{"id":35171,"date":"2022-08-25T11:03:26","date_gmt":"2022-08-25T05:33:26","guid":{"rendered":"https:\/\/apboardsolutions.guru\/?p=35171"},"modified":"2022-08-25T11:03:26","modified_gmt":"2022-08-25T05:33:26","slug":"inter-2nd-year-maths-2a-de-moivres-theorem-solutions-ex-2b","status":"publish","type":"post","link":"https:\/\/apboardsolutions.guru\/inter-2nd-year-maths-2a-de-moivres-theorem-solutions-ex-2b\/","title":{"rendered":"Inter 2nd Year Maths 2A De Moivre\u2019s Theorem Solutions Ex 2(b)"},"content":{"rendered":"

Practicing the Intermediate 2nd Year Maths 2A Textbook Solutions<\/a> Inter 2nd Year Maths 2A De Moivre\u2019s Theorem Solutions Exercise 2(b) will help students to clear their doubts quickly.<\/p>\n

Intermediate 2nd Year Maths 2A De Moivre\u2019s Theorem Solutions Exercise 2(b)<\/h2>\n

I. Find all the values of the following.<\/span><\/p>\n

Question 1.
\n(i) (1 – i\u221a3)1\/3<\/sup>
\nSolution:
\n\"Inter<\/p>\n

(ii) (-i)1\/6<\/sup>
\nSolution:
\n\"Inter
\n\"Inter<\/p>\n

(iii) (1 + i)2\/3<\/sup>
\nSolution:
\n\"Inter<\/p>\n

(iv) (-16)1\/4<\/sup>
\nSolution:
\n\"Inter<\/p>\n

(v) (-32)1\/5<\/sup>
\nSolution:
\n\"Inter<\/p>\n

Question 2.
\nIf A, B, C are angles of a triangle such that x = cis A, y = cis B, z = cis C, then find the value of xyz.
\nSolution:
\n\u2234 A, B, C are angles of a triangle
\n\u21d2 A + B + C = 180\u00b0 ………..(1)
\nx = cis A, y = cis B, Z = cis C
\nxyz = cis (A + B + C)
\n= cos (A + B + C) + i sin (A + B + C)
\n= cos (180\u00b0) + i sin (180\u00b0)
\n= -1 + i(0)
\n= -1
\n\u2234 xyz = -1<\/p>\n

\"Inter<\/p>\n

Question 3.
\n(i) If x = cis \u03b8, then find the value of \\(\\left[x^{6}+\\frac{1}{x^{6}}\\right]\\)
\nSolution:
\n\u2235 x = cos \u03b8 + i sin \u03b8
\n\u21d2 x6<\/sup> = (cos \u03b8 + i sin \u03b8)6<\/sup> = cos 6\u03b8 + i sin 6\u03b8
\n\u21d2 \\(\\frac{1}{x^{6}}\\) = cos 6\u03b8 – i sin 6\u03b8
\n\u2234 \\(x^{6}+\\frac{1}{x^{6}}\\) = 2 cos 6\u03b8<\/p>\n

(ii) Find the cube roots of 8.
\nSolution:
\nLet x3<\/sup> = 8
\n\u21d2 x = 81\/3<\/sup>
\n\u21d2 x = (23<\/sup>)1\/3<\/sup> (1)1\/3<\/sup>
\n\u21d2 x = 2 (1)1\/3<\/sup>
\nSince cube roots of unity are 1, \u03c9, \u03c92<\/sup>
\n\u2234 The cube roots or 8 are 2, 2\u03c9, 2\u03c92<\/sup><\/p>\n

Question 4.
\nIf 1, \u03c9, \u03c92<\/sup> are the cube roots of unity, then prove that
\n(i) \\(\\frac{1}{2+\\omega}-\\frac{1}{1+2 \\omega}=\\frac{1}{1+\\omega}\\)
\nSolution:
\n\u03c9 is a cube root of unity.
\n1 + \u03c9 + \u03c92<\/sup> = 0 and \u03c93<\/sup> = 1
\n\"Inter<\/p>\n

(ii) (2 – \u03c9) (2 – \u03c92<\/sup>) (2 – \u03c910<\/sup>) (2 – \u03c911<\/sup>) = 49
\nSolution:
\n\u2235 1, \u03c9, \u03c92<\/sup> are the cube roots of unity.
\n\u03c93<\/sup> = 1 and 1 + \u03c9 + \u03c92<\/sup> = 0
\n2 – \u03c910<\/sup> = 2 – \u03c99<\/sup> . \u03c9
\n= 2 – (\u03c93<\/sup>)3<\/sup> \u03c9
\n= 2 – (1)3<\/sup> \u03c9
\n= 2 – \u03c9
\nand 2 – \u03c911<\/sup> = 2 – (\u03c93<\/sup>)3<\/sup> . \u03c92<\/sup>
\n= 2 – (1)3<\/sup> \u03c92<\/sup>
\n= 2 – \u03c92<\/sup>
\n(2 – \u03c9) (2 – \u03c92<\/sup>) = 4 – 2\u03c9 – 2\u03c92<\/sup> + \u03c93<\/sup>
\n= 4 – 2(\u03c9 + \u03c92<\/sup>) + 1
\n= 4 – 2(-1) + 1
\n= 4 + 2 + 1
\n= 7
\n\u2234 (2 – \u03c9) (2 – \u03c92<\/sup>) (2 – \u03c910<\/sup>) (2 – \u03c911<\/sup>) = (2 – \u03c9) (2 – \u03c92<\/sup>) (2 – \u03c9) (2 – \u03c92<\/sup>)
\n= ((2 – \u03c9) (2 – \u03c92<\/sup>))2<\/sup>
\n= 72<\/sup>
\n= 49<\/p>\n

(iii) (x + y + z) (x + y\u03c9 + z\u03c92<\/sup>) (x + y\u03c92<\/sup> + z\u03c9) = x3<\/sup> + y3<\/sup> + z3<\/sup> – 3xyz
\nSolution:
\n\u2235 1, \u03c9, \u03c92<\/sup> are the cube roots of unity.
\n\u21d2 1 + \u03c9 + \u03c92<\/sup> = 0 and \u03c93<\/sup> = 1
\nNow consider,
\n(x + y\u03c9 + z\u03c92<\/sup>) (x + y\u03c92<\/sup> + z\u03c9)
\n= x2<\/sup> + xy\u03c92<\/sup> + zx\u03c9 + xy\u03c9 + y2<\/sup>\u03c93<\/sup> + yz\u03c92<\/sup> + zx\u03c92<\/sup> + yz\u03c94<\/sup> + z2<\/sup>\u03c93<\/sup>
\n= x2<\/sup> + y2<\/sup> (1) + z2<\/sup> (1) + xy (\u03c9 + \u03c92<\/sup>) + yz (\u03c94<\/sup> + \u03c92<\/sup>) + zx (\u03c9 + \u03c92<\/sup>)
\n= x2<\/sup> + y2<\/sup> + z2<\/sup> + xy (-1) + yz (\u03c9 + \u03c92<\/sup>) + zx (-1)
\n= x2<\/sup> + y2<\/sup> + z2<\/sup> – xy – yz – zx ……….(1)
\nL.H.S = (x + y + z) (x + y\u03c9 + z\u03c92<\/sup>) (x + y\u03c92<\/sup> + z\u03c9)
\n= (x + y + z) (x2<\/sup> + y2<\/sup> + z2<\/sup> – xy – yz – zx) [by (1)]
\n= x3<\/sup> + y3<\/sup> + z3<\/sup> – 3xyz
\n= R.H.S<\/p>\n

\"Inter<\/p>\n

Question 5.
\nProve that -\u03c9, and -\u03c92<\/sup> are the roots of z2<\/sup> – z + 1 =0, where \u03c9 and \u03c92<\/sup> are the complex cube roots of unity.
\nSolution:
\nSince \u03c9 and \u03c92<\/sup> are the complex cube roots of unity
\n\u2234 1 + \u03c9 + \u03c92<\/sup> = 0 and \u03c92<\/sup> = 1
\nz2<\/sup> – z + 1 = (-\u03c9)2<\/sup> – (-\u03c9) + 1
\n= \u03c92<\/sup> + \u03c9 + 1
\n= 0
\n\u2234 -\u03c9 is a root of the equation z2<\/sup> – z + 1 = 0
\nz2<\/sup> – z + 1 = (-\u03c92<\/sup>)2<\/sup> – (-\u03c92<\/sup>) + 1
\n= \u03c94<\/sup> + \u03c92<\/sup> + 1
\n= \u03c93<\/sup> . \u03c9 + \u03c92<\/sup> + 1
\n= \u03c9 + \u03c92<\/sup>+ 1
\n= 0
\n\u2234 -\u03c92<\/sup> is a root of the equation z2<\/sup> – z + 1 = 0<\/p>\n

Question 6.
\nIf 1, \u03c9, \u03c92<\/sup> are the cube roots of unity, then find the values of the following.
\n(i) (a + b)3<\/sup> + (a\u03c9 + b\u03c92<\/sup>)3<\/sup> + (a\u03c92<\/sup> + b\u03c9)3<\/sup>
\nSolution:
\nSince 1, \u03c9, \u03c92<\/sup> are the cube roots of unity
\n\u2234 1 + \u03c9 + \u03c92<\/sup> = 0 and \u03c93<\/sup> = 1
\nNow (a + b)3<\/sup> = a3<\/sup> + 3a2<\/sup>b + 3ab2<\/sup> + b3<\/sup> ……..(1)
\n(a\u03c9 + b\u03c92<\/sup>)3<\/sup> = [\u03c9(a + b\u03c9)]3<\/sup>
\n= \u03c93<\/sup> (a + b\u03c9)3<\/sup>
\n= (1) (a + b\u03c9)3<\/sup>
\n= a3<\/sup> + 3a2<\/sup>b\u03c9 + 3ab2<\/sup>\u03c92<\/sup> + b3<\/sup>\u03c93<\/sup>
\n= a3<\/sup> + 3a2<\/sup>b\u03c9 + 3ab2<\/sup>\u03c92<\/sup> + b3<\/sup> ……….(2)
\n\u2235 \u03c93<\/sup> = 1
\nand (a\u03c92<\/sup> + b\u03c9)3<\/sup> = [\u03c9(a\u03c9 + b)]3<\/sup>
\n= \u03c93<\/sup> (a\u03c9 + b)3<\/sup>
\n= (1) (a\u03c9 + b)3<\/sup>
\n= a3<\/sup>\u03c93<\/sup> + 3a2<\/sup>b\u03c92<\/sup> + 3ab2<\/sup>\u03c9 + b3<\/sup>
\n= a3<\/sup>(1) + 3a2<\/sup>b\u03c92<\/sup> + 3ab2<\/sup>\u03c9 + b3<\/sup>
\n\u2234 (a\u03c92<\/sup> + b\u03c9)3<\/sup> = a3<\/sup> + 3a2<\/sup>b\u03c92<\/sup> + 3ab2<\/sup>\u03c9 + b3<\/sup> ……….(3)
\nAdding (1), (2) and (3)
\n(a + b)3<\/sup> + (a\u03c9 + b\u03c92<\/sup>)3<\/sup> + (a\u03c92<\/sup> + b\u03c9)3<\/sup> = 3a3<\/sup> + 3a2<\/sup>b (1 + \u03c9 + \u03c92<\/sup>) + 3ab2<\/sup> (1 + \u03c9 + \u03c92<\/sup>) + 3b3<\/sup>
\n= 3(a3<\/sup> + b3<\/sup>) + 3a2<\/sup>b (0) + 3ab2<\/sup> (0)
\n= 3(a3<\/sup> + b3<\/sup>)
\n\u2234 (a + b)3<\/sup> + (a\u03c9 + b\u03c92<\/sup>)3<\/sup> + (a\u03c92<\/sup> + b\u03c9)3<\/sup> = 3 (a3<\/sup> + b3<\/sup>)<\/p>\n

(ii) (a + 2b)2<\/sup> + (a\u03c92<\/sup> + 2b\u03c9)2<\/sup> + (a\u03c9 + 2b\u03c92<\/sup>)2<\/sup>
\nSolution:
\n(a + 2b)2<\/sup> = a2<\/sup> + 4ab + 4b2<\/sup> ……….(1)
\n(a\u03c92<\/sup> + 2b\u03c9)2<\/sup> = a2<\/sup>\u03c94<\/sup> + 4ab\u03c93<\/sup> + 4b2<\/sup>\u03c92<\/sup>
\n= a2<\/sup>\u03c93<\/sup>\u03c9 + 4ab (1) + 4b2<\/sup>\u03c92<\/sup>
\n= a2<\/sup>\u03c9 + 4ab + 4b2<\/sup>\u03c92<\/sup> ………..(2)
\nand (a\u03c9 + 2b\u03c92<\/sup>)2<\/sup> = a2<\/sup>\u03c92<\/sup> + 4ab\u03c93<\/sup> + 4b2<\/sup>\u03c94<\/sup>
\n= a2<\/sup>\u03c92<\/sup> + 4ab (1) + 4b2<\/sup>\u03c93<\/sup>\u03c9
\n= a2<\/sup>\u03c92<\/sup> + 4ab + 4b2<\/sup> (1) \u03c9
\n\u2234 (a\u03c9 + 2b\u03c92<\/sup>)2<\/sup> = a2<\/sup>\u03c92<\/sup> + 4ab + 4b2<\/sup>\u03c9 ……….(3)
\nBy Adding (1), (2) and (3)
\n(a + 2b)2<\/sup> + (a\u03c92<\/sup> + 2b\u03c9)2<\/sup> + (a\u03c9 + 2b\u03c92<\/sup>)2<\/sup>
\n= a2<\/sup> (1 + \u03c9 + \u03c92<\/sup>) + 12ab + 4b2<\/sup> (1 + \u03c9 + \u03c92<\/sup>)
\n= a2<\/sup> (0) + 12ab + 4b2<\/sup> (0)
\n= 12ab
\n\u2234 (a + 2b)2<\/sup> + (a\u03c92<\/sup> + 2b\u03c9)2<\/sup> + (a\u03c9 + 2b\u03c92<\/sup>)2<\/sup> = 12ab<\/p>\n

\"Inter<\/p>\n

(iii) (1 – \u03c9 + \u03c92<\/sup>)3<\/sup>
\nSolution:
\n(1 – \u03c9 + \u03c92<\/sup>)3<\/sup> = (-\u03c9 – \u03c9)3<\/sup>
\n= (-2\u03c9)3<\/sup>
\n= -8\u03c93<\/sup>
\n= -8(1)
\n= -8 (\u2235 1 + \u03c9 + \u03c92<\/sup> = 0)<\/p>\n

(iv) (1 – \u03c9) (1 – \u03c92<\/sup>) (1 – \u03c94<\/sup>) (1 – \u03c98<\/sup>)
\nSolution:
\n1 – \u03c94<\/sup> = 1 – (\u03c93<\/sup>) \u03c9
\n= 1 – (1) \u03c9
\n= 1 – \u03c9
\n1 – \u03c98<\/sup> = 1 – (\u03c93<\/sup>)2<\/sup> \u03c92<\/sup>
\n= 1 – (1) \u03c92<\/sup>
\n= 1 – \u03c92<\/sup>
\n\u2234 (1 – \u03c9) (1 – \u03c92<\/sup>) (1 – \u03c94<\/sup>)(1 – \u03c98<\/sup>) = (1 – \u03c9) (1 – \u03c92<\/sup>) (1 – \u03c9) (1 – \u03c92<\/sup>)
\n= [(1 – \u03c9) (1 – \u03c92<\/sup>)]2<\/sup>
\n= (1 – \u03c9 – \u03c92<\/sup> + \u03c93<\/sup>)2<\/sup>
\n= [1 – (\u03c9 + \u03c92<\/sup>) + 1] (\u2235 1 + \u03c9 + \u03c92<\/sup> = 0)
\n= [1 – (-1) + 1]2<\/sup>
\n= (3)2<\/sup>
\n= 9
\n\u2234 (1 – \u03c9) (1 – \u03c92<\/sup>) (1 – \u03c94<\/sup>) (1 – \u03c98<\/sup>) = 9<\/p>\n

(v) \\(\\left[\\frac{a+b \\omega+c \\omega^{2}}{c+a \\omega+b \\omega^{2}}\\right]+\\left[\\frac{a+b \\omega+c \\omega^{2}}{b+c \\omega+a \\omega^{2}}\\right]\\)
\nSolution:
\n\u2234 1, \u03c9, \u03c92<\/sup> are the cube roots of unity
\n\u21d2 \u03c93<\/sup> = 1 and 1 + \u03c9 + \u03c92<\/sup> = 0 ………..(1)
\n\"Inter<\/p>\n

(vi) (i + \u03c9)3<\/sup> + (1 + \u03c92<\/sup>)3<\/sup>
\nSolution:
\n(i + \u03c9)3<\/sup> + (1 + \u03c92<\/sup>)3<\/sup> = (-\u03c92<\/sup>)3<\/sup> +(-\u03c9)3<\/sup>
\n= -\u03c96<\/sup> – \u03c93<\/sup>
\n= -1 – 1
\n= -2
\n\u2234 (1 + \u03c9)3<\/sup> + (1 + \u03c92<\/sup>)3<\/sup> = -2<\/p>\n

(vii) (1 – \u03c9 + \u03c92<\/sup>)5<\/sup> + (1 + \u03c9 – \u03c92<\/sup>)5<\/sup>
\nSolution:
\n(1 – \u03c9 + \u03c92<\/sup>)5<\/sup> + (1 + \u03c9 – \u03c92<\/sup>)5 = (-\u03c9 – \u03c9)5<\/sup> + (-\u03c92<\/sup> – \u03c92<\/sup>)5<\/sup>
\n= (-2\u03c9)5<\/sup> + (-2\u03c92<\/sup>)5<\/sup>
\n= -32\u03c95<\/sup> – 32\u03c910<\/sup>
\n= -32(\u03c95<\/sup> + \u03c910<\/sup>)
\n= -32(\u03c92<\/sup> + \u03c9)
\n= -32(-1)
\n= 32
\n\u2234 (1 – \u03c9 + \u03c92<\/sup>)5<\/sup> + (1 + \u03c9 – \u03c92<\/sup>)5<\/sup> = 32<\/p>\n

II.<\/span><\/p>\n

Question 1.
\nSolve the following equations.
\n(i) x4<\/sup> – 1 = 0
\nSolution:
\nx4<\/sup> – 1 = 0
\n\u21d2 x4<\/sup> = 1
\n\u21d2 x4<\/sup> = cos 0\u00b0 + i sin 0\u00b0
\n\u21d2 x4<\/sup> = cos 2k\u03c0 + i sin 2k\u03c0
\n\u21d2 x = (cos 2k\u03c0 + i sin 2k\u03c0)1\/4<\/sup>
\n= cos \\(\\frac{k \\pi}{2}\\), k = 0, 1, 2, 3
\ni.e., cos 0\u00b0 + i sin 0\u00b0, cos \\(\\frac{\\pi}{2}\\) + i sin \\(\\frac{\\pi}{2}\\), cos \u03c0 + i sin \u03c0, cos \\(\\frac{3 \\pi}{2}\\) + i sin \\(\\frac{3 \\pi}{2}\\),
\ni.e., 1, i, -1, -i = \u00b11, \u00b1i<\/p>\n

(ii) x5<\/sup> + 1 = 0
\nSolution:
\nx5<\/sup> + 1 = 0
\n\u21d2 x5<\/sup> = -1
\n\u21d2 x5<\/sup> = cos \u03c0 + i sin \u03c0
\n\u21d2 x5<\/sup> = cos(2k + 1) \u03c0 + i sin(2k + 1) \u03c0, k \u2208 z
\n\u21d2 x = [cos(2k + 1) \u03c0 + i sin(2k + 1) \u03c0]1\/5<\/sup>
\n\u21d2 x = cis \\(\\frac{(2 k+1) \\pi}{5}\\), k = 0, 1, 2, 3, 4<\/p>\n

(iii) x9<\/sup> – x5<\/sup> + x4<\/sup> – 1 = 0
\nSolution:
\nx9<\/sup> – x5<\/sup> + x4<\/sup> – 1 = 0
\n\u21d2 x5<\/sup> (x4<\/sup> – 1) + 1 (x4<\/sup> – 1) = 0
\n\u21d2 (x4<\/sup> – 1) (x5<\/sup> + 1) = 0
\n\u21d2 x4<\/sup> – 1 = 0
\nSolving the roots are \u00b11, \u00b1i
\n(see the above problem)
\nx5<\/sup> + 1 = 0
\nSolving the roots are cis \\(\\frac{(2 k+1) \\pi}{5}\\)
\nk = 0, 1, 2, 3, 4 (see the above problem)
\n\u2234 The roots of the given equation are \u00b11, \u00b1i, cis (2k + 1) \\(\\frac{\\pi}{5}\\), k = 0, 1, 2, 3, 4
\ni.e., \u00b11, \u00b1i, cis(\\(\\pm \\frac{\\pi}{5}\\)), cis(\\(\\pm \\frac{3 \\pi}{5}\\))<\/p>\n

(iv) x4<\/sup> + 1 = 0
\nSolution:
\nx4<\/sup> + 1 = 0
\n\u21d2 x4<\/sup> = -1
\n\u21d2 x4<\/sup> = cos \u03c0 + i sin \u03c0
\n\u2234 x4<\/sup> = cos(2k\u03c0 + \u03c0) + i sin(2k\u03c0 + \u03c0),
\n\u2234 x = [cis(2k + 1)\u03c0]1\/4<\/sup>
\n= cis(2k + 1) \\(\\frac{\\pi}{4}\\), where k = 0, 1, 2, 3
\n\u2234 x = \\({cis} \\frac{\\pi}{4}, {cis}\\left(\\frac{3 \\pi}{4}\\right), {cis}\\left(\\frac{5 \\pi}{4}\\right)\\) and \\({cis}\\left(\\frac{7 \\pi}{4}\\right)\\)
\nThese four values of x are the solutions to the given equation.<\/p>\n

\"Inter<\/p>\n

Question 2.
\nFind the common roots of x12<\/sup> – 1 = 0 and x4<\/sup> + x2<\/sup> + 1 = 0
\nSolution:
\nConsider x12<\/sup> – 1 = 0
\n\u21d2 x12<\/sup> = 1
\n\u21d2 x12<\/sup> = (cos 0 + i sin 0)
\n\u21d2 x12<\/sup> = (cos 2k\u03c0 + i sin 2k\u03c0), k is a positive integer
\n\u21d2 x = (cos 2k\u03c0 + i sin 2k\u03c0)1\/2<\/sup>
\n\"Inter
\n\"Inter<\/p>\n

Question 3.
\nFind the number of 15th roots of unity, which are also the 25th roots of unity.
\nSolution:
\nThe number of common roots = H.C.F of {15, 25} = 5<\/p>\n

Question 4.
\nIf the cube roots of unity are 1, \u03c9, \u03c92, then find the roots of the equation (x – 1)3<\/sup> + 8 = 0.
\nSolution:
\nGiven (x – 1)3<\/sup> + 8 = 0
\n\u21d2 (x – 1)3<\/sup> = -8
\n\u21d2 (x – 1)3<\/sup> = (-2)3<\/sup> (1)3<\/sup>
\n\u21d2 (x – 1) = (-2) (1)1\/3<\/sup>
\n\u21d2 x – 1 = -2, -2\u03c9, -2\u03c92<\/sup>
\n\u21d2 x = 1 – 2, 1 – 2\u03c9, 1 – 2\u03c92<\/sup>
\n\u21d2 x = -1, 1 – 2\u03c9, 1 – 2\u03c92<\/sup><\/p>\n

\"Inter<\/p>\n

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

Question 6.
\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)\\) = 12
\nSolution:
\nGiven z2<\/sup> + z + 1 = 0
\n\u21d2 z = \\(\\frac{-1 \\pm \\sqrt{1-4.1 .1}}{2}\\)
\n= \\(\\frac{-1 \\pm i \\sqrt{3}}{2}\\)
\n= \\(\\frac{-1+i \\sqrt{3}}{2}, \\frac{-1-i \\sqrt{3}}{2}\\)
\n= \u03c9, \u03c92<\/sup>
\n\u2234 1 + \u03c9 + \u03c92<\/sup> = 0 and \u03c93<\/sup> = 1
\nIf z = \u03c9 then
\n\"Inter
\n= (\u03c9 + \u03c92<\/sup>)2<\/sup> + (\u03c92<\/sup> + \u03c9)2<\/sup> + (1 + 1)2<\/sup> + (\u03c9 + \u03c92<\/sup>)2<\/sup> + (\u03c92<\/sup> + \u03c9)2<\/sup> + (1 + 1 )2<\/sup>
\n= (-1)2<\/sup> + (-1)2<\/sup> + 4 + (-1)2<\/sup> + (-1)2<\/sup> + 4
\n= 1 + 1 + 4 + 1 + 1 + 4
\n= 12
\nSimilarly If z = \u03c92<\/sup> then
\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)\\) = 12<\/p>\n

III.<\/span><\/p>\n

Question 1.
\nIf 1, \u03b1, \u03b12<\/sup>, \u03b13<\/sup>, ……., \u03b1n-1<\/sup> be the nth roots of unity, then prove that 1p<\/sup> + \u03b1p<\/sup> + (\u03b12<\/sup>)p<\/sup> + (\u03b13<\/sup>)p<\/sup> + …… + (\u03b1n-p<\/sup>)2p<\/sup> = 0
\n= 0; if p \u2260 kn
\n= n; if p \u2260 kn, where p, k \u2208 N
\nSolution:
\nnth roots of unity are 1, \u03b1, \u03b12<\/sup>, ………., \u03b1n-1<\/sup>
\n\"Inter
\n\"Inter
\n\u2234 Each term of the series in (1) is 1
\nHence the sum of the series 1 + \u03b1p + (\u03b12<\/sup>)p + (\u03b13<\/sup>)p + …….. + (\u03b1n-1<\/sup>)p<\/sup>
\n= 1 + 1 + 1 + ……… + 1 (n times)
\n= n(1)
\n= n<\/p>\n

\"Inter<\/p>\n

Question 2.
\nProve that the sum of 99th powers 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 = 0.
\nSolution:
\n\"Inter
\n\"Inter<\/p>\n

Question 3.
\nIf n is a positive integer, show that \\((p+i q)^{1 \/ n}+(p-i q)^{1 \/ n}=2\\left(p^{2}+q^{2}\\right)^{1 \/ 2 n}\\) . \\(\\cos \\left(\\frac{1}{n}, \\tan \\frac{q}{p}\\right)\\)
\nSolution:
\n\"Inter
\n\"Inter<\/p>\n

Question 4.
\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)^{8 \/ 3}\\) is -1.
\nSolution:
\n\"Inter
\n\"Inter<\/p>\n

\"Inter<\/p>\n

Question 5.
\nSolve (x – 1)n<\/sup> = xn<\/sup>, n is a positive integer.
\nSolution:
\nSince x = 0 is not a solution of the given equation, it is equivalent to the equation \\(\\left(\\frac{x-1}{x}\\right)^{n}=1\\)
\nClearly \\(\\left(\\frac{x-1}{x}\\right)^{n}=1\\)
\n\u21d2 \\(\\frac{x-1}{x}\\) is an nth root of unity other than one.
\nSuppose that \u03c9 is an nth root of unity and \u03c9 \u2260 1.
\nThen, \\(\\frac{x-1}{x}\\) = \u03c9
\n\u21d2 x – 1 = x\u03c9
\n\u21d2 (1 – \u03c9) x = 1
\n\u21d2 x = \\(\\frac{1}{1-\\omega}\\), (\u2235 \u03c9 \u2260 1) ……….(1)
\n\"Inter
\n\"Inter<\/p>\n","protected":false},"excerpt":{"rendered":"

Practicing the Intermediate 2nd Year Maths 2A Textbook Solutions Inter 2nd Year Maths 2A De Moivre\u2019s Theorem Solutions Exercise 2(b) will help students to clear their doubts quickly. Intermediate 2nd Year Maths 2A De Moivre\u2019s Theorem Solutions Exercise 2(b) I. Find all the values of the following. Question 1. (i) (1 – i\u221a3)1\/3 Solution: (ii) …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[14],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/35171"}],"collection":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/comments?post=35171"}],"version-history":[{"count":1,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/35171\/revisions"}],"predecessor-version":[{"id":35172,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/35171\/revisions\/35172"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/media?parent=35171"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/categories?post=35171"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/tags?post=35171"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}