# 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: (iii) $$\overline {3i} + \frac{2}{2-i}$$
Solution:  Question 2.
If z = x + iy, find the following in rectangular form.
(i) Re($$\frac {1}{z}$$) (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  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.

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