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:
Inter 2nd Year Maths 2A De Moivre’s Theorem Solutions Ex 2(b) I Q1(i)

(ii) (-i)1/6
Solution:
Inter 2nd Year Maths 2A De Moivre’s Theorem Solutions Ex 2(b) I Q1(ii)
Inter 2nd Year Maths 2A De Moivre’s Theorem Solutions Ex 2(b) I Q1(ii).1

(iii) (1 + i)2/3
Solution:
Inter 2nd Year Maths 2A De Moivre’s Theorem Solutions Ex 2(b) I Q1(iii)

(iv) (-16)1/4
Solution:
Inter 2nd Year Maths 2A De Moivre’s Theorem Solutions Ex 2(b) I Q1(iv)

(v) (-32)1/5
Solution:
Inter 2nd Year Maths 2A De Moivre’s Theorem Solutions Ex 2(b) I Q1(v)

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

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

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 = 81/3
⇒ 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
Inter 2nd Year Maths 2A De Moivre’s Theorem Solutions Ex 2(b) I Q4(i)

(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

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

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 + 3a22 + 3ab2ω + b3
= a3(1) + 3a22 + 3ab2ω + b3
∴ (aω2 + bω)3 = a3 + 3a22 + 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

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

(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)
Inter 2nd Year Maths 2A De Moivre’s Theorem Solutions Ex 2(b) I Q6(v)

(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.

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

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
Inter 2nd Year Maths 2A De Moivre’s Theorem Solutions Ex 2(b) II Q2
Inter 2nd Year Maths 2A De Moivre’s Theorem Solutions Ex 2(b) II Q2.1

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

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

Question 5.
Find the product of all the values of (1 + i)4/5.
Solution:
Inter 2nd Year Maths 2A De Moivre’s Theorem Solutions Ex 2(b) II Q5

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
Inter 2nd Year Maths 2A De Moivre’s Theorem Solutions Ex 2(b) II Q6
= (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, ……., αn-1 be the nth roots of unity, then prove that 1p + αp + (α2)p + (α3)p + …… + (αn-p)2p = 0
= 0; if p ≠ kn
= n; if p ≠ kn, where p, k ∈ N
Solution:
nth roots of unity are 1, α, α2, ………., αn-1
Inter 2nd Year Maths 2A De Moivre’s Theorem Solutions Ex 2(b) III Q1
Inter 2nd Year Maths 2A De Moivre’s Theorem Solutions Ex 2(b) III Q1.1
∴ Each term of the series in (1) is 1
Hence the sum of the series 1 + αp + (α2)p + (α3)p + …….. + (αn-1)p
= 1 + 1 + 1 + ……… + 1 (n times)
= n(1)
= n

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

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:
Inter 2nd Year Maths 2A De Moivre’s Theorem Solutions Ex 2(b) III Q2
Inter 2nd Year Maths 2A De Moivre’s Theorem Solutions Ex 2(b) III Q2.1

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:
Inter 2nd Year Maths 2A De Moivre’s Theorem Solutions Ex 2(b) III Q3
Inter 2nd Year Maths 2A De Moivre’s Theorem Solutions Ex 2(b) III Q3.1

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:
Inter 2nd Year Maths 2A De Moivre’s Theorem Solutions Ex 2(b) III Q4
Inter 2nd Year Maths 2A De Moivre’s Theorem Solutions Ex 2(b) III Q4.1

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

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)
Inter 2nd Year Maths 2A De Moivre’s Theorem Solutions Ex 2(b) III Q5
Inter 2nd Year Maths 2A De Moivre’s Theorem Solutions Ex 2(b) III Q5.1