{"id":67949,"date":"2023-09-07T16:44:00","date_gmt":"2023-09-07T11:14:00","guid":{"rendered":"https:\/\/www.kopykitab.com\/blog\/?p=67949"},"modified":"2023-11-24T10:18:21","modified_gmt":"2023-11-24T04:48:21","slug":"rd-sharma-solutions-class-11-maths-chapter-13-exercise-13-2","status":"publish","type":"post","link":"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-13-exercise-13-2\/","title":{"rendered":"RD Sharma Class 11 Solutions Chapter 13 Exercise 13.2 (Updated for 2024)"},"content":{"rendered":"\n<p><img class=\"alignnone size-full wp-image-122738\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/RD-Sharma-Class-11-Solutions-Chapter-13-Exercise-13.2.jpg\" alt=\"RD Sharma Solutions Class 11 Maths Chapter 13 Exercise 13.2\" width=\"1200\" height=\"675\" srcset=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/RD-Sharma-Class-11-Solutions-Chapter-13-Exercise-13.2.jpg 1200w, https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/RD-Sharma-Class-11-Solutions-Chapter-13-Exercise-13.2-768x432.jpg 768w\" sizes=\"(max-width: 1200px) 100vw, 1200px\" \/><\/p>\n<p><strong>RD Sharma Solutions Class 11 Maths Chapter 13 Exercise 13.2<\/strong>: We are here to help you clear your Class 11 Maths exam with flying colors. For that we would recommend you <a href=\"https:\/\/www.kopykitab.com\/blog\/cbse-class-11-maths-rd-sharma-solutions\/\" target=\"_blank\" rel=\"noopener\">RD Sharma Solutions Class 11 Maths.<\/a> You can easily solve questions with its help and clear your doubts. Maths is made easier, all thanks to the <a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-13-complex-numbers\/\" target=\"_blank\" rel=\"noopener\">RD Sharma Solutions Class 11 Maths Chapter 13<\/a> Exercise 13.2.\u00a0<\/p>\n<div id=\"ez-toc-container\" class=\"ez-toc-v2_0_47_1 counter-hierarchy ez-toc-counter ez-toc-grey ez-toc-container-direction\">\n<div class=\"ez-toc-title-container\">\n<p class=\"ez-toc-title\">Table of Contents<\/p>\n<span class=\"ez-toc-title-toggle\"><a href=\"#\" class=\"ez-toc-pull-right ez-toc-btn ez-toc-btn-xs ez-toc-btn-default ez-toc-toggle\" aria-label=\"ez-toc-toggle-icon-1\"><label for=\"item-69e7674828fd5\" aria-label=\"Table of Content\"><span style=\"display: flex;align-items: center;width: 35px;height: 30px;justify-content: center;direction:ltr;\"><svg style=\"fill: #000000;color:#000000\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" class=\"list-377408\" width=\"20px\" height=\"20px\" viewBox=\"0 0 24 24\" 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href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-13-exercise-13-2\/#download-rd-sharma-solutions-class-11-maths-chapter-13-exercise-132-pdf\" title=\"Download RD Sharma Solutions Class 11 Maths Chapter 13 Exercise 13.2 PDF:\">Download RD Sharma Solutions Class 11 Maths Chapter 13 Exercise 13.2 PDF:<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-2\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-13-exercise-13-2\/#access-answers-of-rd-sharma-solutions-class-11-maths-chapter-13-exercise-132\" title=\"Access answers of RD Sharma Solutions Class 11 Maths Chapter 13 Exercise 13.2\">Access answers of RD Sharma Solutions Class 11 Maths Chapter 13 Exercise 13.2<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-3\" 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href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-13-exercise-13-2\/#why-is-kopykitab%e2%80%99s-rd-sharma-solutions-class-11-maths-chapter-13-exercise-132-the-best-study-material\" title=\"Why is Kopykitab\u2019s RD Sharma Solutions Class 11 Maths Chapter 13 Exercise 13.2 the best study material?\">Why is Kopykitab\u2019s RD Sharma Solutions Class 11 Maths Chapter 13 Exercise 13.2 the best study material?<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-6\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-13-exercise-13-2\/#is-rd-sharma-solutions-for-class-12-maths-chapter-13-exercise-132-for-free\" title=\"Is RD Sharma Solutions for Class 12 Maths Chapter 13 Exercise 13.2 for free?\">Is RD Sharma Solutions for Class 12 Maths Chapter 13 Exercise 13.2 for free?<\/a><\/li><\/ul><\/li><\/ul><\/nav><\/div>\n<h2><span class=\"ez-toc-section\" id=\"download-rd-sharma-solutions-class-11-maths-chapter-13-exercise-132-pdf\"><\/span>Download RD Sharma Solutions Class 11 Maths Chapter 13 Exercise 13.2 PDF:<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p style=\"text-align: justify;\"><a href=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/RD-Sharma-Solutions-Class-11-Maths-Chapter-13-Ex-13.2-1.pdf\" target=\"_blank\" rel=\"noopener\">RD Sharma Class 11 Solutions Chapter 13 Exercise 13.2<\/a><\/p>\n<div id=\"example1\" style=\"text-align: justify;\">\u00a0<\/div>\n<p style=\"text-align: justify;\"><style>\n.pdfobject-container { height: 500px;}<br \/>\n.pdfobject { border: 1px solid #666; }<br \/>\n<\/style><\/p>\n<p style=\"text-align: justify;\"><script src=\"https:\/\/www.kopykitab.com\/_utility\/js\/pdfobject.min.js\"><\/script><br \/><script>PDFObject.embed(\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/RD-Sharma-Solutions-Class-11-Maths-Chapter-13-Ex-13.2-1.pdf\",\"#example1\");<\/script><\/p>\n<h2><span class=\"ez-toc-section\" id=\"access-answers-of-rd-sharma-solutions-class-11-maths-chapter-13-exercise-132\"><\/span>Access answers of RD Sharma Solutions Class 11 Maths Chapter 13 Exercise 13.2<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<article id=\"post-53091\" class=\"post-53091 page type-page status-publish hentry\">\n<p><strong>1. Express the following complex numbers in the standard form a + ib.<\/strong><\/p>\n<p><strong>(i) (1 + i) (1 + 2i)<\/strong><\/p>\n<p><strong>(ii) (3 + 2i) \/ (-2 + i)<\/strong><\/p>\n<p><strong>(iii) 1\/(2 + i)<sup>2<\/sup><\/strong><\/p>\n<p><strong>(iv) (1 \u2013 i) \/ (1 + i)<\/strong><\/p>\n<p><strong>(v) (2 + i)<sup>3<\/sup>\u00a0\/ (2 + 3i)<\/strong><\/p>\n<p><strong>(vi) [(1 + i) (1 +\u221a3i)] \/ (1 \u2013 i)<\/strong><\/p>\n<p><strong>(vii) (2 + 3i) \/ (4 + 5i)<\/strong><\/p>\n<p><strong>(viii) (1 \u2013 i)<sup>3<\/sup>\u00a0\/ (1 \u2013 i<sup>3<\/sup>)<\/strong><\/p>\n<p><strong>(ix) (1 + 2i)<sup>-3<\/sup><\/strong><\/p>\n<p><strong>(x) (3 \u2013 4i) \/ [(4 \u2013 2i) (1 + i)]<\/strong><\/p>\n<p><strong>(xi)<\/strong><\/p>\n<p><img title=\"RD Sharma Solutions for Class 11 Maths Chapter 13 \u2013 Complex Numbers image - 1\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/03\/rd-sharma-solutions-for-class-11-maths-chapter-13-complex-numbers-image-1.gif\" alt=\"RD Sharma Solutions for Class 11 Maths Chapter 13 \u2013 Complex Numbers image - 1\" \/><\/p>\n<p><strong>(xii) (5 +\u221a2i) \/ (1-\u221a2i)<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p><strong>(i)\u00a0<\/strong>(1 + i) (1 + 2i)<\/p>\n<p>Let us simplify and express in the standard form of (a + ib).<\/p>\n<p>(1 + i) (1 + 2i) = (1+i)(1+2i)<\/p>\n<p>= 1(1+2i)+i(1+2i)<\/p>\n<p>= 1+2i+i+2i<sup>2<\/sup><\/p>\n<p>= 1+3i+2(-1) [since, i<sup>2\u00a0<\/sup>= -1]<\/p>\n<p>= 1+3i-2<\/p>\n<p>= -1+3i<\/p>\n<p>\u2234 The values of a and b are -1 and 3.<\/p>\n<p><strong>(ii)\u00a0<\/strong>(3 + 2i) \/ (-2 + i)<\/p>\n<p>Let us simplify and express in the standard form of (a + ib).<\/p>\n<p>(3 + 2i) \/ (-2 + i) = [(3 + 2i) \/ (-2 + i)] \u00d7 (-2-i) \/ (-2-i) [multiply and divide with (-2-i)]<\/p>\n<p>= [3(-2-i) + 2i (-2-i)] \/ [(-2)<sup>2<\/sup>\u00a0\u2013 (i)<sup>2<\/sup>]<\/p>\n<p>= [-6 -3i \u2013 4i -2i<sup>2<\/sup>] \/ (4-i<sup>2<\/sup>)<\/p>\n<p>= [-6 -7i -2(-1)] \/ (4 \u2013 (-1)) [since, i<sup>2\u00a0<\/sup>= -1]<\/p>\n<p>= [-4 -7i] \/ 5<\/p>\n<p>\u2234 The values of a and b are -4\/5 and -7\/5<\/p>\n<p><strong>(iii)\u00a0<\/strong>1\/(2 + i)<sup>2<\/sup><\/p>\n<p>Let us simplify and express in the standard form of (a + ib).<\/p>\n<p>1\/(2 + i)<sup>2<\/sup>\u00a0= 1\/(2<sup>2<\/sup>\u00a0+ i<sup>2<\/sup>\u00a0+ 2(2) (i))<\/p>\n<p>= 1\/ (4 \u2013 1 + 4i) [since, i<sup>2\u00a0<\/sup>= -1]<\/p>\n<p>= 1\/(3 + 4i) [multiply and divide with (3 \u2013 4i)]<\/p>\n<p>= 1\/(3 + 4i) \u00d7 (3 \u2013 4i)\/ (3 \u2013 4i)]<\/p>\n<p>= (3-4i)\/ (3<sup>2<\/sup>\u00a0\u2013 (4i)<sup>2<\/sup>)<\/p>\n<p>= (3-4i)\/ (9 \u2013 16i<sup>2<\/sup>)<\/p>\n<p>= (3-4i)\/ (9 \u2013 16(-1)) [since, i<sup>2\u00a0<\/sup>= -1]<\/p>\n<p>= (3-4i)\/25<\/p>\n<p>\u2234 The values of a and b are 3\/25 and -4\/25<\/p>\n<p><strong>(iv)\u00a0<\/strong>(1 \u2013 i) \/ (1 + i)<\/p>\n<p>Let us simplify and express in the standard form of (a + ib).<\/p>\n<p>(1 \u2013 i) \/ (1 + i) = (1 \u2013 i) \/ (1 + i) \u00d7 (1-i)\/(1-i) [multiply and divide with (1-i)]<\/p>\n<p>= (1<sup>2<\/sup>\u00a0+ i<sup>2<\/sup>\u00a0\u2013 2(1)(i)) \/ (1<sup>2<\/sup>\u00a0\u2013 i<sup>2<\/sup>)<\/p>\n<p>= (1 + (-1) -2i) \/ (1 \u2013 (-1))<\/p>\n<p>= -2i\/2<\/p>\n<p>= -i<\/p>\n<p>\u2234 The values of a and b are 0 and -1<\/p>\n<p><strong>(v)\u00a0<\/strong>(2 + i)<sup>3<\/sup>\u00a0\/ (2 + 3i)<\/p>\n<p>Let us simplify and express in the standard form of (a + ib).<\/p>\n<p>(2 + i)<sup>3<\/sup>\u00a0\/ (2 + 3i) = (2<sup>3<\/sup>\u00a0+ i<sup>3<\/sup>\u00a0+ 3(2)<sup>2<\/sup>(i) + 3(i)<sup>2<\/sup>(2)) \/ (2 + 3i)<\/p>\n<p>= (8 + (i<sup>2<\/sup>.i) + 3(4)(i) + 6i<sup>2<\/sup>) \/ (2 + 3i)<\/p>\n<p>= (8 + (-1)i + 12i + 6(-1)) \/ (2 + 3i)<\/p>\n<p>= (2 + 11i) \/ (2 + 3i)<\/p>\n[multiply and divide with (2-3i)]\n<p>= (2 + 11i)\/(2 + 3i) \u00d7 (2-3i)\/(2-3i)<\/p>\n<p>= [2(2-3i) + 11i(2-3i)] \/ (2<sup>2<\/sup>\u00a0\u2013 (3i)<sup>2<\/sup>)<\/p>\n<p>= (4 \u2013 6i + 22i \u2013 33i<sup>2<\/sup>) \/ (4 \u2013 9i<sup>2<\/sup>)<\/p>\n<p>= (4 + 16i \u2013 33(-1)) \/ (4 \u2013 9(-1)) [since, i<sup>2\u00a0<\/sup>= -1]<\/p>\n<p>= (37 + 16i) \/ 13<\/p>\n<p>\u2234 The values of a and b are 37\/13 and 16\/13.<\/p>\n<p><strong>(vi)\u00a0<\/strong>[(1 + i) (1 +\u221a3i)] \/ (1 \u2013 i)<\/p>\n<p>Let us simplify and express in the standard form of (a + ib).<\/p>\n[(1 + i) (1 +\u221a3i)] \/ (1 \u2013 i) = [1(1+\u221a3i) + i(1+\u221a3i)] \/ (1-i)\n<p>= (1 + \u221a3i + i + \u221a3i<sup>2<\/sup>) \/ (1 \u2013 i)<\/p>\n<p>= (1 + (\u221a3+1)i + \u221a3(-1)) \/ (1-i) [since, i<sup>2\u00a0<\/sup>= -1]<\/p>\n<p>= [(1-\u221a3) + (1+\u221a3)i] \/ (1-i)<\/p>\n[multiply and divide with (1+i)]\n<p>= [(1-\u221a3) + (1+\u221a3)i] \/ (1-i) \u00d7 (1+i)\/(1+i)<\/p>\n<p>= [(1-\u221a3) (1+i) + (1+\u221a3)i(1+i)] \/ (1<sup>2<\/sup>\u00a0\u2013 i<sup>2<\/sup>)<\/p>\n<p>= [1-\u221a3+ (1-\u221a3)i + (1+\u221a3)i + (1+\u221a3)i<sup>2<\/sup>] \/ (1-(-1)) [since, i<sup>2\u00a0<\/sup>= -1]<\/p>\n<p>= [(1-\u221a3)+(1-\u221a3+1+\u221a3)i+(1+\u221a3)(-1)] \/ 2<\/p>\n<p>= (-2\u221a3 + 2i) \/ 2<\/p>\n<p>= -\u221a3 + i<\/p>\n<p>\u2234 The values of a and b are -\u221a3 and 1.<\/p>\n<p><strong>(vii)\u00a0<\/strong>(2 + 3i) \/ (4 + 5i)<\/p>\n<p>Let us simplify and express in the standard form of (a + ib).<\/p>\n<p>(2 + 3i) \/ (4 + 5i) = [multiply and divide with (4-5i)]<\/p>\n<p>= (2 + 3i) \/ (4 + 5i) \u00d7 (4-5i)\/(4-5i)<\/p>\n<p>= [2(4-5i) + 3i(4-5i)] \/ (4<sup>2<\/sup>\u00a0\u2013 (5i)<sup>2<\/sup>)<\/p>\n<p>= [8 \u2013 10i + 12i \u2013 15i<sup>2<\/sup>] \/ (16 \u2013 25i<sup>2<\/sup>)<\/p>\n<p>= [8+2i-15(-1)] \/ (16 \u2013 25(-1)) [since, i<sup>2\u00a0<\/sup>= -1]<\/p>\n<p>= (23 + 2i) \/ 41<\/p>\n<p>\u2234 The values of a and b are 23\/41 and 2\/41.<\/p>\n<p><strong>(viii)\u00a0<\/strong>(1 \u2013 i)<sup>3<\/sup>\u00a0\/ (1 \u2013 i<sup>3<\/sup>)<\/p>\n<p>Let us simplify and express in the standard form of (a + ib).<\/p>\n<p>(1 \u2013 i)<sup>3<\/sup>\u00a0\/ (1 \u2013 i<sup>3<\/sup>) = [1<sup>3<\/sup>\u00a0\u2013 3(1)<sup>2<\/sup>i + 3(1)(i)<sup>2<\/sup>\u00a0\u2013 i<sup>3<\/sup>] \/ (1-i<sup>2<\/sup>.i)<\/p>\n<p>= [1 \u2013 3i + 3(-1)-i<sup>2<\/sup>.i] \/ (1 \u2013 (-1)i) [since, i<sup>2\u00a0<\/sup>= -1]<\/p>\n<p>= [-2 \u2013 3i \u2013 (-1)i] \/ (1+i)<\/p>\n<p>= [-2-4i] \/ (1+i)<\/p>\n[Multiply and divide with (1-i)]\n<p>= [-2-4i] \/ (1+i) \u00d7 (1-i)\/(1-i)<\/p>\n<p>= [-2(1-i)-4i(1-i)] \/ (1<sup>2<\/sup>\u00a0\u2013 i<sup>2<\/sup>)<\/p>\n<p>= [-2+2i-4i+4i<sup>2<\/sup>] \/ (1 \u2013 (-1))<\/p>\n<p>= [-2-2i+4(-1)] \/2<\/p>\n<p>= (-6-2i)\/2<\/p>\n<p>= -3 \u2013 i<\/p>\n<p>\u2234 The values of a and b are -3 and -1.<\/p>\n<p><strong>(ix)\u00a0<\/strong>(1 + 2i)<sup>-3<\/sup><\/p>\n<p>Let us simplify and express in the standard form of (a + ib).<\/p>\n<p>(1 + 2i)<sup>-3<\/sup>\u00a0= 1\/(1 + 2i)<sup>3<\/sup><\/p>\n<p>= 1\/(1<sup>3<\/sup>+3(1)<sup>2<\/sup>\u00a0(2i)+2(1)(2i)<sup>2<\/sup>\u00a0+ (2i)<sup>3<\/sup>)<\/p>\n<p>= 1\/(1+6i+4i<sup>2<\/sup>+8i<sup>3<\/sup>)<\/p>\n<p>= 1\/(1+6i+4(-1)+8i<sup>2<\/sup>.i) [since, i<sup>2\u00a0<\/sup>= -1]<\/p>\n<p>= 1\/(-3+6i+8(-1)i) [since, i<sup>2\u00a0<\/sup>= -1]<\/p>\n<p>= 1\/(-3-2i)<\/p>\n<p>= -1\/(3+2i)<\/p>\n[Multiply and divide with (3-2i)]\n<p>= -1\/(3+2i) \u00d7 (3-2i)\/(3-2i)<\/p>\n<p>= (-3+2i)\/(3<sup>2<\/sup>\u00a0\u2013 (2i)<sup>2<\/sup>)<\/p>\n<p>= (-3+2i) \/ (9-4i<sup>2<\/sup>)<\/p>\n<p>= (-3+2i) \/ (9-4(-1))<\/p>\n<p>= (-3+2i) \/13<\/p>\n<p>\u2234 The values of a and b are -3\/13 and 2\/13.<\/p>\n<p><strong>(x)\u00a0<\/strong>(3 \u2013 4i) \/ [(4 \u2013 2i) (1 + i)]<\/p>\n<p>Let us simplify and express in the standard form of (a + ib).<\/p>\n<p>(3 \u2013 4i) \/ [(4 \u2013 2i) (1 + i)] = (3-4i)\/ [4(1+i)-2i(1+i)]<\/p>\n<p>= (3-4i)\/ [4+4i-2i-2i<sup>2<\/sup>]<\/p>\n<p>= (3-4i)\/ [4+2i-2(-1)] [since, i<sup>2\u00a0<\/sup>= -1]<\/p>\n<p>= (3-4i)\/ (6+2i)<\/p>\n[Multiply and divide with (6-2i)]\n<p>= (3-4i)\/ (6+2i) \u00d7 (6-2i)\/(6-2i)<\/p>\n<p>= [3(6-2i)-4i(6-2i)] \/ (6<sup>2<\/sup>\u00a0\u2013 (2i)<sup>2<\/sup>)<\/p>\n<p>= [18 \u2013 6i \u2013 24i + 8i<sup>2<\/sup>] \/ (36 \u2013 4i<sup>2<\/sup>)<\/p>\n<p>= [18 \u2013 30i + 8 (-1)] \/ (36 \u2013 4 (-1)) [since, i<sup>2\u00a0<\/sup>= -1]<\/p>\n<p>= [10-30i] \/ 40<\/p>\n<p>= (1 \u2013 3i) \/ 4<\/p>\n<p>\u2234 The values of a and b are 1\/4 and -3\/4.<\/p>\n<p><strong>(xi)<\/strong><\/p>\n<p><img title=\"RD Sharma Solutions for Class 11 Maths Chapter 13 \u2013 Complex Numbers\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/03\/rd-sharma-solutions-for-class-11-maths-chapter-13-complex-numbers-image-2.png\" alt=\"RD Sharma Solutions for Class 11 Maths Chapter 13 \u2013 Complex Numbers image - 2\" \/><br \/><img title=\"RD Sharma Solutions for Class 11 Maths Chapter 13 \u2013 Complex Numbers\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/09\/rd-sharma-solutions-for-class-11-maths-chapter-13-complex-numbers-image-3.jpg\" alt=\"RD Sharma Solutions for Class 11 Maths Chapter 13 \u2013 Complex Numbers image - 3\" \/><\/p>\n<p><strong>(xii)\u00a0<\/strong>(5 +\u221a2i) \/ (1-\u221a2i)<\/p>\n<p>Let us simplify and express in the standard form of (a + ib).<\/p>\n<p>(5 +\u221a2i) \/ (1-\u221a2i) = [Multiply and divide with (1+<strong>\u221a<\/strong>2i)]<\/p>\n<p>= (5 +\u221a2i) \/ (1-\u221a2i) \u00d7 (1+<strong>\u221a<\/strong>2i)\/(1+<strong>\u221a<\/strong>2i)<\/p>\n<p>= [5(1+<strong>\u221a<\/strong>2i) +\u00a0<strong>\u221a<\/strong>2i(1+<strong>\u221a<\/strong>2i)] \/ (1<sup>2<\/sup>\u00a0\u2013 (<strong>\u221a<\/strong>2)<sup>2<\/sup>)<\/p>\n<p>= [5+5<strong>\u221a<\/strong>2i +\u00a0<strong>\u221a<\/strong>2i + 2i<sup>2<\/sup>] \/ (1 \u2013 2i<sup>2<\/sup>)<\/p>\n<p>= [5 + 6<strong>\u221a<\/strong>2i + 2(-1)] \/ (1-2(-1)) [since, i<sup>2\u00a0<\/sup>= -1]<\/p>\n<p>= [3+6<strong>\u221a<\/strong>2i]\/3<\/p>\n<p>= 1+ 2<strong>\u221a<\/strong>2i<\/p>\n<p>\u2234 The values of a and b are 1 and 2<strong>\u221a<\/strong>2.<\/p>\n<p><strong>2. Find the real values of x and y, if<\/strong><\/p>\n<p><strong>(i) (x + iy) (2 \u2013 3i) = 4 + i<\/strong><\/p>\n<p><strong>(ii) (3x \u2013 2i y) (2 + i)<sup>2<\/sup>\u00a0= 10(1 + i)<\/strong><\/p>\n<p><img title=\"RD Sharma Solutions for Class 11 Maths Chapter 13 \u2013 Complex Numbers\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/03\/rd-sharma-solutions-for-class-11-maths-chapter-13-complex-numbers-image-4.gif\" alt=\"RD Sharma Solutions for Class 11 Maths Chapter 13 \u2013 Complex Numbers image - 4\" \/><\/p>\n<p><strong>(iv) (1 + i) (x + iy) = 2 \u2013 5i<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p><strong>(i)\u00a0<\/strong>(x + iy) (2 \u2013 3i) = 4 + i<\/p>\n<p>Given:<\/p>\n<p>(x + iy) (2 \u2013 3i) = 4 + i<\/p>\n<p>Let us simplify the expression, and we get<\/p>\n<p>x(2 \u2013 3i) + iy(2 \u2013 3i) = 4 + i<\/p>\n<p>2x \u2013 3xi + 2yi \u2013 3yi<sup>2\u00a0<\/sup>= 4 + i<\/p>\n<p>2x + (-3x+2y)i \u2013 3y (-1) = 4 + i [since, i<sup>2\u00a0<\/sup>= -1]<\/p>\n<p>2x + (-3x+2y)i + 3y = 4 + i [since, i<sup>2\u00a0<\/sup>= -1]<\/p>\n<p>(2x+3y) + i(-3x+2y) = 4 + i<\/p>\n<p>Equating Real and Imaginary parts on both sides, we get<\/p>\n<p>2x+3y = 4\u2026 (i)<\/p>\n<p>And -3x+2y = 1\u2026 (ii)<\/p>\n<p>Multiply (i) by 3 and (ii) by 2 and add<\/p>\n<p>On solving, we get<\/p>\n<p>6x \u2013 6x \u2013 9y + 4y = 12 + 2<\/p>\n<p>13y = 14<\/p>\n<p>y = 14\/13<\/p>\n<p>Substitute the value of y in (i) , and we get<\/p>\n<p>2x+3y = 4<\/p>\n<p>2x + 3(14\/13) = 4<\/p>\n<p>2x = 4 \u2013 (42\/13)<\/p>\n<p>= (52-42)\/13<\/p>\n<p>2x = 10\/13<\/p>\n<p>x = 5\/13<\/p>\n<p>x = 5\/13, y = 14\/13<\/p>\n<p>\u2234 The real values of x and y are 5\/13 and 14\/13.<\/p>\n<p><strong>(ii)\u00a0<\/strong>(3x \u2013 2i y) (2 + i)<sup>2<\/sup>\u00a0= 10(1 + i)<\/p>\n<p>Given:<\/p>\n<p>(3x \u2013 2iy) (2+i)<sup>2\u00a0<\/sup>= 10(1+i)<\/p>\n<p>(3x \u2013 2yi) (2<sup>2<\/sup>+i<sup>2<\/sup>+2(2)(i)) = 10+10i<\/p>\n<p>(3x \u2013 2yi) (4 + (-1)+4i) = 10+10i [since, i<sup>2\u00a0<\/sup>= -1]<\/p>\n<p>(3x \u2013 2yi) (3+4i) = 10+10i<\/p>\n<p>Let us divide with 3+4i into both sides, and we get<\/p>\n<p>(3x \u2013 2yi) = (10+10i)\/(3+4i)<\/p>\n<p>= Now multiply and divide with (3-4i)<\/p>\n<p>= [10(3-4i) + 10i(3-4i)] \/ (3<sup>2<\/sup>\u00a0\u2013 (4i)<sup>2<\/sup>)<\/p>\n<p>= [30-40i+30i-40i<sup>2<\/sup>] \/ (9 \u2013 16i<sup>2<\/sup>)<\/p>\n<p>= [30-10i-40(-1)] \/ (9-16(-1))<\/p>\n<p>= [70-10i]\/25<\/p>\n<p>Now, equating Real and Imaginary parts on both sides, we get<\/p>\n<p>3x = 70\/25 and -2y = -10\/25<\/p>\n<p>x = 70\/75 and y = 1\/5<\/p>\n<p>x = 14\/15 and y = 1\/5<\/p>\n<p>\u2234 The real values of x and y are 14\/15 and 1\/5.<\/p>\n<p><img title=\"RD Sharma Solutions for Class 11 Maths Chapter 13 \u2013 Complex Numbers\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/03\/rd-sharma-solutions-for-class-11-maths-chapter-13-complex-numbers-image-5.png\" alt=\"RD Sharma Solutions for Class 11 Maths Chapter 13 \u2013 Complex Numbers image - 5\" \/><\/p>\n<p>(4+2i) x-3i-3 + (9-7i)y = 10i<\/p>\n<p>(4x+9y-3) + i(2x-7y-3) = 10i<\/p>\n<p>Now, equating Real and Imaginary parts on both sides, we get,<\/p>\n<p>4x+9y-3 = 0 \u2026 (i)<\/p>\n<p>And 2x-7y-3 = 10<\/p>\n<p>2x-7y = 13 \u2026 (ii)<\/p>\n<p>Multiply (i) by 7 and (ii) by 9 and add.<\/p>\n<p>On solving these equations, we get<\/p>\n<p>28x + 18x + 63y \u2013 63y = 117 + 21<\/p>\n<p>46x = 117 + 21<\/p>\n<p>46x = 138<\/p>\n<p>x = 138\/46<\/p>\n<p>= 3<\/p>\n<p>Substitute the value of x in (i), and we get<\/p>\n<p>4x+9y-3 = 0<\/p>\n<p>9y = -9<\/p>\n<p>y = -9\/9<\/p>\n<p>= -1<\/p>\n<p>x = 3 and y = -1<\/p>\n<p>\u2234 The real values of x and y are 3 and -1.<\/p>\n<p><strong>(iv)\u00a0<\/strong>(1 + i) (x + iy) = 2 \u2013 5i<\/p>\n<p>Given:<\/p>\n<p>(1 + i) (x + iy) = 2 \u2013 5i<\/p>\n<p>Divide with (1+i) into both the sides, and we get,<\/p>\n<p>(x + iy) = (2 \u2013 5i)\/(1+i)<\/p>\n<p>Multiply and divide by (1-i).<\/p>\n<p>= (2 \u2013 5i)\/(1+i) \u00d7 (1-i)\/(1-i)<\/p>\n<p>= [2(1-i) \u2013 5i (1-i)] \/ (1<sup>2<\/sup>\u00a0\u2013 i<sup>2<\/sup>)<\/p>\n<p>= [2 \u2013 7i + 5(-1)] \/ 2 [since, i<sup>2\u00a0<\/sup>= -1]<\/p>\n<p>= (-3-7i)\/2<\/p>\n<p>Now, equating Real and Imaginary parts on both sides, we get<\/p>\n<p>x = -3\/2 and y = -7\/2<\/p>\n<p>\u2234 The real values of x and y are -3\/2 and -7\/2.<\/p>\n<p><strong>3. Find the conjugates of the following complex numbers:<\/strong><\/p>\n<p><strong>(i) 4 \u2013 5i<\/strong><\/p>\n<p><strong>(ii) 1 \/ (3 + 5i)<\/strong><\/p>\n<p><strong>(iii) 1 \/ (1 + i)<\/strong><\/p>\n<p><strong>(iv) (3 \u2013 i)<sup>2<\/sup>\u00a0\/ (2 + i)<\/strong><\/p>\n<p><strong>(v) [(1 + i) (2 + i)] \/ (3 + i)<\/strong><\/p>\n<p><strong>(vi) [(3 \u2013 2i) (2 + 3i)] \/ [(1 + 2i) (2 \u2013 i)]<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p><strong>(i)\u00a0<\/strong>4 \u2013 5i<\/p>\n<p>Given:<\/p>\n<p>4 \u2013 5i<\/p>\n<p>We know the conjugate of a complex number (a + ib) is (a \u2013 ib).<\/p>\n<p>So,<\/p>\n<p>\u2234 The conjugate of (4 \u2013 5i) is (4 + 5i)<\/p>\n<p><strong>(ii)\u00a0<\/strong>1 \/ (3 + 5i)<\/p>\n<p>Given:<\/p>\n<p>1 \/ (3 + 5i)<\/p>\n<p>Since the given complex number is not in the standard form of (a + ib),<\/p>\n<p>Let us convert to standard form by multiplying and dividing with (3 \u2013 5i).<\/p>\n<p>We get,<\/p>\n<p><img title=\"RD Sharma Solutions for Class 11 Maths Chapter 13 \u2013 Complex Numbers\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/03\/rd-sharma-solutions-for-class-11-maths-chapter-13-complex-numbers-image-6.png\" alt=\"RD Sharma Solutions for Class 11 Maths Chapter 13 \u2013 Complex Numbers image - 6\" \/><\/p>\n<p>We know the conjugate of a complex number (a + ib) is (a \u2013 ib).<\/p>\n<p>So,<\/p>\n<p>\u2234 The conjugate of (3 \u2013 5i)\/34 is (3 + 5i)\/34<\/p>\n<p><strong>(iii)\u00a0<\/strong>1 \/ (1 + i)<\/p>\n<p>Given:<\/p>\n<p>1 \/ (1 + i)<\/p>\n<p>Since the given complex number is not in the standard form of (a + ib),<\/p>\n<p>Let us convert to standard form by multiplying and dividing with (1 \u2013 i).<\/p>\n<p>We get,<\/p>\n<p><img title=\"RD Sharma Solutions for Class 11 Maths Chapter 13 \u2013 Complex Numbers\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/03\/rd-sharma-solutions-for-class-11-maths-chapter-13-complex-numbers-image-7.png\" alt=\"RD Sharma Solutions for Class 11 Maths Chapter 13 \u2013 Complex Numbers image - 7\" \/><\/p>\n<p>We know the conjugate of a complex number (a + ib) is (a \u2013 ib).<\/p>\n<p>So,<\/p>\n<p>\u2234 The conjugate of (1-i)\/2 is (1+i)\/2<\/p>\n<p><strong>(iv)\u00a0<\/strong>(3 \u2013 i)<sup>2<\/sup>\u00a0\/ (2 + i)<\/p>\n<p>Given:<\/p>\n<p>(3 \u2013 i)<sup>2<\/sup>\u00a0\/ (2 + i)<\/p>\n<p>Since the given complex number is not in the standard form of (a + ib),<\/p>\n<p>Let us convert to standard form.<\/p>\n<p><img title=\"RD Sharma Solutions for Class 11 Maths Chapter 13 \u2013 Complex Numbers\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/03\/rd-sharma-solutions-for-class-11-maths-chapter-13-complex-numbers-image-8.png\" alt=\"RD Sharma Solutions for Class 11 Maths Chapter 13 \u2013 Complex Numbers image - 8\" \/><\/p>\n<p>We know the conjugate of a complex number (a + ib) is (a \u2013 ib).<\/p>\n<p>So,<\/p>\n<p>\u2234 The conjugate of (2 \u2013 4i) is (2 + 4i)<\/p>\n<p><strong>(v)\u00a0<\/strong>[(1 + i) (2 + i)] \/ (3 + i)<\/p>\n<p>Given:<\/p>\n[(1 + i) (2 + i)] \/ (3 + i)\n<p>Since the given complex number is not in the standard form of (a + ib),<\/p>\n<p>Let us convert to standard form.<\/p>\n<p><img title=\"RD Sharma Solutions for Class 11 Maths Chapter 13 \u2013 Complex Numbers\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/03\/rd-sharma-solutions-for-class-11-maths-chapter-13-complex-numbers-image-9.png\" alt=\"RD Sharma Solutions for Class 11 Maths Chapter 13 \u2013 Complex Numbers image - 9\" \/><\/p>\n<p><img title=\"RD Sharma Solutions for Class 11 Maths Chapter 13 \u2013 Complex Numbers\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/03\/rd-sharma-solutions-for-class-11-maths-chapter-13-complex-numbers-image-10.png\" alt=\"RD Sharma Solutions for Class 11 Maths Chapter 13 \u2013 Complex Numbers image - 10\" \/><\/p>\n<p>We know the conjugate of a complex number (a + ib) is (a \u2013 ib).<\/p>\n<p>So,<\/p>\n<p>\u2234 The conjugate of (3 + 4i)\/5 is (3 \u2013 4i)\/5<\/p>\n<p><strong>(vi)\u00a0<\/strong>[(3 \u2013 2i) (2 + 3i)] \/ [(1 + 2i) (2 \u2013 i)]<\/p>\n<p>Given:<\/p>\n[(3 \u2013 2i) (2 + 3i)] \/ [(1 + 2i) (2 \u2013 i)]\n<p>Since the given complex number is not in the standard form of (a + ib),<\/p>\n<p>Let us convert to standard form.<\/p>\n<p><img title=\"RD Sharma Solutions for Class 11 Maths Chapter 13 \u2013 Complex Numbers\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/03\/rd-sharma-solutions-for-class-11-maths-chapter-13-complex-numbers-image-11.png\" alt=\"RD Sharma Solutions for Class 11 Maths Chapter 13 \u2013 Complex Numbers image - 11\" \/><\/p>\n<p>We know the conjugate of a complex number (a + ib) is (a \u2013 ib).<\/p>\n<p>So,<\/p>\n<p>\u2234 The conjugate of (63 \u2013 16i)\/25 is (63 + 16i)\/25<\/p>\n<p><strong>4. Find the multiplicative inverse of the following complex numbers.<\/strong><\/p>\n<p><strong>(i) 1 \u2013 i<\/strong><\/p>\n<p><strong>(ii) (1 + i \u221a3)<sup>2<\/sup><\/strong><\/p>\n<p><strong>(iii) 4 \u2013 3i<\/strong><\/p>\n<p><strong>(iv) \u221a5 + 3i<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p><strong>(i)\u00a0<\/strong>1 \u2013 i<\/p>\n<p>Given:<\/p>\n<p>1 \u2013 i<\/p>\n<p>We know the multiplicative inverse of a complex number (Z) is Z<sup>-1<\/sup>\u00a0or 1\/Z.<\/p>\n<p>So,<\/p>\n<p><img title=\"RD Sharma Solutions for Class 11 Maths Chapter 13 \u2013 Complex Numbers\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/03\/rd-sharma-solutions-for-class-11-maths-chapter-13-complex-numbers-image-12.png\" alt=\"RD Sharma Solutions for Class 11 Maths Chapter 13 \u2013 Complex Numbers image - 12\" \/><\/p>\n<p>\u2234 The multiplicative inverse of (1 \u2013 i) is (1 + i)\/2<\/p>\n<p><strong>(ii)\u00a0<\/strong>(1 + i \u221a3)<sup>2<\/sup><\/p>\n<p>Given:<\/p>\n<p>(1 + i \u221a3)<sup>2<\/sup><\/p>\n<p>Z = (1 + i \u221a3)<sup>2<\/sup><\/p>\n<p>= 1<sup>2<\/sup>\u00a0+ (i \u221a3)<sup>2<\/sup>\u00a0+ 2 (1) (i\u221a3)<\/p>\n<p>= 1 + 3i<sup>2<\/sup>\u00a0+ 2 i\u221a3<\/p>\n<p>= 1 + 3(-1) + 2 i\u221a3 [since, i<sup>2<\/sup>\u00a0= -1]<\/p>\n<p>= 1 \u2013 3 + 2 i\u221a3<\/p>\n<p>= -2 + 2 i\u221a3<\/p>\n<p>We know the multiplicative inverse of a complex number (Z) is Z<sup>-1<\/sup>\u00a0or 1\/Z.<\/p>\n<p>So,<\/p>\n<p>Z = -2 + 2 i\u221a3<\/p>\n<p><img title=\"RD Sharma Solutions for Class 11 Maths Chapter 13 \u2013 Complex Numbers\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/03\/rd-sharma-solutions-for-class-11-maths-chapter-13-complex-numbers-image-13.png\" alt=\"RD Sharma Solutions for Class 11 Maths Chapter 13 \u2013 Complex Numbers image - 13\" \/><\/p>\n<p>\u2234 The multiplicative inverse of (1 + i\u221a3)<sup>2<\/sup>\u00a0is (-1-i\u221a3)\/8<\/p>\n<p><strong>(iii)\u00a0<\/strong>4 \u2013 3i<\/p>\n<p>Given:<\/p>\n<p>4 \u2013 3i<\/p>\n<p>We know the multiplicative inverse of a complex number (Z) is Z<sup>-1<\/sup>\u00a0or 1\/Z.<\/p>\n<p>So,<\/p>\n<p>Z = 4 \u2013 3i<\/p>\n<p><img title=\"RD Sharma Solutions for Class 11 Maths Chapter 13 \u2013 Complex Numbers\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/03\/rd-sharma-solutions-for-class-11-maths-chapter-13-complex-numbers-image-14.png\" alt=\"RD Sharma Solutions for Class 11 Maths Chapter 13 \u2013 Complex Numbers image - 14\" \/><\/p>\n<p>\u2234 The multiplicative inverse of (4 \u2013 3i) is (4 + 3i)\/25<\/p>\n<p><strong>(iv)\u00a0<\/strong>\u221a5 + 3i<\/p>\n<p>Given:<\/p>\n<p>\u221a5 + 3i<\/p>\n<p>We know the multiplicative inverse of a complex number (Z) is Z<sup>-1<\/sup>\u00a0or 1\/Z.<\/p>\n<p>So,<\/p>\n<p>Z = \u221a5 + 3i<\/p>\n<p><img title=\"RD Sharma Solutions for Class 11 Maths Chapter 13 \u2013 Complex Numbers\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/03\/rd-sharma-solutions-for-class-11-maths-chapter-13-complex-numbers-image-15.png\" alt=\"RD Sharma Solutions for Class 11 Maths Chapter 13 \u2013 Complex Numbers image - 15\" \/><\/p>\n<p>\u2234 The multiplicative inverse of (\u221a5 + 3i) is (\u221a5 \u2013 3i)\/14<\/p>\n<p><strong>6. If z<sub>1<\/sub>\u00a0= (2 \u2013 i), z<sub>2<\/sub>\u00a0= (-2 + i), find<\/strong><\/p>\n<p><img title=\"RD Sharma Solutions for Class 11 Maths Chapter 13 \u2013 Complex Numbers\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/03\/rd-sharma-solutions-for-class-11-maths-chapter-13-complex-numbers-image-18.png\" alt=\"RD Sharma Solutions for Class 11 Maths Chapter 13 \u2013 Complex Numbers image - 18\" \/><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given:<\/p>\n<p>z<sub>1<\/sub>\u00a0= (2 \u2013 i) and z<sub>2<\/sub>\u00a0= (-2 + i)<\/p>\n<p><img title=\"RD Sharma Solutions for Class 11 Maths Chapter 13 \u2013 Complex Numbers\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/09\/rd-sharma-solutions-for-class-11-maths-chapter-13-complex-numbers-image-19.jpg\" alt=\"RD Sharma Solutions for Class 11 Maths Chapter 13 \u2013 Complex Numbers image - 19\" \/><\/p>\n<p><img title=\"RD Sharma Solutions for Class 11 Maths Chapter 13 \u2013 Complex Numbers\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/03\/rd-sharma-solutions-for-class-11-maths-chapter-13-complex-numbers-image-20.png\" alt=\"RD Sharma Solutions for Class 11 Maths Chapter 13 \u2013 Complex Numbers image - 20\" \/><\/p>\n<p><strong>7. Find the modulus of [(1 + i)\/(1 \u2013 i)] \u2013 [(1 \u2013 i)\/(1 + i)]<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given:<\/p>\n[(1 + i)\/(1 \u2013 i)] \u2013 [(1 \u2013 i)\/(1 + i)]\n<p>So,<\/p>\n<p>Z = [(1 + i)\/(1 \u2013 i)] \u2013 [(1 \u2013 i)\/(1 + i)]<\/p>\n<p>Let us simplify, and we get<\/p>\n<p>= [(1+i) (1+i) \u2013 (1-i) (1-i)] \/ (1<sup>2<\/sup>\u00a0\u2013 i<sup>2<\/sup>)<\/p>\n<p>= [1<sup>2<\/sup>\u00a0+ i<sup>2<\/sup>\u00a0+ 2(1)(i) \u2013 (1<sup>2<\/sup>\u00a0+ i<sup>2<\/sup>\u00a0\u2013 2(1)(i))] \/ (1 \u2013 (-1)) [Since, i<sup>2<\/sup>\u00a0= -1]<\/p>\n<p>= 4i\/2<\/p>\n<p>= 2i<\/p>\n<p>We know that for a complex number\u00a0Z = (a+ib) it\u2019s magnitude is given by |z| =\u00a0<strong>\u221a<\/strong>(a<sup>2<\/sup>\u00a0+ b<sup>2<\/sup>)<\/p>\n<p>So,<\/p>\n<p>|Z| =\u00a0<strong>\u221a<\/strong>(0<sup>2<\/sup>\u00a0+ 2<sup>2<\/sup>)<\/p>\n<p>= 2<\/p>\n<p>\u2234 The modulus of [(1 + i)\/(1 \u2013 i)] \u2013 [(1 \u2013 i)\/(1 + i)] is 2.<\/p>\n<p><strong>8. If x + iy = (a+ib)\/(a-ib), prove that x<sup>2<\/sup>\u00a0+ y<sup>2<\/sup>\u00a0= 1<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given:<\/p>\n<p>x + iy = (a+ib)\/(a-ib)<\/p>\n<p>We know that for a complex number\u00a0Z = (a+ib) it\u2019s magnitude is given by |z| =\u00a0<strong>\u221a<\/strong>(a<sup>2<\/sup>\u00a0+ b<sup>2<\/sup>)<\/p>\n<p>So,<\/p>\n<p>|a\/b| is |a| \/ |b|<\/p>\n<p>Applying Modulus on both sides, we get,<\/p>\n<p><img title=\"RD Sharma Solutions for Class 11 Maths Chapter 13 \u2013 Complex Numbers\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/03\/rd-sharma-solutions-for-class-11-maths-chapter-13-complex-numbers-image-21.png\" alt=\"RD Sharma Solutions for Class 11 Maths Chapter 13 \u2013 Complex Numbers image - 21\" \/><\/p>\n<p><strong>9. Find the least positive integral value of n for which [(1+i)\/(1-i)]<sup>n<\/sup>\u00a0is real.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given:<\/p>\n[(1+i)\/(1-i)]<sup>n<\/sup>\n<p>Z = [(1+i)\/(1-i)]<sup>n<\/sup><\/p>\n<p>Now, let us multiply and divide by (1+i), and we get<\/p>\n<p><img title=\"RD Sharma Solutions for Class 11 Maths Chapter 13 \u2013 Complex Numbers\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/03\/rd-sharma-solutions-for-class-11-maths-chapter-13-complex-numbers-image-22.png\" alt=\"RD Sharma Solutions for Class 11 Maths Chapter 13 \u2013 Complex Numbers image - 22\" \/><\/p>\n<p>= i [which is not real]<\/p>\n<p>For n = 2, we have<\/p>\n[(1+i)\/(1-i)]<sup>2<\/sup>\u00a0= i<sup>2<\/sup>\n<p>= -1 [which is real]<\/p>\n<p>So, the smallest positive integral \u2018n\u2019 that can make\u00a0[(1+i)\/(1-i)]<sup>n<\/sup>\u00a0real is 2.<\/p>\n<p>\u2234\u00a0The smallest positive integral value of \u2018n\u2019 is 2.<\/p>\n<p><strong>10. Find the real values of \u03b8 for which the complex number (1 + i cos \u03b8) \/ (1 \u2013 2i cos \u03b8) is purely real.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given:<\/p>\n<p>(1 + i cos \u03b8) \/ (1 \u2013 2i cos \u03b8)<\/p>\n<p>Z = (1 + i cos \u03b8) \/ (1 \u2013 2i cos \u03b8)<\/p>\n<p>Let us multiply and divide by (1 + 2i cos \u03b8)<\/p>\n<p><img title=\"RD Sharma Solutions for Class 11 Maths Chapter 13 \u2013 Complex Numbers\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/03\/rd-sharma-solutions-for-class-11-maths-chapter-13-complex-numbers-image-23.png\" alt=\"RD Sharma Solutions for Class 11 Maths Chapter 13 \u2013 Complex Numbers image - 23\" \/><\/p>\n<p>For a complex number to be purely real, the imaginary part should be equal to zero.<\/p>\n<p>So,<\/p>\n<p><img title=\"RD Sharma Solutions for Class 11 Maths Chapter 13 \u2013 Complex Numbers\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/03\/rd-sharma-solutions-for-class-11-maths-chapter-13-complex-numbers-image-24.png\" alt=\"RD Sharma Solutions for Class 11 Maths Chapter 13 \u2013 Complex Numbers image - 24\" \/><\/p>\n<p>3cos \u03b8 = 0 (since,\u00a01 + 4cos<sup>2<\/sup>\u03b8 \u2265 1)<\/p>\n<p>cos \u03b8 = 0<\/p>\n<p>cos \u03b8 = cos \u03c0\/2<\/p>\n<p>\u03b8 = [(2n+1)\u03c0] \/ 2, for n \u2208 Z<\/p>\n<p>= 2n\u03c0\u00a0<strong>\u00b1\u00a0<\/strong>\u03c0\/2, for n \u2208 Z<\/p>\n<p>\u2234\u00a0The values of \u03b8 to get the complex number to be purely real is 2n\u03c0\u00a0<strong>\u00b1\u00a0<\/strong>\u03c0\/2, for n \u2208 Z<\/p>\n<p><strong>11. Find the smallest positive integer value of\u00a0n\u00a0for which (1+i)<sup>\u00a0n<\/sup>\u00a0\/ (1-i)<sup>\u00a0n-2<\/sup>\u00a0is a real number.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given:<\/p>\n<p>(1+i)<sup>\u00a0n<\/sup>\u00a0\/ (1-i)<sup>\u00a0n-2<\/sup><\/p>\n<p>Z = (1+i)<sup>\u00a0n<\/sup>\u00a0\/ (1-i)<sup>\u00a0n-2<\/sup><\/p>\n<p>Let us multiply and divide by (1 \u2013 i)<sup>2<\/sup><\/p>\n<p><img title=\"RD Sharma Solutions for Class 11 Maths Chapter 13 \u2013 Complex Numbers\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/03\/rd-sharma-solutions-for-class-11-maths-chapter-13-complex-numbers-image-25.png\" alt=\"RD Sharma Solutions for Class 11 Maths Chapter 13 \u2013 Complex Numbers image - 25\" \/><\/p>\n<p>For n = 1,<\/p>\n<p>Z = -2i<sup>1+1<\/sup><\/p>\n<p>= -2i<sup>2<\/sup><\/p>\n<p>= 2, which is a real number.<\/p>\n<p>\u2234\u00a0The smallest positive integer value of n is 1.<\/p>\n<p><strong>12. If [(1+i)\/(1-i)]<sup>3<\/sup>\u00a0\u2013 [(1-i)\/(1+i)]<sup>3<\/sup>\u00a0= x + iy, find (x, y)<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given:<\/p>\n[(1+i)\/(1-i)]<sup>3<\/sup>\u00a0\u2013 [(1-i)\/(1+i)]<sup>3<\/sup>\u00a0= x + iy\n<p>Let us rationalize the denominator, we get<\/p>\n<p><img title=\"RD Sharma Solutions for Class 11 Maths Chapter 13 \u2013 Complex Numbers\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/03\/rd-sharma-solutions-for-class-11-maths-chapter-13-complex-numbers-image-26.png\" alt=\"RD Sharma Solutions for Class 11 Maths Chapter 13 \u2013 Complex Numbers image - 26\" \/><\/p>\n<p>i<sup>3<\/sup>\u2013(-i)<sup>3\u00a0<\/sup>= x + iy<\/p>\n<p>2i<sup>3\u00a0<\/sup>= x + iy<\/p>\n<p>2i<sup>2<\/sup>.i = x + iy<\/p>\n<p>2(-1)I = x + iy<\/p>\n<p>-2i = x + iy<\/p>\n<p>Equating Real and Imaginary parts on both sides, we get<\/p>\n<p>x = 0 and y = -2<\/p>\n<p>\u2234\u00a0The values of x and y are 0 and -2.<\/p>\n<p><strong>13. If (1+i)<sup>2<\/sup>\u00a0\/ (2-i) = x + iy, find x + y<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given:<\/p>\n<p>(1+i)<sup>2<\/sup>\u00a0\/ (2-i) = x + iy<\/p>\n<p>Upon expansion, we get<\/p>\n<p><img title=\"RD Sharma Solutions for Class 11 Maths Chapter 13 \u2013 Complex Numbers\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/03\/rd-sharma-solutions-for-class-11-maths-chapter-13-complex-numbers-image-27.png\" alt=\"RD Sharma Solutions for Class 11 Maths Chapter 13 \u2013 Complex Numbers image - 27\" \/><\/p>\n<p>Let us equate real and imaginary parts on both sides, and we get<\/p>\n<p>x = -2\/5 and y = 4\/5<\/p>\n<p>so,<\/p>\n<p>x + y = -2\/5 + 4\/5<\/p>\n<p>= (-2+4)\/5<\/p>\n<p>= 2\/5<\/p>\n<p>\u2234\u00a0The value of (x + y) is 2\/5<\/p>\n<\/article>\n<p>This is the complete blog on RD Sharma Solutions for Class 11 Maths Chapter 13 Exercise 13.2. To Know more about the <a href=\"https:\/\/cbse.nic.in\/\" target=\"_blank\" rel=\"noopener\">CBSE<\/a>\u00a0Class 11 Maths exams, you can ask in the comments.<\/p>\n<h2><span class=\"ez-toc-section\" id=\"faqs-on-rd-sharma-solutions-for-class-11-maths-chapter-13-exercise-132\"><\/span>FAQs on RD Sharma Solutions for Class 11 Maths Chapter 13 Exercise 13.2<span class=\"ez-toc-section-end\"><\/span><\/h2>\n\n\n<div id=\"rank-math-faq\" class=\"rank-math-block\">\n<div class=\"rank-math-list \">\n<div id=\"faq-question-1630662855927\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"where-can-i-download-rd-sharma-class-11-solutions-chapter-13-exercise-132-free-pdf\"><\/span>Where can I download RD Sharma Class 11 Solutions Chapter 13 Exercise 13.2 free PDF?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>You can download RD Sharma Solutions for Class 11 Maths Chapter 13 Exercise 13.2 free PDF from the above article.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1630662928735\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"why-is-kopykitab%e2%80%99s-rd-sharma-solutions-class-11-maths-chapter-13-exercise-132-the-best-study-material\"><\/span>Why is Kopykitab\u2019s RD Sharma Solutions Class 11 Maths Chapter 13 Exercise 13.2 the best study material?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>The RD Sharma Solutions Class 11 Maths Chapter 13 Exercise 13.2 available on Kopykitab\u2019s website has been created by highly qualified experts to assist students in achieving high scores on the board exam. The solutions are well-organized and logical, giving pupils a clear picture of the most important questions.\u00a0<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1630662944509\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"is-rd-sharma-solutions-for-class-12-maths-chapter-13-exercise-132-for-free\"><\/span>Is RD Sharma Solutions for Class 12 Maths Chapter 13 Exercise 13.2 for free?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>Yes, You can get RD Sharma Solutions Class 12 Maths Chapter 13 Exercise 13.2 for free.<\/p>\n\n<\/div>\n<\/div>\n<\/div>\n<\/div>","protected":false},"excerpt":{"rendered":"<p>RD Sharma Solutions Class 11 Maths Chapter 13 Exercise 13.2: We are here to help you clear your Class 11 Maths exam with flying colors. For that we would recommend you RD Sharma Solutions Class 11 Maths. You can easily solve questions with its help and clear your doubts. Maths is made easier, all thanks &#8230; <a title=\"RD Sharma Class 11 Solutions Chapter 13 Exercise 13.2 (Updated for 2024)\" class=\"read-more\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-13-exercise-13-2\/\" aria-label=\"More on RD Sharma Class 11 Solutions Chapter 13 Exercise 13.2 (Updated for 2024)\">Read more<\/a><\/p>\n","protected":false},"author":243,"featured_media":122738,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"fifu_image_url":"","fifu_image_alt":""},"categories":[2985,73411,73410],"tags":[3428,73334,73564],"amp_enabled":true,"_links":{"self":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/67949"}],"collection":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/users\/243"}],"replies":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/comments?post=67949"}],"version-history":[{"count":5,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/67949\/revisions"}],"predecessor-version":[{"id":511865,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/67949\/revisions\/511865"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media\/122738"}],"wp:attachment":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media?parent=67949"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/categories?post=67949"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/tags?post=67949"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}