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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 ω^{3} = 1 and 1 + ω + ω^{2} = 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 = x^{2} + 1

⇒ x^{2} – 2x cos α + 1 = 0

Question 5.

Solve the equation z^{3} + 27 = 0

Solution:

z^{3} + 27 = 0

⇒ z^{3} = -27

⇒ z^{3} = 3^{3}(-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 – 2ω^{2}

Solution:

(z – 1)^{3} + 8 = 0

⇒ (z – 1 )^{3} = -8

The roots are -1, 1 – 2ω, 1 – 2ω^{2}

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, ω^{2}, ……., ω^{8}

Sum of the roots:

1 + ω + ω^{2} + …. + ω^{8} = 0 ⇒ ω + ω^{2} + ω^{3} + …… + ω^{8} = -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 + ω^{2})^{6} + (1 + ω – ω^{2})^{6} = 128

(ii) (1 + ω)(1 + ω^{2})(1 + ω^{4})(1 + ω^{8})……(1 + ω^{2}^{n}) = 1

Solution:

(i) (1 – ω + ω²)^{6} + (1 + ω – ω²)^{6}

= (-ω + ω)^{6} + (-ω – ω²)^{6}

= (-2ω)^{6} + (-2ω²)^{6
}= 2^{6 }[ω^{6} + ω^{12}] [∵ ω^{6 }= 1]

= 64 [1 + 1]

= 64 × 2

= 128

(ii) (1 – ω) (1 + ω²) (1 + ω) (1 + ω^{2}) ….. 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.

(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: x^{4} + 4 = 0

Solution:

Question 7.

If x = a + b, y = aω + bω^{2}, z = aω^{2} + bω, show that

(i) xyz = a^{3} + b^{3}

(ii) x^{3} + y^{3} + z^{3} = 3(a^{3} + b^{3})

Solution:

(i) x = a + A; y = aω + bω^{2}, z = aω^{2} + bω

Now xyz = (a + b) (aω + bω^{2}) (abω^{2} + bω) = (a+A) [aω^{3} + abω^{2} + abω + b^{2}ω^{3}]

= (a + b) (a^{2} – ab + b^{2}) = a^{3} + b^{3}

xyz = a^{3} + b^{3}

(ii) x = a + b, y = aω + bω^{2}, z = aω^{2} + bω

x + y + z = (a + aω + aω^{2}) + (b + bω^{2} + bω)

= a (1 + ω + ω^{2}) + b (1 + ω + ω^{2}) = a(0) + b(0) = 0

Now x + y + z = 0 ⇒ x^{3} + y^{3} + z^{3} = 3xyz

Here xyz = a^{3} + b^{3}

∴ x^{3} + y^{3} + z^{3} = 3 (a^{3} + b^{3})