# The Cauchy–Riemann Equations

*The local differential characterisation of holomorphy, why it forces the real and imaginary parts to be harmonic, and how the harmonic-conjugate relationship prefigures the Kramers–Kronig integral.*

## Derivation

Write `f(z) = u(x, y) + i v(x, y)` where `z = x + iy`. Suppose f is
complex-differentiable at z₀. The derivative

```
f′(z₀) = lim_{h → 0}  [f(z₀ + h) − f(z₀)] / h
```

must give the same value regardless of the direction of h. Take two
specific directions.

**Along the real axis**: `h = Δx`, real.

```
f′(z₀) = lim_{Δx → 0}  [u(x₀+Δx, y₀) − u(x₀, y₀) + i (v(x₀+Δx, y₀) − v(x₀, y₀))] / Δx
        = ∂u/∂x + i ∂v/∂x
```

**Along the imaginary axis**: `h = i Δy`.

```
f′(z₀) = lim_{Δy → 0}  [u(x₀, y₀+Δy) − u(x₀, y₀) + i (v(x₀, y₀+Δy) − v(x₀, y₀))] / (i Δy)
        = −i ∂u/∂y + ∂v/∂y
        = ∂v/∂y − i ∂u/∂y
```

Equating the two expressions for f′(z₀) and matching real and imaginary
parts gives the **Cauchy–Riemann (CR) equations**:

```
∂u/∂x =  ∂v/∂y        (CR-1)
∂u/∂y = −∂v/∂x        (CR-2)
```

The converse also holds with a mild regularity assumption (Looman–
Menchoff; for us it suffices that u, v are C¹): if u, v have continuous
first partials satisfying CR on an open set, then f = u + iv is
holomorphic there.

## Immediate consequences

**Formula for the derivative.** Any of four equivalent forms,

```
f′(z) = ∂u/∂x + i ∂v/∂x
      = ∂v/∂y − i ∂u/∂y
      = ∂u/∂x − i ∂u/∂y
      = ∂v/∂y + i ∂v/∂x
```

which is often useful because it reduces computing f′ to real-variable
partial differentiation.

**Conformality.** The Jacobian of (x, y) ↦ (u, v) under CR is

```
J = | ∂u/∂x  ∂u/∂y | = (∂u/∂x)² + (∂u/∂y)² = |f′(z)|²
    | ∂v/∂x  ∂v/∂y |
```

so wherever f′(z) ≠ 0, the map is locally a rotation-plus-scaling — it
preserves angles. Holomorphic maps are **conformal** away from their
critical points.

**Wirtinger form.** Define

```
∂/∂z   = ½ (∂/∂x − i ∂/∂y)
∂/∂z̄  = ½ (∂/∂x + i ∂/∂y)
```

Then CR is equivalent to the single compact equation

```
∂f/∂z̄ = 0
```

— holomorphic functions are those that depend on z alone and not on z̄.

## Harmonic real and imaginary parts

Differentiate CR-1 with respect to x and CR-2 with respect to y, and
add:

```
∂²u/∂x² + ∂²u/∂y² =  ∂²v/∂x∂y − ∂²v/∂y∂x = 0
```

Similarly, differentiating CR-1 with respect to y and CR-2 with respect
to x and subtracting gives

```
∂²v/∂x² + ∂²v/∂y² = 0
```

So both u and v satisfy **Laplace's equation** `∇² φ = 0` — they are
**harmonic** functions. This is the first major consequence of
holomorphy: the two scalar fields that make up a holomorphic function
are not arbitrary smooth fields, they are solutions of a second-order
elliptic PDE.

Equivalently, every harmonic function on a simply connected domain is
locally the real part of some holomorphic function.

## Harmonic conjugates

Given a harmonic u on a simply connected domain D, a **harmonic
conjugate** v is a harmonic function such that f = u + iv is holomorphic
on D. The CR equations give v by line integration:

```
v(x, y) = ∫  (−∂u/∂y dx + ∂u/∂x dy) + constant
```

The integral is path-independent on a simply connected D precisely
because u is harmonic (the integrand is a closed 1-form). v is unique up
to an additive real constant.

So: **u determines v up to a constant**, and vice versa. Fix the value
of v at one point, and the entire imaginary part of a holomorphic
function is locked by its real part.

This is already a mini-version of K–K. The K–K integrals are the
**global, boundary-value** analogue: if u and v are the boundary values
of a harmonic conjugate pair on the upper half-plane that decays at
infinity, then u and v on the real axis are related by the Hilbert
transform

```
v(x) = (1/π) P ∫_{−∞}^{∞}  u(x′) / (x′ − x) dx′
```

Chapters 11–12 derive exactly this from Cauchy's integral formula.

## Worked example 1: f(z) = z²

Here `u(x, y) = x² − y²`, `v(x, y) = 2xy`.

```
∂u/∂x = 2x,   ∂v/∂y = 2x          ✓ CR-1
∂u/∂y = −2y, ∂v/∂x = 2y,  −∂v/∂x = −2y  ✓ CR-2
```

Harmonicity:

```
∂²u/∂x² + ∂²u/∂y² = 2 − 2 = 0   ✓
∂²v/∂x² + ∂²v/∂y² = 0 + 0 = 0   ✓
```

Derivative: `f′(z) = ∂u/∂x + i ∂v/∂x = 2x + 2iy = 2z` as expected.

## Worked example 2: the Lorentz susceptibility on the real axis

Take the single-resonance susceptibility

```
χ(ω) = ω_p² / (ω₀² − ω² − iγω)
```

with γ > 0. Multiply numerator and denominator by the conjugate:

```
χ(ω) = ω_p² (ω₀² − ω² + iγω) / [(ω₀² − ω²)² + γ² ω²]
```

so

```
χ′(ω) = ω_p² (ω₀² − ω²) / [(ω₀² − ω²)² + γ² ω²]
χ″(ω) = ω_p² γ ω         / [(ω₀² − ω²)² + γ² ω²]
```

χ(ω) extended to complex ω is meromorphic with poles in the lower
half-plane, hence analytic in the UHP. CR are satisfied everywhere
χ is defined; χ′ and χ″ are a harmonic conjugate pair on the UHP whose
boundary values on the real axis are the two curves above. The K–K
relations will recover χ′ from χ″ (and vice versa) by a contour
integral.

## Where CR stop being enough

CR is a **local** condition. Holomorphy, once promoted by Cauchy's
theorem to a global condition, yields the integral formula and thence
the analytic-continuation machinery that K–K exploits. The next chapter
builds the integral side.

## See also

- [03_complex_functions_analyticity.md](03_complex_functions_analyticity.md)
- [05_contour_integration.md](05_contour_integration.md)
- [06_cauchy_theorems.md](06_cauchy_theorems.md)