The Cauchy–Riemann Equations

The complex derivative exists only when the difference quotient gives the same limit from every direction. Demanding agreement between just two directions — along the real axis and along the imaginary axis — already pins down a remarkable pair of equations. They are the Cauchy–Riemann equations, and they are the algebraic heart of the whole subject.

Write f(z) = u(x, y) + i\,v(x, y) as usual, with z = x + iy. We will compute f'(z) two ways and force the answers to match. Both ways use partial derivatives of u and v.

Deriving the equations

Step 1 — approach along the real axis. Let h = \Delta x \to 0. Only x changes, so the quotient becomes a partial derivative in x:

f'(z) = \frac{\partial u}{\partial x} + i\,\frac{\partial v}{\partial x} = u_x + i\,v_x.

Step 2 — approach along the imaginary axis. Let h = i\,\Delta y \to 0. Now only y changes, and dividing by h = i\,\Delta y introduces a 1/i:

f'(z) = \frac{1}{i}\left(\frac{\partial u}{\partial y} + i\,\frac{\partial v}{\partial y}\right) = \frac{u_y + i\,v_y}{i}.

Step 3 — simplify using 1/i = -i. Multiply through by -i:

f'(z) = -i\,(u_y + i\,v_y) = v_y - i\,u_y.

Step 4 — set the two expressions equal. For the derivative to exist, the real-axis answer and the imaginary-axis answer must agree:

u_x + i\,v_x = v_y - i\,u_y.

Step 5 — match real and imaginary parts. Two complex numbers are equal exactly when their real parts match and their imaginary parts match:

\underbrace{u_x = v_y}_{\text{real parts}}, \qquad \underbrace{v_x = -u_y}_{\text{imaginary parts}}.

These are the Cauchy–Riemann equations. Whenever a complex function is differentiable, its real and imaginary parts are locked together by them.

Two worked checks

The squaring map passes. For f(z) = z^2 we found u = x^2 - y^2 and v = 2xy. Differentiate:

u_x = 2x, \quad v_y = 2x, \qquad u_y = -2y, \quad v_x = 2y.

Indeed u_x = v_y (both 2x) and v_x = -u_y (since 2y = -(-2y)). Cauchy–Riemann holds at every point — consistent with f'(z) = 2z.

Conjugation fails. For f(z) = \bar{z} = x - iy we have u = x and v = -y:

u_x = 1, \qquad v_y = -1.

Since u_x = 1 \ne -1 = v_y, the first Cauchy–Riemann equation is violated everywhere — confirming, with no limits at all, that \bar{z} is nowhere differentiable.

For f(z) = u(x, y) + i\,v(x, y) to be complex-differentiable:

It is striking how little we assumed. We only required that two of the infinitely many approach directions — the real and imaginary axes — give the same derivative. That alone forces the Cauchy–Riemann equations. With the mild extra hypothesis that the partials are continuous, those same equations turn out to be sufficient: any (u, v) obeying them, with continuous partials, is genuinely complex-differentiable. So Cauchy–Riemann is not just a necessary symptom of differentiability — it very nearly is differentiability.

Check Cauchy–Riemann live

Pick a function with the function control — z^2 or \bar{z} — and move the point with the x and y sliders. The diagram computes u_x, v_y, u_y, and -v_x at that point and flags whether the two equations u_x = v_y and u_y = -v_x hold (green) or not (red). The squaring map is always green; conjugation is always red.