---
title: "Taylor series of 1/(1-z) about z = 3i"
subject: "Applied Mathematics"
chapter: 1
key: "semester_4/applied_mathematics:taylor-series-1-over-1-z-about-3i"
source: https://www.bctengineer.com/q/semester_4/applied_mathematics/taylor-series-1-over-1-z-about-3i
site: BCT Engineering
---

# Taylor series of 1/(1-z) about z = 3i

Applied Mathematics · Chapter 1: Complex Analysis

- Asked in: 2071 Bhadra
- Video: [Expand in Taylor's Series the function f(z) = 1/z](https://www.youtube.com/watch?v=5wHhr4jmmjU) — Spectrum of Mathematics
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/applied_mathematics/taylor-series-1-over-1-z-about-3i
- Whole chapter: https://www.bctengineer.com/md/semester_4/applied_mathematics/chapter-1

## Question

Find the Taylor's series of f(z) = 1/(1 - z) about z = 3i.

## Short answer

f(z) = Σ_{n=0}^∞ (z − 3i)^n / (1 − 3i)^{n+1}, valid for |z − 3i| < √10. The first terms are (1+3i)/10 + ((1+3i)/10)^2 (z−3i) + ….

## Answer

**Taylor series** about z = a: f(z) = Σ aₙ (z − a)ⁿ with aₙ = f⁽ⁿ⁾(a)/n!.

f(z) = 1/(1 − z) has its only singularity at z = 1. The distance from the centre 3i to 1 is |1 − 3i| = √10, so the series is valid for |z − 3i| < √10.

#### Expansion
Let t = z − 3i, so z = 3i + t and

1 − z = 1 − 3i − t = (1 − 3i)[1 − t/(1 − 3i)].

```text
f(z) = 1/(1 − 3i) · 1/[1 − t/(1 − 3i)]
     = 1/(1 − 3i) · Σ_{n=0}^{∞} [t/(1 − 3i)]ⁿ        (|t| < |1 − 3i| = √10)
     = Σ_{n=0}^{∞} (z − 3i)ⁿ / (1 − 3i)ⁿ⁺¹
```

**Answer: 1/(1 − z) = Σ_{n=0}^{∞} (z − 3i)ⁿ/(1 − 3i)ⁿ⁺¹,  |z − 3i| < √10.**

Since 1/(1 − 3i) = (1 + 3i)/10, the first terms are

f(z) = (1 + 3i)/10 + [(1 + 3i)/10]² (z − 3i) + [(1 + 3i)/10]³ (z − 3i)² + …

(Check: f(3i) = 1/(1 − 3i) = (1 + 3i)/10 ✓.)

## Check yourself

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

1. Why is the series about 3i valid for |z-3i| < √10?
   A. The singularity z = 1 is at distance 1
   B. The singularity is at 3i itself
   C. The distance to z=1 is |1+3i| = 10
   D. The only singularity z = 1 is at distance |1-3i| = √10 from 3i
   Answer: D. R is the distance from the centre to the nearest singularity, which is z = 1. (Key point: Radius)

2. How is 1-z rewritten with t = z-3i so a geometric series applies?
   A. 1 - z = (1-3i)[1 - t/(1-3i)]
   B. 1 - z = (1-3i)[1 + t]
   C. 1 - z = 1 - t
   D. 1 - z = (1+3i)[1 - t/(1+3i)]
   Answer: A. 1 - z = 1 - 3i - t = (1-3i)(1 - t/(1-3i)). (Key point: Key rewrite)

3. What is the general term of the series?
   A. (z-3i)^n / (1-3i)^n
   B. (z-3i)^n / (1-3i)^{n+1}
   C. (z-3i)^n (1-3i)^{n+1}
   D. (z-3i)^n / (1+3i)^{n+1}
   Answer: B. Summing (1/(1-3i)) [t/(1-3i)]^n gives t^n/(1-3i)^{n+1}. (Key point: Result)

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