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>\nTS 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>\nCase – 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>\nCase – 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<\/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>\nQuestion 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<\/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>\nSimilarly,
\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<\/p>\n
<\/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<\/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<\/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<\/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<\/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<\/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<\/p>\n
<\/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
<\/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
<\/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>\nx4<\/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>\nx7<\/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<\/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>\nQuestion 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<\/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<\/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
<\/p>\n
<\/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<\/p>\n
<\/p>\n
<\/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<\/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>\nQuestion 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> – a