Samacheer Kalvi 12th Maths Guide Chapter 2 Complex Numbers Ex 2.4

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:
Samacheer Kalvi 12th Maths Guide Chapter 2 Complex Numbers Ex 2.4 1

(iii) \(\overline {3i} + \frac{2}{2-i}\)
Solution:
Samacheer Kalvi 12th Maths Guide Chapter 2 Complex Numbers Ex 2.4 2

Samacheer Kalvi 12th Maths Guide Chapter 2 Complex Numbers Ex 2.4

Question 2.
If z = x + iy, find the following in rectangular form.
(i) Re(\(\frac {1}{z} \))
Answer:
Samacheer Kalvi 12th Maths Guide Chapter 2 Complex Numbers Ex 2.4 3

(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 z1 = 2 – i and z2 = -4 + 3i, find the inverse of z1, z2 and \(\frac {z_1}{z_2} \)
Solution:
z1 z2 = (z – i) (-4 + 3i)
= -8 + 3 + 4i + 6i
= – 5 + 70i
Samacheer Kalvi 12th Maths Guide Chapter 2 Complex Numbers Ex 2.4 4

Samacheer Kalvi 12th Maths Guide Chapter 2 Complex Numbers Ex 2.4

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:
Samacheer Kalvi 12th Maths Guide Chapter 2 Complex Numbers Ex 2.4 5

Samacheer Kalvi 12th Maths Guide Chapter 2 Complex Numbers Ex 2.4

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:
Samacheer Kalvi 12th Maths Guide Chapter 2 Complex Numbers Ex 2.4 6
= y = Im(z)

Samacheer Kalvi 12th Maths Guide Chapter 2 Complex Numbers Ex 2.4

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 = \(\frac{1}{2}\)
Samacheer Kalvi 12th Maths Guide Chapter 2 Complex Numbers Ex 2.4 7

(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

Samacheer Kalvi 12th Maths Guide Chapter 2 Complex Numbers Ex 2.4

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
Samacheer Kalvi 12th Maths Guide Chapter 2 Complex Numbers Ex 2.4 8
= (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

Samacheer Kalvi 12th Maths Guide Chapter 2 Complex Numbers Ex 2.4

(ii) Samacheer Kalvi 12th Maths Guide Chapter 2 Complex Numbers Ex 2.4 9 is real
Solution:
Samacheer Kalvi 12th Maths Guide Chapter 2 Complex Numbers Ex 2.4 10
= 2 + i
Therefore
Samacheer Kalvi 12th Maths Guide Chapter 2 Complex Numbers Ex 2.4 11
= (2 – i)15 + (2 + i)15 = z
since \(\overline {z}\) = z ⇒ z is purely real.

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