Samacheer Kalvi 12th Maths Solutions Chapter 2 Complex Numbers Ex 2.8

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Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 2 Complex Numbers Ex 2.8

Question 1.
If to ω ≠ 1 is a cube root of unity, then show that \(\frac{a+b \omega+c \omega^{2}}{b+c \omega+a \omega^{2}}+\frac{a+b \omega+c \omega^{2}}{c+a \omega+b \omega^{2}}=1\)
Solution:
Since ω is a cube root of unity, we have ω3 = 1 and 1 + ω + ω2 = 0
Samacheer Kalvi 12th Maths Solutions Chapter 2 Complex Numbers Ex 2.8 Q1

Samacheer Kalvi 12th Maths Solutions Chapter 2 Complex Numbers Ex 2.8

Question 2.
Show that \(\left(\frac{\sqrt{3}}{2}+\frac{i}{2}\right)^{5}+\left(\frac{\sqrt{3}}{2}-\frac{i}{2}\right)^{5}=-\sqrt{3}\)
Solution:
Samacheer Kalvi 12th Maths Solutions Chapter 2 Complex Numbers Ex 2.8 Q2
Samacheer Kalvi 12th Maths Solutions Chapter 2 Complex Numbers Ex 2.8 Q2.1

Question 3.
Find the value of \(\left(\frac{1+\sin \frac{\pi}{10}+i \cos \frac{\pi}{10}}{1+\sin \frac{\pi}{10}-i \cos \frac{\pi}{10}}\right)^{10}\)
Solution:
Samacheer Kalvi 12th Maths Solutions Chapter 2 Complex Numbers Ex 2.8 Q3

Question 4.
If 2 cos α = x + \(\frac{1}{x}\) and 2 cos β = y + \(\frac{1}{y}\), show that
Samacheer Kalvi 12th Maths Solutions Chapter 2 Complex Numbers Ex 2.8 Q4
Solution:
(i) 2 cos α = x + \(\frac{1}{x}\)
⇒ 2 cos α = \(\frac{x^{2}+1}{x}\)
⇒ 2x cos α = x2 + 1
⇒ x2 – 2x cos α + 1 = 0
Samacheer Kalvi 12th Maths Solutions Chapter 2 Complex Numbers Ex 2.8 Q4.1
Samacheer Kalvi 12th Maths Solutions Chapter 2 Complex Numbers Ex 2.8 Q4.2
Samacheer Kalvi 12th Maths Solutions Chapter 2 Complex Numbers Ex 2.8 Q4.3

Samacheer Kalvi 12th Maths Solutions Chapter 2 Complex Numbers Ex 2.8

Question 5.
Solve the equation z3 + 27 = 0
Solution:
z3 + 27 = 0
⇒ z3 = -27
⇒ z3 = 33(-1)
⇒ z = 3(-1)1/3
Samacheer Kalvi 12th Maths Solutions Chapter 2 Complex Numbers Ex 2.8 Q5

Question 6.
If ω ≠ 1 is a cube root of unity, show that the roots of the equation (z – 1)3 + 8 = 0 are -1, 1 – 2ω, 1 – 2ω2
Solution:
(z – 1)3 + 8 = 0
⇒ (z – 1 )3 = -8
Samacheer Kalvi 12th Maths Solutions Chapter 2 Complex Numbers Ex 2.8 Q6
The roots are -1, 1 – 2ω, 1 – 2ω2

Question 7.
Find the value of \(\sum_{k=1}^{8}\left(\cos \frac{2 k \pi}{9}+i \sin \frac{2 k \pi}{9}\right)\)
Solution:
\(\sum_{k=1}^{8}\left(\cos \frac{2 k \pi}{9}+i \sin \frac{2 k \pi}{9}\right)\)
We know that 9th roots of unit are 1, ω, ω2, ……., ω8
Sum of the roots:
1 + ω + ω2 + …. + ω8 = 0 ⇒ ω + ω2 + ω3 + …… + ω8 = -1
Samacheer Kalvi 12th Maths Solutions Chapter 2 Complex Numbers Ex 2.8 Q7
The sum of all the terms \(\sum_{k=1}^{8}\left(\cos \frac{2 k \pi}{9}+i \sin \frac{2 k \pi}{9}\right)\) = -1

Question 8.
If ω ≠ 1 is a cube root of unity, show that
(i) (1 – ω + ω2)6 + (1 + ω – ω2)6 = 128
(ii) (1 + ω)(1 + ω2)(1 + ω4)(1 + ω8)……(1 + ω2n) = 1
Solution:
(i) (1 – ω + ω²)6 + (1 + ω – ω²)6
= (-ω + ω)6 + (-ω – ω²)6
= (-2ω)6 + (-2ω²)6
= 26 6 + ω12] [∵ ω6 = 1]
= 64 [1 + 1]
= 64 × 2
= 128
(ii) (1 – ω) (1 + ω²) (1 + ω) (1 + ω2) ….. 2n factors
(-ω²) (-ω) (-ω²) (-ω) ……. 2n factor
(ω³) (ω³) …… 2n factor
= 1.1 …….. 2n factor
= 1

Question 9.
If z = 2 – 2i, find the rotation of z by θ radians in the counter clockwise direction about the origin when
(i) θ = \(\frac{\pi}{3}\)
(ii) θ = \(\frac{2 \pi}{3}\)
(iii) θ = \(\frac{3 \pi}{2}\)
Solution:
(i) z = 2 – 2i = 2 (1 – i) = r(cos θ + i sin θ)
\(r=\sqrt{x^{2}+y^{2}}=2 \sqrt{1+1}=2 \sqrt{2}\)
\(\alpha=\tan ^{-1}=\left|\frac{y}{x}\right|=\tan ^{-1}|1|=\frac{\pi}{4}\)
(1 – i) lies in IV quadrant
θ = -α = \(-\frac{\pi}{4}\)
\(\Rightarrow z=2 \sqrt{2}\left[\cos \left(\frac{-\pi}{4}\right)+i \sin \left(\frac{-\pi}{4}\right)\right]\)
z is rotated by θ = \(\frac{\pi}{3}\) in the counter clock wise direction.
Samacheer Kalvi 12th Maths Solutions Chapter 2 Complex Numbers Ex 2.8 Q9

(ii) z is rotated by θ = \(\frac{2 \pi}{3}\) in the counter clockwise direction.
Samacheer Kalvi 12th Maths Solutions Chapter 2 Complex Numbers Ex 2.8 Q9.1

(iii) z is rotated by θ = \(\frac{3 \pi}{2}\) in the counter clockwise direction.
Samacheer Kalvi 12th Maths Solutions Chapter 2 Complex Numbers Ex 2.8 Q9.2

Samacheer Kalvi 12th Maths Solutions Chapter 2 Complex Numbers Ex 2.8

Question 10.
Prove that the values of \(\sqrt[4]{-1} \text { are } \pm \frac{1}{\sqrt{2}}(1 \pm i)\)
Solution:
Samacheer Kalvi 12th Maths Solutions Chapter 2 Complex Numbers Ex 2.8 Q10
The roots are \(\pm \frac{1}{\sqrt{2}}(1 \pm i)\)

Samacheer Kalvi 12th Maths Solutions Chapter 2 Complex Numbers Ex 2.8 Additional Problems

Question 1.
Samacheer Kalvi 12th Maths Solutions Chapter 2 Complex Numbers Ex 2.8 1
Solution:
Samacheer Kalvi 12th Maths Solutions Chapter 2 Complex Numbers Ex 2.8 2

Question 2.
Samacheer Kalvi 12th Maths Solutions Chapter 2 Complex Numbers Ex 2.8 3
Solution:
Samacheer Kalvi 12th Maths Solutions Chapter 2 Complex Numbers Ex 2.8 4
Samacheer Kalvi 12th Maths Solutions Chapter 2 Complex Numbers Ex 2.8 5

Question 3.
Prove that: (1 + i)4n and (1 + i)4n + 2 are real and purely imaginary respectively.
Solution:
Let z = 1 + i
Samacheer Kalvi 12th Maths Solutions Chapter 2 Complex Numbers Ex 2.8 6
= 2i (-1)n which is purely imaginary.

Samacheer Kalvi 12th Maths Solutions Chapter 2 Complex Numbers Ex 2.8

Question 4.
Samacheer Kalvi 12th Maths Solutions Chapter 2 Complex Numbers Ex 2.8 7
Solution:
Samacheer Kalvi 12th Maths Solutions Chapter 2 Complex Numbers Ex 2.8 8

Question 5.
If a = cos 2α + i sin 2α, b = cos 2β + i sin 2β and c = cos 2γ + i sin 2γ, prove that
Samacheer Kalvi 12th Maths Solutions Chapter 2 Complex Numbers Ex 2.8 9
Solution:
Samacheer Kalvi 12th Maths Solutions Chapter 2 Complex Numbers Ex 2.8 10

Question 6.
Solve: x4 + 4 = 0
Solution:
Samacheer Kalvi 12th Maths Solutions Chapter 2 Complex Numbers Ex 2.8 11

Question 7.
If x = a + b, y = aω + bω2, z = aω2 + bω, show that
(i) xyz = a3 + b3
(ii) x3 + y3 + z3 = 3(a3 + b3)
Solution:
(i) x = a + A; y = aω + bω2, z = aω2 + bω
Now xyz = (a + b) (aω + bω2) (abω2 + bω) = (a+A) [aω3 + abω2 + abω + b2ω3]
= (a + b) (a2 – ab + b2) = a3 + b3
xyz = a3 + b3

(ii) x = a + b, y = aω + bω2, z = aω2 + bω
x + y + z = (a + aω + aω2) + (b + bω2 + bω)
= a (1 + ω + ω2) + b (1 + ω + ω2) = a(0) + b(0) = 0
Now x + y + z = 0 ⇒ x3 + y3 + z3 = 3xyz
Here xyz = a3 + b3
∴ x3 + y3 + z3 = 3 (a3 + b3)

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