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.

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

Question 2.
If m, n are integers and x = cos α + i sin α, y = cos β + i sin β then prove that
xmyn + \(\frac{1}{x^m y^n}\) = 2 cos (mα + nβ) and
xmyn – \(\frac{1}{x^m y^n}\) = 2i sin (mα + nβ)
Solution:
Given that
xm = (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,
xmyn = 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) xmyn + \(\frac{1}{x^m y^n}\)
= cos (mα + nβ) + i sin (mα + nβ) + cos (mα + nβ) – i sin (mα + nβ)
= 2 cos (mα + nβ)
∴ xmyn + \(\frac{1}{x^m y^n}\) = 2 cos (mα + nβ)

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

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

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

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

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:

TS Inter Second Year Maths 2A De Moivre’s Theorem Important Questions Long Answer Type 3

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 – \(\frac{3}{2}\) = \(\frac{3}{2}\).

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

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:

TS Inter Second Year Maths 2A De Moivre’s Theorem Important Questions Long Answer Type 4

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 \(\frac{n \pi}{2}\).

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

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:

TS Inter Second Year Maths 2A De Moivre’s Theorem Important Questions Long Answer Type 5

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:

TS Inter Second Year Maths 2A De Moivre’s Theorem Important Questions Long Answer Type 6

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

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.

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

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

TS Inter Second Year Maths 2A De Moivre’s Theorem Important Questions Long Answer Type 7

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 \(\frac{13 \pi}{7}\)

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

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 \(\frac{7 \pi}{4}\).

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

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]

TS Inter Second Year Maths 2A De Moivre’s Theorem Important Questions Long Answer Type 8

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,

TS Inter Second Year Maths 2A De Moivre’s Theorem Important Questions Long Answer Type 9

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

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

TS Inter Second Year Maths 2A De Moivre’s Theorem Important Questions Long Answer Type 10

TS Inter Second Year Maths 2A De Moivre’s Theorem Important Questions Long Answer Type 11

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

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

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

TS Inter Second Year Maths 2A De Moivre’s Theorem Important Questions Long Answer Type 12

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 + ………….. + anxn, then show that
(i) a0 – a2 + a4 – a6 + ……. = 2n/2 cos \(\frac{n \pi}{4}\)
(ii) a1 – a3 + a5 + ……………. = 2n/2 sin \(\frac{n \pi}{4}\)
Solution:
Given,
(1 + x)n = a0 + a1x + a2x2 + ……………. + anxn
Put x = i then
(1 + i)n = a0 + a1(i) + a2i2 + a3i3 + a4i4 + a5i5 + …………… + anin
= a0 + a1i – a2 – a3i – a4 + a5i + …………. + anin
= (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
= 2n/2 (cos \(\frac{n \pi}{4}\) + i sin \(\frac{n \pi}{4}\))
From (1),
2n/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 + ………………. = 2n/2 cos \(\frac{n \pi}{4}\)

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

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

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 \(\frac{9\pi}{5}\).

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

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 \(\frac{5 \pi}{3}\).

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

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

TS Inter Second Year Maths 2A De Moivre’s Theorem Important Questions Long Answer Type 13

∴ 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

TS Inter Second Year Maths 2A De Moivre’s Theorem Important Questions Long Answer Type 14

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

TS Inter Second Year Maths 2A De Moivre’s Theorem Important Questions Long Answer Type 15

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 = 41/5 [cos(2k + 1)π + isin(2k + 1)π]1/5
k = 0, 1, 2, 3, 4
= 41/5 [cos (2k + 1) \(\frac{\pi}{5}\) + i sin (2k + 1) \(\frac{\pi}{5}\)]
= 41/5 cis (2k + 1) \(\frac{\pi}{4}\), k = 0, 1, 2, 3, 4

TS Inter Second Year Maths 2A De Moivre’s Theorem Important Questions Long Answer Type 16

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

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

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:

TS Inter Second Year Maths 2A De Moivre’s Theorem Important Questions Long Answer Type 17

= (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.

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

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

TS Inter Second Year Maths 2A De Moivre’s Theorem Important Questions Long Answer Type 18

= 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.

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