## Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 2 Complex Numbers Ex 2.4

Question 1.

Write the following in the rectangular form:

(i) \(\overline { (5+9i)+(2-4i) } \)

Solution:

\(\overline { (5+9i)+(2-4i) } \) = \(\overline {(5+9i)} \) + \(\overline {(2-4i)} \)

= 5 – 9i + 2 + 4i

= 7 – 5i

Samacheer Kalvi 12th Maths Guide

(ii) \(\frac {10-5i}{6+2i} \)

Solution:

(iii) \(\overline {3i} + \frac{2}{2-i}\)

Solution:

Question 2.

If z = x + iy, find the following in rectangular form.

(i) Re(\(\frac {1}{z} \))

Answer:

(ii) Re(i\(\overline {z}\)) = Re i(x – iy) = Re (ix + y)

= y

(iii) Im (3z + 4\(\overline {z}\) – 4i)

= Im (3x + 3 iy + 4x- 4iy – 4i)

= Im (7x + i (-y – 4) = -y – 4

Question 3.

If z_{1} = 2 – i and z_{2} = -4 + 3i, find the inverse of z_{1}, z_{2} and \(\frac {z_1}{z_2} \)

Solution:

z_{1} z_{2} = (z – i) (-4 + 3i)

= -8 + 3 + 4i + 6i

= – 5 + 70i

Question 4.

The complex numbers u, v, and w are related by \(\frac {1}{u}\) = \(\frac {1}{v}\) + \(\frac {1}{w}\). If v = 3 – 4i and w = 4 + 3i, find u in rectangular form.

Solution:

Question 5.

Prove the following properties:

(i) z is real if and only if z = \(\overline {z}\)

Solution:

Let z = x + iy

Then \(\overline {z}\) = x – iy

If z is real y = 0 then z – x and so z = x

∴ z = \(\overline {z}\)

conversely Let z = \(\overline {z}\) ⇒ x + iy = x – iy

⇒ 2 iy = 0

⇒ y = 0

∴ z is real.

(ii) Re(z) = \(\frac{z+\bar{z}}{2}\) and Im(z) = \(\frac{z-\bar{z}}{2i}\)

Solution:

= y = Im(z)

Question 6.

Find the least value of the positive integer n for which (√3 + i)^{n} (i) real, (ii) purely imaginary.

Solution:

Given (√3 + i)^{n}

Let z = √3 + i ⇒ x = √3, y = 1

∴ r = \(\sqrt{3+1}\) = 2

√3 = 2 cos θ ⇒ cos θ = \(\frac{√3}{2}\) θ is I quadrant

1 = 2 sin θ ⇒ sin θ = \(\frac{1}{2}\)

(i) to be purely real

cos \(\frac{nπ}{6}\) = cos π , sin \(\frac{nπ}{6}\) = sin π

⇒ \(\frac{nπ}{6}\) = π ⇒ n = 6

(ii) to be purely imaginary cos\(\frac{nπ}{6}\)= 0,

sin \(\frac{nπ}{6}\) = 1

\(\frac{nπ}{6}\) = \(\frac{π}{2}\) ⇒ n = 3

and so sin \(\frac{3π}{6}\) = sin \(\frac{π}{2}\) = 1

∴ If n = 3, it is purely imaginary

Question 7.

Show that

(i) (2 + i√3)^{10} – (2 – i√3)^{10} is purely imaginary.

Solution:

Let z = (2 + i√3)^{10} – (2 – i√3)^{10
}Let Z

= (2 – i√3)^{10} – (2 + i√3)^{10}

= -[(z + i√3)^{10} – (2 – i√3)^{10}]

= -z

since \(\overline {z}\) = -z ⇒ z is purely imaginary

(ii) is real

Solution:

= 2 + i

Therefore

= (2 – i)^{15} + (2 + i)^{15} = z

since \(\overline {z}\) = z ⇒ z is purely real.