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. i^{9} + i^{19}

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)

= – 3i^{2} = (- 3)(- 1) = 3

= a + ib, where a = 3, b = 0.

2. i^{9} + i^{19} = i.i^{8} + i.i^{18} = i(i^{2})^{4} + i(i^{2})^{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 + 7i^{2})

= (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 + 6i^{2} – 4i^{3} + i^{4}

= 1 – 4i + 6(- 1) – 4i(i^{2}) + (i^{2})^{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) + 27i^{3}

= \(\frac{1}{27}\) + i + 9(- 1) + 27ii^{2}

= \(\frac{1}{27}\) + 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: