---
title: "Taylor series expansion"
subject: "Mathematics III"
chapter: 2
key: "semester_3/mathematics_iii:taylor-series-expansion-three"
source: https://www.bctengineer.com/q/semester_3/mathematics_iii/taylor-series-expansion-three
site: BCT Engineering
---

# Taylor series expansion

Mathematics III · Chapter 2: Line, Surface and Volume Integrals

- Asked in: 2081 Chaitra (new course), 2 marks
- Video: [COMPLEX INTEGRATION | LECTURE 05 | Taylor Series | PRADEEP GIRI SIR](https://www.youtube.com/watch?v=6wo62OVQBzQ) — Pradeep Giri Academy
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/mathematics_iii/taylor-series-expansion-three
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-2

## Question

Expand the function f(z) = 1/z about z = i in Taylor's series.

## Short answer

Let w=z-i. 1/z = 1/(i+w) = -i/(1-iw) = -i Σ_{n=0}^∞ (iw)^n = -i Σ_{n=0}^∞ i^n (z-i)^n, converging for |z-i|<1.

## Answer

Let w = z − i, so z = i + w. The only singularity of 1/z is z = 0, at distance 1 from i. So the series converges for |z − i| < 1.

```text
1/z = 1/(i + w) = (1/i) · 1/(1 + w/i)
    = −i · 1/(1 − i w)          [since 1/i = −i and w/i = −iw]
    = −i Σ_{n=0}^∞ (i w)ⁿ        for |w| < 1
```

**Answer:**

f(z) = −i Σ_{n=0}^∞ iⁿ (z − i)ⁿ = −i + (z − i) + i (z − i)² − (z − i)³ − ..., for |z − i| < 1.

Check: f'(i) = −1/i² = 1, which matches the coefficient of (z − i).

## Check yourself

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

1. The series about i converges for:
   A. |z - i| < 1
   B. |z-i|<2
   C. |z|<1
   D. all z
   Answer: A. Singularity at 0 is distance 1 from i. (Key point: Radius)

2. 1/(i+w) = 
   A. i/(1 - iw)
   B. -i/(1 - i w)
   C. -i/(1+iw)
   D. 1/(1 - w)
   Answer: B. 1/i = -i and w/i = -iw. (Key point: Rewrite)

3. Coefficient of (z-i)^0 is:
   A. i
   B. 1
   C. -i
   D. -1
   Answer: C. f(i) = 1/i = -i. (Key point: Coefficient)

## Related in Mathematics III

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