonempty open regions $D = D_1 \cup D_2$ with $D_1 \cap D_2 = \emptyset$), then $f$ could take one constant value on $D_1$ and another on $D_2$:

$f(z) = \begin{cases} k, &z \in D_1\\\\ -k, &z \in D_2 \end{cases}$

This function would have ∣f(z)∣ = k everywhere, but wouldn’t be constant. Furthermore, in real analysis, one can have non-constant functions with constant magnitude — for instance, $f(x) = e^{ix} = \cos x + i\sin x$ on $\mathbb{R}$ satisfies $|f(x)| = 1$ for all $x$ but is not constant. However, this doesn’t violate anything because $e^{ix}$ is not analytic as a function of a complex variable on any open subset of $\mathbb{C}$. It is only analytic as a function of a real variable.

Analytic Functions with Zero Derivative Are Constant

Theorem. Let $f(z) = u(x, y) + iv(x, y)$ be a complex function that is differentiable (analytic) at every point in a domain $D \subseteq \mathbb{C}$. If $f'(z) = 0$ for all $z \in D$, then $f$ must be constant on $D$.

Proof.

Recall that for a complex function f(z) = u(x, y) + iv(x, y) where z = x + iy , the complex derivative is defined as: $f'(z) = \lim_{h \to 0} \frac{f(z + h) - f(z)}{h}$

When this limit exists for all z ∈ D, f is analytic on D. For an analytic function, the complex derivative can be expressed using partial derivatives: $f'(z) = \frac{\partial u}{\partial x}+i\frac{\partial v}{\partial x}$ This comes from considering the limit as h→0 along the real axis (h = t ∈ ℝ): $f'(z) = \lim_{t \to 0} \frac{f(z + t) - f(z)}{t}$

Alternatively, using the Cauchy-Riemann equations, we can also write: $f'(z) = \frac{\partial v}{\partial y}-i\frac{\partial u}{\partial y}$

By assumption, $f'(z) = 0$ for all $z \in D$. For a complex number to be zero, both its real and imaginary parts must vanish. From the first expression: $\frac{\partial u}{\partial x} = 0, \qquad \frac{\partial v}{\partial x} = 0 \quad \text{throughout } D.$

From the second expression: $\frac{\partial v}{\partial y} = 0, \qquad \frac{\partial u}{\partial y} = 0 \quad \text{throughout } D.$

Therefore, all four first-order partial derivatives of $u$ and $v$ are zero throughout $D$.

For a differentiable function $u(x, y)$, the directional derivative in the direction of any unit vector $\vec{\mathbf{v}} = (a, b)$ is: $D_{\vec{\mathbf{v}}} u = \nabla u \cdot \vec{\mathbf{v}} = \frac{\partial u}{\partial x} a + \frac{\partial u}{\partial y} b = 0 \cdot a + 0 \cdot b = 0.$

This holds for every direction $\vec{\mathbf{v}}$, and the same applies to $v(x, y)$. Both $u$ and $v$ have zero directional derivatives in all directions at every point in $D$. As it was established previously, this implies that $u$ and $v$ are each constant on the connected domain $D$. Therefore, $f$ is constant. $\blacksquare$

Examples

• $f(z) = \bar{z}, f(x + iy) = x - iy$ for all $x + iy \in \mathbb{C}$.

1. Identify u and v: u(x, y) = x, v(x, y) = -y.
2. Compute Partial Derivatives. $\frac{\partial u}{\partial x} = 1, \frac{\partial u}{\partial y} = 0, \frac{\partial v}{\partial x} = 0, \frac{\partial v}{\partial y}= -1$. All partial derivatives exist and are continuous everywhere.
3. Check Cauchy-Riemann Equations: $\frac{\partial u}{\partial x} = 1 \ne 0 = \frac{\partial v}{\partial y}$ at any point f, so $f(z)=\bar{z}$ is not differentiable at any point in $\mathbb{C}$.

• f(z) = $e^xcos(y) + ie^xsin(y) = e^x(cos(y)+isin(y)) = e^xe^{iy} = e^{x+iy} = e^z$. Let’s verify that it satisfies the conditions for complex differentiability:

1. Identify u and v: $u(x, y) = e^x \cos y$, $v(x, y) = e^x \sin y$.
2. Compute Partial Derivatives. $\frac{\partial u}{\partial x} = e^xcos(y), \frac{\partial u}{\partial y} = -e^xsin(y), \frac{\partial v}{\partial x} = e^xsin(y), \frac{\partial v}{\partial y}= e^xcos(y)$. All partial derivatives exist and are continuous everywhere (exponentials and trigonometric functions are smooth).
3. Check Cauchy-Riemann Equations: $\frac{\partial u}{\partial x} = e^xcos(y) = \frac{\partial v}{\partial y}✓, \frac{\partial u}{\partial y} = -e^xsin(y) = -\frac{\partial v}{\partial x}✓$. The Cauchy-Riemann equations are satisfied at all points (x,y) .
4. Since all partial derivatives are continuous on ℝ² and the Cauchy-Riemann equations hold everywhere, $f(z) = e^z$ is complex differentiable everywhere — it is an entire function.
5. As previously noted, this derivation also provides expressions for the complex derivative itself: (Real Axis) f’(z) = $\frac{\partial u}{\partial x} + i\frac{\partial v}{\partial x} = e^xcos(y) + ie^xcos(y) = e^x(cos(y)+isin(y)) = e^xe^{iy} = e^{x+iy} = e^z.$

Thus, $\frac{d}{dz} e^z = e^z$, recovering the familiar formula.

• f(z) = $z^2 = x^2 - y^2 + i(2xy)$ where z = x + iy.

1. Identify Real and Imaginary Parts: u(x, y) = x² - y² (real part), v(x, y) = 2xy (imaginary part).
2. Compute the First-Order Partial Derivatives. $\frac{\partial u}{\partial x} = 2x, \frac{\partial u}{\partial y} = -2y, \frac{\partial v}{\partial x} = 2y, \frac{\partial v}{\partial y}= 2x$. All partial derivatives exist and are continuous everywhere in ℝ², since they are polynomials.
3. Check the Cauchy-Riemann Equations. The Cauchy-Riemann equations are: $\frac{\partial u}{\partial x} = 2x = \frac{\partial v}{\partial y}✓, \frac{\partial u}{\partial y} = -2y = -\frac{\partial v}{\partial x}✓$. Both equations hold for all $(x, y) \in \mathbb{R}^2$.
4. Conclusion: $f(z)$ is complex differentiable at every point in $\mathbb{C}$ — it is entire.
5. Compute the Derivative: f’(z) = $\frac{\partial u}{\partial x} + i\frac{\partial v}{\partial x} = 2x + i(2y) =2(x+iy)=2z.$
6. Observation: $f(z) = (x + iy)^2 = z^2$ is simply a polynomial, and hence automatically entire.

• $f(z) = r^2(\cos 2\theta + i\sin 2\theta)$ for $z = re^{i\theta} \neq 0$

1. Express f(z) in Cartesian Coordinates. With $x = r\cos\theta$, $y = r\sin\theta$, we have $r^2 = x^2 + y^2$, $\cos \theta = \frac{x}{\sqrt{x^2 + y^2}}, \sin \theta = \frac{y}{\sqrt{x^2 + y^2}}, \cos 2\theta = \cos^2\theta - \sin^2\theta = \frac{x^2 - y^2}{x^2 + y^2},$ and $\sin 2\theta = 2\sin\theta\cos\theta = \frac{2xy}{x^2 + y^2}.$
   Therefore, $f(z) = (x^2 + y^2)\left(\frac{x^2 - y^2}{x^2 + y^2} + i\frac{2xy}{x^2 + y^2}\right) = (x^2 - y^2) + i(2xy)$ for $z \neq 0$.
   But $z^2 = (x + iy)^2 = (x^2 - y^2) + i(2xy)$, so $f(z) = z^2$ for $z \neq 0$. Since $\lim_{z \to 0} f(z) = 0$, we can extend $f$ continuously by defining $f(0) = 0$.
2. For $z \neq 0$, the real and imaginary parts are $u(x, y) = x^2 - y^2$ and $v(x, y) = 2xy$. As shown in Example 3, these satisfy the Cauchy–Riemann equations everywhere with continuous partial derivatives. Hence $f$ is complex differentiable everywhere (for $z \neq 0$, and at $z = 0$ by direct verification or extension).

• $f(z) = z^2 + \bar{z}$

1. Express the function f in terms of x and y. $f(z) = (x + iy)^2 + (x - iy) = (x^2 - y^2 + 2ixy) + (x - iy) =[\text{Group real and imaginary parts}] (x^2 - y^2 + x) + i(2xy - y).$
2. Identify the real and imaginary parts. $u(x, y) = x^2 - y^2 + x$, $v(x, y) = 2xy - y$. These are polynomials (smooth), so all partial derivatives exist and are continuous everywhere.
3. Compute the First-Order Partial Derivatives. $\frac{\partial u}{\partial x} = 2x + 1, \frac{\partial u}{\partial y} = -2y, \frac{\partial v}{\partial x} = 2y, \frac{\partial v}{\partial y}= 2x-1$. All partials are continuous on ℝ².
4. Check the Cauchy-Riemann Equations. $\frac{\partial u}{\partial x} = 2x + 1 \neq 2x - 1 = \frac{\partial v}{\partial y}.$. The equation $2x + 1 = 2x - 1$ simplifies to $1 = -1$, which is never true. The second equation $\frac{\partial u}{\partial y} = -2y = -\frac{\partial v}{\partial x}$ does hold for all $(x, y)$, but both equations must be satisfied simultaneously.
5. Since the Cauchy-Riemann equations are not satisfied at any point, we conclude: $f(z) = z^2 + \bar{z}$ is not complex differentiable at any point in $\mathbb{C}$. Thus, f is nowhere analytic.

   This illustrates that the presence of $\bar{z}$ in a formula typically destroys analyticity.

• $f(z) = x^3 - 3xy^2 + i(3x^2y - y^3)$

1. Identify $u(x, y)$ and $v(x, y), u(x, y) = x^3 - 3xy^2, v(x, y) = 3x^2y - y^3$
2. Compute Partial Derivatives. $\frac{\partial u}{\partial x} = 3x^2 - 3y^2, \frac{\partial u}{\partial y} = -6xy, \frac{\partial v}{\partial x} = 6xy, \frac{\partial v}{\partial y}= 3x^2 - 3y^2$. All partial derivatives exist and are continuous everywhere in ℝ², since they are polynomials.
3. Check the Cauchy-Riemann Equations. $ \frac{\partial u}{\partial x} = 3x^2 - 3y^2 = \frac{\partial v}{\partial y} = 3x^2 - 3y^2, \frac{\partial u}{\partial y} = -6xy = -\frac{\partial v}{\partial x}$
4. Since both Cauchy-Riemann equations are satisfied and the partial derivatives of u and v are continuous (as they are polynomials), the function $f(z) = x^3 - 3xy^2 + i(3x^2y - y^3)$ is complex differentiable at every point in ℂ, so it is entire.
5. Moreover, its derivative is: $f'(x) = \frac{\partial u}{\partial x} + i\frac{\partial v}{\partial x} = 3x^2 - 3y^2 + i(6xy) = 3(x^2 - y^2 + 2xyi) = 3(x+iy)^2 = 3z^2.$ Thus, f’(z) = 3z².
6. Notice that $z^3 = (x + iy)^3 = x^3 + 3x^2(iy) + 3x(iy)^2 + (iy)^3 = x^3 - 3xy^2 + i(3x^2 y - y^3)$ which is analytic everywhere in the complex plane (i.e., entire), and its derivative is: f’(z) = 3z².
7. This function is a special case of De Moivre’s Theorem. For any real angle $\theta$ and integer n, $(\cos(\theta) + i\sin(\theta))^n = \cos(n\theta) + i\sin(n\theta).$ Every nonzero complex number z can be expressed as z = $r\bigl(\cos\theta + i\sin\theta\bigr)$. Applying De Moivre’s Theorem with n = 3, we get $z^3 = \bigl[r(\cos(\theta) + i\sin(\theta))\bigr]^3 = r^3\bigl(\cos(3\theta) + i\sin(3\theta)\bigr) =[\text{So in polar coordinates:}] r^3e^{i3\theta}$ which again confirms it’s analytic for all $z \in \mathbb{C}$.

The Jacobian Matrix and Complex Differentiability

To understand the geometry of complex functions, we first view a complex function $f(z) = u(x,y) + iv(x,y)$ as a standard 2D vector mapping from $\mathbb{R}^2 \to \mathbb{R}^2$, where $(x, y) \mapsto (u(x, y), v(x, y))$.

If the partial derivatives of $u$ and $v$ exist and are continuous at a point $(x_0, y_0)$, the function is real-differentiable. This means the change in $f$ near that point can be approximated by a linear transformation (represented by the Jacobian matrix) plus an error term (a negligible error) that vanishes rapidly.

The linear transformation is represented by the Jacobian matrix: $J_f(x_0, y_0) = \begin{pmatrix} \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} \\ \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y} \end{pmatrix}$. This matrix encapsulates how $f$ transforms small changes in the input near $(x_0, y_0)$. For any small displacement vector $(\Delta x, \Delta y)^T$, the resulting change in the output is approximated by multiplying this vector by the Jacobian, $J_F \cdot (\Delta x, \Delta y)^T$.

Every real-differentiable function has such a Jacobian, but it can represent any linear map (rotations, shears, scalings, etc.).

Complex Differentiability: A Much Stronger Condition

A function is complex differentiable at $z_0 = x_0 + iy_0$ if the limit $f'(z_0) = \lim_{h \to 0}\frac{f(z_0+h)-f(z_0)}{h}$ exists as a complex number. This automatically implies real differentiability, but the converse is false. Complex differentiability imposes extra constraints because the limit must be independent of the direction in which $h \to 0$ (path-independence).

Complex differentiability is a much stronger condition because it forces the partial derivatives to satisfy the Cauchy-Riemann equations: $\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}, \quad \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}$

Let $a = \frac{\partial u}{\partial x}$ and $b = \frac{\partial v}{\partial x}$. Substituting the Cauchy-Riemann equations into our general Jacobian forces it into a highly specific form or structure: $J_f(x_0, y_0) = \begin{pmatrix} a & -b \\ b & a \end{pmatrix}$

This matrix is not arbitrary. It is mathematically identical to multiplication by the complex number $a + ib$. To see this, compare how the matrix acts on a vector $(x, y)^T$ with how the complex number $(a+ib)$ multiplies by $(x+iy)$:

• Matrix Multiplication: $\begin{pmatrix} a & -b \\ b & a \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} ax - by \\ bx + ay \end{pmatrix}$.
• Complex Multiplication: $(a + ib)(x + iy) = (ax - by) + i(bx + ay)$

The two results are identical. Moreover, the complex derivative at $z_0$ is $f'(z_0) = \frac{\partial u}{\partial x} + i\frac{\partial v}{\partial x} = a + ib$, we reach a profound conclusion: the linear transformation represented by the Jacobian matrix acts exactly like multiplication by the complex derivative $f'(z_0)$. The local linear approximation is not just any real-linear map; it is complex-linear.

Complex differentiability at $z_0$ means the local behavior of $f$ is indistinguishable from multiplying by a single complex number. The real first-order approximation is not just affine, it is strictly complex-affine; f(z) is approximated by $f(z) \approx f(z_0) + f'(z_0)(z - z_0)$ where $f'(z_0) = \frac{\partial u}{\partial x}(x_0, y_0) + i\frac{\partial v}{\partial x}(x_0, y_0)$.

Geometric Interpretation: The “Amplitwist”

Because the local derivative acts entirely through complex multiplication, it has a beautiful and strict geometric interpretation.

When you multiply any complex number by $f'(z_0) = Re^{i\theta}$, two geometric transformations happen simultaneously:

• Scaling (Amplification): The magnitude of the input is scaled by the modulus $|f'(z_0)| = \sqrt{a^2 + b^2}$.
• Rotation (Twist): The angle of the input is rotated by the argument $\arg(f'(z_0))$.

This combined local effect is often called an “amplitwist”. It tells us that a complex differentiable function can locally translate, rotate, and scale the plane, but it cannot shear, reflect, or stretch the plane unevenly. This perfect preservation of local shape and intersection angles is why these mappings are called conformal.

Algebraic Proof of the Amplitwist

We can easily verify this geometric behavior using the properties of complex arithmetic, which perfectly mirrors our Jacobian matrix.

Assume $f'(z_0) \neq 0$. Let $\alpha = \alpha_1 + i\alpha_2$ represent a very small displacement vector originating from $z_0$. The total derivative tells us how this small vector is transformed: $\text{Transformed Vector} = f'(z_0) \cdot \alpha$

Using the standard rules for the magnitude and argument of complex multiplication, we immediately get:

• The Scaling Factor: $|f'(z_0) \cdot \alpha| = |f'(z_0)| |\alpha|$. The length of our small vector $\alpha$ is uniformly scaled by the constant $|f'(z_0)|$, regardless of which direction $\alpha$ points.
• The Rotation Angle: $\arg(f'(z_0) \cdot \alpha) = \arg(f'(z_0)) + \arg(\alpha)$. The direction of our small vector $\alpha$ is rotated by the exact angle $\arg(f'(z_0))$.

Notice how complex arithmetic bypasses the need to expand long polynomials of partial derivatives. The geometric truth of the “amplitwist” falls out naturally: small circles centered at $z_0$ map to small circles centered at $f(z_0)$, scaled by the absolute value of the derivative, and rotated by its argument.