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

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:

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:

Question 4.
If 2 cos α = x + \(\frac{1}{x}\) and 2 cos β = y + \(\frac{1}{y}\), show that

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

Question 5.
Solve the equation z³ + 27 = 0
Solution:
z³ + 27 = 0
⇒ z³ = -27
⇒ z³ = 3³(-1)
⇒ z = 3(-1)^1/3

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

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

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) ω is a cube root of unity ω³ = 1; 1 + ω + ω² = 0
(1 – ω + ω²)⁶ + (1 + ω – ω²)⁶
= (-ω – ω)⁶ + (-ω² – ω²)⁶
= (-2ω)⁶ + (-2ω²)⁶
= (-2)⁶(ω⁶ + ω¹²)
= (64)(1 + 1)
= 128
(ii) (1 + ω)(1 + ω²)(1 + ω⁴)(1 + ω⁸) …… (1 + ω²ⁿ)
= (1 + ω)(1 + ω²)(1 + ω⁴)(1 + ω⁸) ……. 2n factors
= (-ω²)(-ω)(-ω²)(-ω) …… 2n factors
= ω³. ω³
= 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.

(ii) z is rotated by θ = \(\frac{2 \pi}{3}\) in the counter clockwise direction.

(iii) z is rotated by θ = \(\frac{3 \pi}{2}\) in the counter clockwise direction.

Question 10.
Prove that the values of \(\sqrt[4]{-1} \text { are } \pm \frac{1}{\sqrt{2}}(1 \pm i)\)
Solution:

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.

Solution:

Question 2.

Solution:

Question 3.
Prove that: (1 + i)⁴ⁿ and (1 + i)^{4n + 2} are real and purely imaginary respectively.
Solution:
Let z = 1 + i

= 2i (-1)ⁿ which is purely imaginary.

Question 4.

Solution:

Question 5.
If a = cos 2α + i sin 2α, b = cos 2β + i sin 2β and c = cos 2γ + i sin 2γ, prove that

Solution:

Question 6.
Solve: x⁴ + 4 = 0
Solution:

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)