---
title: "Conditions for analytic; log z analytic"
subject: "Applied Mathematics"
chapter: 1
key: "semester_4/applied_mathematics:analytic-log-z"
source: https://www.bctengineer.com/q/semester_4/applied_mathematics/analytic-log-z
site: BCT Engineering
---

# Conditions for analytic; log z analytic

Applied Mathematics · Chapter 1: Complex Analysis

- Asked in: 2068 Bhadra
- Video: [How to verify log z is analytic function || Complex Analysis || 18mat41 || Dr Prashant Patil](https://www.youtube.com/watch?v=ZdJp7zSuAYA) — Dr Prashant Patil
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/applied_mathematics/analytic-log-z
- Whole chapter: https://www.bctengineer.com/md/semester_4/applied_mathematics/chapter-1

## Question

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.

## 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.

## Answer

#### 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.

```text
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.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Necessary condition for analyticity:
   A. C–R equations hold
   B. f is real
   C. u and v are constant
   D. f is bounded
   Answer: A. C–R. (Key point: Necessary)

2. For log z = ln r + iθ, u and v are:
   A. u = r, v = θ
   B. u = ln r, v = θ
   C. u = θ, v = ln r
   D. u = ln θ, v = r
   Answer: B. By definition. (Key point: Components)

3. The polar C–R check u_r = (1/r) v_θ gives:
   A. 1/r = 1/r
   B. 0 = 1
   C. r = θ
   D. 1 = 1/r
   Answer: A. u_r = 1/r, v_θ = 1. (Key point: Check)

4. f′(z) =
   A. 1/z
   B. log z
   C. z
   D. e^z
   Answer: A. e^{−iθ}/r = 1/z. (Key point: Derivative)

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