---
title: "Taylor series of 1/(z²+4) about z = i"
subject: "Applied Mathematics"
chapter: 1
key: "semester_4/applied_mathematics:taylor-series-1-over-z2-plus-4-about-i"
source: https://www.bctengineer.com/q/semester_4/applied_mathematics/taylor-series-1-over-z2-plus-4-about-i
site: BCT Engineering
---

# Taylor series of 1/(z²+4) about z = i

Applied Mathematics · Chapter 1: Complex Analysis

- Asked in: 2073 Magh, 1+4 marks; 2070 Bhadra, 5 marks; 2070 Magh, 5 marks; 2069 Bhadra (asked 4 times)
- Video: [19. Taylor's and Laurent's Series | Problem#1 | Most Important | Complete Concept](https://www.youtube.com/watch?v=OZAZImytcXE) — MKS TUTORIALS by Manoj Sir
- 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-z2-plus-4-about-i
- Whole chapter: https://www.bctengineer.com/md/semester_4/applied_mathematics/chapter-1

## Question

Find the Taylor's series expansion of f(z) = 1/(z² + 4) about the point z = i.

## Short answer

Use partial fractions 1/(z²+4)=1/(4i)[1/(z−2i)−1/(z+2i)] and set t=z−i. The Taylor series valid for |z−i|<1 is Σ (1/4)[(−i)ⁿ+iⁿ/3^(n+1)](z−i)ⁿ.

## Answer

**Taylor's series** of f(z) about z = a (f analytic in |z − a| < R):  f(z) = Σ aₙ(z − a)ⁿ, with aₙ = f⁽ⁿ⁾(a)/n!.

Here f(z) = 1/(z² + 4) has poles at z = ±2i. The nearest one to z = i is 2i, at distance 1. So the series is valid for |z − i| < 1.

#### Partial fractions
1/(z² + 4) = (1/4i) [1/(z − 2i) − 1/(z + 2i)]

Let t = z − i, so z = i + t.

- z − 2i = t − i = −i(1 + it), so 1/(z − 2i) = i · 1/(1 + it) = i Σ (−it)ⁿ = i Σ (−i)ⁿ tⁿ  (valid for |t| < 1)
- z + 2i = t + 3i = 3i(1 − it/3), so 1/(z + 2i) = (−i/3) · 1/(1 − it/3) = (−i/3) Σ (i/3)ⁿ tⁿ  (valid for |t| < 3)

#### Combine
```text
f(z) = (1/4i) [ i Σ (−i)ⁿ tⁿ + (i/3) Σ (i/3)ⁿ tⁿ ]
     = (1/4) Σ (−i)ⁿ tⁿ + (1/12) Σ (i/3)ⁿ tⁿ
```

**Answer:**

f(z) = Σ_{n=0}^{∞} (1/4) [ (−i)ⁿ + iⁿ/3ⁿ⁺¹ ] (z − i)ⁿ,  |z − i| < 1.

#### First few terms
```text
a₀ = (1/4)(1 + 1/3)        = 1/3
a₁ = (1/4)(−i + i/9)       = −2i/9
a₂ = (1/4)(−1 − 1/27)      = −7/27
a₃ = (1/4)(i − i/81)       = 20i/81
```
f(z) = 1/3 − (2i/9)(z − i) − (7/27)(z − i)² + (20i/81)(z − i)³ + …

Check: f(i) = 1/(−1 + 4) = 1/3, and f′(i) = −2i/(3)² = −2i/9, which match a₀ and a₁.

## Check yourself

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

1. Why is the Taylor series of 1/(z²+4) about z = i valid for |z − i| < 1?
   A. The nearest singularity 2i is at distance 1 from i
   B. The function is entire
   C. Singularities are at ±i
   D. The nearest singularity is at 0
   Answer: A. Poles at ±2i; 2i is distance 1 from i (−2i is distance 3). (Key point: Radius)

2. 1/(z²+4) is split as:
   A. (1/4i)[1/(z − 2) − 1/(z + 2)]
   B. (1/2i)[1/(z − 2i) − 1/(z + 2i)]
   C. (1/4i)[1/(z − 2i) − 1/(z + 2i)]
   D. (1/4)[1/(z − 2i) + 1/(z + 2i)]
   Answer: C. 1/(z−2i) − 1/(z+2i) = 4i/(z²+4), so the factor is 1/4i. (Key point: Partial fractions)

3. What substitution is used to expand about z = i?
   A. t = z + i
   B. t = z/i
   C. t = z − 2i
   D. t = z − i
   Answer: D. Writing z = i + t gives powers of (z − i). (Key point: Shift)

4. What are a0 and a1 of the series?
   A. a0 = 1/3, a1 = 2i/9
   B. a0 = 3, a1 = −2i
   C. a0 = 1/3, a1 = −2i/9
   D. a0 = 1/4, a1 = −i/4
   Answer: C. a0 = f(i) = 1/3 and a1 = f′(i) = −2i/9. (Key point: Coefficients)

## Close variants set in other exams

- [Taylor series of (2z³+1)/(z²+z) about z = 1](https://www.bctengineer.com/question.md?key=semester_4%2Fapplied_mathematics%3Ataylor-series-2z3-about-1) — Applied Mathematics, chapter 1, asked 5×

## Related in Applied Mathematics

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- [State and prove Taylor's series](https://www.bctengineer.com/question.md?key=semester_4%2Fapplied_mathematics%3Ataylor-series-prove) — Applied Mathematics, chapter 1, asked 2×
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