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Derive Cauchy-Riemann equations (Cartesian)
Short answer
An analytic function is differentiable in a neighborhood of each point. Differentiating f(z)=u+iv along the real and imaginary axes and equating the two expressions for f′(z) gives u_x=v_y and u_y=−v_x, the Cauchy–Riemann equations in Cartesian form.
Define an analytic function of a complex variable. Derive the Cauchy-Riemann equations in Cartesian form for f(z) = u + iv.
Analytic function
A function f(z) is analytic at a point z₀ if it is differentiable at z₀ and at every point in some neighbourhood of z₀. It is analytic in a region if it is differentiable at every point of that region.
Cauchy–Riemann equations
Theorem. If f(z) = u(x, y) + i v(x, y) is analytic in a region, then at every point of it
∂u/∂x = ∂v/∂y and ∂u/∂y = −∂v/∂x.
Derivation
Since f is differentiable at z = x + iy, the limit
f′(z) = lim_{δz→0} [f(z + δz) − f(z)]/δz
exists and is the same along every path by which δz → 0. Here δz = δx + i δy.
Path 1: along the real axis (δy = 0, so δz = δx → 0):
f′(z) = lim [u(x + δx, y) − u(x, y)]/δx + i [v(x + δx, y) − v(x, y)]/δx = ∂u/∂x + i ∂v/∂x. ... (1)
Path 2: along the imaginary axis (δx = 0, so δz = i δy → 0):
f′(z) = lim [u(x, y + δy) − u(x, y)]/(i δy) + i [v(x, y + δy) − v(x, y)]/(i δy) = (1/i) ∂u/∂y + ∂v/∂y = ∂v/∂y − i ∂u/∂y. ... (2)
The two values must be equal. Equate real and imaginary parts of (1) and (2):
∂u/∂x = ∂v/∂y
∂v/∂x = −∂u/∂y
These are the Cauchy–Riemann (C–R) equations. ∎
Also f′(z) = u_x + i v_x = v_y − i u_y.
Sufficient condition. If u, v, and their first partial derivatives are continuous in a region and satisfy the C–R equations there, then f(z) = u + iv is analytic in that region.
✓ Answer checked by Claude Opus 5.5 (high effort)
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