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 = r^{2}(cos^{2} θ + sin^{2} θ) = r^{2}

∴ 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

1^{2} + 1^{2} = r^{2}

∴ 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

r^{2} = (- 1)^{2} + 1^{2} = 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

r^{2} = (- 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 r^{2} = (- 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

r^{2} = 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 r^{2} = 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 \(\frac{π}{2}\).