---
title: "Complex Differentiability and the Cauchy-Riemann Equations"
description: "Demand a derivative independent of the direction of approach, and derive the two partial differential equations that demand forces."
canonical: https://lightmysky.com/learn/mathematics/complex-differentiability-and-the-cauchy-riemann-equations-mt_nZNmZcl9Ss
source: https://lightmysky.com/learn/mathematics/complex-differentiability-and-the-cauchy-riemann-equations-mt_nZNmZcl9Ss.md
retrieved: 2026-09-12
---

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# Complex Differentiability and the Cauchy-Riemann Equations

Demand a derivative independent of the direction of approach, and derive the two partial differential equations that demand forces.

Subject: Mathematics · Area: Complex Analysis · Ages 19 to 20
Page: https://lightmysky.com/learn/mathematics/complex-differentiability-and-the-cauchy-riemann-equations-mt_nZNmZcl9Ss

## Ready when they can

- Derive the Cauchy-Riemann equations by comparing approach along the two axes
- Show a specific function fails to be differentiable anywhere despite being smooth as a real map
- State the extra continuity condition that makes the equations sufficient

## Lesson: One derivative from every direction

Complex differentiability demands a derivative independent of the approach direction. Approach along the real axis gives u_x plus i v_x, while approach along the imaginary axis gives v_y minus i u_y. One derivative forces them equal, and that equality is the Cauchy-Riemann pair: u_x is v_y and u_y is negative v_x. Practice the partials until they are boring: for u is x squared minus y squared, u_x at (3, 1) is 6, and for v is 2xy, v_y at (3, 1) is 6.

**Example.** Smooth as a real map does not mean differentiable as a complex one. Conjugation splits into u is x and v is negative y, so u_x is 1 against v_y negative 1 everywhere: differentiable nowhere. The square modulus u is x squared plus y squared with v is 0 satisfies the pair only at the origin, so |z| squared is differentiable only at 0. By contrast z squared passes everywhere: at (1, 1), u_y and negative v_x are both negative 2.

The equations alone are only necessary. Continuity of the partials near the point upgrades them to sufficient, and that hypothesis cannot be dropped. Polynomials have continuous partials everywhere, which is why z squared passes without drama. Without continuity, the equations can hold at a point while differentiability fails.

**Tip.** Apply the full checklist per function: split into u and v, differentiate, test the pair, verify continuity. Conjugation fails at step three everywhere, and |z| squared passes step three only at 0. Methodical checking beats inspired guessing.

**Recap.** Equate the two axial approaches to get the equations, test them point by point, and check continuity before claiming differentiability.

## Practice

18 questions on this page, each with its working shown.

## Needs first

- [Complex Functions and the Complex Plane as a Domain](https://lightmysky.com/learn/mathematics/complex-functions-and-the-complex-plane-as-a-domain-mt_hMo851VTEs)
- [Partial Derivatives](https://lightmysky.com/learn/mathematics/partial-derivatives-mt_tu11fd9Xk9)

## Opens up

- [Contour Integrals Along Parametrised Paths](https://lightmysky.com/learn/mathematics/contour-integrals-along-parametrised-paths-mt_CaJ7ruAv8P)
- [Harmonic Functions and What Analyticity Forces](https://lightmysky.com/learn/mathematics/harmonic-functions-and-what-analyticity-forces-mt_jqajP39XI4)
- [Conformal Maps and Mobius Transformations](https://lightmysky.com/learn/mathematics/conformal-maps-and-mobius-transformations-mt_WtHCbAT4GI)