Practicing the Intermediate 2nd Year Maths 2A Textbook Solutions Inter 2nd Year Maths 2A Complex Numbers Solutions Exercise 1(a) will help students to clear their doubts quickly.

## Intermediate 2nd Year Maths 2A Complex Numbers Solutions Exercise 1(a)

I.

Question 1.

If z_{1} = (2, -1), z_{2} = (6, 3), find z_{1} – z_{2}.

Solution:

z_{1} = (2, -1), z_{2} = (6, 3)

∴ z_{1} – z_{2} = (2 – 6, -1 – 3) = (-4, -4)

Question 2.

If z_{1} = (3, 5) and z_{2} = (2, 6), find z_{1} . z_{2}

Solution:

Given z_{1} = (3, 5) = 3 + 5i

and z_{2} = (2, 6) = 2 + 6i

z_{1} . z_{2} = (3 + 5i) . (2 + 6i)

= 6 + 10i + 18i + 30i^{2}

= 6 + 28i + 30(-1) [since i^{2} = -1]

= -24 + 28i

= (-24, 28)

Question 3.

Write the additive inverse of the following complex numbers.

(i) (√3, 5)

(ii) (-6, 5) + (10, -4)

(iii) (2, 1) (-4, 6)

Solution:

The additive inverse of (a, b) is (-a, -b)

(i) The additive inverse of (√3, 5) is (-√3, -5)

(ii) (-6, 5) + (10, -4)

= (-6 + 10, 5 + (-4))

= (4, 1)

∴ The additive inverse of (4, 1) is (-4, -1)

(iii) (2, 1) . (-4, 6)

= ((2 × -4 – 1 × 6), (1 × -4 + 2 × 6))

= (-8 – 6, -4 + 12)

= (-14, 8)

∴ The additive inverse of (-14, 8) is (14, -8)

II.

Question 1.

If z_{1} = (6, 3); z_{2} = (2, -1), find z_{1}/z_{2}.

Solution:

Given z_{1} = (6, 3) = 6 + 3i

and z_{2} = (2, -1) = 2 – i

Question 2.

If z = (cos θ, sin θ), find (z – \(\frac{1}{z}\))

Solution:

Given z = (cos θ, sin θ) = cos θ + i sin θ

⇒ \(\frac{1}{z}\) = cos θ – i sin θ

∴ z – \(\frac{1}{z}\) = (cos θ + i sin θ) – (cos θ – i sin θ)

= 2 i sin θ

= 0 + i (2 sin θ)

= (0, 2 sin θ)

Question 3.

Write the multiplicative inverse of the following complex numbers.

(i) (3, 4)

(ii) (sin θ, cos θ)

(iii) (7, 24)

(iv) (-2, 1)

Solution:

The multiplicative inverse of the complex number (a, b) is \(\left(\frac{a}{a^{2}+b^{2}}, \frac{-b}{a^{2}+b^{2}}\right)\)

(i) Multiplicative inverse of (3, 4) = \(\left(\frac{3}{3^{2}+4^{2}} \cdot \frac{-4}{3^{2}+4^{2}}\right)\) = \(\left(\frac{3}{25}, \frac{-4}{25}\right)\)

(ii) Multiplicative inverse of (sin θ, cos θ) = \(\left(\frac{\sin \theta}{\sin ^{2} \theta+\cos ^{2} \theta}, \frac{-\cos \theta}{\sin ^{2} \theta+\cos ^{2} \theta}\right)\) = (sin θ, -cos θ)

(iii) Multiplicative inverse of (7, 24) = \(\left(\frac{7}{7^{2}+24^{2}}, \frac{-24}{7^{2}+24^{2}}\right)\) = \(\left(\frac{7}{625}, \frac{-24}{625}\right)\)

(iv) Multiplicative inverse of (-2, 1) = \(\left(\frac{-2}{(-2)^{2}+(1)^{2}}, \frac{-1}{(-2)^{2}+(1)^{2}}\right)\) = \(\left(-\frac{2}{5},-\frac{1}{5}\right)\)