# GSEB Solutions Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations Ex 5.2

Gujarat Board GSEB Textbook Solutions Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations Ex 5.2 Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations Ex 5.2

Find the modulus and the argument of each of the complex numbers in questions 1 to 2:
1. z = – 1 – i$$\sqrt{3}$$
2. z = – $$\sqrt{3}$$ + i
Solutions to questions 1 to 2:
1. z = – 1 – i$$\sqrt{3}$$
âˆ´ Let – 1 – i$$\sqrt{3}$$ = r(cos Î¸ + isinÎ¸)
âˆ´ – 1 = rcos Î¸ and $$\sqrt{-3}$$ = r sin Î¸
4 = 1 + 3 = r2(cos2 Î¸ + sin2 Î¸) = r2
âˆ´ r = $$\sqrt{4}$$ = 2
tan Î¸ = $$\frac{\sqrt{3}}{-1}$$ = – $$\sqrt{3}$$.
where Î¸ lies in third quadrant.
âˆ´ Î¸ = – 180Â° + 60Â° = – 120Â° = – $$\frac{2Ï€}{3}$$
|z| = 2, arg z = – $$\frac{2Ï€}{3}$$

2. z = – $$\sqrt{3}$$ + i = r(cos Î¸ + 180Â°)
âˆ´ rcos Î¸ = – $$\sqrt{3}$$, r sin Î¸ = 1
Squaring and adding ($$\sqrt{3}$$)2 + 1 = 4
âˆ´ r = 2, we get

tan Î¸ = $$\frac{-1}{\sqrt{3}}$$ â‡’ Î¸ lies in II Quadrant
âˆ´ Î¸ = 180Â° – 30Â° = 150Â° = $$\frac{5Ï€}{6}$$
âˆ´ |z| = 2, arg z = $$\frac{5Ï€}{6}$$.

Convert each of the complex numbers given in questions 3 to 8 in the polar form:
3. 1 – i
4. – 1 + i
5. – 1 – i
6. – 3
7. $$\sqrt{3}$$ + i
8. i
Solutions to questions 3 to 8:
3. If 1 – i = r(cos Î¸ + isin Î¸)
âˆ´ rcos Î¸ = 1, rsin Î¸ = – 1
12 + 12 = r2
âˆ´ r = $$\sqrt{2}$$
and tan Î¸ = $$\frac{- 1}{1}$$ = – 1.
âˆ´ Î¸ lies in IV quadrant, since sin Î¸ is negative and cos Î¸ is positive.
âˆ´ Î¸ = – 45Â° = – $$\frac{Ï€}{4}$$.
âˆ´ Polar form of 1 – i is
$$\sqrt{2}$$[cos (- $$\frac{Ï€}{4}$$) + isin (- $$\frac{Ï€}{4}$$)]

4. z = – 1 + i = r(cos Î¸ + isin Î¸)
âˆ´ rcos Î¸ = – 1, r sin Î¸ = 1
r2 = (- 1)2 + 12 = 2 is r = $$\sqrt{2}$$.
Here, sin Î¸ is +ve and cos Î¸ is – ve. Therefore, Î¸ lies in the second quadrant.
i.e; Î¸ = Ï€ – $$\frac{Ï€}{4}$$ = $$\frac{3Ï€}{4}$$.
âˆ´ z = $$\sqrt{2}$$(cos $$\frac{3Ï€}{4}$$ + isin $$\frac{3Ï€}{4}$$).

5. z = – 3 = r(cos Î¸ + i sin Î¸)
âˆ´ rcos Î¸ = – 1, r sin Î¸ = – 1
r2 = (- 1)2 + (- 1)2 = z
âˆ´ r = $$\sqrt{2}$$.

6. z = – 3 = r(cos Î¸ + i sin Î¸)
âˆ´ rcos Î¸ = – 3, rsin Î¸ = 0.
Squaring and adding, we get r2 = (- 3)2
âˆ´ r = 3.
tan Î¸ = 0 â‡’ Î¸ = Ï€ [âˆµ cos Ï€ = 0]
âˆ´ – 3 = 3(cos Ï€ + isin Ï€).

7. r = $$\sqrt{3}$$ + i = r(cos Ï€ + isinÏ€)
âˆ´ rcos Î¸ = $$\sqrt{3}$$, r sin Î¸ = 1
Also, tan Î¸ = $$\frac{1}{\sqrt{3}}$$. Also, sin Î¸ and cos Î¸ both are positive.
âˆ´ Î¸ = 30Â° = $$\frac{Ï€}{6}$$.
Polar form of z is 2(cos $$\frac{Ï€}{6}$$ + isin $$\frac{Ï€}{6}$$).
Now, sin Î¸ = 1, cos Î¸ = 0 at Î¸ = $$\frac{Ï€}{2}$$.
âˆ´ Polar form of z is cos $$\frac{Ï€}{2}$$ + isin $$\frac{Ï€}{2}$$.