Complex Differentiability and the Cauchy-Riemann Equations
Define the complex derivative as a limit over all approach directions, derive the Cauchy-Riemann equations, and state the continuity condition that makes them sufficient.
What a learner can do afterwards
- Define the complex difference quotient, derive the Cauchy-Riemann equations, and recover the derivative when differentiability is established
- Distinguish differentiability sets from holomorphic neighbourhoods, including real-smooth failures
- State the sufficient continuity hypothesis precisely and show counterexamples where CR holds without differentiability
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The complex derivative of f at z0 is the limit of [f(z0 + h) - f(z0)] / h as h tends to 0 through nonzero complex values, with f defined on an open neighbourhood of z0. The limit must be the same for every way h can approach 0, not just along the real or imaginary axis. This all-path demand is what makes complex differentiability stronger than real differentiability of the two component functions.
Derive the Cauchy-Riemann equations. Write f = u + iv and let h = t be real. The quotient tends to u_x + i v_x. Now let h = ik be imaginary. The quotient tends to (u_y + i v_y) / i, which is v_y - i u_y. Equating real and imaginary parts gives u_x = v_y and u_y = -v_x. Once differentiability is established, f'(z0) = u_x + i v_x.
The equations are necessary but not sufficient by themselves. The sufficient hypothesis adds that the four partial derivatives exist in a neighbourhood of z0 and are continuous at z0, together with the equations holding at z0. Failure of the equations proves non-differentiability. Failure of continuity alone does not prove non-differentiability; a direct difference quotient can still settle the question.
For z squared, u = x² - y² and v = 2xy. The four partials are u_x = 2x, u_y = -2y, v_x = 2y, v_y = 2x. Both equations hold everywhere. Polynomial partials are continuous, so f'(z) = 2x + i 2y = 2z.
For f(z) = conj(z)²/z for z ≠ 0 and f(0) = 0, real increments give f(t) = t, so u_x = 1 and v_x = 0. Imaginary increments give f(it) = it, so u_y = 0 and v_y = 1. Both CR equations hold: u_x = v_y = 1 and u_y = -v_x = 0. The quotient Q(h) = (conj(h)/h)² equals 1 on real h and -1 on h = t(1+i). Two paths disagree, so no derivative at 0. Also |f(z)| = |z|, so f is continuous at 0 despite failing differentiability.
An open neighbourhood of a point contains a disk of positive radius around it. A function is holomorphic on an open set if differentiable at every point of that set. Holomorphic at a point means differentiable throughout some open neighbourhood of it. The conjugate function has u = x and v = -y, so u_x = 1 and v_y = -1 everywhere: nowhere differentiable, hence nowhere holomorphic.
The function |z| squared satisfies the equations only at 0. At 0 the quotient |h| squared over h has modulus |h|, which tends to 0, so the derivative is 0 at 0 only. It is nowhere holomorphic.
Define the derivative over all paths, derive the equations by equating axial approaches, add continuous partials for sufficiency, and distinguish pointwise from holomorphic.
All ten extra tasks and three readings are original authorship created for this topic. The previous enrichment carried unsupported CC0 credits on custom summaries that overstated source scope; those are replaced with original null-license entries. The ten PG records and three textbook records were read as design studies only. The counterexample f(z) = conj(z)²/z and the all-path bound |h|²/h are standard mathematical examples, not source text.
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