Gujarat Board GSEB Textbook Solutions Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations Ex 5.1 Textbook Questions and Answers.

Gujarat Board Textbook Solutions Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations Ex 5.1

Express each of the complex numbers given in questions 1 to 10 in the form a + ib:
1. (5i) (- (frac{3}{5}) i)
2. i⁹ + i¹⁹
3. i^-39
4. 3(7 + i7) + i(7 + i7)
5. (1 – i) – (- 1 + i6)
6. ((frac{1}{5}) + i(frac{2}{5})) – (4 + i(frac{5}{2}))
7. [((frac{1}{3}) + i(frac{7}{3}))] + (4 + i(frac{1}{3})) – (- (frac{4}{3}) + i)
8. (i – 4)⁴
9. ((frac{1}{3}) + 3i)³
10. (- 2 – (frac{1}{3}) i)³
Solutions to questions 1 to 10:
1. (5i) (- (frac{3i}{5})) = (- 5 × (frac{3}{5})) × (i × i)
= – 3i² = (- 3)(- 1) = 3
= a + ib, where a = 3, b = 0.

2. i⁹ + i¹⁹ = i.i⁸ + i.i¹⁸ = i(i²)⁴ + i(i²)⁹
= i(- 1)⁴ + 1(- 1)⁹ = i – i = 0.
= a + ib, where a = 0, b = 0.

3. i^-39 = (frac{1}{i^{39}}) = (frac{1}{i.i^{38}}) = (frac{1}{ileft(i^{2}right)^{19}}) = (frac{1}{i(-1)^{19}}) = – (frac{1}{i})
= – (frac{1}{i}) × (frac{i}{i}) = – (frac{i}{i^{2}}) = (frac{- i}{- 1}) = i
= a + ib, where a = 0, b = 1.

4. 3(7 + i.7) + i(7 + i.7) = (21 + 21i) + (7i + 7i²)
= (21 + 21i) + (7i – 7) = 14 + 28i
= a + ib, where a = 14, b = 28.

5. (1 – i) – (- 1 + i.6) = (1 – i) + (1 – 6i) = 1 + 1 = – i – 6i
= 2 – 7i
= (a + ib), where a = 2, b = – 7.

6. ((frac{1}{5}) + i.(frac{2}{5})) – (4 + i.(frac{5}{2})) = ((frac{1}{5}) + (frac{2}{5})i) + (- 4 – (frac{5}{2})i)
= (frac{1}{5}) – 4 + (frac{2}{5})i – (frac{5}{2})i = – (frac{19}{5}) – (- (frac{2}{5}) + (frac{5}{2}))i
= – (frac{19}{5}) – (frac{21}{10})i = a + ib
where a = (frac{- 19}{5}) and b = (frac{- 21}{10}).

7. ((frac{1}{3}) + i(frac{7}{3})) + (4 + i(frac{1}{3})) – (- (frac{4}{3}) + i)
= ((frac{1}{3}) + (frac{7}{3})i) + (4 + (frac{1}{3})i) + ((frac{4}{3}) – i)
= ((frac{1}{3}) + 4 + (frac{4}{3})) + i((frac{7}{3}) + (frac{1}{3}) – 1) = (frac{17 }{3}) + i.(frac{5}{3})
= (a + ib), where a = (frac{17}{3}), b = (frac{5}{3}).

8. (1 – i)⁴ = 1 – 4i + 6i² – 4i³ + i⁴
= 1 – 4i + 6(- 1) – 4i(i²) + (i²)²
= 1 – 4i – 6 – 4i(- 1) + (- 1)²
= 1 – 4i – 6 + 4i + 1
= (1 – 6 + 1) = – 4 = (a + ib),
where a = – 4, b = 0.

9. ((frac{1}{3}) + 3i)³ = ((frac{1}{3}))³ + 3.((frac{1}{3}))² (3i) + 3. ((frac{1}{3}))(3i)² + (3i)³
= (frac{1}{27}) + i + 9(- 1) + 27i³
= (frac{1}{27}) + i + 9(- 1) + 27ii²
= (frac{1}{27}) + i – 9 + 27i(- 1) = (frac{1}{27}) + i – 9 – 27i
= ((frac{1}{27}) – 9) + (1 – 27)i
= – (frac{242}{27}) = – 26i
= a + ib, where a = (frac{- 242}{27}), b = – 26.

10.

= a + ib, where a = – (frac{22}{3}), b = – (frac{107}{27}).

Find the multiplicative inverse of each of the complex numbers given in questions 11 to 13:
11. 4 – 3i
12. (sqrt{5}) + 3i
13. – i
Solutions to questions 11 to 13:
11. Multiplicative inverse of 4 – 3i

12. Multiplicative inverse of (sqrt{5}) + 3i

13. Multiplicative inverse of – i
= (frac{1}{- i}) = (frac{- 1}{i}) × (frac{i}{i}) = (frac{-i}{i^{2}}) = (frac{- i}{- 1}) = i.

14. Express the following expression in the form a + ib:
(frac{(3+i sqrt{5})(3-i sqrt{5})}{(sqrt{3}+sqrt{2 i})-(sqrt{3}-i sqrt{2})})
Solution: