Cauchy-Riemann equations in polar form

Analytic function

f(z) is analytic at z₀ if it is differentiable at z₀ and at every point of some neighbourhood of z₀.

C–R equations in polar form

Let z = re^{iθ} and f(z) = u(r, θ) + i v(r, θ). Then the C–R equations are

∂u/∂r = (1/r) ∂v/∂θ and ∂v/∂r = −(1/r) ∂u/∂θ.

Derivation

Use f′(z) = lim [f(z + δz) − f(z)]/δz and approach z in two ways. Here z = re^{iθ}.

Radial direction (θ fixed, r changes by δr): δz = e^{iθ} δr, so

f′(z) = (∂f/∂r)/e^{iθ} = e^{−iθ} (u_r + i v_r). ... (1)

Angular direction (r fixed, θ changes by δθ): δz = i r e^{iθ} δθ, so

f′(z) = (∂f/∂θ)/(i r e^{iθ}) = (e^{−iθ}/(ir)) (u_θ + i v_θ) = (e^{−iθ}/r)(v_θ − i u_θ). ... (2)

Equate (1) and (2) and cancel e^{−iθ}:

u_r + i v_r = (1/r) v_θ − i (1/r) u_θ

Equating real and imaginary parts:

u_r = (1/r) v_θ
v_r = −(1/r) u_θ

Link with the Cartesian form

Using x = r cos θ, y = r sin θ and u_x = v_y, u_y = −v_x:

• v_θ = −r sin θ · v_x + r cos θ · v_y = r (u_y sin θ + u_x cos θ) = r u_r
• u_θ = −r sin θ · u_x + r cos θ · u_y = −r (v_y sin θ + v_x cos θ) = −r v_r

This gives the same polar equations. Also f′(z) = e^{−iθ}(u_r + i v_r).