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Conditions for analytic; log z analytic

Short answer

Necessary conditions are Cauchy-Riemann equations and continuity of partial derivatives. f(z)=log z with u=ln r, v=θ satisfies these for r>0, so analytic except at origin (and branch cut); f′(z)=1/z.

State necessary conditions for a function f(z) to be analytic. Show that the function f(z) = log z is analytic everywhere except at the origin.

Necessary conditions

If f(z) = u(x, y) + i v(x, y) is analytic in a region, then in that region u and v satisfy the Cauchy–Riemann equations:

∂u/∂x = ∂v/∂y, ∂u/∂y = −∂v/∂x

(in polar form: u_r = (1/r) v_θ, v_r = −(1/r) u_θ). Also, u and v are harmonic. If, in addition, the first partial derivatives are continuous, the C–R equations are sufficient.

log z is analytic except at the origin

Let z = re^{iθ}. The principal value is

log z = ln r + iθ, −π < θ ≤ π (or any single branch of width 2π).

So u = ln r and v = θ, defined for r > 0.

u_r = 1/r      u_θ = 0
v_r = 0        v_θ = 1
  • u_r = 1/r = (1/r) v_θ ✓
  • v_r = 0 = −(1/r) u_θ ✓

The partial derivatives are continuous for r > 0. So the C–R equations hold and f is analytic at every z ≠ 0 (excluding points on the branch cut, along the negative real axis, for the principal branch).

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

At z = 0, ln r is undefined (r → 0 gives −∞) and arg z is undefined, so log z is neither defined nor differentiable there.

Conclusion: log z is analytic everywhere except at the origin (and on the branch cut), with derivative 1/z.

✓ Answer checked by Claude Opus 5.5 (high effort)

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