Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 2 Complex Numbers Ex 2.4 Textbook Questions and Answers, Notes.

## 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

(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\(\bar{z}\)) = Re[i(\(\overline{x+i y}\))]

= Re(ix + y)

= y

(iii) Im(3z + 4\(\bar{z}\) – 4i)

= Im (3(x + iy) + 4(x – iy) – 4i)

= Im (3x + 3iy + 4x – 4iy – 4i)

= Im (3x + 4 + i (3y – 4y – 4)

= Im (3x + 4x + i(-y – 4))

= Im [7x + i(-y – 4)]

= -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} = 2 – i, z_{2} = -4 + 3i

z_{1} z_{2} = (2 – i) (-4 + 3i)

= -8 + 3 + 4i + 6i

= -5 + 10i

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:

v = 3 – 4i, w = 4 + 3i

Question 5.

Prove the following properties:

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

Solution:

z is real iff z = \(\bar{z}\)

Let z = x + iy

z = \(\bar{z}\)

⇒ x + iy = x – iy

⇒ 2iy = 0

⇒ y = 0

⇒ z is real.

z is real iff z = \(\bar{z}\)

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

Solution:

let z = x + iy

\(\overline {z}\) = x – iy

Hence proved

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}

= (√3)² + 2i √3 + (i)²

= 3 + 2i √3 – 1

= 2 + 2i √3

= 2(1 + i√3)

put n = 3 or 4 or 5

then real part is not possible

which is purely real ∴ n = 6

(ii) (√3 + i)^{n}

which is purely imaginary

∴ n = 3

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

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

(ii) \(\left(\frac{19-7 i}{9+i}\right)^{12}\) + \(\left(\frac{20-5 i}{7-6 i}\right)^{12}\) is real

Solution:

∴ z is real.