---
title: "Show 1/z⁴ is analytic except z = 0"
subject: "Applied Mathematics"
chapter: 1
key: "semester_4/applied_mathematics:analytic-1-over-z4"
source: https://www.bctengineer.com/q/semester_4/applied_mathematics/analytic-1-over-z4
site: BCT Engineering
---

# Show 1/z⁴ is analytic except z = 0

Applied Mathematics · Chapter 1: Complex Analysis

- Asked in: 2070 Magh, 5 marks
- Video: [#12 || Problem#4 ||  𝒇(𝒛)=𝒛^𝒏,  is analytic || Derivative || Complex Functions || 18MAT41||](https://www.youtube.com/watch?v=zXCYVGBiBMQ) — Math E Connect
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/applied_mathematics/analytic-1-over-z4
- Whole chapter: https://www.bctengineer.com/md/semester_4/applied_mathematics/chapter-1

## Question

Define analytic function. Show that the function f(z) = 1/z⁴ is analytic except at z = 0.

## Short answer

A function is analytic if differentiable in a neighbourhood. f(z)=1/z⁴ is analytic for z≠0, with f′(z)=−4/z⁵; it fails at z=0 because it is undefined.

## Answer

#### Analytic function
f(z) is **analytic** at a point if it is differentiable at that point and at all points in some neighbourhood of it.

#### Method 1: C–R equations in polar form
Let z = re^{iθ}. Then f(z) = z⁻⁴ = r⁻⁴ e^{−4iθ} = r⁻⁴ cos 4θ − i r⁻⁴ sin 4θ.

u = r⁻⁴ cos 4θ,  v = −r⁻⁴ sin 4θ.

```text
u_r = −4 r⁻⁵ cos 4θ        u_θ = −4 r⁻⁴ sin 4θ
v_r =  4 r⁻⁵ sin 4θ        v_θ = −4 r⁻⁴ cos 4θ
```

Check the polar C–R equations (valid for r ≠ 0):

- u_r = −4r⁻⁵ cos 4θ and (1/r) v_θ = −4r⁻⁵ cos 4θ. Equal ✓
- v_r = 4r⁻⁵ sin 4θ and −(1/r) u_θ = 4r⁻⁵ sin 4θ. Equal ✓

All partial derivatives are continuous for r ≠ 0, so the C–R equations are satisfied and sufficient conditions hold. Hence f is analytic at every z ≠ 0.

#### Method 2: direct differentiation
For z ≠ 0,

f′(z) = lim_{h→0} [ (z + h)⁻⁴ − z⁻⁴ ] / h = lim [ z⁴ − (z + h)⁴ ] / [ h z⁴ (z + h)⁴ ] = −4z³/(z⁴ · z⁴) = −4/z⁵.

The limit exists at every z ≠ 0.

At z = 0, f is not even defined, so it is not differentiable there.

**Conclusion: f(z) = 1/z⁴ is analytic everywhere except at z = 0, and f′(z) = −4/z⁵.**

## Check yourself

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

1. Analytic at a point means:
   A. Continuous
   B. Differentiable there and in a neighbourhood
   C. Real
   D. Defined
   Answer: B. Standard. (Key point: Definition)

2. z⁻⁴ in polar form has u =
   A. r⁻⁴ sin 4θ
   B. r⁴ cos 4θ
   C. r⁻⁴ cos 4θ
   D. r⁻⁴ cos θ
   Answer: C. z⁻⁴ = r⁻⁴ e^{−4iθ}. (Key point: Polar form)

3. f′(z) is:
   A. −4/z⁵
   B. −4z³
   C. 1/z³
   D. 4/z⁵
   Answer: A. Differentiate z⁻⁴. (Key point: Derivative)

4. At z=0 the function is not analytic because:
   A. u is zero
   B. It is not defined/differentiable there
   C. C–R fails everywhere
   D. f′ is zero
   Answer: B. 1/z⁴ blows up. (Key point: Origin)

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