---
title: "Find a bilinear transformation"
subject: "Mathematics III"
chapter: 2
key: "semester_3/mathematics_iii:find-a-bilinear-transformation-two"
source: https://www.bctengineer.com/q/semester_3/mathematics_iii/find-a-bilinear-transformation-two
site: BCT Engineering
---

# Find a bilinear transformation

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

- Asked in: 2082 Kartik (new course), 4 marks
- Video: [#3 || Problem #1 || Bilinear Transformation || z=1, i, -1 and w=i, 0, -i || 18MAT41 ||](https://www.youtube.com/watch?v=-BWPDj5xH3E) — Math E Connect
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/mathematics_iii/find-a-bilinear-transformation-two
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-2

## Question

Find the bilinear transform that maps the points 1, i, −1 in the z-plane into the points 0, 1, ∞ into the w plane.

## Short answer

Zero at z=1 and pole at z=-1 give w = k(z-1)/(z+1). Substituting z=i, w=1 yields k=-i, so w = -i(z-1)/(z+1) = i(1-z)/(1+z).

## Answer

Let the transformation be w = (az + b)/(cz + d).

- z = 1 maps to w = 0, so the numerator has a factor (z − 1).
- z = −1 maps to w = ∞, so the denominator has a factor (z + 1).

Hence w = k (z − 1)/(z + 1).

Use z = i → w = 1:

```text
1 = k (i − 1)/(i + 1)
(i − 1)/(i + 1) = (i − 1)(1 − i)/((1 + i)(1 − i)) = (i − i² − 1 + i)/2 = 2i/2 = i
so k·i = 1  → k = 1/i = −i
```

```text
w = −i (z − 1)/(z + 1) = i (1 − z)/(1 + z)
```

**Answer: w = i(1 − z)/(1 + z).** Check: z = 1 → 0; z = −1 → ∞; z = i → i(1 − i)/(1 + i) = i(−i) = 1 ✓.

## Check yourself

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

1. z=1 -> w=0 gives numerator factor:
   A. (z + 1)
   B. (z - i)
   C. (z - 1)
   D. (z - 0)
   Answer: C. Zero of w. (Key point: Zero)

2. z=-1 -> w=infinity gives denominator factor:
   A. (z - 1)
   B. (z + i)
   C. (z - i)
   D. (z + 1)
   Answer: D. Pole of w. (Key point: Pole)

3. From z=i -> w=1, k =
   A. -i
   B. i
   C. 1
   D. -1
   Answer: A. (i-1)/(i+1) = i so k = 1/i = -i. (Key point: Constant)

4. The transformation is:
   A. w = i(1+z)/(1-z)
   B. w = i(1 - z)/(1 + z)
   C. w = (z-1)/(z+1)
   D. w = -i(1-z)/(1+z)
   Answer: B. -i(z-1)/(z+1) = i(1-z)/(1+z). (Key point: Result)

## Related in Mathematics III

- [Find a bilinear transformation](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Afind-a-bilinear-transformation) — Mathematics III, chapter 2

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