Question -
Answer -
(i) (1 + i) (1 + 2i)
Let us simplify andexpress in the standard form of (a + ib),
(1 + i) (1 + 2i) =(1+i)(1+2i)
= 1(1+2i)+i(1+2i)
= 1+2i+i+2i2
= 1+3i+2(-1) [since, i2 =-1]
= 1+3i-2
= -1+3i
∴ The values of a,b are -1, 3.
(ii) (3 + 2i) / (-2 + i)
Let us simplify andexpress in the standard form of (a + ib),
(3 + 2i) / (-2 + i) =[(3 + 2i) / (-2 + i)] × (-2-i) / (-2-i) [multiply and divide with (-2-i)]
= [3(-2-i) + 2i(-2-i)] / [(-2)2 – (i)2]
= [-6 -3i – 4i -2i2]/ (4-i2)
= [-6 -7i -2(-1)] / (4– (-1)) [since, i2 = -1]
= [-4 -7i] / 5
∴ The values of a,b are -4/5, -7/5
(iii) 1/(2 + i)2
Let us simplify andexpress in the standard form of (a + ib),
1/(2 + i)2 =1/(22 + i2 + 2(2) (i))
= 1/ (4 – 1 + 4i)[since, i2 = -1]
= 1/(3 + 4i) [multiplyand divide with (3 – 4i)]
= 1/(3 + 4i) × (3 –4i)/ (3 – 4i)]
= (3-4i)/ (32 –(4i)2)
= (3-4i)/ (9 – 16i2)
= (3-4i)/ (9 – 16(-1))[since, i2 = -1]
= (3-4i)/25
∴ The values of a,b are 3/25, -4/25
(iv) (1 – i) / (1 + i)
Let us simplify andexpress in the standard form of (a + ib),
(1 – i) / (1 + i) = (1– i) / (1 + i) × (1-i)/(1-i) [multiply and divide with (1-i)]
= (12 +i2 – 2(1)(i)) / (12 – i2)
= (1 + (-1) -2i) / (1– (-1))
= -2i/2
= -i
∴ The values of a,b are 0, -1
(v) (2 + i)3 /(2 + 3i)
Let us simplify andexpress in the standard form of (a + ib),
(2 + i)3 /(2 + 3i) = (23 + i3 + 3(2)2(i) +3(i)2(2)) / (2 + 3i)
= (8 + (i2.i)+ 3(4)(i) + 6i2) / (2 + 3i)
= (8 + (-1)i + 12i +6(-1)) / (2 + 3i)
= (2 + 11i) / (2 + 3i)
[multiplyand divide with (2-3i)]
= (2 + 11i)/(2 + 3i) ×(2-3i)/(2-3i)
= [2(2-3i) +11i(2-3i)] / (22 – (3i)2)
= (4 – 6i + 22i – 33i2)/ (4 – 9i2)
= (4 + 16i – 33(-1)) /(4 – 9(-1)) [since, i2 = -1]
= (37 + 16i) / 13
∴ The values of a,b are 37/13, 16/13
(vi) [(1 + i) (1 +√3i)] /(1 – i)
Let us simplify andexpress in the standard form of (a + ib),
[(1 +i) (1 +√3i)] / (1 – i) = [1(1+√3i) + i(1+√3i)] / (1-i)
= (1 + √3i + i + √3i2)/ (1 – i)
= (1 + (√3+1)i +√3(-1)) / (1-i) [since, i2 = -1]
= [(1-√3) + (1+√3)i] /(1-i)
[multiplyand divide with (1+i)]
= [(1-√3) + (1+√3)i] /(1-i) × (1+i)/(1+i)
= [(1-√3) (1+i) +(1+√3)i(1+i)] / (12 – i2)
= [1-√3+ (1-√3)i +(1+√3)i + (1+√3)i2] / (1-(-1)) [since, i2 = -1]
= [(1-√3)+(1-√3+1+√3)i+(1+√3)(-1)]/ 2
= (-2√3 + 2i) / 2
= -√3 + i
∴ The values of a,b are -√3, 1
(vii) (2 + 3i) / (4 + 5i)
Let us simplify andexpress in the standard form of (a + ib),
(2 + 3i) / (4 + 5i) =[multiply and divide with (4-5i)]
= (2 + 3i) / (4 + 5i)× (4-5i)/(4-5i)
= [2(4-5i) + 3i(4-5i)]/ (42 – (5i)2)
= [8 – 10i + 12i – 15i2]/ (16 – 25i2)
= [8+2i-15(-1)] / (16– 25(-1)) [since, i2 = -1]
= (23 + 2i) / 41
∴ The values of a,b are 23/41, 2/41
(viii) (1 – i)3 /(1 – i3)
Let us simplify andexpress in the standard form of (a + ib),
(1 – i)3 /(1 – i3) = [13 – 3(1)2i + 3(1)(i)2 –i3] / (1-i2.i)
= [1 – 3i + 3(-1)-i2.i]/ (1 – (-1)i) [since, i2 = -1]
= [-2 – 3i – (-1)i] /(1+i)
= [-2-4i] / (1+i)
[Multiplyand divide with (1-i)]
= [-2-4i] / (1+i) ×(1-i)/(1-i)
= [-2(1-i)-4i(1-i)] /(12 – i2)
= [-2+2i-4i+4i2]/ (1 – (-1))
= [-2-2i+4(-1)] /2
= (-6-2i)/2
= -3 – i
∴ The values of a,b are -3, -1
(ix) (1 + 2i)-3
Let us simplify andexpress in the standard form of (a + ib),
(1 + 2i)-3 =1/(1 + 2i)3
= 1/(13+3(1)2 (2i)+2(1)(2i)2 +(2i)3)
= 1/(1+6i+4i2+8i3)
= 1/(1+6i+4(-1)+8i2.i)[since, i2 = -1]
= 1/(-3+6i+8(-1)i)[since, i2 = -1]
= 1/(-3-2i)
= -1/(3+2i)
[Multiplyand divide with (3-2i)]
= -1/(3+2i) ×(3-2i)/(3-2i)
= (-3+2i)/(32 –(2i)2)
= (-3+2i) / (9-4i2)
= (-3+2i) / (9-4(-1))
= (-3+2i) /13
∴ The values of a,b are -3/13, 2/13
(x) (3 – 4i) / [(4 – 2i)(1 + i)]
Let us simplify andexpress in the standard form of (a + ib),
(3 – 4i) / [(4 – 2i)(1 + i)] = (3-4i)/ [4(1+i)-2i(1+i)]
= (3-4i)/ [4+4i-2i-2i2]
= (3-4i)/ [4+2i-2(-1)][since, i2 = -1]
= (3-4i)/ (6+2i)
[Multiplyand divide with (6-2i)]
= (3-4i)/ (6+2i) ×(6-2i)/(6-2i)
= [3(6-2i)-4i(6-2i)] /(62 – (2i)2)
= [18 – 6i – 24i + 8i2]/ (36 – 4i2)
= [18 – 30i + 8 (-1)]/ (36 – 4 (-1)) [since, i2 = -1]
= [10-30i] / 40
= (1 – 3i) / 4
∴ The values of a,b are 1/4, -3/4
(xi) 
(xii) (5 +√2i) / (1-√2i)
Let us simplify andexpress in the standard form of (a + ib),
(5 +√2i) / (1-√2i) =[Multiply and divide with (1+√2i)]
= (5 +√2i) / (1-√2i) ×(1+√2i)/(1+√2i)
= [5(1+√2i) + √2i(1+√2i)] / (12 –(√2)2)
= [5+5√2i + √2i + 2i2] / (1 – 2i2)
= [5 + 6√2i + 2(-1)] / (1-2(-1)) [since, i2 =-1]
= [3+6√2i]/3
= 1+ 2√2i
∴ The values of a,b are 1, 2√2