# Inter 2nd Year Maths 2A Complex Numbers Solutions Ex 1(a)

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 z1 = (2, -1), z2 = (6, 3), find z1 – z2.
Solution:
z1 = (2, -1), z2 = (6, 3)
∴ z1 – z2 = (2 – 6, -1 – 3) = (-4, -4)

Question 2.
If z1 = (3, 5) and z2 = (2, 6), find z1 . z2
Solution:
Given z1 = (3, 5) = 3 + 5i
and z2 = (2, 6) = 2 + 6i
z1 . z2 = (3 + 5i) . (2 + 6i)
= 6 + 10i + 18i + 30i2
= 6 + 28i + 30(-1) [since i2 = -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 z1 = (6, 3); z2 = (2, -1), find z1/z2.
Solution:
Given z1 = (6, 3) = 6 + 3i
and z2 = (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) = $$\left(\frac{-2}{(-2)^{2}+(1)^{2}}, \frac{-1}{(-2)^{2}+(1)^{2}}\right)$$ = $$\left(-\frac{2}{5},-\frac{1}{5}\right)$$