Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 2 Complex Numbers Ex 2.8 Textbook Questions and Answers, Notes.

## 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

Solution:

L.H.S

Question 2.

Show that

Solution:

Question 3.

Solution:

Aliter method:

Question 4.

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

(i) \(\frac{x}{y}+\frac{y}{x}=2 \cos (\alpha-\beta)\)

Solution:

Given 2 cos α = x + \(\frac{1}{x}\) and cos β = y + \(\frac{1}{y}\)

simplifying x² – 2x cos α + 1 = 0

if x = cos α + i sin α, then \(\frac{1}{x}\) = cos α – i sin α

similarly y = cos β + i sin β and \(\frac{1}{y}\) = cos β – i sin β

Hence proved

(ii) xy = (cos α + i sin α)(cos β + i sin β )

Solution:

xy = (cos α + i sin α) (cos β + i sin β)

= cos (α + β) + i sin (α + β)

[∵ arg (z_{1}z_{2}) = arg z_{1} + arg z_{2}

\(\frac{1}{xy}\) = cos (α + β) – i sin (α + β)

∴ xy – \(\frac{1}{xy}\) = cos (α + β) + i sin (α + β) – cos (α + β) + i sin (α + β)

= 2i sin (α + β)

Hence proved

(iii) \(\frac{x^{m}}{y^{n}}-\frac{y^{n}}{x^{m}}\) = 2 i sin (mα – nβ)

Solution:

Hence Proved

(iv) x^{m }y^{n} + \(\frac { 1 }{ x^m y^n }\) = 2 cos(mα – nβ)

Solution:

= (cos mα + sin mα) (cos nβ + i sin nβ)

= cos (mα + nβ) + i sin (mα + nβ)

Hence proved

Question 5.

Solve the equation z³ + 27 = 0.

Solution:

z³ + 27 = 0

z³ = – 27 = 27 (-1)

= 27 [cos(π + 2kπ) + i sin(π + 2kπ)], k ∈ z

∴ z = (27)^{1/3}[cos (2k + 1)π + i sin (2k+1)π]^{1/3 }k ∈ z

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:

Given ω ≠ 1 is a active root of unity

(z – 1)³ + 8 = 0

(z- 1)³ = -8

z – 1 = (-8)^{1/3} (1)^{1/3}

= (-2) (1, ω, ω²)

z – 1 = (-2, -2ω, -2ω²)

= z – 1 = -2

z = -2 + 1 = -1

z – 1 = -2ω

z = 1 – 2ω

z – 1 = -2ω²

z = 1 – 2ω²

Question 7.

Solution:

\(\sum_{k=1}^{8}\) (cos \(\frac { 2kπ }{ 9 }\) + i sin \(\frac { 2kπ }{ 9 }\))

The sum all nth root of unity is

1 + ω + ω² + …….. + ω^{n-1} = 0

From the given polar from , it is clear that the complex number is 1 + i0 (unity)

Question 8.

If ω ≠ 1 is a cube root of unity, show that

(i) (1 – ω + ω²)^{6} + (1 + ω – ω²)^{6} = 128.

Solution:

ω is a cube root of unity ω^{3} = 1; 1 + ω + ω^{2} = 0

(1 – ω + ω^{2})^{6} + (1 + ω – ω^{2})^{6}

= (-ω – ω)^{6} + (-ω^{2} – ω^{2})^{6}

= (-2ω)^{6} + (-2ω^{2})^{6}

= (-2)^{6}(ω^{6} + ω^{12})

= (64)(1 + 1)

= 128

(ii) (1 + ω) (1 + ω²) (1 + ω^{4}) (1 + ω^{8})….. (1 + ω^{2n}) = 1

Solution:

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

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

= (-ω^{2})(-ω)(-ω^{2})(-ω) …… 2n factors

= ω^{3}. ω^{3}

= 1

Hence proved.

Question 9.

If z = 2 – 2i, find the rotation of z by θ radians in the counterclockwise direction about the origin when

Solution:

Let 2 – 2i

Modules = |z| = \(\sqrt{2^2+2^2}\) = 2√2

Argument θ = tan^{-1}(\(\frac{-2}{2}\)) = tan^{-1}(-1) = –\(\frac{π}{4}\)

(i) when ‘z’ is rotated in the counter clockwise direction about the origin when θ = \(\frac{π}{3}\) i.,e argument of new position

(ii) θ = \(\frac{2π}{3}\)

Solution:

(iii) θ = \(\frac{3π}{3}\)

Solution: