TS Inter Second Year Maths 2A De Moivre’s Theorem Important Questions Long Answer Type

Students must practice these Maths 2A Important Questions TS Inter Second Year Maths 2A De Moivre’s Theorem Important Questions Long Answer Type to help strengthen their preparations for exams.

TS Inter Second Year Maths 2A De Moivre’s Theorem Important Questions Long Answer Type

Question 1.
State and prove De Moivre’s theorem for integral index. [May ‘03]
Solution:
Theorem : (De Moivre’s for integral index)
If n is an integer then
(cos θ + i sin θ)ⁿ = cos nθ + i sinnθ
Proof:
Case – 1:
Suppose ‘n’ is a positive integer
Let, S(n) be the statement that
(cos θ + i sin θ)ⁿ = cos nθ + i sin nθ
If n = 1 then
L.H.S: (cos θ + i sin θ)ⁿ
= (cos θ + i sin θ)¹
= cos θ + i sin θ
R.H.S: cos nθ +i sin nθ
= cos (1θ) + i sin (1θ)
= cos θ + i sin θ
∴ L.H.S = R.H.S
∴ S(1) is true.
Assume that S(k) is true
∴ (cos θ + i sin θ)^k = cos kθ + i sin kθ
Now,
(cos θ + i sin θ)^k+1 = (cos θ + i sin θ)^k (cos θ + i sin θ) .
= (cos kθ + i sin kθ) (cos θ + i sin θ)
= cos kθ cos θ + i cos kθ sin θ + i sin kθ cos θ – sin kθ sin θ
= (cos kθ cos θ – sin kθ sin θ) + i(sin kθ cos θ + cos kθ sin θ)
= cos (kθ + θ) + i sin(kθ + θ)
= cos(k + 1)θ + i sin(k + 1)θ
∴ S(k + 1) is true.
By the principle of mathematical induction S(n) is true, ∀ n ∈ N.

Case – 2 :
If n = 0 then
L.H.S: (cos θ + i sin θ)° = 1
R.H.S: cos nθ + i sin nθ = cos(0θ) + isin (0θ)
= cos θ + i sin θ
= 1 + i(0) = 1
∴ LH.S = R.H.S
∴ (cos θ + i sin θ)ⁿ = cos nθ + i sin nθ.

Case – 3 :
Suppose ‘n’ is a negative integer
Let, n = – m where m ∈ Z⁺
L.H.S = (cos θ + i sin θ)
= (cos θ + i sin θ)^-m
= \(\frac{1}{(\cos \theta+i \sin \theta)^m}\)
= \(\frac{1}{\cos m \theta+i \sin m \theta}\) [∵ from case (1)]
= \(\frac{1}{\cos m \theta+i \sin m \theta} \times \frac{\cos m \theta-i \sin m \theta}{\cos m \theta-i \sin m \theta}\)
= \(\frac{\cos m \theta-i \sin m \theta}{\cos ^2 m \theta+\sin ^2 m \theta}\)
= \(\frac{\cos m \theta-i \sin m \theta}{1}\)
= cos mθ – i sin mθ
= cos (- m)θ + i sin(- m)θ
= cos nθ + i sin nθ
= R.H.S
∴ (cos θ + i sin θ)ⁿ = cos nθ + i sin nθ, ∀ n ∈ Z.

Question 2.
If m, n are integers and x = cos α + i sin α, y = cos β + i sin β then prove that
x^myⁿ + \(\frac{1}{x^m y^n}\) = 2 cos (mα + nβ) and
x^myⁿ – \(\frac{1}{x^m y^n}\) = 2i sin (mα + nβ)
Solution:
Given that
x^m = (cos α + i sin α)^m
= cos mα + i sin mα = cis mα
yⁿ = (cos β + isin β)ⁿ
= cos nβ + i sin nβ = cis nβ
Now,
x^myⁿ = cis mα . cis nβ
= cis (mα + nβ)
= cos (mα + nβ) + i sin (mα + nβ)
\(\frac{1}{x^m y^n}=\frac{1}{\cos (m \alpha+n \beta)+i \sin (m \alpha+n \beta)}\)
= cos (mα + nβ) – i sin (mα + nβ)

(i) x^myⁿ + \(\frac{1}{x^m y^n}\)
= cos (mα + nβ) + i sin (mα + nβ) + cos (mα + nβ) – i sin (mα + nβ)
= 2 cos (mα + nβ)
∴ x^myⁿ + \(\frac{1}{x^m y^n}\) = 2 cos (mα + nβ)

(ii) x^myⁿ – \(\frac{1}{x^m y^n}\)
= cos (mα + nβ) + i sin (mα + nβ) – cos (mα + nβ) + i sin (mα + nβ)
= 2i sin (mα + nβ)
∴ x^myⁿ – \(\frac{1}{x^m y^n}\) = 2i sin (mα + nβ)

Question 3.
If n is a positive integer, show that (1 + i)ⁿ + (1 – i)ⁿ = \(2^{\frac{n+2}{2}} \cos \left(\frac{n \pi}{4}\right)\). [AP – Mar. 2015, Mar.’99]
Solution:

Let, 1 + i = r (cos θ + i sin θ)
then, r cos θ = 1, r sin θ = 1
∴ r = \(\sqrt{\mathrm{x}^2+\mathrm{y}^2}=\sqrt{1^2+1^2}=\sqrt{1+1}=\sqrt{2}\)
Hence.
√2 cos θ = 1,
cos θ = \(\frac{1}{\sqrt{2}}\)

√2 sin θ = 1
sin θ = \(\frac{1}{\sqrt{2}}\)

∴ θ lies in Q₁.
∴ θ = \(\frac{\pi}{4}\)

Similarly,
1 – i = √2 (cos \(\frac{\pi}{4}\) – i sin \(\frac{\pi}{4}\))
L.H.S:
(1 + i)ⁿ + (1 – i)ⁿ

Question 4.
If n is an integer then show that (1 + cos θ + i sin θ)ⁿ + (1 + cos θ – i sin θ)ⁿ = 2ⁿ⁺¹ cosⁿ \(\left(\frac{\theta}{2}\right)\) – i cos \(\left(\frac{n \theta}{2}\right)\) . [May ‘01, ‘97, March ’10. Mar. ’93, TS & AP – Mar. 2017]
Solution:

Question 5.
If cos α + cos β + cos γ = 0 = sin α + sin β + sin γ, prove that cos² α + cos² β + cos² γ = \(\frac{3}{2}\) = sin² α + sin² β + sin² γ. [AP – Mar., May 2016; TS – Mar. ‘18. ‘15. May ‘15, May ‘09, March ‘03, ‘96, March ‘13 (old)]
Solution:
Given that,
cos α + cos β + cos γ = 0 = sin α + sin β + sin γ
Let, x = cos α + i sin α
y = cos β + i sin β
z = cos γ + i sin γ
Now,
x + y + z = cos α + i sin α + cos β + i sin β + cos γ + i sin γ
= (cos α + cos β + cos γ) + i(sin α + sin β + sin γ)
= 0 + i(0) = 0
∴ x + y + z = 0
Squaring on both sides,
(x + y + z)² = 0
x² + y² + z² + 2 (xy + yz + zx) = 0
x² + y² + z² = – 2 (xy + yz + zx)
= \(\frac{-2 x y z}{x y z}\) (xy + y + zx)
= – 2xyz \(\left(\frac{1}{z}+\frac{1}{x}+\frac{1}{y}\right)\)
= – 2xyz (cos γ – i sin γ + cos α – i sin α + cos β – i sin β)
= – 2xyz [(cos α + cos β + cos γ) – i (sin α + sin β + sin γ)]
= – 2xyz [0 – i . 0]
= – 2xyz(0) = 0
∴ x² + y² + z² = 0
(cos α + i sin α)² + (cos β + i sin β)² + (cos γ + i sin γ)² = 0
cos 2α + i sin2α + cos 2β + i sin 2β + cos 2γ + i sin 2γ = 0
(cos 2α + cos 2β + cos 2γ) + i(sin 2α + sin 2β + sin 2γ) = 0
Comparing real parts on both sides, we get
(i) cos 2α + cos 2β + cos 2γ = 0
2 cos² α – 1 + 2 cos² β – 1 + 2 cos² γ – 1 = 0
2 (cos² α + cos² β + cos² γ) = 3
cos² α + cos² β + cos² γ = \(\frac{3}{2}\)

(ii) 1 – sin² α + 1 – sin² β + 1 – sin² γ = \(\frac{3}{2}\)
3 – (sin² α + sin² β + sin² γ) = \(\frac{3}{2}\)
sin² α + sin² β + sin² γ = 3 – \(\frac{3}{2}\) = \(\frac{3}{2}\).

Question 6.
If n is an integer then show that (1 + i)²ⁿ + (1 – i)²ⁿ = 2ⁿ⁺¹ cos \(\frac{n \pi}{2}\) [May ’14. ’02, ’98, ’93, March ’09]
Solution:

Let, 1 + i = r (cos θ + i sin θ)
then r cos θ = 1, r sin θ = 1
r = \(\sqrt{\mathrm{x}^2+\mathrm{y}^2}=\sqrt{1^2+1^2}=\sqrt{1+1}=\sqrt{2}\)
Hence,
√2 cos θ = 1,
cos θ = \(\frac{1}{\sqrt{2}}\)
√2 sin θ = 1
sin θ = \(\frac{1}{\sqrt{2}}\)
∴ θ lies in Q₁.
∴ θ = \(\frac{\pi}{4}\)
∴ 1 + i = √2 (cos \(\frac{\pi}{4}\) + i cos \(\frac{\pi}{4}\))
Similarly,
1 – i = √2 (cos \(\frac{\pi}{4}\) – i sin \(\frac{\pi}{4}\))
L.H.S:
(1 + i)²ⁿ + (1 – i)²ⁿ = [√2 (cos \(\frac{\pi}{4}\) + i sin \(\frac{\pi}{4}\))]²ⁿ + [√2 (cos \(\frac{\pi}{4}\) – i sin \(\frac{\pi}{4}\))]²ⁿ
= 2ⁿ (cos \(\frac{n \pi}{2}\) + i sin \(\frac{n \pi}{2}\)) + 2ⁿ (cos \(\frac{n \pi}{2}\) – i sin \(\frac{n \pi}{2}\))
= 2ⁿ [cos \(\frac{n \pi}{2}\) + i sin \(\frac{n \pi}{2}\) + cos \(\frac{n \pi}{2}\) – i sin \(\frac{n \pi}{2}\)]
= 2ⁿ . 2 cos \(\frac{n \pi}{2}\)
= 2ⁿ⁺¹ cos \(\frac{n \pi}{2}\)
= R.H.S
∴ (1 + i)²ⁿ + (1 – i)²ⁿ = 2ⁿ⁺¹ cos \(\frac{n \pi}{2}\).

Question 7.
If α, β are the roots of the equation x² – 2x + 4 = 0 then for any n ∈ N show that αⁿ + βⁿ = 2ⁿ⁺¹ cos \(\frac{n \pi}{3}\) [TS – May 2016; March ’14, ’11, May ’88, AP – Mar. 2019]
Solution:

Given quadratic equation is x² – 2x + 4 = 0
Comparing ax² + bx + c = 0 we get, a = 1, b = – 2, c = 4
∴ x = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\)
= \(\frac{-(-2) \pm \sqrt{4-16}}{2(1)}\)
= \(\frac{2 \pm \sqrt{-12}}{2}\)
= \(\frac{2 \pm i 2 \sqrt{3}}{2}\)
= 1 ± √3i
Since, α, β are the roots of the equation
x² – 2x + 4 = 0 then
α = 1 + i√3, β = 1 – i√3
Let,
α = 1 + i√3 = r (cos θ + i sin θ)
then r cos θ = 1, r sin θ = √3
r = \(\sqrt{x^2+y^2}=\sqrt{(1)^2+(\sqrt{3})^2}\)
= \(\sqrt{1+3}=\sqrt{4}\) = 2
Hence,
2 cos θ = 1,
cos θ = \(\frac{1}{2}\)
2 sin θ = √3
sin θ = \(\frac{\sqrt{3}}{2}\)
∴ θ lies in the Q₁.
∴ θ = \(\frac{\pi}{3}\)
α = 1 + i√3 = 2(cos \(\frac{\pi}{3}\) + i sin \(\frac{\pi}{3}\))
Similarly,
β = 1 – i√3 = 2(cos \(\frac{\pi}{3}\) – i sin \(\frac{\pi}{3}\))
L.H.S:

Question 8.
If cos α + cos β + cos γ = 0 = sin α + sin β + sin γ then show that
(i) cos 3α + cos 3β + cos 3γ = 2 cos (α + β + γ)
(ii) sin 3α + sin 3β + sin 3γ = 2 sin (α + β + γ)
(iii) cos (α + β) + cos (β + γ) + cos (γ + α) = 0
Soluton:
Given that,
cos α + cos β + cos γ = 0 = sin α + sin β + sin γ
Let,
x = cos α + i sin α
y = cos β + i sin β
z = cos γ + i sin γ
Now,
x + y + z = cos α + i sin α + cos β + i sin β + cos γ + i sin γ
= (cos α + cos β + cos γ) + i(sin α + sin β + sin γ)
= 0 + i (0) = 0
∴ x + y + z = 0
⇒ x³ + y³ + z³ = 3xyz
(cos α + i sin α) + (cos β + i sin β) + (cos γ + i sin γ) = 3 (cos α + i sin α) (cos β + i sin β) (cos γ + i sin γ)
cos 3α + i sin 3α + cos 3β + i sin 3β + cos 3γ + i sin 3γ = 3 cis α cis β cis γ
(cos 3α + cos 3β + cos 3γ) + i(sin 3α + sin 3β + sin 3γ) = 3 cis (α + β + γ)
= 3 [cos (α + β + γ) + i sin (α + β + γ)]

(i) Comparing the real parts on both sides we get,
cos 3α + i sin 3α + cos 3β = 3 cos (α + β + γ)

(ii) Comparing the imaginary parts on both sides we get,
sin 3α + sin 3β + sin 3γ = 3 sin (α + β + γ)

(iii) Let,
x = cos α + i sin α
⇒ \(\frac{1}{x}\) = cos α – i sin α
y = cos β + i sin β
⇒ \(\frac{1}{y}\) = cos β – i sin β
z = cos γ + i sin γ
⇒ \(\frac{1}{z}\) = cos γ – i sin γ
Now,
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\) = cos α – i sin α + cos β – i sin β + cos γ – i sin γ
= (cos α + cos β + cos γ) – i(sin α + sin β + sin γ)
= 0 + i(0) = 0
∴ \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\) = 0
\(\frac{y z+x z+x y}{x y z}\) = 0
(cos α + i sin α) (cos β + i sin β) + (cos β + i sin β) (cos γ + i sin γ) + (cos γ + i sin γ) (cos α + i sin α) = 0
cis α cis β + cis β . cis γ + cis γ . cis α = 0
cis (α + β) + cis (β + γ) + cis (γ + α) = 0
cos (α + β) + i sin (α + β) + cos (β + γ) + isin (β + γ) + cos (γ + α) + isin (γ + α) = 0
[cos (α + β) + cos (β + γ) + cos (γ + α)] + i[sin (α + β) + sin (β + γ) + sin (γ + α)] = 0
Comparing real parts on both sides we get,
cos (α + β) + cos (β + γ) + cos (γ + α) = 0
Comparing imaginary parts on both sides
we get,
sin (α + β) + sin (β + γ) + sin (γ + α) = 0.

Question 9.
If n is an Integer and z = cis θ, (θ ≠ (2n + 1)\(\frac{\pi}{2}\)), then show that \(\frac{z^{2 n}-1}{z^{2 n}+1}\) = i tan nθ. [Mar. ’12. ’19(TS)]
Solution:
Given that,
z = cis θ = cos θ + i sin θ

Question 10.
Find all the roots of the equation x¹¹ – x⁷ + x⁴ – 1 = 0 [Board Paper].
Solution:
Given equation is x¹¹ – x⁷ + x⁴ – 1 = 0
x⁷ (x⁴ – 1) . 1 (x⁴ – 1) = 0
(x⁴ – 1) (x⁷ + 1) = 0
x⁴ – 1 = 0 or x⁷ + 1 = 0

x⁴ – 1 = 0:
x⁴ = 1
x = (1)^1/4
= (cos 0 + i sin 0)^1/4
= [cos (2kπ + 0) + isin(2kπ + 0)]^1/4
k = 0, 1, 2, 3
= [cos (2kπ) + i sin(2kπ]^1/4
= cos (\(\frac{k\pi}{2}\)) + sin (\(\frac{k\pi}{2}\))
= cis (\(\frac{k\pi}{2}\)), k = 0, 1, 2, 3
∴ The values of x are
cis (0), cis (\(\frac{\pi}{2}\)), cis π, cis (\(\frac{3\pi}{2}\)) : 1, i, – 1, – i

x⁷ + 1 = 0:
x⁷ = – 1
x = (- 1)^1/7
= [cos π + i sin π]
= [cos (2kπ + π) + i sin (2kπ + π)]^1/7, k = 0, 1, 2, 3, 4, 5, 6
= [cos (2k + 1)π + i sin (2k + 1)π]^1/7
= [cos (2k + 1)\(\frac{\pi}{7}\) + i sin (2k + 1)\(\frac{\pi}{7}\)]
= cis (2k + 1)\(\frac{\pi}{7}\), k = 0, 1, 2, 3, 4, 5, 6
The values of x are
cis \(\frac{\pi}{7}\), cis \(\frac{3 \pi}{7}\), cis \(\frac{5 \pi}{7}\), cis π, cis \(\frac{9 \pi}{7}\), cis \(\frac{11 \pi}{7}\), cis \(\frac{13 \pi}{7}\)
∴ The roots of the equation are ± 1, ± i, cis \(\frac{\pi}{7}\), cis \(\frac{3 \pi}{7}\), cis \(\frac{5 \pi}{7}\), cis π, cis \(\frac{9 \pi}{7}\), cis \(\frac{11 \pi}{7}\), cis \(\frac{13 \pi}{7}\)

Question 11.
Solve the equation x⁴ – 1 = 0.
Solution:
Given equation is x⁴ – 1 = 0
x⁴ = 1
x = (1)^1/4
= [cos o + i sin o]^1/4
= [cos (2kπ + 0) + i sin(2kπ + 0)]^1/4
k = 0, 1, 2, 3
= [cos 2kπ + i sin 2kπ]^1/4
= cos \(\frac{k \pi}{2}\) + i sin \(\frac{k \pi}{2}\)
= cis \(\frac{k \pi}{2}\), k = 0, 1, 2, 3
lf k = 0 then x = cos 0 + i sin 0 = 1 + i . 0 = 1
If k = 1 then x = cos \(\frac{\pi}{2}\) + i sin \(\frac{\pi}{2}\) = 0 + i . 1 = i
If k = 2 then x = cos π + i sin π
= – 1 + i .0 = – 1
If k = 3 then x cos \(\frac{3 \pi}{2}\) + i sin \(\frac{3 \pi}{2}\)
= o + i (- 1) = – i
∴ The roots of the given equation are ± 1, ± i.

Question 12.
Solve the equation x⁴ + 1 = 0. [May ’97]
Solution:
Given equation is x⁴ + 1 = 0
x⁴ = – 1
x = (- 1)^1/4
= [cos π + i sin π]^1/4
= [cos (2kπ + π) + i sin (2kπ + π)]^1/4, k = 0, 1, 2, 3
= [cos (2k + 1)π + i sin(2k + 1)π]^1/4
= cos(2k + 1) \(\frac{\pi}{4}\) + i sin(2k + 1) \(\frac{\pi}{4}\)
= cis (2k + 1) \(\frac{\pi}{4}\), k = 0, 1, 2, 3
If k = 0, x = cis \(\frac{\pi}{4}\)
If k = 1, x = cis \(\frac{3 \pi}{4}\)
If k = 2, x = cis \(\frac{5 \pi}{4}\)
If k = 3, x = cis \(\frac{7 \pi}{4}\)
∴ The roots of the given equation are cis \(\frac{\pi}{4}\), cis \(\frac{3 \pi}{4}\), cis \(\frac{5 \pi}{4}\), cis \(\frac{7 \pi}{4}\).

Question 13.
If n is a positive integer, show that (p + iq)^1/n + (p – iq)^1/n = 2(p² + q²)^1/2n . cos \(\left(\frac{1}{n} \tan ^{-1} \frac{q}{p}\right)\) [AP – Mar. 18, May ’15; Mar. ’01]
Solution:
Let p + iq = r (cos θ + i sin θ)
then r cos θ = p, r sin θ = q
r = \(\sqrt{\mathrm{x}^2+\mathrm{y}^2}=\sqrt{\mathrm{p}^2+\mathrm{q}^2}\)
Hence,
\(\sqrt{p^2+q^2}\) cos θ = p, \(\sqrt{p^2+q^2}\) sin θ = q
cos θ = \(\frac{p}{\sqrt{p^2+q^2}}\)
sin θ = \(\frac{q}{\sqrt{p^2+q^2}}\)
∴ tan θ = \(\frac{\sin \theta}{\cos \theta}=\frac{\frac{q}{\sqrt{p^2+q^2}}}{\frac{p}{\sqrt{p^2+q^2}}}=\frac{q}{p}\)
θ = \(\tan ^{-1}\left(\frac{q}{p}\right)\)
∴ p + iq = \(\sqrt{p^2+q^2}\) [cos θ + i sin θ]
Similarly,
p – iq = \(\sqrt{p^2+q^2}\) [cos θ – i sin θ]

Question 14.
Show that one value of \(\left[\frac{1+\sin \frac{\pi}{8}+i \cos \frac{\pi}{8}}{1+\sin \frac{\pi}{8}-i \cos \frac{\pi}{8}}\right]^{\frac{8}{3}}\) is – 1. [TS – Mar. 2016; May ’12, May ’10]
Solution:
Given,

Question 15.
Solve (x – i)ⁿ = xⁿ, n is a positive integer. [March ’02]
Solution:
Given equation is (x – 1)ⁿ = xⁿ
Taking nth root of each side of (x – 1)ⁿ = xⁿ
we have
x – 1 = x(1)^1/n
= x [cos 0 – i sin 0]^1/n
= x [cos (2kπ + 0) + i sin (2kπ + 0)]^1/n
k = 0, 1, ………….., (n – 1)
= x [cos 2kπ + i sin 2kπ]^1/n

Some More Maths 2A De Moivre’s Theorem Important Questions

Question 1.
Find the value of (1 + i√3)³.
Solution:

Let 1 + i√3 = r (cos θ + i sin θ)
then r cos θ = 1, r sin θ = √3
r = \(\sqrt{\mathrm{x}^2+\mathrm{y}^2}=\sqrt{(1)^2+(\sqrt{3})^2}\)
= \(\sqrt{1+3}=\sqrt{4}\) = 2
Hence,
2 cos θ = 1, 2 sin θ = √3
cos θ = \(\frac{1}{2}\), sin θ = \(\frac{\sqrt{3}}{2}\)
∴ θ lies in the Q₁.
∴ θ = \(\frac{\pi}{3}\)
∴ 1 + i√3 = 2 (cos \(\frac{\pi}{3}\) + i sin \(\frac{\pi}{3}\))
Now,
(1 + i√3)³ = [2 (cos \(\frac{\pi}{3}\) + i sin \(\frac{\pi}{3}\))]³
= 8 (cos π + i sin π)
= 8 (- 1 + i . 0) = – 8.

Question 2.
If (1 + x)ⁿ = a₀ + a₁x + a₂x² + ………….. + a_nxⁿ, then show that
(i) a₀ – a₂ + a₄ – a₆ + ……. = 2^n/2 cos \(\frac{n \pi}{4}\)
(ii) a₁ – a₃ + a₅ + ……………. = 2^n/2 sin \(\frac{n \pi}{4}\)
Solution:
Given,
(1 + x)ⁿ = a₀ + a₁x + a₂x² + ……………. + a_nxⁿ
Put x = i then
(1 + i)ⁿ = a₀ + a₁(i) + a₂i² + a₃i³ + a₄i⁴ + a₅i⁵ + …………… + a_niⁿ
= a₀ + a₁i – a₂ – a₃i – a₄ + a₅i + …………. + a_niⁿ
= (a₀ – a₂ + a₄ – a₆ + …………. ) + i(a₁ – a₃ + a₅ – …………) ……………..(1)
Let, 1 + i = r (cos θ + i sin θ)
then r cos θ = 1, r sin θ = 1
r = \(\sqrt{\mathrm{x}^2+\mathrm{y}^2}=\sqrt{1^2+1^2}=\sqrt{1+1}=\sqrt{2}\)
Hence,
√2 cos θ = i
cos θ = \(\frac{1}{\sqrt{2}}\)
√2 sin θ = 1
sin θ = \(\frac{1}{\sqrt{2}}\)
∴ θ lies in the Q₁.
∴ θ = \(\frac{\pi}{4}\)
∴ (1 + i) = √2 (cos \(\frac{\pi}{4}\) + i sin \(\frac{\pi}{4}\))
(1 + i)ⁿ = [√2 (cos \(\frac{\pi}{4}\) + i sin \(\frac{\pi}{4}\))]ⁿ
= 2^n/2 (cos \(\frac{n \pi}{4}\) + i sin \(\frac{n \pi}{4}\))
From (1),
2^n/2 (cos \(\frac{n \pi}{4}\) + i sin \(\frac{n \pi}{4}\)) = (a₀ – a₂ + a₄ – a₆ + ……………….) + i (a₁ – a₃ + a₅ – ………….)

(i) Comparing real parts on both sides we get,
a₀ – a₂ + a₄ – a₆ + ………………. = 2^n/2 cos \(\frac{n \pi}{4}\)

(ii) Comparing imaginary parts on bothsides we get,
a₁ – a₃ + a₅ – …………… = 2^n/2 sin \(\frac{n \pi}{4}\).

Question 3.
Solve the equation x⁵ + 1 = 0.
Solution:
Given equation is x⁵ + 1 = 0
x⁵ = – 1
x = (- 1)^1/5
= [cos π + i sin π]^1/5
= [cos(2kπ + π) + i sin (2kπ + π)]^1/5, k = 0, 1, 2, 3, 4
= [cos(2k + 1)π + i sin (2k + 1)π]^1/5
= cis (2k + 1) \(\frac{\pi}{5}\), k = 0, 1, 2, 3, 4
If k = 0,
⇒ x = cis \(\frac{\pi}{5}\)

If k = 1,
⇒ x = cis \(\frac{3 \pi}{5}\)

If k = 2,
⇒ x = cis π

If k = 3,
⇒ x = cis \(\frac{7 \pi}{5}\)

If k = 4,
⇒ x = cis \(\frac{9 \pi}{5}\)
∴ The roots of the given equation are cis \(\frac{\pi}{5}\), cis \(\frac{\pi}{5}\), cis π, cis \(\frac{7 \pi}{5}\), cis \(\frac{9 \pi}{5}\).

Question 4.
Solve the equation x⁹ – x⁵ + x⁴ – 1 = 0.
Solution:
Given equation is x⁹ – x⁵ + x⁴ – 1 = 0
x⁵ (x⁴ – 1) + 1 (x⁴ – 1) = 0
(x⁴ – 1) (x⁵ + 1) = 0
x⁴ – 1 = 0 or x⁵ + 1 = 0

x⁴ – 1 = 0:
x⁴ = 1
x = (1)^1/4
= [cos 0 + i sin 0]^1/4
= [cos (2kπ + 0) + i sin (2kπ + 0)]^1/4, k = 0, 1, 2, 3
= [cos (2kπ) + i sin(2kπ)]^1/4
= cos \(\frac{k \pi}{2}\) + i sin \(\frac{k \pi}{2}\)
= cis \(\frac{k \pi}{2}\), k = 0, 1, 2, 3
If k = 0
⇒ x = cos 0 + i sin 0
= 1 + i . 0 = 1
If k = 1
⇒ x= cos \(\frac{\pi}{2}\) + i sin \(\frac{\pi}{2}\)
= 0 + i . 1 = 1
If k = 2
⇒ x = cos π + i sin π
= – 1 + i . 0 = – 1
If k = 3x
⇒ x = cos \(\frac{3 \pi}{2}\) + i sin \(\frac{3 \pi}{2}\)
= 1 + i(- 1) = – i
∴ The values of x are ± 1, ± 1.

x⁵ + 1 = 0:
x⁵ = (- 1)
x = (- 1)^1/5
= [cos π + i sin π]^1/5
= [cos (2kπ + π) + i sin(2kπ + π)]^1/5, k = 0, 1, 2, 3, 4
= [cos(2k + 1)π + i sin(2k + 1)π]^1/5
= cos (2k + 1) \(\frac{\pi}{5}\) + i sin(2k + 1) \(\frac{\pi}{5}\)
= cis (2k + 1) \(\frac{\pi}{5}\), k = 0, 1, 2, 3, 4
If k = 0
⇒ x = cis \(\frac{\pi}{5}\)
If k = 1
⇒ x = cis \(\frac{3 \pi}{5}\)
If k = 2
⇒ x = cis π
If k = 3
⇒ x = cis \(\frac{7 \pi}{5}\)
If x = 4
⇒ x = cis \(\frac{9 \pi}{5}\)
∴ The values of x are cis \(\frac{\pi}{5}\), cis \(\frac{3 \pi}{5}\), cis π, cis \(\frac{7 \pi}{5}\), cis \(\frac{9\pi}{5}\).
∴ The roots of the given equation are ± 1, ± i, cis \(\frac{\pi}{5}\), cis \(\frac{3 \pi}{5}\), cis π, cis \(\frac{7 \pi}{5}\), cis \(\frac{9\pi}{5}\).

Question 5.
Find the common roots of x¹² – 1 = 0 and x + x + 1 = 0.
Solution:
Given equation is x¹² – 1 = 0
x¹² = 1
x = (1)¹²
= [cos 0 + i sin 0]^1/12
= [cos (2kπ + 0) + i sin (2kπ + 0)]^1/12, k = 0, 1, ……………, 11
= [cos 2kπ + i sin 2kπ]^1/12
= cos \(\frac{k \pi}{6}\) + i sin \(\frac{k \pi}{6}\)
= cis \(\frac{k \pi}{6}\), k = 0, 1, 2, …………., 11
If k = 0
⇒ x = cis 0
If k = 1
⇒ x = cis \(\frac{\pi}{6}\)
If k = 2
⇒ x = cis \(\frac{\pi}{3}\)
If k = 3
⇒ x = cis \(\frac{\pi}{2}\)
If k = 4
⇒ x = cis \(\frac{2 \pi}{3}\)
If k = 5
⇒ x = cis \(\frac{5 \pi}{6}\)
If k = 6
⇒ x = cis π
If k = 7
⇒ x = cis \(\frac{7 \pi}{6}\)
If k = 8
⇒ x = cis \(\frac{4 \pi}{3}\)
If k = 9
⇒ x = cis \(\frac{3 \pi}{2}\)
If k = 10
⇒ x = cis \(\frac{5 \pi}{3}\)
If k = 11
⇒ x = cis \(\frac{11 \pi}{6}\)
∴ Therootsof x¹² – 1 = 0 are cis 0, cis \(\frac{\pi}{6}\), cis \(\frac{\pi}{3}\), cis \(\frac{\pi}{2}\), cis \(\frac{2 \pi}{3}\), cis \(\frac{5 \pi}{6}\), cis π, cis \(\frac{7 \pi}{6}\), cis \(\frac{4 \pi}{3}\), cis \(\frac{3 \pi}{2}\), cis \(\frac{5 \pi}{3}\), cis \(\frac{11 \pi}{6}\)

Given equation is x⁴ + x² + 1 = 0
Multiplying on both sides with x² – 1
we get,
(x² – 1) (x⁴ + x² + 1) = 0
x⁶ – 1 = 0
x⁶ = 1
x = (1)^1/6
= [cos 0 + i sin 0]^1/6
= [cos (2kπ + 0) + i sin (2kπ + 0)]^1/6, k = 0, 1, 2, 3, 4, 5
= [cos 2kπ + i sin 2kπ]^1/6
= cos \(\frac{k \pi}{3}\) + i sin \(\frac{k \pi}{3}\)
= cis \(\frac{k \pi}{3}\), k = 0, 1, 2, 3, 4, 5
If k = 0
⇒ x = cis 0
If k = 1
⇒ x = cis \(\frac{\pi}{3}\)
If k = 2
⇒ x = cis \(\frac{2 \pi}{3}\)
If k = 3
⇒ x = cis π
If k = 4
⇒ x = cis \(\frac{4 \pi}{3}\)
If k = 5
⇒ x = cis \(\frac{5 \pi}{3}\)
The values of given equation are cis 0, cis \(\frac{\pi}{3}\), cis \(\frac{2 \pi}{3}\), cis π, cis \(\frac{4 \pi}{3}\), cts \(\frac{5 \pi}{3}\).
∴ The common roots of the given equations are cis \(\frac{\pi}{3}\), cis \(\frac{2 \pi}{3}\), cis \(\frac{4 \pi}{3}\), cis \(\frac{5 \pi}{3}\).

Question 6.
Find the number of 15th roots of unity, which are also 25th roots of unity.
Solution:
15th roots of unity:
Let, x = \(\sqrt[15]{1}=(1)^{1 / 15}\)
= [cos 0 + i sin 0]^1/15
= [cos (2nπ + 0) + i sin(2nπ + 0)]^1/15
n = 0, 1, 2, …………., 14
= [cos 2nπ + i sin 2nπ]^1/15
= \(\cos \frac{2 n \pi}{15}+i \sin \frac{2 n \pi}{15}\)
= cis \(\frac{2n \pi}{15}\), n = 0, 1, 2, …………….., 14

25th roots of unity:
Let, x = \(\sqrt[25]{1}\)
= (1)^1/25
= [cos 0 + i sin 0]^1/25
= [cos (2mπ + 0) + i sin (2mπ + 0)]^1/25
m = 0, 1, 2, ………….., 24
= [cos 2mπ + i sin 2mπ]^1/25
= cos \(\frac{2 \mathrm{~m} \pi}{25}\) + i sin \(\frac{2 \mathrm{~m} \pi}{25}\)
= cis \(\frac{2 \mathrm{~m} \pi}{25}\), m = 0, 1, 2, ……………., 24

∴ The common roots are cis 0, cis \(\frac{2 \pi}{5}\), cis \(\frac{4 \pi}{5}\), cis \(\frac{6 \pi}{5}\), cis \(\frac{8 \pi}{5}\).
∴ The number of common roots = 5 (or)
The GCM of 15 and 25 is

Question 7.
Find the product of all the values of (1 + i)^4/5.
Solution:

Let, 1 + i = r (cos θ + i sin θ)
then r cos θ = 1, r sin θ = 1
r = \(\sqrt{x^2+y^2}=\sqrt{1^2+1^2}=\sqrt{1+1}=\sqrt{2}\)
Hence,
√2 cos θ = i,
cos θ = \(\frac{1}{\sqrt{2}}\)
√2 sin θ = 1
sin θ = \(\frac{1}{\sqrt{2}}\)
∴ θ lies in the Q₁.
∴ θ = \(\frac{\pi}{4}\)
∴ 1 + i = √2 (cos \(\frac{\pi}{4}\) + i sin \(\frac{\pi}{4}\))
Now,
(1 + i) = [√2 (cos \(\frac{\pi}{4}\) + i sin \(\frac{\pi}{4}\))]⁴
= 4 (cos π + i sin π)
= 4[cos(2kπ + π) + i sin (2kπ + π)]
= 4[cos(2k + 1)π + isin(2k + 1)π]
(1 + i)^4/5 = 4^1/5 [cos(2k + 1)π + isin(2k + 1)π]^1/5
k = 0, 1, 2, 3, 4
= 4^1/5 [cos (2k + 1) \(\frac{\pi}{5}\) + i sin (2k + 1) \(\frac{\pi}{5}\)]
= 4^1/5 cis (2k + 1) \(\frac{\pi}{4}\), k = 0, 1, 2, 3, 4

= 4 cis (5π)
= 4 cis π
= 4 (cos π + i sin π)
= 4 [- 1 + i . 0] = – 4.

Question 8.
If z² + z + 1 = 0, where z is a complex number, prove that
\(\left(z+\frac{1}{z}\right)^2+\left(z^2+\frac{1}{z^2}\right)^2+\left(z^3+\frac{1}{z^3}\right)^2\) + \(\left(z^4+\frac{1}{z^4}\right)^2+\left(z^5+\frac{1}{z^5}\right)^2+\left(z^6+\frac{1}{z^6}\right)^2\) = 12.
Solution:
Given equation is z² + z + 1 = 0
Comparing with ax² + bx + c we get, a = 1, b = 1, c = 1
The roots of the given equation is
z = \(\frac{-\mathrm{b} \pm \sqrt{\mathrm{b}^2-4 \mathrm{ac}}}{2 \mathrm{a}}=\frac{-1 \pm \sqrt{1-4 \cdot 1 \cdot 1}}{2(1)}\)
= \(\frac{-1 \pm \sqrt{1-4}}{2} \Rightarrow \frac{-1 \pm \sqrt{3} i}{2}\) = ω, ω²
Let, z = ω
L.H.S:

= (ω + ω²)² + (ω² + ω) + 4 + (ω + ω²) + (ω² + ω)² + 4
= (- 1)² + (- 1)² + 4 + (- 1)² + (- 1)² + 4
= 1 + 1 + 4 + 1 + 1 + 4
= 12 = R.H.S.

Question 9.
Prove the sum of 99th power of the roots of the equation x⁷ – 1 = 0 is zero and hence deduce the roots of x⁶ + x⁵ + x⁴ + x³ + x² + x + 1 = o.
Solution:
Given equation is x⁷ – 1 = 0
x⁷ = 1
x = (1)^1/7
= [cos 0 + i sin 0]^1/7
= [cos (2kπ + 0) + i sin(2kπ + 0)]^1/7
k = 0, 1, 2, 3, …………….., 6
= [cos 2kπ + i sin 2kπ]^1/7
= cos \(\frac{2 \mathrm{k} \pi}{7}\) + i sin \(\frac{2 \mathrm{k} \pi}{7}\)
= cis \(\frac{2 \mathrm{k} \pi}{7}\), k = 0, 1, 2, …………, 6
If k = 0
⇒ cis 0 = x
If k = 1
⇒ cis \(\frac{2 \pi}{7}\) = x
If k = 2
⇒ cis \(\frac{4 \pi}{7}\) = x
If k = 3
⇒ cis \(\frac{6 \pi}{7}\) = x
If k = 4
⇒ cis \(\frac{8 \pi}{7}\) = x
If k = 5
⇒ cis \(\frac{10 \pi}{7}\) = x
If k = 6
⇒ cis \(\frac{12 \pi}{7}\) = x
∴ All the values of x⁷ – 1 = 0 are cis 0, cis \(\frac{2 \pi}{7}\), cis \(\frac{4 \pi}{7}\), cis \(\frac{6 \pi}{7}\), cis \(\frac{8 \pi}{7}\), cis \(\frac{10 \pi}{7}\), cis \(\frac{12 \pi}{7}\)
The 99th power of the roots of the equation x⁷ – 1 = 0 are

= 1 + ω + ω² + ω³ + ω⁴ + ω⁵ + ω⁶
= 0 (from theorem)
Given equation is
x⁶ + x⁵ + x⁴ + x³ + x² + x + 1 = 0
Multiplying on both sides with (x – 1)
(x – 1) (x⁶ + x⁵ + x⁴ + x³ + x² + x + 1) = 0
x⁷ + x⁶ + x⁵ + x⁴ + x³ + x² + x – x⁶ – x⁵ – x⁴ – x³ – x² – x – 1 = 0
x⁷ – 1 = 0
x = 1 is also a root of x⁷ – 1 = 0.
∴ The roots of the equation x⁶ + x⁵ + x⁴ + x³ + x² + x + 1 = 0 are x = cis \(\frac{2k \pi}{7}\), k = 1, 2, 3, ………., 6.