Question

Imaginary part of \( \frac{(1 - i)^3}{(2 - i)(3 - 2i)} \) is:

• A. \( \frac{22}{65} \)
• B. \( \frac{6}{65} \)
• C. \( -\frac{6}{65} \)
• D. \( -\frac{22}{65} \)

Hint:

When dealing with complex numbers, use the conjugate of the denominator to simplify complex fractions. Expanding and simplifying step by step ensures accuracy.

Answer 1

The Correct Option is D

We are asked to find the imaginary part of the expression: \[ \frac{(1 - i)^3}{(2 - i)(3 - 2i)}. \] Step 1: Expanding \( (1 - i)^3 \) We start by expanding \( (1 - i)^3 \). Use the binomial expansion for \( (a - b)^3 \): \[ (1 - i)^3 = 1^3 - 3(1^2)(i) + 3(1)(i^2) - i^3. \] Now calculate the terms: \[ 1^3 = 1, \quad 3(1^2)(i) = 3i, \quad 3(1)(i^2) = 3(-1) = -3, \quad -i^3 = -(-i) = i. \] Thus: \[ (1 - i)^3 = 1 - 3i - 3 + i = -2 - 2i. \] Step 2: Expanding \( (2 - i)(3 - 2i) \) Next, we expand \( (2 - i)(3 - 2i) \) using distributive property: \[ (2 - i)(3 - 2i) = 2(3) + 2(-2i) - i(3) - i(-2i). \] Now calculate the terms: \[ 2(3) = 6, \quad 2(-2i) = -4i, \quad -i(3) = -3i, \quad -i(-2i) = 2i^2 = -2. \] Thus: \[ (2 - i)(3 - 2i) = 6 - 4i - 3i - 2 = 4 - 7i. \] Step 3: Dividing the two expressions Now, we divide the two expressions: \[ \frac{-2 - 2i}{4 - 7i}. \] To simplify this, multiply both the numerator and denominator by the conjugate of the denominator \( 4 + 7i \): \[ \frac{-2 - 2i}{4 - 7i} \times \frac{4 + 7i}{4 + 7i} = \frac{(-2 - 2i)(4 + 7i)}{(4 - 7i)(4 + 7i)}. \] First, simplify the denominator: \[ (4 - 7i)(4 + 7i) = 4^2 - (7i)^2 = 16 - (-49) = 16 + 49 = 65. \] Now, expand the numerator: \[ (-2 - 2i)(4 + 7i) = -2(4) - 2(7i) - 2i(4) - 2i(7i) = -8 - 14i - 8i + 14 = 6 - 22i. \] Thus, the expression becomes: \[ \frac{6 - 22i}{65}. \] Step 4: Identifying the imaginary part The expression is \( \frac{6}{65} - \frac{22i}{65} \), so the imaginary part is: \[ -\frac{22}{65}. \] Thus, the imaginary part of \( \frac{(1 - i)^3}{(2 - i)(3 - 2i)} \) is \( -\frac{22}{65} \).