{"id":4156,"date":"2023-12-20T10:47:12","date_gmt":"2023-12-20T05:17:12","guid":{"rendered":"https:\/\/wordpress-505192-1602719.cloudwaysapps.com\/?p=4156"},"modified":"2023-12-20T17:24:23","modified_gmt":"2023-12-20T11:54:23","slug":"samacheer-kalvi-12th-maths-solutions-chapter-2-ex-2-8","status":"publish","type":"post","link":"https:\/\/samacheerguru.com\/samacheer-kalvi-12th-maths-solutions-chapter-2-ex-2-8\/","title":{"rendered":"Samacheer Kalvi 12th Maths Solutions Chapter 2 Complex Numbers Ex 2.8"},"content":{"rendered":"

You can Download Samacheer Kalvi 12th Maths Book Solutions<\/a> Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.<\/p>\n

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 2 Complex Numbers Ex 2.8<\/h2>\n

Question 1.
\nIf to \u03c9 \u2260 1 is a cube root of unity, then show that \\(\\frac{a+b \\omega+c \\omega^{2}}{b+c \\omega+a \\omega^{2}}+\\frac{a+b \\omega+c \\omega^{2}}{c+a \\omega+b \\omega^{2}}=1\\)
\nSolution:
\nSince \u03c9 is a cube root of unity, we have \u03c93<\/span><\/sup><\/span> = 1 and 1 + \u03c9 + \u03c92<\/span><\/sup><\/span> = 0
\n\"Samacheer<\/p>\n

\"Samacheer<\/p>\n

Question 2.
\nShow that \\(\\left(\\frac{\\sqrt{3}}{2}+\\frac{i}{2}\\right)^{5}+\\left(\\frac{\\sqrt{3}}{2}-\\frac{i}{2}\\right)^{5}=-\\sqrt{3}\\)
\nSolution:
\n\"Samacheer
\n\"Samacheer<\/p>\n

Question 3.
\nFind the value of \\(\\left(\\frac{1+\\sin \\frac{\\pi}{10}+i \\cos \\frac{\\pi}{10}}{1+\\sin \\frac{\\pi}{10}-i \\cos \\frac{\\pi}{10}}\\right)^{10}\\)
\nSolution:
\n\"Samacheer<\/p>\n

Question 4.
\nIf 2 cos \u03b1 = x + \\(\\frac{1}{x}\\) and 2 cos \u03b2 = y + \\(\\frac{1}{y}\\), show that
\n\"Samacheer
\nSolution:
\n(i) 2 cos \u03b1 = x + \\(\\frac{1}{x}\\)
\n\u21d2 2 cos \u03b1 = \\(\\frac{x^{2}+1}{x}\\)
\n\u21d2 2x cos \u03b1 = x2<\/span><\/sup><\/span> + 1
\n\u21d2 x2<\/span><\/sup><\/span> – 2x cos \u03b1 + 1 = 0
\n\"Samacheer
\n\"Samacheer
\n\"Samacheer<\/p>\n

\"Samacheer<\/p>\n

Question 5.
\nSolve the equation z3<\/span><\/sup><\/span> + 27 = 0
\nSolution:
\nz3<\/span><\/sup><\/span> + 27 = 0
\n\u21d2 z3<\/span><\/sup><\/span> = -27
\n\u21d2 z3<\/span><\/sup><\/span> = 33<\/span><\/sup><\/span>(-1)
\n\u21d2 z = 3(-1)1\/3<\/span><\/sup><\/span>
\n\"Samacheer<\/p>\n

Question 6.
\nIf \u03c9 \u2260 1 is a cube root of unity, show that the roots of the equation (z – 1)3<\/span><\/sup><\/span> + 8 = 0 are -1, 1 – 2\u03c9, 1 – 2\u03c92<\/span><\/sup><\/span>
\nSolution:
\n(z – 1)3<\/span><\/sup><\/span> + 8 = 0
\n\u21d2 (z – 1 )3<\/span><\/sup><\/span> = -8
\n\"Samacheer
\nThe roots are -1, 1 – 2\u03c9, 1 – 2\u03c92<\/span><\/sup><\/span><\/p>\n

Question 7.
\nFind the value of \\(\\sum_{k=1}^{8}\\left(\\cos \\frac{2 k \\pi}{9}+i \\sin \\frac{2 k \\pi}{9}\\right)\\)
\nSolution:
\n\\(\\sum_{k=1}^{8}\\left(\\cos \\frac{2 k \\pi}{9}+i \\sin \\frac{2 k \\pi}{9}\\right)\\)
\nWe know that 9th roots of unit are 1, \u03c9, \u03c92<\/span><\/sup><\/span>, ……., \u03c98<\/span><\/sup><\/span>
\nSum of the roots:
\n1 + \u03c9 + \u03c92<\/span><\/sup><\/span> + …. + \u03c98<\/span><\/sup><\/span> = 0 \u21d2 \u03c9 + \u03c92<\/span><\/sup><\/span> + \u03c93<\/span><\/sup><\/span> + …… + \u03c98<\/span><\/sup><\/span> = -1
\n\"Samacheer
\nThe sum of all the terms \\(\\sum_{k=1}^{8}\\left(\\cos \\frac{2 k \\pi}{9}+i \\sin \\frac{2 k \\pi}{9}\\right)\\) = -1<\/p>\n

Question 8.
\nIf \u03c9 \u2260 1 is a cube root of unity, show that
\n(i) (1 – \u03c9 + \u03c92<\/span><\/sup><\/span>)6<\/span><\/sup><\/span> + (1 + \u03c9 – \u03c92<\/span><\/sup><\/span>)6<\/span><\/sup><\/span> = 128
\n(ii) (1 + \u03c9)(1 + \u03c92<\/span><\/sup><\/span>)(1 + \u03c94<\/span><\/sup><\/span>)(1 + \u03c98<\/span><\/sup><\/span>)……(1 + \u03c92<\/span><\/sup><\/span>n<\/span><\/sup><\/span>) = 1
\nSolution:
\n(i) \u03c9 is a cube root of unity \u03c93<\/span><\/sup><\/span> = 1; 1 + \u03c9 + \u03c92<\/span><\/sup><\/span> = 0
\n(1 – \u03c9 + \u03c92<\/span><\/sup><\/span>)6<\/span><\/sup><\/span> + (1 + \u03c9 – \u03c92<\/span><\/sup><\/span>)6<\/span><\/sup><\/span>
\n= (-\u03c9 – \u03c9)6<\/span><\/sup><\/span> + (-\u03c92<\/span><\/sup><\/span> – \u03c92<\/span><\/sup><\/span>)6<\/span><\/sup><\/span>
\n= (-2\u03c9)6<\/span><\/sup><\/span> + (-2\u03c92<\/span><\/sup><\/span>)6<\/span><\/sup><\/span>
\n= (-2)6<\/span><\/sup><\/span>(\u03c96<\/span><\/sup><\/span> + \u03c912<\/span><\/sup><\/span>)
\n= (64)(1 + 1)
\n= 128
\n(ii) (1 + \u03c9)(1 + \u03c92<\/span><\/sup><\/span>)(1 + \u03c94<\/span><\/sup><\/span>)(1 + \u03c98<\/span><\/sup><\/span>) …… (1 + \u03c92n<\/span><\/sup><\/span>)
\n= (1 + \u03c9)(1 + \u03c92<\/span><\/sup><\/span>)(1 + \u03c94<\/span><\/sup><\/span>)(1 + \u03c98<\/span><\/sup><\/span>) ……. 2n factors
\n= (-\u03c92<\/span><\/sup><\/span>)(-\u03c9)(-\u03c92<\/span><\/sup><\/span>)(-\u03c9) …… 2n factors
\n= \u03c93<\/span><\/sup><\/span>. \u03c93<\/span><\/sup><\/span>
\n= 1<\/p>\n

Question 9.
\nIf z = 2 – 2i, find the rotation of z by \u03b8 radians in the counter clockwise direction about the origin when
\n(i) \u03b8 = \\(\\frac{\\pi}{3}\\)
\n(ii) \u03b8 = \\(\\frac{2 \\pi}{3}\\)
\n(iii) \u03b8 = \\(\\frac{3 \\pi}{2}\\)
\nSolution:
\n(i) z = 2 – 2i = 2 (1 – i) = r(cos \u03b8 + i sin \u03b8)
\n\\(r=\\sqrt{x^{2}+y^{2}}=2 \\sqrt{1+1}=2 \\sqrt{2}\\)
\n\\(\\alpha=\\tan ^{-1}=\\left|\\frac{y}{x}\\right|=\\tan ^{-1}|1|=\\frac{\\pi}{4}\\)
\n(1 – i) lies in IV quadrant
\n\u03b8 = -\u03b1 = \\(-\\frac{\\pi}{4}\\)
\n\\(\\Rightarrow z=2 \\sqrt{2}\\left[\\cos \\left(\\frac{-\\pi}{4}\\right)+i \\sin \\left(\\frac{-\\pi}{4}\\right)\\right]\\)
\nz is rotated by \u03b8 = \\(\\frac{\\pi}{3}\\) in the counter clock wise direction.
\n\"Samacheer
\n(ii) z is rotated by \u03b8 = \\(\\frac{2 \\pi}{3}\\) in the counter clockwise direction.
\n\"Samacheer
\n(iii) z is rotated by \u03b8 = \\(\\frac{3 \\pi}{2}\\) in the counter clockwise direction.
\n\"Samacheer<\/p>\n

\"Samacheer<\/p>\n

Question 10.
\nProve that the values of \\(\\sqrt[4]{-1} \\text { are } \\pm \\frac{1}{\\sqrt{2}}(1 \\pm i)\\)
\nSolution:
\n\"Samacheer
\nThe roots are \\(\\pm \\frac{1}{\\sqrt{2}}(1 \\pm i)\\)<\/p>\n

Samacheer Kalvi 12th Maths Solutions Chapter 2 Complex Numbers Ex 2.8 Additional Problems<\/h3>\n

Question 1.
\n\"Samacheer
\nSolution:
\n\"Samacheer<\/p>\n

Question 2.
\n\"Samacheer
\nSolution:
\n\"Samacheer
\n\"Samacheer<\/p>\n

Question 3.
\nProve that: (1 + i)4n<\/sup> and (1 + i)4n + 2<\/sup> are real and purely imaginary respectively.
\nSolution:
\nLet z = 1 + i
\n\"Samacheer
\n= 2i (-1)n<\/sup> which is purely imaginary.<\/p>\n

\"Samacheer<\/p>\n

Question 4.
\n\"Samacheer
\nSolution:
\n\"Samacheer<\/p>\n

Question 5.
\nIf a = cos 2\u03b1 + i sin 2\u03b1, b = cos 2\u03b2 + i sin 2\u03b2 and c = cos 2\u03b3 + i sin 2\u03b3, prove that
\n\"Samacheer
\nSolution:
\n\"Samacheer<\/p>\n

Question 6.
\nSolve: x4<\/sup> + 4 = 0
\nSolution:
\n\"Samacheer<\/p>\n

Question 7.
\nIf x = a + b, y = a\u03c9 + b\u03c92<\/sup>, z = a\u03c92<\/sup> + b\u03c9, show that
\n(i) xyz = a3<\/sup> + b3<\/sup>
\n(ii) x3<\/sup> + y3<\/sup> + z3<\/sup> = 3(a3<\/sup> + b3<\/sup>)
\nSolution:
\n(i) x = a + A; y = a\u03c9 + b\u03c92<\/sup>, z = a\u03c92<\/sup> + b\u03c9
\nNow xyz = (a + b) (a\u03c9 + b\u03c92<\/sup>) (ab\u03c92<\/sup> + b\u03c9) = (a+A) [a\u03c93<\/sup> + ab\u03c92<\/sup> + ab\u03c9 + b2<\/sup>\u03c93<\/sup>]
\n= (a + b) (a2<\/sup> – ab + b2<\/sup>) = a3<\/sup> + b3<\/sup>
\nxyz = a3<\/sup> + b3<\/sup><\/p>\n

(ii) x = a + b, y = a\u03c9 + b\u03c92<\/sup>, z = a\u03c92<\/sup> + b\u03c9
\nx + y + z = (a + a\u03c9 + a\u03c92<\/sup>) + (b + b\u03c92<\/sup> + b\u03c9)
\n= a (1 + \u03c9 + \u03c92<\/sup>) + b (1 + \u03c9 + \u03c92<\/sup>) = a(0) + b(0) = 0
\nNow x + y + z = 0 \u21d2 x3<\/sup> + y3<\/sup> + z3<\/sup> = 3xyz
\nHere xyz = a3<\/sup> + b3<\/sup>
\n\u2234 x3<\/sup> + y3<\/sup> + z3<\/sup> = 3 (a3<\/sup> + b3<\/sup>)<\/p>\n","protected":false},"excerpt":{"rendered":"

You can Download Samacheer Kalvi 12th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations. Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 2 Complex Numbers Ex 2.8 Question 1. If to \u03c9 \u2260 1 is a cube root of unity, then show […]<\/p>\n","protected":false},"author":2,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":"","footnotes":""},"categories":[8],"tags":[],"blocksy_meta":"","jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/samacheerguru.com\/wp-json\/wp\/v2\/posts\/4156"}],"collection":[{"href":"https:\/\/samacheerguru.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/samacheerguru.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/samacheerguru.com\/wp-json\/wp\/v2\/users\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/samacheerguru.com\/wp-json\/wp\/v2\/comments?post=4156"}],"version-history":[{"count":1,"href":"https:\/\/samacheerguru.com\/wp-json\/wp\/v2\/posts\/4156\/revisions"}],"predecessor-version":[{"id":55291,"href":"https:\/\/samacheerguru.com\/wp-json\/wp\/v2\/posts\/4156\/revisions\/55291"}],"wp:attachment":[{"href":"https:\/\/samacheerguru.com\/wp-json\/wp\/v2\/media?parent=4156"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/samacheerguru.com\/wp-json\/wp\/v2\/categories?post=4156"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/samacheerguru.com\/wp-json\/wp\/v2\/tags?post=4156"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}