{"id":35597,"date":"2022-11-23T16:36:07","date_gmt":"2022-11-23T11:06:07","guid":{"rendered":"https:\/\/tsboardsolutions.com\/?p=35597"},"modified":"2022-11-24T12:30:56","modified_gmt":"2022-11-24T07:00:56","slug":"maths-2a-de-moivres-theorem-important-questions-long-answer-type","status":"publish","type":"post","link":"https:\/\/tsboardsolutions.com\/maths-2a-de-moivres-theorem-important-questions-long-answer-type\/","title":{"rendered":"TS Inter Second Year Maths 2A De Moivre\u2019s Theorem Important Questions Long Answer Type"},"content":{"rendered":"

Students must practice these Maths 2A Important Questions<\/a> TS Inter Second Year Maths 2A De Moivre\u2019s Theorem Important Questions Long Answer Type to help strengthen their preparations for exams.<\/p>\n

TS Inter Second Year Maths 2A De Moivre\u2019s Theorem Important Questions Long Answer Type<\/h2>\n

Question 1.
\nState and prove De Moivre\u2019s theorem for integral index. [May \u201803]
\nSolution:
\nTheorem : (De Moivre\u2019s for integral index)
\nIf n is an integer then
\n(cos \u03b8 + i sin \u03b8)n<\/sup> = cos n\u03b8 + i sinn\u03b8
\nProof:
\nCase – 1:
\nSuppose \u2018n\u2019 is a positive integer
\nLet, S(n) be the statement that
\n(cos \u03b8 + i sin \u03b8)n<\/sup> = cos n\u03b8 + i sin n\u03b8
\nIf n = 1 then
\nL.H.S: (cos \u03b8 + i sin \u03b8)n<\/sup>
\n= (cos \u03b8 + i sin \u03b8)1<\/sup>
\n= cos \u03b8 + i sin \u03b8
\nR.H.S: cos n\u03b8 +i sin n\u03b8
\n= cos (1\u03b8) + i sin (1\u03b8)
\n= cos \u03b8 + i sin \u03b8
\n\u2234 L.H.S = R.H.S
\n\u2234 S(1) is true.
\nAssume that S(k) is true
\n\u2234 (cos \u03b8 + i sin \u03b8)k<\/sup> = cos k\u03b8 + i sin k\u03b8
\nNow,
\n(cos \u03b8 + i sin \u03b8)k+1<\/sup> = (cos \u03b8 + i sin \u03b8)k<\/sup> (cos \u03b8 + i sin \u03b8) .
\n= (cos k\u03b8 + i sin k\u03b8) (cos \u03b8 + i sin \u03b8)
\n= cos k\u03b8 cos \u03b8 + i cos k\u03b8 sin \u03b8 + i sin k\u03b8 cos \u03b8 – sin k\u03b8 sin \u03b8
\n= (cos k\u03b8 cos \u03b8 – sin k\u03b8 sin \u03b8) + i(sin k\u03b8 cos \u03b8 + cos k\u03b8 sin \u03b8)
\n= cos (k\u03b8 + \u03b8) + i sin(k\u03b8 + \u03b8)
\n= cos(k + 1)\u03b8 + i sin(k + 1)\u03b8
\n\u2234 S(k + 1) is true.
\nBy the principle of mathematical induction S(n) is true, \u2200 n \u2208 N.<\/p>\n

Case – 2 :
\nIf n = 0 then
\nL.H.S: (cos \u03b8 + i sin \u03b8)\u00b0 = 1
\nR.H.S: cos n\u03b8 + i sin n\u03b8 = cos(0\u03b8) + isin (0\u03b8)
\n= cos \u03b8 + i sin \u03b8
\n= 1 + i(0) = 1
\n\u2234 LH.S = R.H.S
\n\u2234 (cos \u03b8 + i sin \u03b8)n<\/sup> = cos n\u03b8 + i sin n\u03b8.<\/p>\n

Case – 3 :
\nSuppose \u2018n\u2019 is a negative integer
\nLet, n = – m where m \u2208 Z+<\/sup>
\nL.H.S = (cos \u03b8 + i sin \u03b8)
\n= (cos \u03b8 + i sin \u03b8)-m<\/sup>
\n= \\(\\frac{1}{(\\cos \\theta+i \\sin \\theta)^m}\\)
\n= \\(\\frac{1}{\\cos m \\theta+i \\sin m \\theta}\\) [\u2235 from case (1)]
\n= \\(\\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}\\)
\n= \\(\\frac{\\cos m \\theta-i \\sin m \\theta}{\\cos ^2 m \\theta+\\sin ^2 m \\theta}\\)
\n= \\(\\frac{\\cos m \\theta-i \\sin m \\theta}{1}\\)
\n= cos m\u03b8 – i sin m\u03b8
\n= cos (- m)\u03b8 + i sin(- m)\u03b8
\n= cos n\u03b8 + i sin n\u03b8
\n= R.H.S
\n\u2234 (cos \u03b8 + i sin \u03b8)n<\/sup> = cos n\u03b8 + i sin n\u03b8, \u2200 n \u2208 Z.<\/p>\n

\"TS<\/p>\n

Question 2.
\nIf m, n are integers and x = cos \u03b1 + i sin \u03b1, y = cos \u03b2 + i sin \u03b2 then prove that
\nxm<\/sup>yn<\/sup> + \\(\\frac{1}{x^m y^n}\\) = 2 cos (m\u03b1 + n\u03b2) and
\nxm<\/sup>yn<\/sup> – \\(\\frac{1}{x^m y^n}\\) = 2i sin (m\u03b1 + n\u03b2)
\nSolution:
\nGiven that
\nxm<\/sup> = (cos \u03b1 + i sin \u03b1)m<\/sup>
\n= cos m\u03b1 + i sin m\u03b1 = cis m\u03b1
\nyn<\/sup> = (cos \u03b2 + isin \u03b2)n<\/sup>
\n= cos n\u03b2 + i sin n\u03b2 = cis n\u03b2
\nNow,
\nxm<\/sup>yn<\/sup> = cis m\u03b1 . cis n\u03b2
\n= cis (m\u03b1 + n\u03b2)
\n= cos (m\u03b1 + n\u03b2) + i sin (m\u03b1 + n\u03b2)
\n\\(\\frac{1}{x^m y^n}=\\frac{1}{\\cos (m \\alpha+n \\beta)+i \\sin (m \\alpha+n \\beta)}\\)
\n= cos (m\u03b1 + n\u03b2) – i sin (m\u03b1 + n\u03b2)<\/p>\n

(i) xm<\/sup>yn<\/sup> + \\(\\frac{1}{x^m y^n}\\)
\n= cos (m\u03b1 + n\u03b2) + i sin (m\u03b1 + n\u03b2) + cos (m\u03b1 + n\u03b2) – i sin (m\u03b1 + n\u03b2)
\n= 2 cos (m\u03b1 + n\u03b2)
\n\u2234 xm<\/sup>yn<\/sup> + \\(\\frac{1}{x^m y^n}\\) = 2 cos (m\u03b1 + n\u03b2)<\/p>\n

(ii) xm<\/sup>yn<\/sup> – \\(\\frac{1}{x^m y^n}\\)
\n= cos (m\u03b1 + n\u03b2) + i sin (m\u03b1 + n\u03b2) – cos (m\u03b1 + n\u03b2) + i sin (m\u03b1 + n\u03b2)
\n= 2i sin (m\u03b1 + n\u03b2)
\n\u2234 xm<\/sup>yn<\/sup> – \\(\\frac{1}{x^m y^n}\\) = 2i sin (m\u03b1 + n\u03b2)<\/p>\n

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

\"TS<\/p>\n

Let, 1 + i = r (cos \u03b8 + i sin \u03b8)
\nthen, r cos \u03b8 = 1, r sin \u03b8 = 1
\n\u2234 r = \\(\\sqrt{\\mathrm{x}^2+\\mathrm{y}^2}=\\sqrt{1^2+1^2}=\\sqrt{1+1}=\\sqrt{2}\\)
\nHence.
\n\u221a2 cos \u03b8 = 1,
\ncos \u03b8 = \\(\\frac{1}{\\sqrt{2}}\\)<\/p>\n

\u221a2 sin \u03b8 = 1
\nsin \u03b8 = \\(\\frac{1}{\\sqrt{2}}\\)<\/p>\n

\u2234 \u03b8 lies in Q1<\/sub>.
\n\u2234 \u03b8 = \\(\\frac{\\pi}{4}\\)<\/p>\n

Similarly,
\n1 – i = \u221a2 (cos \\(\\frac{\\pi}{4}\\) – i sin \\(\\frac{\\pi}{4}\\))
\nL.H.S:
\n(1 + i)n<\/sup> + (1 – i)n<\/sup><\/p>\n

\"TS<\/p>\n

\"TS<\/p>\n

Question 4.
\nIf n is an integer then show that (1 + cos \u03b8 + i sin \u03b8)n<\/sup> + (1 + cos \u03b8 – i sin \u03b8)n<\/sup> = 2n+1<\/sup> cosn<\/sup> \\(\\left(\\frac{\\theta}{2}\\right)\\) – i cos \\(\\left(\\frac{n \\theta}{2}\\right)\\) . [May \u201801, \u201897, March ’10. Mar. ’93, TS & AP – Mar. 2017]
\nSolution:<\/p>\n

\"TS<\/p>\n

Question 5.
\nIf cos \u03b1 + cos \u03b2 + cos \u03b3 = 0 = sin \u03b1 + sin \u03b2 + sin \u03b3, prove that cos2<\/sup> \u03b1 + cos2<\/sup> \u03b2 + cos2<\/sup> \u03b3 = \\(\\frac{3}{2}\\) = sin2<\/sup> \u03b1 + sin2<\/sup> \u03b2 + sin2<\/sup> \u03b3. [AP – Mar., May 2016; TS – Mar. \u201818. \u201815. May \u201815, May \u201809, March \u201803, \u201896, March \u201813 (old)]
\nSolution:
\nGiven that,
\ncos \u03b1 + cos \u03b2 + cos \u03b3 = 0 = sin \u03b1 + sin \u03b2 + sin \u03b3
\nLet, x = cos \u03b1 + i sin \u03b1
\ny = cos \u03b2 + i sin \u03b2
\nz = cos \u03b3 + i sin \u03b3
\nNow,
\nx + y + z = cos \u03b1 + i sin \u03b1 + cos \u03b2 + i sin \u03b2 + cos \u03b3 + i sin \u03b3
\n= (cos \u03b1 + cos \u03b2 + cos \u03b3) + i(sin \u03b1 + sin \u03b2 + sin \u03b3)
\n= 0 + i(0) = 0
\n\u2234 x + y + z = 0
\nSquaring on both sides,
\n(x + y + z)2<\/sup> = 0
\nx2<\/sup> + y2<\/sup> + z2<\/sup> + 2 (xy + yz + zx) = 0
\nx2<\/sup> + y2<\/sup> + z2<\/sup> = – 2 (xy + yz + zx)
\n= \\(\\frac{-2 x y z}{x y z}\\) (xy + y + zx)
\n= – 2xyz \\(\\left(\\frac{1}{z}+\\frac{1}{x}+\\frac{1}{y}\\right)\\)
\n= – 2xyz (cos \u03b3 – i sin \u03b3 + cos \u03b1 – i sin \u03b1 + cos \u03b2 – i sin \u03b2)
\n= – 2xyz [(cos \u03b1 + cos \u03b2 + cos \u03b3) – i (sin \u03b1 + sin \u03b2 + sin \u03b3)]
\n= – 2xyz [0 – i . 0]
\n= – 2xyz(0) = 0
\n\u2234 x2<\/sup> + y2<\/sup> + z2<\/sup> = 0
\n(cos \u03b1 + i sin \u03b1)2<\/sup> + (cos \u03b2 + i sin \u03b2)2<\/sup> + (cos \u03b3 + i sin \u03b3)2<\/sup> = 0
\ncos 2\u03b1 + i sin2\u03b1 + cos 2\u03b2 + i sin 2\u03b2 + cos 2\u03b3 + i sin 2\u03b3 = 0
\n(cos 2\u03b1 + cos 2\u03b2 + cos 2\u03b3) + i(sin 2\u03b1 + sin 2\u03b2 + sin 2\u03b3) = 0
\nComparing real parts on both sides, we get
\n(i) cos 2\u03b1 + cos 2\u03b2 + cos 2\u03b3 = 0
\n2 cos2<\/sup> \u03b1 – 1 + 2 cos2<\/sup> \u03b2 – 1 + 2 cos2<\/sup> \u03b3 – 1 = 0
\n2 (cos2<\/sup> \u03b1 + cos2<\/sup> \u03b2 + cos2<\/sup> \u03b3) = 3
\ncos2<\/sup> \u03b1 + cos2<\/sup> \u03b2 + cos2<\/sup> \u03b3 = \\(\\frac{3}{2}\\)<\/p>\n

(ii) 1 – sin2<\/sup> \u03b1 + 1 – sin2<\/sup> \u03b2 + 1 – sin2<\/sup> \u03b3 = \\(\\frac{3}{2}\\)
\n3 – (sin2<\/sup> \u03b1 + sin2<\/sup> \u03b2 + sin2<\/sup> \u03b3) = \\(\\frac{3}{2}\\)
\nsin2<\/sup> \u03b1 + sin2<\/sup> \u03b2 + sin2<\/sup> \u03b3 = 3 – \\(\\frac{3}{2}\\) = \\(\\frac{3}{2}\\).<\/p>\n

\"TS<\/p>\n

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

\"TS<\/p>\n

Let, 1 + i = r (cos \u03b8 + i sin \u03b8)
\nthen r cos \u03b8 = 1, r sin \u03b8 = 1
\nr = \\(\\sqrt{\\mathrm{x}^2+\\mathrm{y}^2}=\\sqrt{1^2+1^2}=\\sqrt{1+1}=\\sqrt{2}\\)
\nHence,
\n\u221a2 cos \u03b8 = 1,
\ncos \u03b8 = \\(\\frac{1}{\\sqrt{2}}\\)
\n\u221a2 sin \u03b8 = 1
\nsin \u03b8 = \\(\\frac{1}{\\sqrt{2}}\\)
\n\u2234 \u03b8 lies in Q1<\/sub>.
\n\u2234 \u03b8 = \\(\\frac{\\pi}{4}\\)
\n\u2234 1 + i = \u221a2 (cos \\(\\frac{\\pi}{4}\\) + i cos \\(\\frac{\\pi}{4}\\))
\nSimilarly,
\n1 – i = \u221a2 (cos \\(\\frac{\\pi}{4}\\) – i sin \\(\\frac{\\pi}{4}\\))
\nL.H.S:
\n(1 + i)2n<\/sup> + (1 – i)2n<\/sup> = [\u221a2 (cos \\(\\frac{\\pi}{4}\\) + i sin \\(\\frac{\\pi}{4}\\))]2n<\/sup> + [\u221a2 (cos \\(\\frac{\\pi}{4}\\) – i sin \\(\\frac{\\pi}{4}\\))]2n<\/sup>
\n= 2n<\/sup> (cos \\(\\frac{n \\pi}{2}\\) + i sin \\(\\frac{n \\pi}{2}\\)) + 2n<\/sup> (cos \\(\\frac{n \\pi}{2}\\) – i sin \\(\\frac{n \\pi}{2}\\))
\n= 2n<\/sup> [cos \\(\\frac{n \\pi}{2}\\) + i sin \\(\\frac{n \\pi}{2}\\) + cos \\(\\frac{n \\pi}{2}\\) – i sin \\(\\frac{n \\pi}{2}\\)]
\n= 2n<\/sup> . 2 cos \\(\\frac{n \\pi}{2}\\)
\n= 2n+1<\/sup> cos \\(\\frac{n \\pi}{2}\\)
\n= R.H.S
\n\u2234 (1 + i)2n<\/sup> + (1 – i)2n<\/sup> = 2n+1<\/sup> cos \\(\\frac{n \\pi}{2}\\).<\/p>\n

\"TS<\/p>\n

Question 7.
\nIf \u03b1, \u03b2 are the roots of the equation x2<\/sup> – 2x + 4 = 0 then for any n \u2208 N show that \u03b1n<\/sup> + \u03b2n<\/sup> = 2n+1<\/sup> cos \\(\\frac{n \\pi}{3}\\) [TS – May 2016; March ’14, ’11, May ’88, AP – Mar. 2019]
\nSolution:<\/p>\n

\"TS<\/p>\n

Given quadratic equation is x2<\/sup> – 2x + 4 = 0
\nComparing ax2<\/sup> + bx + c = 0 we get, a = 1, b = – 2, c = 4
\n\u2234 x = \\(\\frac{-b \\pm \\sqrt{b^2-4 a c}}{2 a}\\)
\n= \\(\\frac{-(-2) \\pm \\sqrt{4-16}}{2(1)}\\)
\n= \\(\\frac{2 \\pm \\sqrt{-12}}{2}\\)
\n= \\(\\frac{2 \\pm i 2 \\sqrt{3}}{2}\\)
\n= 1 \u00b1 \u221a3i
\nSince, \u03b1, \u03b2 are the roots of the equation
\nx2<\/sup> – 2x + 4 = 0 then
\n\u03b1 = 1 + i\u221a3, \u03b2 = 1 – i\u221a3
\nLet,
\n\u03b1 = 1 + i\u221a3 = r (cos \u03b8 + i sin \u03b8)
\nthen r cos \u03b8 = 1, r sin \u03b8 = \u221a3
\nr = \\(\\sqrt{x^2+y^2}=\\sqrt{(1)^2+(\\sqrt{3})^2}\\)
\n= \\(\\sqrt{1+3}=\\sqrt{4}\\) = 2
\nHence,
\n2 cos \u03b8 = 1,
\ncos \u03b8 = \\(\\frac{1}{2}\\)
\n2 sin \u03b8 = \u221a3
\nsin \u03b8 = \\(\\frac{\\sqrt{3}}{2}\\)
\n\u2234 \u03b8 lies in the Q1<\/sub>.
\n\u2234 \u03b8 = \\(\\frac{\\pi}{3}\\)
\n\u03b1 = 1 + i\u221a3 = 2(cos \\(\\frac{\\pi}{3}\\) + i sin \\(\\frac{\\pi}{3}\\))
\nSimilarly,
\n\u03b2 = 1 – i\u221a3 = 2(cos \\(\\frac{\\pi}{3}\\) – i sin \\(\\frac{\\pi}{3}\\))
\nL.H.S:<\/p>\n

\"TS<\/p>\n

\"TS<\/p>\n

Question 8.
\nIf cos \u03b1 + cos \u03b2 + cos \u03b3 = 0 = sin \u03b1 + sin \u03b2 + sin \u03b3 then show that
\n(i) cos 3\u03b1 + cos 3\u03b2 + cos 3\u03b3 = 2 cos (\u03b1 + \u03b2 + \u03b3)
\n(ii) sin 3\u03b1 + sin 3\u03b2 + sin 3\u03b3 = 2 sin (\u03b1 + \u03b2 + \u03b3)
\n(iii) cos (\u03b1 + \u03b2) + cos (\u03b2 + \u03b3) + cos (\u03b3 + \u03b1) = 0
\nSoluton:
\nGiven that,
\ncos \u03b1 + cos \u03b2 + cos \u03b3 = 0 = sin \u03b1 + sin \u03b2 + sin \u03b3
\nLet,
\nx = cos \u03b1 + i sin \u03b1
\ny = cos \u03b2 + i sin \u03b2
\nz = cos \u03b3 + i sin \u03b3
\nNow,
\nx + y + z = cos \u03b1 + i sin \u03b1 + cos \u03b2 + i sin \u03b2 + cos \u03b3 + i sin \u03b3
\n= (cos \u03b1 + cos \u03b2 + cos \u03b3) + i(sin \u03b1 + sin \u03b2 + sin \u03b3)
\n= 0 + i (0) = 0
\n\u2234 x + y + z = 0
\n\u21d2 x3<\/sup> + y3<\/sup> + z3<\/sup> = 3xyz
\n(cos \u03b1 + i sin \u03b1) + (cos \u03b2 + i sin \u03b2) + (cos \u03b3 + i sin \u03b3) = 3 (cos \u03b1 + i sin \u03b1) (cos \u03b2 + i sin \u03b2) (cos \u03b3 + i sin \u03b3)
\ncos 3\u03b1 + i sin 3\u03b1 + cos 3\u03b2 + i sin 3\u03b2 + cos 3\u03b3 + i sin 3\u03b3 = 3 cis \u03b1 cis \u03b2 cis \u03b3
\n(cos 3\u03b1 + cos 3\u03b2 + cos 3\u03b3) + i(sin 3\u03b1 + sin 3\u03b2 + sin 3\u03b3) = 3 cis (\u03b1 + \u03b2 + \u03b3)
\n= 3 [cos (\u03b1 + \u03b2 + \u03b3) + i sin (\u03b1 + \u03b2 + \u03b3)]<\/p>\n

(i) Comparing the real parts on both sides we get,
\ncos 3\u03b1 + i sin 3\u03b1 + cos 3\u03b2 = 3 cos (\u03b1 + \u03b2 + \u03b3)<\/p>\n

(ii) Comparing the imaginary parts on both sides we get,
\nsin 3\u03b1 + sin 3\u03b2 + sin 3\u03b3 = 3 sin (\u03b1 + \u03b2 + \u03b3)<\/p>\n

(iii) Let,
\nx = cos \u03b1 + i sin \u03b1
\n\u21d2 \\(\\frac{1}{x}\\) = cos \u03b1 – i sin \u03b1
\ny = cos \u03b2 + i sin \u03b2
\n\u21d2 \\(\\frac{1}{y}\\) = cos \u03b2 – i sin \u03b2
\nz = cos \u03b3 + i sin \u03b3
\n\u21d2 \\(\\frac{1}{z}\\) = cos \u03b3 – i sin \u03b3
\nNow,
\n\\(\\frac{1}{x}+\\frac{1}{y}+\\frac{1}{z}\\) = cos \u03b1 – i sin \u03b1 + cos \u03b2 – i sin \u03b2 + cos \u03b3 – i sin \u03b3
\n= (cos \u03b1 + cos \u03b2 + cos \u03b3) – i(sin \u03b1 + sin \u03b2 + sin \u03b3)
\n= 0 + i(0) = 0
\n\u2234 \\(\\frac{1}{x}+\\frac{1}{y}+\\frac{1}{z}\\) = 0
\n\\(\\frac{y z+x z+x y}{x y z}\\) = 0
\n(cos \u03b1 + i sin \u03b1) (cos \u03b2 + i sin \u03b2) + (cos \u03b2 + i sin \u03b2) (cos \u03b3 + i sin \u03b3) + (cos \u03b3 + i sin \u03b3) (cos \u03b1 + i sin \u03b1) = 0
\ncis \u03b1 cis \u03b2 + cis \u03b2 . cis \u03b3 + cis \u03b3 . cis \u03b1 = 0
\ncis (\u03b1 + \u03b2) + cis (\u03b2 + \u03b3) + cis (\u03b3 + \u03b1) = 0
\ncos (\u03b1 + \u03b2) + i sin (\u03b1 + \u03b2) + cos (\u03b2 + \u03b3) + isin (\u03b2 + \u03b3) + cos (\u03b3 + \u03b1) + isin (\u03b3 + \u03b1) = 0
\n[cos (\u03b1 + \u03b2) + cos (\u03b2 + \u03b3) + cos (\u03b3 + \u03b1)] + i[sin (\u03b1 + \u03b2) + sin (\u03b2 + \u03b3) + sin (\u03b3 + \u03b1)] = 0
\nComparing real parts on both sides we get,
\ncos (\u03b1 + \u03b2) + cos (\u03b2 + \u03b3) + cos (\u03b3 + \u03b1) = 0
\nComparing imaginary parts on both sides
\nwe get,
\nsin (\u03b1 + \u03b2) + sin (\u03b2 + \u03b3) + sin (\u03b3 + \u03b1) = 0.<\/p>\n

\"TS<\/p>\n

Question 9.
\nIf n is an Integer and z = cis \u03b8, (\u03b8 \u2260 (2n + 1)\\(\\frac{\\pi}{2}\\)), then show that \\(\\frac{z^{2 n}-1}{z^{2 n}+1}\\) = i tan n\u03b8. [Mar. ’12. ’19(TS)]
\nSolution:
\nGiven that,
\nz = cis \u03b8 = cos \u03b8 + i sin \u03b8<\/p>\n

\"TS<\/p>\n

Question 10.
\nFind all the roots of the equation x11<\/sup> – x7<\/sup> + x4<\/sup> – 1 = 0 [Board Paper].
\nSolution:
\nGiven equation is x11<\/sup> – x7<\/sup> + x4<\/sup> – 1 = 0
\nx7<\/sup> (x4<\/sup> – 1) . 1 (x4<\/sup> – 1) = 0
\n(x4<\/sup> – 1) (x7<\/sup> + 1) = 0
\nx4<\/sup> – 1 = 0 or x7<\/sup> + 1 = 0<\/p>\n

x4<\/sup> – 1 = 0:
\nx4<\/sup> = 1
\nx = (1)1\/4<\/sup>
\n= (cos 0 + i sin 0)1\/4<\/sup>
\n= [cos (2k\u03c0 + 0) + isin(2k\u03c0 + 0)]1\/4<\/sup>
\nk = 0, 1, 2, 3
\n= [cos (2k\u03c0) + i sin(2k\u03c0]1\/4<\/sup>
\n= cos (\\(\\frac{k\\pi}{2}\\)) + sin (\\(\\frac{k\\pi}{2}\\))
\n= cis (\\(\\frac{k\\pi}{2}\\)), k = 0, 1, 2, 3
\n\u2234 The values of x are
\ncis (0), cis (\\(\\frac{\\pi}{2}\\)), cis \u03c0, cis (\\(\\frac{3\\pi}{2}\\)) : 1, i, – 1, – i<\/p>\n

x7<\/sup> + 1 = 0:
\nx7<\/sup> = – 1
\nx = (- 1)1\/7<\/sup>
\n= [cos \u03c0 + i sin \u03c0]
\n= [cos (2k\u03c0 + \u03c0) + i sin (2k\u03c0 + \u03c0)]1\/7<\/sup>, k = 0, 1, 2, 3, 4, 5, 6
\n= [cos (2k + 1)\u03c0 + i sin (2k + 1)\u03c0]1\/7<\/sup>
\n= [cos (2k + 1)\\(\\frac{\\pi}{7}\\) + i sin (2k + 1)\\(\\frac{\\pi}{7}\\)]
\n= cis (2k + 1)\\(\\frac{\\pi}{7}\\), k = 0, 1, 2, 3, 4, 5, 6
\nThe values of x are
\ncis \\(\\frac{\\pi}{7}\\), cis \\(\\frac{3 \\pi}{7}\\), cis \\(\\frac{5 \\pi}{7}\\), cis \u03c0, cis \\(\\frac{9 \\pi}{7}\\), cis \\(\\frac{11 \\pi}{7}\\), cis \\(\\frac{13 \\pi}{7}\\)
\n\u2234 The roots of the equation are \u00b1 1, \u00b1 i, cis \\(\\frac{\\pi}{7}\\), cis \\(\\frac{3 \\pi}{7}\\), cis \\(\\frac{5 \\pi}{7}\\), cis \u03c0, cis \\(\\frac{9 \\pi}{7}\\), cis \\(\\frac{11 \\pi}{7}\\), cis \\(\\frac{13 \\pi}{7}\\)<\/p>\n

\"TS<\/p>\n

Question 11.
\nSolve the equation x4<\/sup> – 1 = 0.
\nSolution:
\nGiven equation is x4<\/sup> – 1 = 0
\nx4<\/sup> = 1
\nx = (1)1\/4<\/sup>
\n= [cos o + i sin o]1\/4<\/sup>
\n= [cos (2k\u03c0 + 0) + i sin(2k\u03c0 + 0)]1\/4<\/sup>
\nk = 0, 1, 2, 3
\n= [cos 2k\u03c0 + i sin 2k\u03c0]1\/4<\/sup>
\n= cos \\(\\frac{k \\pi}{2}\\) + i sin \\(\\frac{k \\pi}{2}\\)
\n= cis \\(\\frac{k \\pi}{2}\\), k = 0, 1, 2, 3
\nlf k = 0 then x = cos 0 + i sin 0 = 1 + i . 0 = 1
\nIf k = 1 then x = cos \\(\\frac{\\pi}{2}\\) + i sin \\(\\frac{\\pi}{2}\\) = 0 + i . 1 = i
\nIf k = 2 then x = cos \u03c0 + i sin \u03c0
\n= – 1 + i .0 = – 1
\nIf k = 3 then x cos \\(\\frac{3 \\pi}{2}\\) + i sin \\(\\frac{3 \\pi}{2}\\)
\n= o + i (- 1) = – i
\n\u2234 The roots of the given equation are \u00b1 1, \u00b1 i.<\/p>\n

Question 12.
\nSolve the equation x4<\/sup> + 1 = 0. [May ’97]
\nSolution:
\nGiven equation is x4<\/sup> + 1 = 0
\nx4<\/sup> = – 1
\nx = (- 1)1\/4<\/sup>
\n= [cos \u03c0 + i sin \u03c0]1\/4<\/sup>
\n= [cos (2k\u03c0 + \u03c0) + i sin (2k\u03c0 + \u03c0)]1\/4<\/sup>, k = 0, 1, 2, 3
\n= [cos (2k + 1)\u03c0 + i sin(2k + 1)\u03c0]1\/4<\/sup>
\n= cos(2k + 1) \\(\\frac{\\pi}{4}\\) + i sin(2k + 1) \\(\\frac{\\pi}{4}\\)
\n= cis (2k + 1) \\(\\frac{\\pi}{4}\\), k = 0, 1, 2, 3
\nIf k = 0, x = cis \\(\\frac{\\pi}{4}\\)
\nIf k = 1, x = cis \\(\\frac{3 \\pi}{4}\\)
\nIf k = 2, x = cis \\(\\frac{5 \\pi}{4}\\)
\nIf k = 3, x = cis \\(\\frac{7 \\pi}{4}\\)
\n\u2234 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}\\).<\/p>\n

\"TS<\/p>\n

Question 13.
\nIf n is a positive integer, show that (p + iq)1\/n<\/sup> + (p – iq)1\/n<\/sup> = 2(p2<\/sup> + q2<\/sup>)1\/2n<\/sup> . cos \\(\\left(\\frac{1}{n} \\tan ^{-1} \\frac{q}{p}\\right)\\) [AP – Mar. 18, May ’15; Mar. ’01]
\nSolution:
\nLet p + iq = r (cos \u03b8 + i sin \u03b8)
\nthen r cos \u03b8 = p, r sin \u03b8 = q
\nr = \\(\\sqrt{\\mathrm{x}^2+\\mathrm{y}^2}=\\sqrt{\\mathrm{p}^2+\\mathrm{q}^2}\\)
\nHence,
\n\\(\\sqrt{p^2+q^2}\\) cos \u03b8 = p, \\(\\sqrt{p^2+q^2}\\) sin \u03b8 = q
\ncos \u03b8 = \\(\\frac{p}{\\sqrt{p^2+q^2}}\\)
\nsin \u03b8 = \\(\\frac{q}{\\sqrt{p^2+q^2}}\\)
\n\u2234 tan \u03b8 = \\(\\frac{\\sin \\theta}{\\cos \\theta}=\\frac{\\frac{q}{\\sqrt{p^2+q^2}}}{\\frac{p}{\\sqrt{p^2+q^2}}}=\\frac{q}{p}\\)
\n\u03b8 = \\(\\tan ^{-1}\\left(\\frac{q}{p}\\right)\\)
\n\u2234 p + iq = \\(\\sqrt{p^2+q^2}\\) [cos \u03b8 + i sin \u03b8]
\nSimilarly,
\np – iq = \\(\\sqrt{p^2+q^2}\\) [cos \u03b8 – i sin \u03b8]<\/p>\n

\"TS<\/p>\n

Question 14.
\nShow 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]
\nSolution:
\nGiven,<\/p>\n

\"TS<\/p>\n

\"TS<\/p>\n

Question 15.
\nSolve (x – i)n<\/sup> = xn<\/sup>, n is a positive integer. [March ’02]
\nSolution:
\nGiven equation is (x – 1)n<\/sup> = xn<\/sup>
\nTaking nth root of each side of (x – 1)n<\/sup> = xn<\/sup>
\nwe have
\nx – 1 = x(1)1\/n<\/sup>
\n= x [cos 0 – i sin 0]1\/n<\/sup>
\n= x [cos (2k\u03c0 + 0) + i sin (2k\u03c0 + 0)]1\/n<\/sup>
\nk = 0, 1, ………….., (n – 1)
\n= x [cos 2k\u03c0 + i sin 2k\u03c0]1\/n<\/sup><\/p>\n

\"TS<\/p>\n

\"TS<\/p>\n

\"TS<\/p>\n

Some More Maths 2A De Moivre\u2019s Theorem Important Questions<\/h3>\n

Question 1.
\nFind the value of (1 + i\u221a3)3<\/sup>.
\nSolution:<\/p>\n

\"TS<\/p>\n

Let 1 + i\u221a3 = r (cos \u03b8 + i sin \u03b8)
\nthen r cos \u03b8 = 1, r sin \u03b8 = \u221a3
\nr = \\(\\sqrt{\\mathrm{x}^2+\\mathrm{y}^2}=\\sqrt{(1)^2+(\\sqrt{3})^2}\\)
\n= \\(\\sqrt{1+3}=\\sqrt{4}\\) = 2
\nHence,
\n2 cos \u03b8 = 1, 2 sin \u03b8 = \u221a3
\ncos \u03b8 = \\(\\frac{1}{2}\\), sin \u03b8 = \\(\\frac{\\sqrt{3}}{2}\\)
\n\u2234 \u03b8 lies in the Q1<\/sub>.
\n\u2234 \u03b8 = \\(\\frac{\\pi}{3}\\)
\n\u2234 1 + i\u221a3 = 2 (cos \\(\\frac{\\pi}{3}\\) + i sin \\(\\frac{\\pi}{3}\\))
\nNow,
\n(1 + i\u221a3)3<\/sup> = [2 (cos \\(\\frac{\\pi}{3}\\) + i sin \\(\\frac{\\pi}{3}\\))]3<\/sup>
\n= 8 (cos \u03c0 + i sin \u03c0)
\n= 8 (- 1 + i . 0) = – 8.<\/p>\n

Question 2.
\nIf (1 + x)n<\/sup> = a0<\/sub> + a1<\/sub>x + a2<\/sub>x2<\/sup> + ………….. + an<\/sub>xn<\/sup>, then show that
\n(i) a0<\/sub> – a2<\/sub> + a4<\/sub> – a6<\/sub> + ……. = 2n\/2<\/sup> cos \\(\\frac{n \\pi}{4}\\)
\n(ii) a1<\/sub> – a3<\/sub> + a5<\/sub> + ……………. = 2n\/2<\/sup> sin \\(\\frac{n \\pi}{4}\\)
\nSolution:
\nGiven,
\n(1 + x)n<\/sup> = a0<\/sub> + a1<\/sub>x + a2<\/sub>x2<\/sup> + ……………. + an<\/sub>xn<\/sup>
\nPut x = i then
\n(1 + i)n<\/sup> = a0<\/sub> + a1<\/sub>(i) + a2<\/sub>i2<\/sup> + a3<\/sub>i3<\/sup> + a4<\/sub>i4<\/sup> + a5<\/sub>i5<\/sup> + …………… + an<\/sub>in<\/sup>
\n= a0<\/sub> + a1<\/sub>i – a2<\/sub> – a3<\/sub>i – a4<\/sub> + a5<\/sub>i + …………. + an<\/sub>in<\/sup>
\n= (a0<\/sub> – a2<\/sub> + a4<\/sub> – a6<\/sub> + …………. ) + i(a1<\/sub> – a3<\/sub> + a5<\/sub> – …………) ……………..(1)
\nLet, 1 + i = r (cos \u03b8 + i sin \u03b8)
\nthen r cos \u03b8 = 1, r sin \u03b8 = 1
\nr = \\(\\sqrt{\\mathrm{x}^2+\\mathrm{y}^2}=\\sqrt{1^2+1^2}=\\sqrt{1+1}=\\sqrt{2}\\)
\nHence,
\n\u221a2 cos \u03b8 = i
\ncos \u03b8 = \\(\\frac{1}{\\sqrt{2}}\\)
\n\u221a2 sin \u03b8 = 1
\nsin \u03b8 = \\(\\frac{1}{\\sqrt{2}}\\)
\n\u2234 \u03b8 lies in the Q1<\/sub>.
\n\u2234 \u03b8 = \\(\\frac{\\pi}{4}\\)
\n\u2234 (1 + i) = \u221a2 (cos \\(\\frac{\\pi}{4}\\) + i sin \\(\\frac{\\pi}{4}\\))
\n(1 + i)n<\/sup> = [\u221a2 (cos \\(\\frac{\\pi}{4}\\) + i sin \\(\\frac{\\pi}{4}\\))]n<\/sup>
\n= 2n\/2<\/sup> (cos \\(\\frac{n \\pi}{4}\\) + i sin \\(\\frac{n \\pi}{4}\\))
\nFrom (1),
\n2n\/2<\/sup> (cos \\(\\frac{n \\pi}{4}\\) + i sin \\(\\frac{n \\pi}{4}\\)) = (a0<\/sub> – a2<\/sub> + a4<\/sub> – a6<\/sub> + ……………….) + i (a1<\/sub> – a3<\/sub> + a5<\/sub> – ………….)<\/p>\n

(i) Comparing real parts on both sides we get,
\na0<\/sub> – a2<\/sub> + a4<\/sub> – a6<\/sub> + ………………. = 2n\/2<\/sup> cos \\(\\frac{n \\pi}{4}\\)<\/p>\n

(ii) Comparing imaginary parts on bothsides we get,
\na1<\/sub> – a3<\/sub> + a5<\/sub> – …………… = 2n\/2<\/sup> sin \\(\\frac{n \\pi}{4}\\).<\/p>\n

\"TS<\/p>\n

Question 3.
\nSolve the equation x5<\/sup> + 1 = 0.
\nSolution:
\nGiven equation is x5<\/sup> + 1 = 0
\nx5<\/sup> = – 1
\nx = (- 1)1\/5<\/sup>
\n= [cos \u03c0 + i sin \u03c0]1\/5<\/sup>
\n= [cos(2k\u03c0 + \u03c0) + i sin (2k\u03c0 + \u03c0)]1\/5<\/sup>, k = 0, 1, 2, 3, 4
\n= [cos(2k + 1)\u03c0 + i sin (2k + 1)\u03c0]1\/5<\/sup>
\n= cis (2k + 1) \\(\\frac{\\pi}{5}\\), k = 0, 1, 2, 3, 4
\nIf k = 0,
\n\u21d2 x = cis \\(\\frac{\\pi}{5}\\)<\/p>\n

If k = 1,
\n\u21d2 x = cis \\(\\frac{3 \\pi}{5}\\)<\/p>\n

If k = 2,
\n\u21d2 x = cis \u03c0<\/p>\n

If k = 3,
\n\u21d2 x = cis \\(\\frac{7 \\pi}{5}\\)<\/p>\n

If k = 4,
\n\u21d2 x = cis \\(\\frac{9 \\pi}{5}\\)
\n\u2234 The roots of the given equation are cis \\(\\frac{\\pi}{5}\\), cis \\(\\frac{\\pi}{5}\\), cis \u03c0, cis \\(\\frac{7 \\pi}{5}\\), cis \\(\\frac{9 \\pi}{5}\\).<\/p>\n

Question 4.
\nSolve the equation x9<\/sup> – x5<\/sup> + x4<\/sup> – 1 = 0.
\nSolution:
\nGiven equation is x9<\/sup> – x5<\/sup> + x4<\/sup> – 1 = 0
\nx5<\/sup> (x4<\/sup> – 1) + 1 (x4<\/sup> – 1) = 0
\n(x4<\/sup> – 1) (x5<\/sup> + 1) = 0
\nx4<\/sup> – 1 = 0 or x5<\/sup> + 1 = 0<\/p>\n

x4<\/sup> – 1 = 0:
\nx4<\/sup> = 1
\nx = (1)1\/4<\/sup>
\n= [cos 0 + i sin 0]1\/4<\/sup>
\n= [cos (2k\u03c0 + 0) + i sin (2k\u03c0 + 0)]1\/4<\/sup>, k = 0, 1, 2, 3
\n= [cos (2k\u03c0) + i sin(2k\u03c0)]1\/4<\/sup>
\n= cos \\(\\frac{k \\pi}{2}\\) + i sin \\(\\frac{k \\pi}{2}\\)
\n= cis \\(\\frac{k \\pi}{2}\\), k = 0, 1, 2, 3
\nIf k = 0
\n\u21d2 x = cos 0 + i sin 0
\n= 1 + i . 0 = 1
\nIf k = 1
\n\u21d2 x= cos \\(\\frac{\\pi}{2}\\) + i sin \\(\\frac{\\pi}{2}\\)
\n= 0 + i . 1 = 1
\nIf k = 2
\n\u21d2 x = cos \u03c0 + i sin \u03c0
\n= – 1 + i . 0 = – 1
\nIf k = 3x
\n\u21d2 x = cos \\(\\frac{3 \\pi}{2}\\) + i sin \\(\\frac{3 \\pi}{2}\\)
\n= 1 + i(- 1) = – i
\n\u2234 The values of x are \u00b1 1, \u00b1 1.<\/p>\n

x5<\/sup> + 1 = 0:
\nx5<\/sup> = (- 1)
\nx = (- 1)1\/5<\/sup>
\n= [cos \u03c0 + i sin \u03c0]1\/5<\/sup>
\n= [cos (2k\u03c0 + \u03c0) + i sin(2k\u03c0 + \u03c0)]1\/5<\/sup>, k = 0, 1, 2, 3, 4
\n= [cos(2k + 1)\u03c0 + i sin(2k + 1)\u03c0]1\/5<\/sup>
\n= cos (2k + 1) \\(\\frac{\\pi}{5}\\) + i sin(2k + 1) \\(\\frac{\\pi}{5}\\)
\n= cis (2k + 1) \\(\\frac{\\pi}{5}\\), k = 0, 1, 2, 3, 4
\nIf k = 0
\n\u21d2 x = cis \\(\\frac{\\pi}{5}\\)
\nIf k = 1
\n\u21d2 x = cis \\(\\frac{3 \\pi}{5}\\)
\nIf k = 2
\n\u21d2 x = cis \u03c0
\nIf k = 3
\n\u21d2 x = cis \\(\\frac{7 \\pi}{5}\\)
\nIf x = 4
\n\u21d2 x = cis \\(\\frac{9 \\pi}{5}\\)
\n\u2234 The values of x are cis \\(\\frac{\\pi}{5}\\), cis \\(\\frac{3 \\pi}{5}\\), cis \u03c0, cis \\(\\frac{7 \\pi}{5}\\), cis \\(\\frac{9\\pi}{5}\\).
\n\u2234 The roots of the given equation are \u00b1 1, \u00b1 i, cis \\(\\frac{\\pi}{5}\\), cis \\(\\frac{3 \\pi}{5}\\), cis \u03c0, cis \\(\\frac{7 \\pi}{5}\\), cis \\(\\frac{9\\pi}{5}\\).<\/p>\n

\"TS<\/p>\n

Question 5.
\nFind the common roots of x12<\/sup> – 1 = 0 and x + x + 1 = 0.
\nSolution:
\nGiven equation is x12<\/sup> – 1 = 0
\nx12<\/sup> = 1
\nx = (1)12<\/sup>
\n= [cos 0 + i sin 0]1\/12<\/sup>
\n= [cos (2k\u03c0 + 0) + i sin (2k\u03c0 + 0)]1\/12<\/sup>, k = 0, 1, ……………, 11
\n= [cos 2k\u03c0 + i sin 2k\u03c0]1\/12<\/sup>
\n= cos \\(\\frac{k \\pi}{6}\\) + i sin \\(\\frac{k \\pi}{6}\\)
\n= cis \\(\\frac{k \\pi}{6}\\), k = 0, 1, 2, …………., 11
\nIf k = 0
\n\u21d2 x = cis 0
\nIf k = 1
\n\u21d2 x = cis \\(\\frac{\\pi}{6}\\)
\nIf k = 2
\n\u21d2 x = cis \\(\\frac{\\pi}{3}\\)
\nIf k = 3
\n\u21d2 x = cis \\(\\frac{\\pi}{2}\\)
\nIf k = 4
\n\u21d2 x = cis \\(\\frac{2 \\pi}{3}\\)
\nIf k = 5
\n\u21d2 x = cis \\(\\frac{5 \\pi}{6}\\)
\nIf k = 6
\n\u21d2 x = cis \u03c0
\nIf k = 7
\n\u21d2 x = cis \\(\\frac{7 \\pi}{6}\\)
\nIf k = 8
\n\u21d2 x = cis \\(\\frac{4 \\pi}{3}\\)
\nIf k = 9
\n\u21d2 x = cis \\(\\frac{3 \\pi}{2}\\)
\nIf k = 10
\n\u21d2 x = cis \\(\\frac{5 \\pi}{3}\\)
\nIf k = 11
\n\u21d2 x = cis \\(\\frac{11 \\pi}{6}\\)
\n\u2234 Therootsof x12<\/sup> – 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 \u03c0, 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}\\)<\/p>\n

Given equation is x4<\/sup> + x2<\/sup> + 1 = 0
\nMultiplying on both sides with x2<\/sup> – 1
\nwe get,
\n(x2<\/sup> – 1) (x4<\/sup> + x2<\/sup> + 1) = 0
\nx6<\/sup> – 1 = 0
\nx6<\/sup> = 1
\nx = (1)1\/6<\/sup>
\n= [cos 0 + i sin 0]1\/6<\/sup>
\n= [cos (2k\u03c0 + 0) + i sin (2k\u03c0 + 0)]1\/6<\/sup>, k = 0, 1, 2, 3, 4, 5
\n= [cos 2k\u03c0 + i sin 2k\u03c0]1\/6<\/sup>
\n= cos \\(\\frac{k \\pi}{3}\\) + i sin \\(\\frac{k \\pi}{3}\\)
\n= cis \\(\\frac{k \\pi}{3}\\), k = 0, 1, 2, 3, 4, 5
\nIf k = 0
\n\u21d2 x = cis 0
\nIf k = 1
\n\u21d2 x = cis \\(\\frac{\\pi}{3}\\)
\nIf k = 2
\n\u21d2 x = cis \\(\\frac{2 \\pi}{3}\\)
\nIf k = 3
\n\u21d2 x = cis \u03c0
\nIf k = 4
\n\u21d2 x = cis \\(\\frac{4 \\pi}{3}\\)
\nIf k = 5
\n\u21d2 x = cis \\(\\frac{5 \\pi}{3}\\)
\nThe values of given equation are cis 0, cis \\(\\frac{\\pi}{3}\\), cis \\(\\frac{2 \\pi}{3}\\), cis \u03c0, cis \\(\\frac{4 \\pi}{3}\\), cts \\(\\frac{5 \\pi}{3}\\).
\n\u2234 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}\\).<\/p>\n

\"TS<\/p>\n

Question 6.
\nFind the number of 15th roots of unity, which are also 25th roots of unity.
\nSolution:
\n15th roots of unity:
\nLet, x = \\(\\sqrt[15]{1}=(1)^{1 \/ 15}\\)
\n= [cos 0 + i sin 0]1\/15<\/sup>
\n= [cos (2n\u03c0 + 0) + i sin(2n\u03c0 + 0)]1\/15<\/sup>
\nn = 0, 1, 2, …………., 14
\n= [cos 2n\u03c0 + i sin 2n\u03c0]1\/15<\/sup>
\n= \\(\\cos \\frac{2 n \\pi}{15}+i \\sin \\frac{2 n \\pi}{15}\\)
\n= cis \\(\\frac{2n \\pi}{15}\\), n = 0, 1, 2, …………….., 14<\/p>\n

25th roots of unity:
\nLet, x = \\(\\sqrt[25]{1}\\)
\n= (1)1\/25<\/sup>
\n= [cos 0 + i sin 0]1\/25<\/sup>
\n= [cos (2m\u03c0 + 0) + i sin (2m\u03c0 + 0)]1\/25<\/sup>
\nm = 0, 1, 2, ………….., 24
\n= [cos 2m\u03c0 + i sin 2m\u03c0]1\/25<\/sup>
\n= cos \\(\\frac{2 \\mathrm{~m} \\pi}{25}\\) + i sin \\(\\frac{2 \\mathrm{~m} \\pi}{25}\\)
\n= cis \\(\\frac{2 \\mathrm{~m} \\pi}{25}\\), m = 0, 1, 2, ……………., 24<\/p>\n

\"TS<\/p>\n

\u2234 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}\\).
\n\u2234 The number of common roots = 5 (or)
\nThe GCM of 15 and 25 is<\/p>\n

\"TS<\/p>\n

Question 7.
\nFind the product of all the values of (1 + i)4\/5<\/sup>.
\nSolution:<\/p>\n

\"TS<\/p>\n

Let, 1 + i = r (cos \u03b8 + i sin \u03b8)
\nthen r cos \u03b8 = 1, r sin \u03b8 = 1
\nr = \\(\\sqrt{x^2+y^2}=\\sqrt{1^2+1^2}=\\sqrt{1+1}=\\sqrt{2}\\)
\nHence,
\n\u221a2 cos \u03b8 = i,
\ncos \u03b8 = \\(\\frac{1}{\\sqrt{2}}\\)
\n\u221a2 sin \u03b8 = 1
\nsin \u03b8 = \\(\\frac{1}{\\sqrt{2}}\\)
\n\u2234 \u03b8 lies in the Q1<\/sub>.
\n\u2234 \u03b8 = \\(\\frac{\\pi}{4}\\)
\n\u2234 1 + i = \u221a2 (cos \\(\\frac{\\pi}{4}\\) + i sin \\(\\frac{\\pi}{4}\\))
\nNow,
\n(1 + i) = [\u221a2 (cos \\(\\frac{\\pi}{4}\\) + i sin \\(\\frac{\\pi}{4}\\))]4<\/sup>
\n= 4 (cos \u03c0 + i sin \u03c0)
\n= 4[cos(2k\u03c0 + \u03c0) + i sin (2k\u03c0 + \u03c0)]
\n= 4[cos(2k + 1)\u03c0 + isin(2k + 1)\u03c0]
\n(1 + i)4\/5<\/sup> = 41\/5<\/sup> [cos(2k + 1)\u03c0 + isin(2k + 1)\u03c0]1\/5<\/sup>
\nk = 0, 1, 2, 3, 4
\n= 41\/5<\/sup> [cos (2k + 1) \\(\\frac{\\pi}{5}\\) + i sin (2k + 1) \\(\\frac{\\pi}{5}\\)]
\n= 41\/5<\/sup> cis (2k + 1) \\(\\frac{\\pi}{4}\\), k = 0, 1, 2, 3, 4<\/p>\n

\"TS<\/p>\n

= 4 cis (5\u03c0)
\n= 4 cis \u03c0
\n= 4 (cos \u03c0 + i sin \u03c0)
\n= 4 [- 1 + i . 0] = – 4.<\/p>\n

\"TS<\/p>\n

Question 8.
\nIf z2<\/sup> + z + 1 = 0, where z is a complex number, prove that
\n\\(\\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.
\nSolution:
\nGiven equation is z2<\/sup> + z + 1 = 0
\nComparing with ax2<\/sup> + bx + c we get, a = 1, b = 1, c = 1
\nThe roots of the given equation is
\nz = \\(\\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)}\\)
\n= \\(\\frac{-1 \\pm \\sqrt{1-4}}{2} \\Rightarrow \\frac{-1 \\pm \\sqrt{3} i}{2}\\) = \u03c9, \u03c92<\/sup>
\nLet, z = \u03c9
\nL.H.S:<\/p>\n

\"TS<\/p>\n

= (\u03c9 + \u03c92<\/sup>)2<\/sup> + (\u03c92<\/sup> + \u03c9) + 4 + (\u03c9 + \u03c92<\/sup>) + (\u03c92<\/sup> + \u03c9)2<\/sup> + 4
\n= (- 1)2<\/sup> + (- 1)2<\/sup> + 4 + (- 1)2<\/sup> + (- 1)2<\/sup> + 4
\n= 1 + 1 + 4 + 1 + 1 + 4
\n= 12 = R.H.S.<\/p>\n

\"TS<\/p>\n

Question 9.
\nProve the sum of 99th power of the roots of the equation x7<\/sup> – 1 = 0 is zero and hence deduce the roots of x6<\/sup> + x5<\/sup> + x4<\/sup> + x3<\/sup> + x2<\/sup> + x + 1 = o.
\nSolution:
\nGiven equation is x7<\/sup> – 1 = 0
\nx7<\/sup> = 1
\nx = (1)1\/7<\/sup>
\n= [cos 0 + i sin 0]1\/7<\/sup>
\n= [cos (2k\u03c0 + 0) + i sin(2k\u03c0 + 0)]1\/7<\/sup>
\nk = 0, 1, 2, 3, …………….., 6
\n= [cos 2k\u03c0 + i sin 2k\u03c0]1\/7<\/sup>
\n= cos \\(\\frac{2 \\mathrm{k} \\pi}{7}\\) + i sin \\(\\frac{2 \\mathrm{k} \\pi}{7}\\)
\n= cis \\(\\frac{2 \\mathrm{k} \\pi}{7}\\), k = 0, 1, 2, …………, 6
\nIf k = 0
\n\u21d2 cis 0 = x
\nIf k = 1
\n\u21d2 cis \\(\\frac{2 \\pi}{7}\\) = x
\nIf k = 2
\n\u21d2 cis \\(\\frac{4 \\pi}{7}\\) = x
\nIf k = 3
\n\u21d2 cis \\(\\frac{6 \\pi}{7}\\) = x
\nIf k = 4
\n\u21d2 cis \\(\\frac{8 \\pi}{7}\\) = x
\nIf k = 5
\n\u21d2 cis \\(\\frac{10 \\pi}{7}\\) = x
\nIf k = 6
\n\u21d2 cis \\(\\frac{12 \\pi}{7}\\) = x
\n\u2234 All the values of x7<\/sup> – 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}\\)
\nThe 99th power of the roots of the equation x7<\/sup> – 1 = 0 are<\/p>\n

\"TS<\/p>\n

= 1 + \u03c9 + \u03c92<\/sup> + \u03c93<\/sup> + \u03c94<\/sup> + \u03c95<\/sup> + \u03c96<\/sup>
\n= 0 (from theorem)
\nGiven equation is
\nx6<\/sup> + x5<\/sup> + x4<\/sup> + x3<\/sup> + x2<\/sup> + x + 1 = 0
\nMultiplying on both sides with (x – 1)
\n(x – 1) (x6<\/sup> + x5<\/sup> + x4<\/sup> + x3<\/sup> + x2<\/sup> + x + 1) = 0
\nx7<\/sup> + x6<\/sup> + x5<\/sup> + x4<\/sup> + x3<\/sup> + x2<\/sup> + x – x6<\/sup> – x5<\/sup> – x4<\/sup> – x3<\/sup> – x2<\/sup> – x – 1 = 0
\nx7<\/sup> – 1 = 0
\nx = 1 is also a root of x7<\/sup> – 1 = 0.
\n\u2234 The roots of the equation x6<\/sup> + x5<\/sup> + x4<\/sup> + x3<\/sup> + x2<\/sup> + x + 1 = 0 are x = cis \\(\\frac{2k \\pi}{7}\\), k = 1, 2, 3, ………., 6.<\/p>\n","protected":false},"excerpt":{"rendered":"

Students must practice these Maths 2A Important Questions TS Inter Second Year Maths 2A De Moivre\u2019s Theorem Important Questions Long Answer Type to help strengthen their preparations for exams. TS Inter Second Year Maths 2A De Moivre\u2019s Theorem Important Questions Long Answer Type Question 1. State and prove De Moivre\u2019s theorem for integral index. [May … Read more<\/a><\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":[],"categories":[26],"tags":[],"yoast_head":"\nTS Inter Second Year Maths 2A De Moivre\u2019s Theorem Important Questions Long Answer Type - TS Board Solutions<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/tsboardsolutions.com\/maths-2a-de-moivres-theorem-important-questions-long-answer-type\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"TS Inter Second Year Maths 2A De Moivre\u2019s Theorem Important Questions Long Answer Type - TS Board Solutions\" \/>\n<meta property=\"og:description\" content=\"Students must practice these Maths 2A Important Questions TS Inter Second Year Maths 2A De Moivre\u2019s Theorem Important Questions Long Answer Type to help strengthen their preparations for exams. 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