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¸

Squaring and adding, we get

4 = 1 + 3 = r^{2}(cos^{2} Î¸ + sin^{2} Î¸) = r^{2}

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

Squaring and adding, we get

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

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

Squaring and adding, we get

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

Squaring and adding, we get

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

Squaring and adding, we get

r^{2} = 3 + 1 = 4, r = 2.

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 \(\frac{Ï€}{6}\)).

8. z = i = r(cos Î¸ + isin Î¸)

âˆ´ rcos Î¸ = 0, r.sin Î¸ = 1.

Squaring and adding, we get r^{2} = 1.

âˆ´ r = 1.

Now, sin Î¸ = 1, cos Î¸ = 0 at Î¸ = \(\frac{Ï€}{2}\).

âˆ´ Polar form of z is cos \(\frac{Ï€}{2}\) + isin \(\frac{Ï€}{2}\).