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 = 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

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

∴ 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

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)