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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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