---
title: "Find v from u = (x-1)³ - 3xy² + 3y²"
subject: "Applied Mathematics"
chapter: 1
key: "semester_4/applied_mathematics:analytic-function-x-1-cubed"
source: https://www.bctengineer.com/q/semester_4/applied_mathematics/analytic-function-x-1-cubed
site: BCT Engineering
---

# Find v from u = (x-1)³ - 3xy² + 3y²

Applied Mathematics · Chapter 1: Complex Analysis

- Asked in: 2072 Asoj, 5 marks
- Video: [Find v if u=(x-1)^3-3xy^2+3y^2 | Harmonic conjugate | complex analysis engineering mathematics, bsc](https://www.youtube.com/watch?v=b-bPEDzHmZw) — Swati Theng Mathematics
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/applied_mathematics/analytic-function-x-1-cubed
- Whole chapter: https://www.bctengineer.com/md/semester_4/applied_mathematics/chapter-1

## Question

If u = (x-1)³ - 3xy² + 3y², determine V so that u + iv is an analytic function of x + iy.

## Short answer

v=3(x−1)²y−y³+constant, making f(z)=(z−1)³+c.

## Answer

u = (x − 1)³ − 3xy² + 3y².

#### Check that u is harmonic
```text
u_x  = 3(x − 1)² − 3y²     u_xx = 6(x − 1)
u_y  = −6xy + 6y           u_yy = −6x + 6
```

u_xx + u_yy = 6x − 6 − 6x + 6 = 0 ✓.

#### Find v by C–R equations (v_y = u_x, v_x = −u_y)
Integrate v_y = 3(x − 1)² − 3y² with respect to y:

v = 3(x − 1)² y − y³ + g(x)

Differentiate with respect to x: v_x = 6(x − 1)y + g′(x).

This must equal −u_y = 6xy − 6y = 6(x − 1)y. So g′(x) = 0 and g = constant.

**Answer: v = 3(x − 1)² y − y³ + c**

#### The analytic function
f′(z) = u_x(z, 0) − i u_y(z, 0) = 3(z − 1)², so f(z) = (z − 1)³ + c.

Check: (z − 1)³ = (x − 1 + iy)³ has real part (x − 1)³ − 3(x − 1)y² = (x − 1)³ − 3xy² + 3y² = u ✓, and imaginary part 3(x − 1)²y − y³ = v ✓.

## Check yourself

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

1. Which relation gives v_y?
   A. v_y = u_y
   B. v_y = −u_y
   C. v_y = −u_x
   D. v_y = u_x
   Answer: D. C–R equations. (Key point: C–R)

2. Integrating v_y = 3(x−1)² − 3y² gives:
   A. v = 6(x−1)y
   B. v = (x−1)³ − y³
   C. v = 3(x−1)²y − y³ + g(x)
   D. v = 3(x−1)²y + y³
   Answer: C. Integrate w.r.t. y. (Key point: Integration)

3. Why is g(x) constant?
   A. y = 0
   B. u is zero
   C. g is given
   D. v_x must equal −u_y = 6(x−1)y, so g′ = 0
   Answer: D. Matching v_x. (Key point: g(x))

4. The analytic function is:
   A. (z+1)³ + c
   B. z³ + c
   C. (z−1)³ + c
   D. i(z−1)³ + c
   Answer: C. f′ = 3(z−1)². (Key point: f)

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