# 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 Ã— $$\frac{3}{5}$$) Ã— (i Ã— 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 = $$\frac{1}{i^{39}}$$ = $$\frac{1}{i.i^{38}}$$ = $$\frac{1}{i\left(i^{2}\right)^{19}}$$ = $$\frac{1}{i(-1)^{19}}$$ = – $$\frac{1}{i}$$
= – $$\frac{1}{i}$$ Ã— $$\frac{i}{i}$$ = – $$\frac{i}{i^{2}}$$ = $$\frac{- i}{- 1}$$ = 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
= $$\frac{1}{- i}$$ = $$\frac{- 1}{i}$$ Ã— $$\frac{i}{i}$$ = $$\frac{-i}{i^{2}}$$ = $$\frac{- i}{- 1}$$ = 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: