Inter 2nd Year Maths 2A De Moivre’s Theorem Solutions Ex 2(b)

Practicing the Intermediate 2nd Year Maths 2A Textbook Solutions Inter 2nd Year Maths 2A De Moivre’s Theorem Solutions Exercise 2(b) will help students to clear their doubts quickly.

Intermediate 2nd Year Maths 2A De Moivre’s Theorem Solutions Exercise 2(b)

I. Find all the values of the following.

Question 1.
(i) (1 – i√3)^1/3
Solution:

(ii) (-i)^1/6
Solution:

(iii) (1 + i)^2/3
Solution:

(iv) (-16)^1/4
Solution:

(v) (-32)^1/5
Solution:

Question 2.
If A, B, C are angles of a triangle such that x = cis A, y = cis B, z = cis C, then find the value of xyz.
Solution:
∴ A, B, C are angles of a triangle
⇒ A + B + C = 180° ………..(1)
x = cis A, y = cis B, Z = cis C
xyz = cis (A + B + C)
= cos (A + B + C) + i sin (A + B + C)
= cos (180°) + i sin (180°)
= -1 + i(0)
= -1
∴ xyz = -1

Question 3.
(i) If x = cis θ, then find the value of \(\left[x^{6}+\frac{1}{x^{6}}\right]\)
Solution:
∵ x = cos θ + i sin θ
⇒ x⁶ = (cos θ + i sin θ)⁶ = cos 6θ + i sin 6θ
⇒ \(\frac{1}{x^{6}}\) = cos 6θ – i sin 6θ
∴ \(x^{6}+\frac{1}{x^{6}}\) = 2 cos 6θ

(ii) Find the cube roots of 8.
Solution:
Let x³ = 8
⇒ x = 8^1/3
⇒ x = (2³)^1/3 (1)^1/3
⇒ x = 2 (1)^1/3
Since cube roots of unity are 1, ω, ω²
∴ The cube roots or 8 are 2, 2ω, 2ω²

Question 4.
If 1, ω, ω² are the cube roots of unity, then prove that
(i) \(\frac{1}{2+\omega}-\frac{1}{1+2 \omega}=\frac{1}{1+\omega}\)
Solution:
ω is a cube root of unity.
1 + ω + ω² = 0 and ω³ = 1

(ii) (2 – ω) (2 – ω²) (2 – ω¹⁰) (2 – ω¹¹) = 49
Solution:
∵ 1, ω, ω² are the cube roots of unity.
ω³ = 1 and 1 + ω + ω² = 0
2 – ω¹⁰ = 2 – ω⁹ . ω
= 2 – (ω³)³ ω
= 2 – (1)³ ω
= 2 – ω
and 2 – ω¹¹ = 2 – (ω³)³ . ω²
= 2 – (1)³ ω²
= 2 – ω²
(2 – ω) (2 – ω²) = 4 – 2ω – 2ω² + ω³
= 4 – 2(ω + ω²) + 1
= 4 – 2(-1) + 1
= 4 + 2 + 1
= 7
∴ (2 – ω) (2 – ω²) (2 – ω¹⁰) (2 – ω¹¹) = (2 – ω) (2 – ω²) (2 – ω) (2 – ω²)
= ((2 – ω) (2 – ω²))²
= 7²
= 49

(iii) (x + y + z) (x + yω + zω²) (x + yω² + zω) = x³ + y³ + z³ – 3xyz
Solution:
∵ 1, ω, ω² are the cube roots of unity.
⇒ 1 + ω + ω² = 0 and ω³ = 1
Now consider,
(x + yω + zω²) (x + yω² + zω)
= x² + xyω² + zxω + xyω + y²ω³ + yzω² + zxω² + yzω⁴ + z²ω³
= x² + y² (1) + z² (1) + xy (ω + ω²) + yz (ω⁴ + ω²) + zx (ω + ω²)
= x² + y² + z² + xy (-1) + yz (ω + ω²) + zx (-1)
= x² + y² + z² – xy – yz – zx ……….(1)
L.H.S = (x + y + z) (x + yω + zω²) (x + yω² + zω)
= (x + y + z) (x² + y² + z² – xy – yz – zx) [by (1)]
= x³ + y³ + z³ – 3xyz
= R.H.S

Question 5.
Prove that -ω, and -ω² are the roots of z² – z + 1 =0, where ω and ω² are the complex cube roots of unity.
Solution:
Since ω and ω² are the complex cube roots of unity
∴ 1 + ω + ω² = 0 and ω² = 1
z² – z + 1 = (-ω)² – (-ω) + 1
= ω² + ω + 1
= 0
∴ -ω is a root of the equation z² – z + 1 = 0
z² – z + 1 = (-ω²)² – (-ω²) + 1
= ω⁴ + ω² + 1
= ω³ . ω + ω² + 1
= ω + ω²+ 1
= 0
∴ -ω² is a root of the equation z² – z + 1 = 0

Question 6.
If 1, ω, ω² are the cube roots of unity, then find the values of the following.
(i) (a + b)³ + (aω + bω²)³ + (aω² + bω)³
Solution:
Since 1, ω, ω² are the cube roots of unity
∴ 1 + ω + ω² = 0 and ω³ = 1
Now (a + b)³ = a³ + 3a²b + 3ab² + b³ ……..(1)
(aω + bω²)³ = [ω(a + bω)]³
= ω³ (a + bω)³
= (1) (a + bω)³
= a³ + 3a²bω + 3ab²ω² + b³ω³
= a³ + 3a²bω + 3ab²ω² + b³ ……….(2)
∵ ω³ = 1
and (aω² + bω)³ = [ω(aω + b)]³
= ω³ (aω + b)³
= (1) (aω + b)³
= a³ω³ + 3a²bω² + 3ab²ω + b³
= a³(1) + 3a²bω² + 3ab²ω + b³
∴ (aω² + bω)³ = a³ + 3a²bω² + 3ab²ω + b³ ……….(3)
Adding (1), (2) and (3)
(a + b)³ + (aω + bω²)³ + (aω² + bω)³ = 3a³ + 3a²b (1 + ω + ω²) + 3ab² (1 + ω + ω²) + 3b³
= 3(a³ + b³) + 3a²b (0) + 3ab² (0)
= 3(a³ + b³)
∴ (a + b)³ + (aω + bω²)³ + (aω² + bω)³ = 3 (a³ + b³)

(ii) (a + 2b)² + (aω² + 2bω)² + (aω + 2bω²)²
Solution:
(a + 2b)² = a² + 4ab + 4b² ……….(1)
(aω² + 2bω)² = a²ω⁴ + 4abω³ + 4b²ω²
= a²ω³ω + 4ab (1) + 4b²ω²
= a²ω + 4ab + 4b²ω² ………..(2)
and (aω + 2bω²)² = a²ω² + 4abω³ + 4b²ω⁴
= a²ω² + 4ab (1) + 4b²ω³ω
= a²ω² + 4ab + 4b² (1) ω
∴ (aω + 2bω²)² = a²ω² + 4ab + 4b²ω ……….(3)
By Adding (1), (2) and (3)
(a + 2b)² + (aω² + 2bω)² + (aω + 2bω²)²
= a² (1 + ω + ω²) + 12ab + 4b² (1 + ω + ω²)
= a² (0) + 12ab + 4b² (0)
= 12ab
∴ (a + 2b)² + (aω² + 2bω)² + (aω + 2bω²)² = 12ab

(iii) (1 – ω + ω²)³
Solution:
(1 – ω + ω²)³ = (-ω – ω)³
= (-2ω)³
= -8ω³
= -8(1)
= -8 (∵ 1 + ω + ω² = 0)

(iv) (1 – ω) (1 – ω²) (1 – ω⁴) (1 – ω⁸)
Solution:
1 – ω⁴ = 1 – (ω³) ω
= 1 – (1) ω
= 1 – ω
1 – ω⁸ = 1 – (ω³)² ω²
= 1 – (1) ω²
= 1 – ω²
∴ (1 – ω) (1 – ω²) (1 – ω⁴)(1 – ω⁸) = (1 – ω) (1 – ω²) (1 – ω) (1 – ω²)
= [(1 – ω) (1 – ω²)]²
= (1 – ω – ω² + ω³)²
= [1 – (ω + ω²) + 1] (∵ 1 + ω + ω² = 0)
= [1 – (-1) + 1]²
= (3)²
= 9
∴ (1 – ω) (1 – ω²) (1 – ω⁴) (1 – ω⁸) = 9

(v) \(\left[\frac{a+b \omega+c \omega^{2}}{c+a \omega+b \omega^{2}}\right]+\left[\frac{a+b \omega+c \omega^{2}}{b+c \omega+a \omega^{2}}\right]\)
Solution:
∴ 1, ω, ω² are the cube roots of unity
⇒ ω³ = 1 and 1 + ω + ω² = 0 ………..(1)

(vi) (i + ω)³ + (1 + ω²)³
Solution:
(i + ω)³ + (1 + ω²)³ = (-ω²)³ +(-ω)³
= -ω⁶ – ω³
= -1 – 1
= -2
∴ (1 + ω)³ + (1 + ω²)³ = -2

(vii) (1 – ω + ω²)⁵ + (1 + ω – ω²)⁵
Solution:
(1 – ω + ω²)⁵ + (1 + ω – ω²)5 = (-ω – ω)⁵ + (-ω² – ω²)⁵
= (-2ω)⁵ + (-2ω²)⁵
= -32ω⁵ – 32ω¹⁰
= -32(ω⁵ + ω¹⁰)
= -32(ω² + ω)
= -32(-1)
= 32
∴ (1 – ω + ω²)⁵ + (1 + ω – ω²)⁵ = 32

II.

Question 1.
Solve the following equations.
(i) x⁴ – 1 = 0
Solution:
x⁴ – 1 = 0
⇒ x⁴ = 1
⇒ x⁴ = cos 0° + i sin 0°
⇒ x⁴ = cos 2kπ + i sin 2kπ
⇒ x = (cos 2kπ + i sin 2kπ)^1/4
= cos \(\frac{k \pi}{2}\), k = 0, 1, 2, 3
i.e., cos 0° + i sin 0°, cos \(\frac{\pi}{2}\) + i sin \(\frac{\pi}{2}\), cos π + i sin π, cos \(\frac{3 \pi}{2}\) + i sin \(\frac{3 \pi}{2}\),
i.e., 1, i, -1, -i = ±1, ±i

(ii) x⁵ + 1 = 0
Solution:
x⁵ + 1 = 0
⇒ x⁵ = -1
⇒ x⁵ = cos π + i sin π
⇒ x⁵ = cos(2k + 1) π + i sin(2k + 1) π, k ∈ z
⇒ x = [cos(2k + 1) π + i sin(2k + 1) π]^1/5
⇒ x = cis \(\frac{(2 k+1) \pi}{5}\), k = 0, 1, 2, 3, 4

(iii) x⁹ – x⁵ + x⁴ – 1 = 0
Solution:
x⁹ – x⁵ + x⁴ – 1 = 0
⇒ x⁵ (x⁴ – 1) + 1 (x⁴ – 1) = 0
⇒ (x⁴ – 1) (x⁵ + 1) = 0
⇒ x⁴ – 1 = 0
Solving the roots are ±1, ±i
(see the above problem)
x⁵ + 1 = 0
Solving the roots are cis \(\frac{(2 k+1) \pi}{5}\)
k = 0, 1, 2, 3, 4 (see the above problem)
∴ The roots of the given equation are ±1, ±i, cis (2k + 1) \(\frac{\pi}{5}\), k = 0, 1, 2, 3, 4
i.e., ±1, ±i, cis(\(\pm \frac{\pi}{5}\)), cis(\(\pm \frac{3 \pi}{5}\))

(iv) x⁴ + 1 = 0
Solution:
x⁴ + 1 = 0
⇒ x⁴ = -1
⇒ x⁴ = cos π + i sin π
∴ x⁴ = cos(2kπ + π) + i sin(2kπ + π),
∴ x = [cis(2k + 1)π]^1/4
= cis(2k + 1) \(\frac{\pi}{4}\), where k = 0, 1, 2, 3
∴ x = \({cis} \frac{\pi}{4}, {cis}\left(\frac{3 \pi}{4}\right), {cis}\left(\frac{5 \pi}{4}\right)\) and \({cis}\left(\frac{7 \pi}{4}\right)\)
These four values of x are the solutions to the given equation.

Question 2.
Find the common roots of x¹² – 1 = 0 and x⁴ + x² + 1 = 0
Solution:
Consider x¹² – 1 = 0
⇒ x¹² = 1
⇒ x¹² = (cos 0 + i sin 0)
⇒ x¹² = (cos 2kπ + i sin 2kπ), k is a positive integer
⇒ x = (cos 2kπ + i sin 2kπ)^1/2

Question 3.
Find the number of 15th roots of unity, which are also the 25th roots of unity.
Solution:
The number of common roots = H.C.F of {15, 25} = 5

Question 4.
If the cube roots of unity are 1, ω, ω2, then find the roots of the equation (x – 1)³ + 8 = 0.
Solution:
Given (x – 1)³ + 8 = 0
⇒ (x – 1)³ = -8
⇒ (x – 1)³ = (-2)³ (1)³
⇒ (x – 1) = (-2) (1)^1/3
⇒ x – 1 = -2, -2ω, -2ω²
⇒ x = 1 – 2, 1 – 2ω, 1 – 2ω²
⇒ x = -1, 1 – 2ω, 1 – 2ω²

Question 5.
Find the product of all the values of (1 + i)4/5.
Solution:

Question 6.
If z² + z + 1 =0, where z is a complex number, prove that
\(\left(z+\frac{1}{z}\right)^{2}+\left(z^{2}+\frac{1}{z^{2}}\right)^{2}+\left(z^{3}+\frac{1}{z^{3}}\right)^{2}\) + \(\left(z^{4}+\frac{1}{z^{4}}\right)^{2}+\left(z^{5}+\frac{1}{z^{5}}\right)^{2}+\left(z^{6}+\frac{1}{z^{6}}\right)\) = 12
Solution:
Given z² + z + 1 = 0
⇒ z = \(\frac{-1 \pm \sqrt{1-4.1 .1}}{2}\)
= \(\frac{-1 \pm i \sqrt{3}}{2}\)
= \(\frac{-1+i \sqrt{3}}{2}, \frac{-1-i \sqrt{3}}{2}\)
= ω, ω²
∴ 1 + ω + ω² = 0 and ω³ = 1
If z = ω then

= (ω + ω²)² + (ω² + ω)² + (1 + 1)² + (ω + ω²)² + (ω² + ω)² + (1 + 1 )²
= (-1)² + (-1)² + 4 + (-1)² + (-1)² + 4
= 1 + 1 + 4 + 1 + 1 + 4
= 12
Similarly If z = ω² then
\(\left(z+\frac{1}{z}\right)^{2}+\left(z^{2}+\frac{1}{z^{2}}\right)^{2}+\left(z^{3}+\frac{1}{z^{3}}\right)^{2}\) + \(\left(z^{4}+\frac{1}{z^{4}}\right)^{2}+\left(z^{5}+\frac{1}{z^{5}}\right)^{2}+\left(z^{6}+\frac{1}{z^{6}}\right)\) = 12

III.

Question 1.
If 1, α, α², α³, ……., α^n-1 be the nth roots of unity, then prove that 1^p + α^p + (α²)^p + (α³)^p + …… + (α^n-p)^2p = 0
= 0; if p ≠ kn
= n; if p ≠ kn, where p, k ∈ N
Solution:
nth roots of unity are 1, α, α², ………., α^n-1


∴ Each term of the series in (1) is 1
Hence the sum of the series 1 + αp + (α²)p + (α³)p + …….. + (α^n-1)^p
= 1 + 1 + 1 + ……… + 1 (n times)
= n(1)
= n

Question 2.
Prove that the sum of 99th powers of the roots of the equation x⁷ – 1 = 0 is zero and hence deduce the roots of x⁶ + x⁵ + x⁴ + x³ + x² + x + 1 = 0.
Solution:

Question 3.
If n is a positive integer, show that \((p+i q)^{1 / n}+(p-i q)^{1 / n}=2\left(p^{2}+q^{2}\right)^{1 / 2 n}\) . \(\cos \left(\frac{1}{n}, \tan \frac{q}{p}\right)\)
Solution:

Question 4.
Show that one value of \(\left(\frac{1+\sin \frac{\pi}{8}+i \cos \frac{\pi}{8}}{1+\sin \frac{\pi}{8}-i \cos \frac{\pi}{8}}\right)^{8 / 3}\) is -1.
Solution:

Question 5.
Solve (x – 1)ⁿ = xⁿ, n is a positive integer.
Solution:
Since x = 0 is not a solution of the given equation, it is equivalent to the equation \(\left(\frac{x-1}{x}\right)^{n}=1\)
Clearly \(\left(\frac{x-1}{x}\right)^{n}=1\)
⇒ \(\frac{x-1}{x}\) is an nth root of unity other than one.
Suppose that ω is an nth root of unity and ω ≠ 1.
Then, \(\frac{x-1}{x}\) = ω
⇒ x – 1 = xω
⇒ (1 – ω) x = 1
⇒ x = \(\frac{1}{1-\omega}\), (∵ ω ≠ 1) ……….(1)