RD Sharma Solutions for Class 11 Maths Chapter 13 Complex

RD Sharma Solutions for Class 11 Maths Chapter 13 Complex Numbers 1. Evaluate the following: (i) i 457 (ii) i 528 (iii) 1/ i58 (iv) i 37 + 1/i 67 (v) [i 41 + 1/i 257] (vi) (i 77 + i 70 + i 87 + i 414)3 (vii) i 30 + i 40 + i 60 (viii) i 49 + i 68 + i 89 + i 110 Solution: (i) i 457 Let us simplify we get, i457 = i (456 + 1) = i 4(114) × i = (1)114 × i = i [since i4 = 1] (ii) i 528 Let us simplify we get, i 528 = i 4(132) = (1)132 = 1 [since i4 = 1] (iii) 1/ i58 Let us simplify we get, 1/ i58 = 1/ i 56+2 = 1/ i 56 × i2 = 1/ (i4)14 × i2 = 1/ i2 [since, i4 = 1] = 1/-1 [since, i2 = -1] = -1 (iv) i 37 + 1/i 67 Let us simplify we get, i 37 + 1/i 67 = i36+1 + 1/ i64+3 = i + 1/i3 [since, i4 = 1] = i + i/i4 =i+i = 2i (v) [i 41 + 1/i 257] Let us simplify we get, [i 41 + 1/i 257] = [i40+1 + 1/ i256+1] = [i + 1/i]9 [since, 1/i = -1] = [i – i] =0 (vi) (i 77 + i 70 + i 87 + i 414)3 Let us simplify we get, (i 77 + i 70 + i 87 + i 414)3 = (i(76 + 1) + i(68 + 2) + i(84 + 3) + i(412 + 2) ) 3 = (i + i2 + i3 + i2)3 [since i3 = – i, i2 = – 1] = (i + (– 1) + (– i) + (– 1)) 3 = (– 2)3 =–8 (vii) i 30 + i 40 + i 60 Let us simplify we get, i 30 + i 40 + i 60 = i(28 + 2) + i40 + i60 = (i4)7 i2 + (i4)10 + (i4)15 = i2 + 110 + 115 =–1+1+1 =1 (viii) i 49 + i 68 + i 89 + i 110 Let us simplify we get, i 49 + i 68 + i 89 + i 110 = i(48 + 1) + i68 + i(88 + 1) + i(116 + 2) = (i4)12×i + (i4)17 + (i4)22×i + (i4)29×i2 =i+1+i–1 = 2i 2. Show that 1 + i10 + i20 + i30 is a real number? Solution: Given: 1 + i10 + i20 + i30 = 1 + i(8 + 2) + i20 + i(28 + 2) = 1 + (i4)2 × i2 + (i4)5 + (i4)7 × i2 = 1 – 1 + 1 – 1 [since, i4 = 1, i2 = – 1] =0 Hence , 1 + i10 + i20 + i30 is a real number. 3. Find the values of the following expressions: (i) i49 + i68 + i89 + i110 (ii) i30 + i80 + i120 (iii) i + i2 + i3 + i4 (iv) i5 + i10 + i15 (v) [i592 + i590 + i588 + i586 + i584] / [i582 + i580 + i578 + i576 + i574] (vi) 1 + i2 + i4 + i6 + i8 + … + i20 (vii) (1 + i)6 + (1 – i)3 Solution: (i) i49 + i68 + i89 + i110 Let us simplify we get, i49 + i68 + i89 + i110 = i (48 + 1) + i68 + i(88 + 1) + i(108 + 2) = (i4)12 × i + (i4)17 + (i4)22 × i + (i4)27 × i2 = i + 1 + i – 1 [since i4 = 1, i2 = – 1] = 2i ∴ i49 + i68 + i89 + i110 = 2i (ii) i30 + i80 + i120 Let us simplify we get, i30 + i80 + i120 = i(28 + 2) + i80 + i120 = (i4)7 × i2 + (i4)20 + (i4)30 = – 1 + 1 + 1 [since i4 = 1, i2 = – 1] =1 ∴ i30 + i80 + i120 = 1 (iii) i + i2 + i3 + i4 Let us simplify we get, i + i2 + i3 + i4 = i + i2 + i2×i + i4 = i – 1 + (– 1) × i + 1 [since i4 = 1, i2 = – 1] =i–1–i+1 =0 ∴ i + i2 + i3 + i4 = 0 (iv) i5 + i10 + i15 Let us simplify we get, i5 + i10 + i15 = i(4 + 1) + i(8 + 2) + i(12 + 3) = (i4)1×i + (i4)2×i2 + (i4)3×i3 = (i4)1×i + (i4)2×i2 + (i4)3×i2×i = 1×i + 1 × (– 1) + 1 × (– 1)×i =i–1–i =–1 ∴ i5 + i10 + i15 = -1 (v) [i592 + i590 + i588 + i586 + i584] / [i582 + i580 + i578 + i576 + i574] Let us simplify we get, [i592 + i590 + i588 + i586 + i584] / [i582 + i580 + i578 + i576 + i574] = [i10 (i582 + i580 + i578 + i576 + i574) / (i582 + i580 + i578 + i576 + i574)] = i10 = i8 i2 = (i4)2 i2 = (1)2 (-1) [since i4 = 1, i2 = -1] = -1 ∴ [i592 + i590 + i588 + i586 + i584] / [i582 + i580 + i578 + i576 + i574] = -1 (vi) 1 + i2 + i4 + i6 + i8 + … + i20 Let us simplify we get, 1 + i2 + i4 + i6 + i8 + … + i20 = 1 + (– 1) + 1 + (– 1) + 1 + … + 1 =1 ∴ 1 + i2 + i4 + i6 + i8 + … + i20 = 1 (vii) (1 + i)6 + (1 – i)3 Let us simplify we get, (1 + i)6 + (1 – i)3 = {(1 + i)2 }3 + (1 – i)2 (1 – i) = {1 + i2 + 2i}3 + (1 + i2 – 2i)(1 – i) = {1 – 1 + 2i}3 + (1 – 1 – 2i)(1 – i) = (2i)3 + (– 2i)(1 – i) = 8i3 + (– 2i) + 2i2 = – 8i – 2i – 2 [since i3 = – i, i2 = – 1] = – 10 i – 2 = – 2(1 + 5i) = – 2 – 10i ∴ (1 + i)6 + (1 – i)3 = – 2 – 10i EXERCISE 13.2 PAGE NO: 13.31 1. Express the following complex numbers in the standard form a + ib: (i) (1 + i) (1 + 2i) (ii) (3 + 2i) / (-2 + i) (iii) 1/(2 + i)2 (iv) (1 – i) / (1 + i) (v) (2 + i)3 / (2 + 3i) (vi) [(1 + i) (1 +√3i)] / (1 – i) (vii) (2 + 3i) / (4 + 5i) (viii) (1 – i)3 / (1 – i3) (ix) (1 + 2i)-3 (x) (3 – 4i) / [(4 – 2i) (1 + i)] (xi) (xii) (5 +√2i) / (1-√2i) Solution: (i) (1 + i) (1 + 2i) Let us simplify and express 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 and express 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 and express 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) [multiply and 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 and express 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 and express 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) [multiply and 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 and express 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) [multiply and 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 and express 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 and express 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) [Multiply and 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 and express 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) [Multiply and 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 and express 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) [Multiply and 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 and express 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 2. Find the real values of x and y, if (i) (x + iy) (2 – 3i) = 4 + i (ii) (3x – 2i y) (2 + i)2 = 10(1 + i) (iv) (1 + i) (x + iy) = 2 – 5i Solution: (i) (x + iy) (2 – 3i) = 4 + i Given: (x + iy) (2 – 3i) = 4 + i Let us simplify the expression we get, x(2 – 3i) + iy(2 – 3i) = 4 + i 2x – 3xi + 2yi – 3yi2 = 4 + i 2x + (-3x+2y)i – 3y (-1) = 4 + i [since, i2 = -1] 2x + (-3x+2y)i + 3y = 4 + i [since, i2 = -1] (2x+3y) + i(-3x+2y) = 4 + i Equating Real and Imaginary parts on both sides, we get 2x+3y = 4… (i) And -3x+2y = 1… (ii) Multiply (i) by 3 and (ii) by 2 and add On solving we get, 6x – 6x – 9y + 4y = 12 + 2 13y = 14 y = 14/13 Substitute the value of y in (i) we get, 2x+3y = 4 2x + 3(14/13) = 4 2x = 4 – (42/13) = (52-42)/13 2x = 10/13 x = 5/13 x = 5/13, y = 14/13 ∴ The real values of x and y are 5/13, 14/13 (ii) (3x – 2i y) (2 + i)2 = 10(1 + i) Given: (3x – 2iy) (2+i)2 = 10(1+i) (3x – 2yi) (22+i2+2(2)(i)) = 10+10i (3x – 2yi) (4 + (-1)+4i) = 10+10i [since, i2 = -1] (3x – 2yi) (3+4i) = 10+10i Let us divide with 3+4i on both sides we get, (3x – 2yi) = (10+10i)/(3+4i) = Now multiply and divide with (3-4i) = [10(3-4i) + 10i(3-4i)] / (32 – (4i)2) = [30-40i+30i-40i2] / (9 – 16i2) = [30-10i-40(-1)] / (9-16(-1)) = [70-10i]/25 Now, equating Real and Imaginary parts on both sides we get 3x = 70/25 and -2y = -10/25 x = 70/75 and y = 1/5 x = 14/15 and y = 1/5 ∴ The real values of x and y are 14/15, 1/5 (4+2i) x-3i-3 + (9-7i)y = 10i (4x+9y-3) + i(2x-7y-3) = 10i Now, equating Real and Imaginary parts on both sides we get, 4x+9y-3 = 0 … (i) And 2x-7y-3 = 10 2x-7y = 13 … (ii) Multiply (i) by 7 and (ii) by 9 and add On solving these equations we get 28x + 18x + 63y – 63y = 117 + 21 46x = 117 + 21 46x = 138 x = 138/46 =3 Substitute the value of x in (i) we get, 4x+9y-3 = 0 9y = -9 y = -9/9 = -1 x = 3 and y = -1 ∴ The real values of x and y are 3 and -1 (iv) (1 + i) (x + iy) = 2 – 5i Given: (1 + i) (x + iy) = 2 – 5i Divide with (1+i) on both the sides we get, (x + iy) = (2 – 5i)/(1+i) Multiply and divide by (1-i) = (2 – 5i)/(1+i) × (1-i)/(1-i) = [2(1-i) – 5i (1-i)] / (12 – i2) = [2 – 7i + 5(-1)] / 2 [since, i2 = -1] = (-3-7i)/2 Now, equating Real and Imaginary parts on both sides we get x = -3/2 and y = -7/2 ∴ Thee real values of x and y are -3/2, -7/2 3. Find the conjugates of the following complex numbers: (i) 4 – 5i (ii) 1 / (3 + 5i) (iii) 1 / (1 + i) (iv) (3 – i)2 / (2 + i) (v) [(1 + i) (2 + i)] / (3 + i) (vi) [(3 – 2i) (2 + 3i)] / [(1 + 2i) (2 – i)] Solution: (i) 4 – 5i Given: 4 – 5i We know the conjugate of a complex number (a + ib) is (a – ib) So, ∴ The conjugate of (4 – 5i) is (4 + 5i) (ii) 1 / (3 + 5i) Given: 1 / (3 + 5i) Since the given complex number is not in the standard form of (a + ib) Let us convert to standard form by multiplying and dividing with (3 – 5i) We get, We know the conjugate of a complex number (a + ib) is (a – ib) So, ∴ The conjugate of (3 – 5i)/34 is (3 + 5i)/34 (iii) 1 / (1 + i) Given: 1 / (1 + i) Since the given complex number is not in the standard form of (a + ib) Let us convert to standard form by multiplying and dividing with (1 – i) We get, We know the conjugate of a complex number (a + ib) is (a – ib) So, ∴ The conjugate of (1-i)/2 is (1+i)/2 (iv) (3 – i)2 / (2 + i) Given: (3 – i)2 / (2 + i) Since the given complex number is not in the standard form of (a + ib) Let us convert to standard form, We know the conjugate of a complex number (a + ib) is (a – ib) So, ∴ The conjugate of (2 – 4i) is (2 + 4i) (v) [(1 + i) (2 + i)] / (3 + i) Given: [(1 + i) (2 + i)] / (3 + i) Since the given complex number is not in the standard form of (a + ib) Let us convert to standard form, We know the conjugate of a complex number (a + ib) is (a – ib) So, ∴ The conjugate of (3 + 4i)/5 is (3 – 4i)/5 (vi) [(3 – 2i) (2 + 3i)] / [(1 + 2i) (2 – i)] Given: [(3 – 2i) (2 + 3i)] / [(1 + 2i) (2 – i)] Since the given complex number is not in the standard form of (a + ib) Let us convert to standard form, We know the conjugate of a complex number (a + ib) is (a – ib) So, ∴ The conjugate of (63 – 16i)/25 is (63 + 16i)/25 4. Find the multiplicative inverse of the following complex numbers: (i) 1 – i (ii) (1 + i √3)2 (iii) 4 – 3i (iv) √5 + 3i Solution: (i) 1 – i Given: 1–i We know the multiplicative inverse of a complex number (Z) is Z-1 or 1/Z So, ∴ The multiplicative inverse of (1 – i) is (1 + i)/2 (ii) (1 + i √3)2 Given: (1 + i √3)2 Z = (1 + i √3)2 = 12 + (i √3)2 + 2 (1) (i√3) = 1 + 3i2 + 2 i√3 = 1 + 3(-1) + 2 i√3 [since, i2 = -1] = 1 – 3 + 2 i√3 = -2 + 2 i√3 We know the multiplicative inverse of a complex number (Z) is Z-1 or 1/Z So, Z = -2 + 2 i√3 ∴ The multiplicative inverse of (1 + i√3)2 is (-1-i√3)/8 (iii) 4 – 3i Given: 4 – 3i We know the multiplicative inverse of a complex number (Z) is Z-1 or 1/Z So, Z = 4 – 3i ∴ The multiplicative inverse of (4 – 3i) is (4 + 3i)/25 (iv) √5 + 3i Given: √5 + 3i We know the multiplicative inverse of a complex number (Z) is Z-1 or 1/Z So, Z = √5 + 3i ∴ The multiplicative inverse of (√5 + 3i) is (√5 – 3i)/14 6. If z1 = (2 – i), z2 = (-2 + i), find Solution: Given: z1 = (2 – i) and z2 = (-2 + i) 7. Find the modulus of [(1 + i)/(1 – i)] – [(1 – i)/(1 + i)] Solution: Given: [(1 + i)/(1 – i)] – [(1 – i)/(1 + i)] So, Z = [(1 + i)/(1 – i)] – [(1 – i)/(1 + i)] Let us simplify, we get = [(1+i) (1+i) – (1-i) (1-i)] / (12 – i2) = [12 + i2 + 2(1)(i) – (12 + i2 – 2(1)(i))] / (1 – (-1)) [Since, i2 = -1] = 4i/2 = 2i We know that for a complex number Z = (a+ib) it’s magnitude is given by |z| = √(a2 + b2) So, |Z| = √(02 + 22) =2 ∴ The modulus of [(1 + i)/(1 – i)] – [(1 – i)/(1 + i)] is 2. 8. If x + iy = (a+ib)/(a-ib), prove that x2 + y2 = 1 Solution: Given: x + iy = (a+ib)/(a-ib) We know that for a complex number Z = (a+ib) it’s magnitude is given by |z| = √(a2 + b2) So, |a/b| is |a| / |b| Applying Modulus on both sides we get, 9. Find the least positive integral value of n for which [(1+i)/(1-i)]n is real. Solution: Given: [(1+i)/(1-i)]n Z = [(1+i)/(1-i)]n Now let us multiply and divide by (1+i), we get = i [which is not real] For n = 2, we have [(1+i)/(1-i)]2 = i2 = -1 [which is real] So, the smallest positive integral ‘n’ that can make [(1+i)/(1-i)]n real is 2. ∴ The smallest positive integral value of ‘n’ is 2. 10. Find the real values of θ for which the complex number (1 + i cos θ) / (1 – 2i cos θ) is purely real. Solution: Given: (1 + i cos θ) / (1 – 2i cos θ) Z = (1 + i cos θ) / (1 – 2i cos θ) Let us multiply and divide by (1 + 2i cos θ) For a complex number to be purely real, the imaginary part should be equal to zero. So, 3cos θ = 0 (since, 1 + 4cos2θ ≥ 1) cos θ = 0 cos θ = cos π/2 θ = [(2n+1)π] / 2, for n ∈ Z = 2nπ ± π/2, for n ∈ Z ∴ The values of θ to get the complex number to be purely real is 2nπ ± π/2, for n ∈ Z 11. Find the smallest positive integer value of n for which (1+i) n / (1-i) n-2 is a real number. Solution: Given: (1+i) n / (1-i) n-2 Z = (1+i) n / (1-i) n-2 Let us multiply and divide by (1 – i)2 For n = 1, Z = -2i1+1 = -2i2 = 2, which is a real number. ∴ The smallest positive integer value of n is 1. 12. If [(1+i)/(1-i)]3 – [(1-i)/(1+i)]3 = x + iy, find (x, y) Solution: Given: [(1+i)/(1-i)]3 – [(1-i)/(1+i)]3 = x + iy Let us rationalize the denominator, we get i3–(-i)3 = x + iy 2i3 = x + iy 2i2.i = x + iy 2(-1)I = x + iy -2i = x + iy Equating Real and Imaginary parts on both sides we get x = 0 and y = -2 ∴ The values of x and y are 0 and -2. 13. If (1+i)2 / (2-i) = x + iy, find x + y Solution: Given: (1+i)2 / (2-i) = x + iy Upon expansion we get, Let us equate real and imaginary parts on both sides we get, x = -2/5 and y = 4/5 so, x + y = -2/5 + 4/5 = (-2+4)/5 = 2/5 ∴ The value of (x + y) is 2/5 EXERCISE 13.3 PAGE NO: 13.39 1. Find the square root of the following complex numbers. (i) – 5 + 12i (ii) -7 – 24i (iii) 1 – i (iv) – 8 – 6i (v) 8 – 15i (vi) -11 – 60√-1 (vii) 1 + 4√-3 (viii) 4i (ix) -i Solution: (i) – 5 + 12i Given: – 5 + 12i We know, Z = a + ib So, √(a + ib) = √(-5+12i) Here, b > 0 Let us simplify now, ∴ Square root of (– 5 + 12i) is ±[2 + 3i] (ii) -7 – 24i Given: -7 – 24i We know, Z = -7 – 24i So, √(a + ib) = √(-7 – 24i) Here, b < 0 Let us simplify now, ∴ Square root of (-7 – 24i) is ± [3 – 4i] (iii) 1 – i Given: 1–i We know, Z = (1 – i) So, √(a + ib) = √(1 – i) Here, b < 0 Let us simplify now, ∴ Square root of (1 – i) is ± [(√(√2+1)/2) – i (√(√2-1)/2)] (iv) -8 -6i Given: -8 -6i We know, Z = -8 -6i So, √(a + ib) = -8 -6i Here, b < 0 Let us simplify now, = [11/2 – i 91/2] = ± [1 – 3i] ∴ Square root of (-8 -6i) is ± [1 – 3i] (v) 8 – 15i Given: 8 – 15i We know, Z = 8 – 15i So, √(a + ib) = 8 – 15i Here, b < 0 Let us simplify now, ∴ Square root of (8 – 15i) is ± 1/√2 (5 – 3i) (vi) -11 – 60√-1 Given: -11 – 60√-1 We know, Z = -11 – 60√-1 So, √(a + ib) = -11 – 60√-1 = -11 – 60i Here, b < 0 Let us simplify now, ∴ Square root of (-11 – 60√-1) is ± (5 – 6i) (vii) 1 + 4√-3 Given: 1 + 4√-3 We know, Z = 1 + 4√-3 So, √(a + ib) = 1 + 4√-3 = 1 + 4(√3) (√-1) = 1 + 4√3i Here, b > 0 Let us simplify now, ∴ Square root of (1 + 4√-3) is ± (2 + √3i) (viii) 4i Given: 4i We know, Z = 4i So, √(a + ib) = 4i Here, b > 0 Let us simplify now, ∴ Square root of 4i is ± √2 (1 + i) (ix) –i Given: -i We know, Z = -i So, √(a + ib) = -i Here, b < 0 Let us simplify now, ∴ Square root of –i is ± 1/√2 (1 – i) EXERCISE 13.4 PAGE NO: 13.57 1. Find the modulus and arguments of the following complex numbers and hence express each of them in the polar form: (i) 1 + i (ii) √3 + i (iii) 1 – i (iv) (1 – i) / (1 + i) (v) 1/(1 + i) (vi) (1 + 2i) / (1 – 3i) (vii) sin 120o – i cos 120o (viii) -16 / (1 + i√3) Solution: We know that the polar form of a complex number Z = x + iy is given by Z = |Z| (cos θ + i sin θ) Where, |Z| = modulus of complex number = √(x2 + y2) θ = arg (z) = argument of complex number = tan-1 (|y| / |x|) (i) 1 + i Given: Z = 1 + i So now, |Z| = √(x2 + y2) = √(12 + 12) = √(1 + 1) = √2 θ = tan-1 (|y| / |x|) = tan-1 (1 / 1) = tan-1 1 Since x > 0, y > 0 complex number lies in 1st quadrant and the value of θ is 00≤θ≤900. θ = π/4 Z = √2 (cos (π/4) + i sin (π/4)) ∴ Polar form of (1 + i) is √2 (cos (π/4) + i sin (π/4)) (ii) √3 + i Given: Z = √3 + i So now, |Z| = √(x2 + y2) = √((√3)2 + 12) = √(3 + 1) = √4 =2 θ = tan-1 (|y| / |x|) = tan-1 (1 / √3) Since x > 0, y > 0 complex number lies in 1st quadrant and the value of θ is 00≤θ≤900. θ = π/6 Z = 2 (cos (π/6) + i sin (π/6)) ∴ Polar form of (√3 + i) is 2 (cos (π/6) + i sin (π/6)) (iii) 1 – i Given: Z = 1 – i So now, |Z| = √(x2 + y2) = √(12 + (-1)2) = √(1 + 1) = √2 θ = tan-1 (|y| / |x|) = tan-1 (1 / 1) = tan-1 1 Since x > 0, y < 0 complex number lies in 4th quadrant and the value of θ is -900≤θ≤00. θ = -π/4 Z = √2 (cos (-π/4) + i sin (-π/4)) = √2 (cos (π/4) – i sin (π/4)) ∴ Polar form of (1 – i) is √2 (cos (π/4) – i sin (π/4)) (iv) (1 – i) / (1 + i) Given: Z = (1 – i) / (1 + i) Let us multiply and divide by (1 – i), we get =0–i So now, |Z| = √(x2 + y2) = √(02 + (-1)2) = √(0 + 1) = √1 θ = tan-1 (|y| / |x|) = tan-1 (1 / 0) = tan-1 ∞ Since x ≥ 0, y < 0 complex number lies in 4th quadrant and the value of θ is -900≤θ≤00. θ = -π/2 Z = 1 (cos (-π/2) + i sin (-π/2)) = 1 (cos (π/2) – i sin (π/2)) ∴ Polar form of (1 – i) / (1 + i) is 1 (cos (π/2) – i sin (π/2)) (v) 1/(1 + i) Given: Z = 1 / (1 + i) Let us multiply and divide by (1 – i), we get So now, |Z| = √(x2 + y2) = √((1/2)2 + (-1/2)2) = √(1/4 + 1/4) = √(2/4) = 1/√2 θ = tan-1 (|y| / |x|) = tan-1 ((1/2) / (1/2)) = tan-1 1 Since x > 0, y < 0 complex number lies in 4th quadrant and the value of θ is -900≤θ≤00. θ = -π/4 Z = 1/√2 (cos (-π/4) + i sin (-π/4)) = 1/√2 (cos (π/4) – i sin (π/4)) ∴ Polar form of 1/(1 + i) is 1/√2 (cos (π/4) – i sin (π/4)) (vi) (1 + 2i) / (1 – 3i) Given: Z = (1 + 2i) / (1 – 3i) Let us multiply and divide by (1 + 3i), we get So now, |Z| = √(x2 + y2) = √((-1/2)2 + (1/2)2) = √(1/4 + 1/4) = √(2/4) = 1/√2 θ = tan-1 (|y| / |x|) = tan-1 ((1/2) / (1/2)) = tan-1 1 Since x < 0, y > 0 complex number lies in 2nd quadrant and the value of θ is 900≤θ≤1800. θ = 3π/4 Z = 1/√2 (cos (3π/4) + i sin (3π/4)) ∴ Polar form of (1 + 2i) / (1 – 3i) is 1/√2 (cos (3π/4) + i sin (3π/4)) (vii) sin 120o – i cos 120o Given: Z = sin 120o – i cos 120o = √3/2 – i (-1/2) = √3/2 + i (1/2) So now, |Z| = √(x2 + y2) = √((√3/2)2 + (1/2)2) = √(3/4 + 1/4) = √(4/4) = √1 =1 θ = tan-1 (|y| / |x|) = tan-1 ((1/2) / (√3/2)) = tan-1 (1/√3) Since x > 0, y > 0 complex number lies in 1st quadrant and the value of θ is 00≤θ≤900. θ = π/6 Z = 1 (cos (π/6) + i sin (π/6)) ∴ Polar form of √3/2 + i (1/2) is 1 (cos (π/6) + i sin (π/6)) (viii) -16 / (1 + i√3) Given: Z = -16 / (1 + i√3) Let us multiply and divide by (1 – i√3), we get So now, |Z| = √(x2 + y2) = √((-4)2 + (4√3)2) = √(16 + 48) = √(64) =8 θ = tan-1 (|y| / |x|) = tan-1 ((4√3) / 4) = tan-1 (√3) Since x < 0, y > 0 complex number lies in 2nd quadrant and the value of θ is 900≤θ≤1800. θ = 2π/3 Z = 8 (cos (2π/3) + i sin (2π/3)) ∴ Polar form of -16 / (1 + i√3) is 8 (cos (2π/3) + i sin (2π/3)) 2. Write (i25)3 in polar form. Solution: Given: Z = (i25)3 = i75 = i74. i = (i2)37. i = (-1)37. i = (-1). i =–i =0–i So now, |Z| = √(x2 + y2) = √(02 + (-1)2) = √(0 + 1) = √1 θ = tan-1 (|y| / |x|) = tan-1 (1 / 0) = tan-1 ∞ Since x ≥ 0, y < 0 complex number lies in 4th quadrant and the value of θ is -900≤θ≤00. θ = -π/2 Z = 1 (cos (-π/2) + i sin (-π/2)) = 1 (cos (π/2) – i sin (π/2)) ∴ Polar form of (i25)3 is 1 (cos (π/2) – i sin (π/2)) 3. Express the following complex numbers in the form r (cos θ + i sin θ): (i) 1 + i tan α (ii) tan α – i (iii) 1 – sin α + i cos α (iv) (1 – i) / (cos π/3 + i sin π/3) Solution: (i) 1 + i tan α Given: Z = 1 + i tan α We know that the polar form of a complex number Z = x + iy is given by Z = |Z| (cos θ + i sin θ) Where, |Z| = modulus of complex number = √(x2 + y2) θ = arg (z) = argument of complex number = tan-1 (|y| / |x|) We also know that tan α is a periodic function with period π. So α is lying in the interval [0, π/2) ∪ (π/2, π]. Let us consider case 1: α ∈ [0, π/2) So now, |Z| = r = √(x2 + y2) = √(12 + tan2 α) = √( sec2 α) = |sec α| since, sec α is positive in the interval [0, π/2) θ = tan-1 (|y| / |x|) = tan-1 (tan α / 1) = tan-1 (tan α) = α since, tan α is positive in the interval [0, π/2) ∴ Polar form is Z = sec α (cos α + i sin α) Let us consider case 2: α ∈ (π/2, π] So now, |Z| = r = √(x2 + y2) = √(12 + tan2 α) = √( sec2 α) = |sec α| = – sec α since, sec α is negative in the interval (π/2, π] θ = tan-1 (|y| / |x|) = tan-1 (tan α / 1) = tan-1 (tan α) = -π + α since, tan α is negative in the interval (π/2, π] θ = -π + α [since, θ lies in 4th quadrant] Z = -sec α (cos (α – π) + i sin (α – π)) ∴ Polar form is Z = -sec α (cos (α – π) + i sin (α – π)) (ii) tan α – i Given: Z = tan α – i We know that the polar form of a complex number Z = x + iy is given by Z = |Z| (cos θ + i sin θ) Where, |Z| = modulus of complex number = √(x2 + y2) θ = arg (z) = argument of complex number = tan-1 (|y| / |x|) We also know that tan α is a periodic function with period π. So α is lying in the interval [0, π/2) ∪ (π/2, π]. Let us consider case 1: α ∈ [0, π/2) So now, |Z| = r = √(x2 + y2) = √(tan2 α + 12) = √( sec2 α) = |sec α| since, sec α is positive in the interval [0, π/2) = sec α θ = tan-1 (|y| / |x|) = tan-1 (1/tan α) = tan-1 (cot α) since, cot α is positive in the interval [0, π/2) = α – π/2 [since, θ lies in 4th quadrant] Z = sec α (cos (α – π/2) + i sin (α – π/2)) ∴ Polar form is Z = sec α (cos (α – π/2) + i sin (α – π/2)) Let us consider case 2: α ∈ (π/2, π] So now, |Z| = r = √(x2 + y2) = √(tan2 α + 12) = √( sec2 α) = |sec α| = – sec α since, sec α is negative in the interval (π/2, π] θ = tan-1 (|y| / |x|) = tan-1 (1/tan α) = tan-1 (cot α) = π/2 + α since, cot α is negative in the interval (π/2, π] θ = π/2 + α [since, θ lies in 3th quadrant] Z = -sec α (cos (π/2 + α) + i sin (π/2 + α)) ∴ Polar form is Z = -sec α (cos (π/2 + α) + i sin (π/2 + α)) (iii) 1 – sin α + i cos α Given: Z = 1 – sin α + i cos α By using the formulas, Sin2 θ + cos2 θ = 1 Sin 2θ = 2 sin θ cos θ Cos 2θ = cos2 θ – sin2 θ So, z= (sin2(α/2) + cos2(α/2) – 2 sin(α/2) cos(α/2)) + i (cos2(α/2) – sin2(α/2)) = (cos(α/2) – sin(α/2))2 + i (cos2(α/2) – sin2(α/2)) We know that the polar form of a complex number Z = x + iy is given by Z = |Z| (cos θ + i sin θ) Where, |Z| = modulus of complex number = √(x2 + y2) θ = arg (z) = argument of complex number = tan-1 (|y| / |x|) Now, We know that sine and cosine functions are periodic with period 2π Here we have 3 intervals: 0 ≤ α ≤ π/2 π/2 ≤ α ≤ 3π/2 3π/2 ≤ α ≤ 2π Let us consider case 1: In the interval 0 ≤ α ≤ π/2 Cos (α/2) > sin (α/2) and also 0 < π/4 + α/2 < π/2 So, ∴ Polar form is Z = √2 (cos (α/2) – sin (α/2)) (cos (π/4 + α/2) + i sin (π/4 + α/2)) Let us consider case 2: In the interval π/2 ≤ α ≤ 3π/2 Cos (α/2) < sin (α/2) and also π/2 < π/4 + α/2 < π So, Since, (1 – sin α) > 0 and cos α < 0 [Z lies in 4th quadrant] = α/2 – 3π/4 ∴ Polar form is Z = –√2 (cos (α/2) – sin (α/2)) (cos (α/2 – 3π/4) + i sin (α/2 – 3π/4)) Let us consider case 3: In the interval 3π/2 ≤ α ≤ 2π Cos (α/2) < sin (α/2) and also π < π/4 + α/2 < 5π/4 So, θ = tan-1 (tan (π/4 + α/2)) = π – (π/4 + α/2) [since, θ lies in 1st quadrant and tan’s period is π] = α/2 – 3π/4 ∴ Polar form is Z = –√2 (cos (α/2) – sin (α/2)) (cos (α/2 – 3π/4) + i sin (α/2 – 3π/4)) (iv) (1 – i) / (cos π/3 + i sin π/3) Given: Z = (1 – i) / (cos π/3 + i sin π/3) Let us multiply and divide by (1 – i√3), we get We know that the polar form of a complex number Z = x + iy is given by Z = |Z| (cos θ + i sin θ) Where, |Z| = modulus of complex number = √(x2 + y2) θ = arg (z) = argument of complex number = tan-1 (|y| / |x|) Now, Since x < 0, y < 0 complex number lies in 3rd quadrant and the value of θ is 1800≤θ≤-900. = tan-1 (2 + √3) = -7π/12 Z = √2 (cos (-7π/12) + i sin (-7π/12)) = √2 (cos (7π/12) – i sin (7π/12)) ∴ Polar form of (1 – i) / (cos π/3 + i sin π/3) is √2 (cos (7π/12) – i sin (7π/12)) 4. If z1 and z2 are two complex number such that |z1| = |z2| and arg (z1) + arg (z2) = π, then show that Solution: Given: |z1| = |z2| and arg (z1) + arg (z2) = π Let us assume arg (z1) = θ arg (z2) = π – θ We know that in the polar form, z = |z| (cos θ + i sin θ) z1 = |z1| (cos θ + i sin θ) …………. (i) z2 = |z2| (cos (π – θ) + i sin (π – θ)) = |z2| (-cos θ + i sin θ) = – |z2| (cos θ – i sin θ) Now let us find the conjugate of = – |z2| (cos θ + i sin θ) …… (ii) (since, |Z2¯¯¯¯¯¯|=|Z2|) Now, z1 / = [|z1| (cos θ + i sin θ)] / [-|z2| (cos θ + i sin θ)] = – |z1| / |z2| [since, |z1| = |z2|] = -1 When we cross multiply we get, z1 = – Hence proved. 5. If z1, z2 and z3, z4 are two pairs of conjugate complex numbers, prove that arg (z1/z4) + arg (z2/z3) = 0 Solution: Given: Hence proved. 6. Express sin π/5 + i (1 – cos π/5) in polar form. Solution: Given: Z = sin π/5 + i (1 – cos π/5) By using the formula, sin 2θ = 2 sin θ cos θ 1- cos 2θ = 2 sin2 θ So, Z = 2 sin π/10 cos π/10 + i (2 sin2 π/10) = 2 sin π/10 (cos π/10 + i sin π/10) ∴ The polar form of sin π/5 + i (1 – cos π/5) is 2 sin π/10 (cos π/10 + i sin π/10)