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})
∴ Let – 1 – i(sqrt{3}) = r(cos θ + isinθ)
∴ – 1 = rcos θ and (sqrt{-3}) = r sin θ
Squaring and adding, we get
4 = 1 + 3 = r²(cos² θ + sin² θ) = r²
∴ r = (sqrt{4}) = 2
tan θ = (frac{sqrt{3}}{-1}) = – (sqrt{3}).
where θ lies in third quadrant.
∴ θ = – 180° + 60° = – 120° = – (frac{2π}{3})
|z| = 2, arg z = – (frac{2π}{3})

2. z = – (sqrt{3}) + i = r(cos θ + 180°)
∴ rcos θ = – (sqrt{3}), r sin θ = 1
Squaring and adding ((sqrt{3}))² + 1 = 4
∴ r = 2, we get

tan θ = (frac{-1}{sqrt{3}}) ⇒ θ lies in II Quadrant
∴ θ = 180° – 30° = 150° = (frac{5π}{6})
∴ |z| = 2, arg z = (frac{5π}{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 θ + isin θ)
∴ rcos θ = 1, rsin θ = – 1
Squaring and adding, we get
1² + 1² = r²
∴ r = (sqrt{2})
and tan θ = (frac{- 1}{1}) = – 1.
∴ θ lies in IV quadrant, since sin θ is negative and cos θ is positive.
∴ θ = – 45° = – (frac{π}{4}).
∴ Polar form of 1 – i is
(sqrt{2})[cos (- (frac{π}{4})) + isin (- (frac{π}{4}))]

4. z = – 1 + i = r(cos θ + isin θ)
∴ rcos θ = – 1, r sin θ = 1
Squaring and adding, we get
r² = (- 1)² + 1² = 2 is r = (sqrt{2}).
Here, sin θ is +ve and cos θ is – ve. Therefore, θ lies in the second quadrant.
i.e; θ = π – (frac{π}{4}) = (frac{3π}{4}).
∴ z = (sqrt{2})(cos (frac{3π}{4}) + isin (frac{3π}{4})).

5. z = – 3 = r(cos θ + i sin θ)
∴ rcos θ = – 1, r sin θ = – 1
Squaring and adding, we get
r² = (- 1)² + (- 1)² = z
∴ r = (sqrt{2}).

6. z = – 3 = r(cos θ + i sin θ)
∴ rcos θ = – 3, rsin θ = 0.
Squaring and adding, we get r² = (- 3)²
∴ r = 3.
tan θ = 0 ⇒ θ = π [∵ cos π = 0]
∴ – 3 = 3(cos π + isin π).

7. r = (sqrt{3}) + i = r(cos π + isinπ)
∴ rcos θ = (sqrt{3}), r sin θ = 1
Squaring and adding, we get
r² = 3 + 1 = 4, r = 2.
Also, tan θ = (frac{1}{sqrt{3}}). Also, sin θ and cos θ both are positive.
∴ θ = 30° = (frac{π}{6}).
Polar form of z is 2(cos (frac{π}{6}) + isin (frac{π}{6})).

8. z = i = r(cos θ + isin θ)
∴ rcos θ = 0, r.sin θ = 1.
Squaring and adding, we get r² = 1.
∴ r = 1.
Now, sin θ = 1, cos θ = 0 at θ = (frac{π}{2}).
∴ Polar form of z is cos (frac{π}{2}) + isin (frac{π}{2}).