QUESTION TWO (a)The binary operation * on \( \mathbb{R} \) is defined by \( a * b=(a-b)^{2}-2 a b \). (i) Show that the operation * is commutative. (ii) Find \( \left(-1^{*} 3\right)^{* 2} \) (b) (i) Express \( Z=\frac{4+2 i}{(1-2 i)^{2}} \) in the form a+bi where a and b are real numbers (ii) Express 0.253 in the form \( \frac{a}{b} \) where a and b are integers

Let's tackle each part of **Question Two** step by step. --- ## **Problem 2(a)(i): Show that the operation * is commutative.** **Given:** The binary operation \( * \) on \( \mathbb{R} \) is defined by: \[ a * b = (a - b)^2 - 2ab \] **To Show:** \( a * b = b * a \) for all real numbers \( a \) and \( b \). **Solution:** 1. **Compute \( a * b \):** \[ a * b = (a - b)^2 - 2ab \] 2. **Compute \( b * a \):** \[ b * a = (b - a)^2 - 2ba \] Notice that \( (b - a)^2 = (a - b)^2 \) and \( 2ba = 2ab \) since multiplication is commutative in \( \mathbb{R} \). 3. **Compare \( a * b \) and \( b * a \):** \[ a * b = (a - b)^2 - 2ab = (b - a)^2 - 2ab = b * a \] **Conclusion:** Since \( a * b = b * a \), the operation \( * \) is **commutative**. --- ## **Problem 2(a)(ii): Find \( (-1 * 3) * 2 \).** **Given:** \( a * b = (a - b)^2 - 2ab \) **Solution:** 1. **First, compute \( -1 * 3 \):** \[ -1 * 3 = (-1 - 3)^2 - 2(-1)(3) \] \[ = (-4)^2 - 2(-3) \] \[ = 16 + 6 \] \[ = 22 \] 2. **Next, compute \( 22 * 2 \):** \[ 22 * 2 = (22 - 2)^2 - 2(22)(2) \] \[ = (20)^2 - 88 \] \[ = 400 - 88 \] \[ = 312 \] **Conclusion:** \( (-1 * 3) * 2 = 312 \) --- ## **Problem 2(b)(i): Express \( Z = \frac{4 + 2i}{(1 - 2i)^2} \) in the form \( a + bi \) where \( a \) and \( b \) are real numbers.** **Solution:** 1. **First, compute \( (1 - 2i)^2 \):** \[ (1 - 2i)^2 = 1^2 - 2 \cdot 1 \cdot 2i + (2i)^2 \] \[ = 1 - 4i + 4i^2 \] Recall that \( i^2 = -1 \): \[ = 1 - 4i - 4 \] \[ = -3 - 4i \] 2. **Express \( Z \) with the computed denominator:** \[ Z = \frac{4 + 2i}{-3 - 4i} \] 3. **Multiply numerator and denominator by the conjugate of the denominator to rationalize:** \[ \text{Conjugate of } (-3 - 4i) \text{ is } (-3 + 4i) \] \[ Z = \frac{(4 + 2i)(-3 + 4i)}{(-3 - 4i)(-3 + 4i)} \] 4. **Compute the denominator:** \[ (-3)^2 - (4i)^2 = 9 - 16i^2 = 9 - 16(-1) = 9 + 16 = 25 \] 5. **Compute the numerator:** \[ (4)(-3) + 4(4i) + 2i(-3) + 2i(4i) = -12 + 16i - 6i + 8i^2 \] \[ = -12 + 10i + 8(-1) = -12 + 10i - 8 = -20 + 10i \] 6. **Combine the results:** \[ Z = \frac{-20 + 10i}{25} = \frac{-20}{25} + \frac{10}{25}i = -\frac{4}{5} + \frac{2}{5}i \] **Conclusion:** \[ Z = -\frac{4}{5} + \frac{2}{5}i \] Or, equivalently: \[ Z = -0.8 + 0.4i \] --- ## **Problem 2(b)(ii): Express \( 0.253 \) in the form \( \frac{a}{b} \) where \( a \) and \( b \) are integers.** **Solution:** 1. **Recognize that \( 0.253 \) is the same as \( \frac{253}{1000} \):** \[ 0.253 = \frac{253}{1000} \] 2. **Simplify the fraction if possible:** - **Factor the numerator and denominator:** \[ 253 = 11 \times 23 \quad (\text{since } 11 \times 23 = 253) \] \[ 1000 = 2^3 \times 5^3 \] - **Determine the greatest common divisor (GCD) of 253 and 1000:** - The prime factors of 253 are 11 and 23. - The prime factors of 1000 are 2 and 5. - They have no common prime factors, so the GCD is 1. 3. **Since the GCD is 1, the fraction is already in its simplest form:** \[ \frac{253}{1000} \] **Conclusion:** \[ 0.253 = \frac{253}{1000} \] ---