GSEB Solutions Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations Ex 5.1

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 = $\frac{1}{i^{39}}$ = $\frac{1}{i.i^{38}}$ = $\frac{1}{i\left(i^{2}\right)^{19}}$ = $\frac{1}{i(-1)^{19}}$ = – $\frac{1}{i}$
= – $\frac{1}{i}$ × $\frac{i}{i}$ = – $\frac{i}{i^{2}}$ = $\frac{- i}{- 1}$ = 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: