{"id":67951,"date":"2023-09-16T17:35:00","date_gmt":"2023-09-16T12:05:00","guid":{"rendered":"https:\/\/www.kopykitab.com\/blog\/?p=67951"},"modified":"2023-12-01T10:48:19","modified_gmt":"2023-12-01T05:18:19","slug":"rd-sharma-solutions-class-11-maths-chapter-13-exercise-13-4","status":"publish","type":"post","link":"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-13-exercise-13-4\/","title":{"rendered":"RD Sharma Class 11 Solutions Chapter 13 Exercise 13.4 (Updated for 2024)"},"content":{"rendered":"\n<p><img class=\"alignnone size-full wp-image-122803\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/RD-Sharma-Class-11-Solutions-Chapter-13-Exercise-13.4.jpg\" alt=\"RD Sharma Solutions Class 11 Maths Chapter 13 Exercise 13.4\" width=\"1200\" height=\"675\" srcset=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/RD-Sharma-Class-11-Solutions-Chapter-13-Exercise-13.4.jpg 1200w, https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/RD-Sharma-Class-11-Solutions-Chapter-13-Exercise-13.4-768x432.jpg 768w\" sizes=\"(max-width: 1200px) 100vw, 1200px\" \/><\/p>\n<p>RD Sharma Solutions Class 11 Maths Chapter 13 Exercise 13.4: You can download the Free PDF of <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> without any hassle from this blog. Subject matter experts have made the solutions easy to understand. To know more about 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 13.4, read the whole blog.<\/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-69da6d1ddda86\" 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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class=\"ez-toc-section\" id=\"download-rd-sharma-solutions-class-11-maths-chapter-13-exercise-134-pdf\"><\/span>Download RD Sharma Solutions Class 11 Maths Chapter 13 Exercise 13.4 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-13.4-1.pdf\" target=\"_blank\" rel=\"noopener\">RD Sharma Solutions Class 11 Maths Chapter 13 Exercise 13.4<\/a><\/p>\n<div id=\"example1\" style=\"text-align: justify;\">&nbsp;<\/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-13.4-1.pdf\",\"#example1\");<\/script><\/p>\n<h2><span class=\"ez-toc-section\" id=\"rd-sharma-solutions-for-class-11-maths-chapter-13-exercise-134\"><\/span>RD Sharma Solutions for Class 11 Maths Chapter 13 Exercise 13.4<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p><strong>1. Find the modulus and argument of the following complex numbers and hence express each of them in the polar form:<\/strong><\/p>\n<p><strong>(i) 1 + i<\/strong><\/p>\n<p><strong>(ii) \u221a3 + i<\/strong><\/p>\n<p><strong>(iii) 1 \u2013 i<\/strong><\/p>\n<p><strong>(iv) (1 \u2013 i) \/ (1 + i)<\/strong><\/p>\n<p><strong>(v) 1\/(1 + i)<\/strong><\/p>\n<p><strong>(vi) (1 + 2i) \/ (1 \u2013 3i)<\/strong><\/p>\n<p><strong>(vii) sin 120<sup>o<\/sup>&nbsp;\u2013 i cos 120<sup>o<\/sup><\/strong><\/p>\n<p><strong>(viii) -16 \/ (1 + i\u221a3)<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>We know that the polar form of a complex number Z = x + iy is given by Z = |Z| (cos \u03b8 + i sin \u03b8)<\/p>\n<p>Where,<\/p>\n<p>|Z| = modulus of complex number =&nbsp;<strong>\u221a<\/strong>(x<sup>2<\/sup>&nbsp;+ y<sup>2<\/sup>)<\/p>\n<p>\u03b8 = arg (z) = argument of complex number = tan<sup>-1<\/sup>&nbsp;(|y| \/ |x|)<\/p>\n<p><strong>(i)&nbsp;<\/strong>1 + i<\/p>\n<p>Given: Z = 1 + i<\/p>\n<p>So now,<\/p>\n<p>|Z| =&nbsp;<strong>\u221a<\/strong>(x<sup>2<\/sup>&nbsp;+ y<sup>2<\/sup>)<\/p>\n<p>=&nbsp;<strong>\u221a<\/strong>(1<sup>2<\/sup>&nbsp;+ 1<sup>2<\/sup>)<\/p>\n<p>=&nbsp;<strong>\u221a<\/strong>(1 + 1)<\/p>\n<p>=&nbsp;<strong>\u221a<\/strong>2<\/p>\n<p>\u03b8 = tan<sup>-1<\/sup>&nbsp;(|y| \/ |x|)<\/p>\n<p>= tan<sup>-1<\/sup>&nbsp;(1 \/ 1)<\/p>\n<p>= tan<sup>-1<\/sup>&nbsp;1<\/p>\n<p>Since x &gt; 0, y &gt; 0 complex number lies in 1<sup>st<\/sup>&nbsp;quadrant and the value of \u03b8 is 0<sup>0<\/sup>\u2264\u03b8\u226490<sup>0<\/sup>.<\/p>\n<p>\u03b8 = \u03c0\/4<\/p>\n<p>Z =&nbsp;<strong>\u221a<\/strong>2 (cos (\u03c0\/4) + i sin (\u03c0\/4))<\/p>\n<p>\u2234 Polar form of (1 + i) is&nbsp;<strong>\u221a<\/strong>2 (cos (\u03c0\/4) + i sin (\u03c0\/4))<\/p>\n<p><strong>(ii)<\/strong>&nbsp;\u221a3 + i<\/p>\n<p>Given: Z = \u221a3 + i<\/p>\n<p>So now,<\/p>\n<p>|Z| =&nbsp;<strong>\u221a<\/strong>(x<sup>2<\/sup>&nbsp;+ y<sup>2<\/sup>)<\/p>\n<p>=&nbsp;<strong>\u221a<\/strong>((\u221a3)<sup>2<\/sup>&nbsp;+ 1<sup>2<\/sup>)<\/p>\n<p>=&nbsp;<strong>\u221a<\/strong>(3 + 1)<\/p>\n<p>=&nbsp;<strong>\u221a<\/strong>4<\/p>\n<p>= 2<\/p>\n<p>\u03b8 = tan<sup>-1<\/sup>&nbsp;(|y| \/ |x|)<\/p>\n<p>= tan<sup>-1<\/sup>&nbsp;(1 \/ \u221a3)<\/p>\n<p>Since x &gt; 0, y &gt; 0 complex number lies in 1<sup>st<\/sup>&nbsp;quadrant and the value of \u03b8 is 0<sup>0<\/sup>\u2264\u03b8\u226490<sup>0<\/sup>.<\/p>\n<p>\u03b8 = \u03c0\/6<\/p>\n<p>Z = 2 (cos (\u03c0\/6) + i sin (\u03c0\/6))<\/p>\n<p>\u2234 Polar form of (\u221a3 + i) is 2 (cos (\u03c0\/6) + i sin (\u03c0\/6))<\/p>\n<p><strong>(iii)<\/strong>&nbsp;1 \u2013 i<\/p>\n<p>Given: Z = 1 \u2013 i<\/p>\n<p>So now,<\/p>\n<p>|Z| =&nbsp;<strong>\u221a<\/strong>(x<sup>2<\/sup>&nbsp;+ y<sup>2<\/sup>)<\/p>\n<p>=&nbsp;<strong>\u221a<\/strong>(1<sup>2<\/sup>&nbsp;+ (-1)<sup>2<\/sup>)<\/p>\n<p>=&nbsp;<strong>\u221a<\/strong>(1 + 1)<\/p>\n<p>=&nbsp;<strong>\u221a<\/strong>2<\/p>\n<p>\u03b8 = tan<sup>-1<\/sup>&nbsp;(|y| \/ |x|)<\/p>\n<p>= tan<sup>-1<\/sup>&nbsp;(1 \/ 1)<\/p>\n<p>= tan<sup>-1<\/sup>&nbsp;1<\/p>\n<p>Since x &gt; 0, y &lt; 0 complex number lies in 4<sup>th<\/sup>&nbsp;quadrant and the value of \u03b8 is -90<sup>0<\/sup>\u2264\u03b8\u22640<sup>0<\/sup>.<\/p>\n<p>\u03b8 = -\u03c0\/4<\/p>\n<p>Z =&nbsp;<strong>\u221a<\/strong>2 (cos (-\u03c0\/4) + i sin (-\u03c0\/4))<\/p>\n<p>=&nbsp;<strong>\u221a<\/strong>2 (cos (\u03c0\/4) \u2013 i sin (\u03c0\/4))<\/p>\n<p>\u2234 The polar form of (1 \u2013 i) is <strong>\u221a<\/strong>2 (cos (\u03c0\/4) \u2013 i sin (\u03c0\/4))<\/p>\n<p><strong>(iv)&nbsp;<\/strong>(1 \u2013 i) \/ (1 + i)<\/p>\n<p>Given: Z = (1 \u2013 i) \/ (1 + i)<\/p>\n<p>Let us multiply and divide by (1 \u2013 i), 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-41.png\" alt=\"RD Sharma Solutions for Class 11 Maths Chapter 13 \u2013 Complex Numbers image - 41\"><\/p>\n<p>= 0 \u2013 i<\/p>\n<p>So now,<\/p>\n<p>|Z| =&nbsp;<strong>\u221a<\/strong>(x<sup>2<\/sup>&nbsp;+ y<sup>2<\/sup>)<\/p>\n<p>=&nbsp;<strong>\u221a<\/strong>(0<sup>2<\/sup>&nbsp;+ (-1)<sup>2<\/sup>)<\/p>\n<p>=&nbsp;<strong>\u221a<\/strong>(0 + 1)<\/p>\n<p>=&nbsp;<strong>\u221a<\/strong>1<\/p>\n<p>\u03b8 = tan<sup>-1<\/sup>&nbsp;(|y| \/ |x|)<\/p>\n<p>= tan<sup>-1<\/sup>&nbsp;(1 \/ 0)<\/p>\n<p>= tan<sup>-1<\/sup>&nbsp;\u221e<\/p>\n<p>Since x \u2265 0, y &lt; 0 complex number lies in 4<sup>th<\/sup>&nbsp;quadrant and the value of \u03b8 is -90<sup>0<\/sup>\u2264\u03b8\u22640<sup>0<\/sup>.<\/p>\n<p>\u03b8 = -\u03c0\/2<\/p>\n<p>Z = 1 (cos (-\u03c0\/2) + i sin (-\u03c0\/2))<\/p>\n<p>= 1 (cos (\u03c0\/2) \u2013 i sin (\u03c0\/2))<\/p>\n<p>\u2234 The polar form of (1 \u2013 i) \/ (1 + i) is 1 (cos (\u03c0\/2) \u2013 i sin (\u03c0\/2))<\/p>\n<p><strong>(v)&nbsp;<\/strong>1\/(1 + i)<\/p>\n<p>Given: Z = 1 \/ (1 + i)<\/p>\n<p>Let us multiply and divide by (1 \u2013 i), 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-42.png\" alt=\"RD Sharma Solutions for Class 11 Maths Chapter 13 \u2013 Complex Numbers image - 42\"><\/p>\n<p>So now,<\/p>\n<p>|Z| =&nbsp;<strong>\u221a<\/strong>(x<sup>2<\/sup>&nbsp;+ y<sup>2<\/sup>)<\/p>\n<p>=&nbsp;<strong>\u221a<\/strong>((1\/2)<sup>2<\/sup>&nbsp;+ (-1\/2)<sup>2<\/sup>)<\/p>\n<p>=&nbsp;<strong>\u221a<\/strong>(1\/4 + 1\/4)<\/p>\n<p>=&nbsp;<strong>\u221a<\/strong>(2\/4)<\/p>\n<p>= 1\/<strong>\u221a<\/strong>2<\/p>\n<p>\u03b8 = tan<sup>-1<\/sup>&nbsp;(|y| \/ |x|)<\/p>\n<p>= tan<sup>-1<\/sup>&nbsp;((1\/2) \/ (1\/2))<\/p>\n<p>= tan<sup>-1<\/sup>&nbsp;1<\/p>\n<p>Since x &gt; 0, y &lt; 0 complex number lies in 4<sup>th<\/sup>&nbsp;quadrant and the value of \u03b8 is -90<sup>0<\/sup>\u2264\u03b8\u22640<sup>0<\/sup>.<\/p>\n<p>\u03b8 = -\u03c0\/4<\/p>\n<p>Z = 1\/<strong>\u221a<\/strong>2 (cos (-\u03c0\/4) + i sin (-\u03c0\/4))<\/p>\n<p>= 1\/<strong>\u221a<\/strong>2 (cos (\u03c0\/4) \u2013 i sin (\u03c0\/4))<\/p>\n<p>\u2234 Polar form of 1\/(1 + i) is 1\/<strong>\u221a<\/strong>2 (cos (\u03c0\/4) \u2013 i sin (\u03c0\/4))<\/p>\n<p><strong>(vi)&nbsp;<\/strong>(1 + 2i) \/ (1 \u2013 3i)<\/p>\n<p>Given: Z = (1 + 2i) \/ (1 \u2013 3i)<\/p>\n<p>Let us multiply and divide by (1 + 3i), 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-43.png\" alt=\"RD Sharma Solutions for Class 11 Maths Chapter 13 \u2013 Complex Numbers image - 43\"><\/p>\n<p>So now,<\/p>\n<p>|Z| =&nbsp;<strong>\u221a<\/strong>(x<sup>2<\/sup>&nbsp;+ y<sup>2<\/sup>)<\/p>\n<p>=&nbsp;<strong>\u221a<\/strong>((-1\/2)<sup>2<\/sup>&nbsp;+ (1\/2)<sup>2<\/sup>)<\/p>\n<p>=&nbsp;<strong>\u221a<\/strong>(1\/4 + 1\/4)<\/p>\n<p>=&nbsp;<strong>\u221a<\/strong>(2\/4)<\/p>\n<p>= 1\/<strong>\u221a<\/strong>2<\/p>\n<p>\u03b8 = tan<sup>-1<\/sup>&nbsp;(|y| \/ |x|)<\/p>\n<p>= tan<sup>-1<\/sup>&nbsp;((1\/2) \/ (1\/2))<\/p>\n<p>= tan<sup>-1<\/sup>&nbsp;1<\/p>\n<p>Since x &lt; 0, y &gt; 0 complex number lies in 2<sup>nd<\/sup>&nbsp;quadrant and the value of \u03b8 is 90<sup>0<\/sup>\u2264\u03b8\u2264180<sup>0<\/sup>.<\/p>\n<p>\u03b8 = 3\u03c0\/4<\/p>\n<p>Z = 1\/<strong>\u221a<\/strong>2 (cos (3\u03c0\/4) + i sin (3\u03c0\/4))<\/p>\n<p>\u2234 Polar form of (1 + 2i) \/ (1 \u2013 3i) is 1\/<strong>\u221a<\/strong>2 (cos (3\u03c0\/4) + i sin (3\u03c0\/4))<\/p>\n<p><strong>(vii)&nbsp;<\/strong>sin 120<sup>o<\/sup>&nbsp;\u2013 i cos 120<sup>o<\/sup><\/p>\n<p>Given: Z = sin 120<sup>o<\/sup>&nbsp;\u2013 i cos 120<sup>o<\/sup><\/p>\n<p>=&nbsp;<strong>\u221a<\/strong>3\/2 \u2013 i (-1\/2)<\/p>\n<p>=&nbsp;<strong>\u221a<\/strong>3\/2 + i (1\/2)<\/p>\n<p>So now,<\/p>\n<p>|Z| =&nbsp;<strong>\u221a<\/strong>(x<sup>2<\/sup>&nbsp;+ y<sup>2<\/sup>)<\/p>\n<p>=&nbsp;<strong>\u221a<\/strong>((<strong>\u221a<\/strong>3\/2)<sup>2<\/sup>&nbsp;+ (1\/2)<sup>2<\/sup>)<\/p>\n<p>=&nbsp;<strong>\u221a<\/strong>(3\/4 + 1\/4)<\/p>\n<p>=&nbsp;<strong>\u221a<\/strong>(4\/4)<\/p>\n<p>=&nbsp;<strong>\u221a<\/strong>1<\/p>\n<p>= 1<\/p>\n<p>\u03b8 = tan<sup>-1<\/sup>&nbsp;(|y| \/ |x|)<\/p>\n<p>= tan<sup>-1<\/sup>&nbsp;((1\/2) \/ (<strong>\u221a<\/strong>3\/2))<\/p>\n<p>= tan<sup>-1<\/sup>&nbsp;(1\/<strong>\u221a<\/strong>3)<\/p>\n<p>Since x &gt; 0, y &gt; 0 complex number lies in 1<sup>st<\/sup>&nbsp;quadrant and the value of \u03b8 is 0<sup>0<\/sup>\u2264\u03b8\u226490<sup>0<\/sup>.<\/p>\n<p>\u03b8 = \u03c0\/6<\/p>\n<p>Z = 1 (cos (\u03c0\/6) + i sin (\u03c0\/6))<\/p>\n<p>\u2234 Polar form of&nbsp;<strong>\u221a<\/strong>3\/2 + i (1\/2) is 1 (cos (\u03c0\/6) + i sin (\u03c0\/6))<\/p>\n<p><strong>(viii)&nbsp;<\/strong>-16 \/ (1 + i\u221a3)<\/p>\n<p>Given: Z = -16 \/ (1 + i\u221a3)<\/p>\n<p>Let us multiply and divide by (1 \u2013 i<strong>\u221a<\/strong>3), 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-44.png\" alt=\"RD Sharma Solutions for Class 11 Maths Chapter 13 \u2013 Complex Numbers image - 44\"><\/p>\n<p>So now,<\/p>\n<p>|Z| =&nbsp;<strong>\u221a<\/strong>(x<sup>2<\/sup>&nbsp;+ y<sup>2<\/sup>)<\/p>\n<p>=&nbsp;<strong>\u221a<\/strong>((-4)<sup>2<\/sup>&nbsp;+ (4<strong>\u221a<\/strong>3)<sup>2<\/sup>)<\/p>\n<p>=&nbsp;<strong>\u221a<\/strong>(16 + 48)<\/p>\n<p>=&nbsp;<strong>\u221a<\/strong>(64)<\/p>\n<p>= 8<\/p>\n<p>\u03b8 = tan<sup>-1<\/sup>&nbsp;(|y| \/ |x|)<\/p>\n<p>= tan<sup>-1<\/sup>&nbsp;((4<strong>\u221a<\/strong>3) \/ 4)<\/p>\n<p>= tan<sup>-1<\/sup>&nbsp;(<strong>\u221a<\/strong>3)<\/p>\n<p>Since x &lt; 0, y &gt; 0 complex number lies in 2<sup>nd<\/sup>&nbsp;quadrant and the value of \u03b8 is 90<sup>0<\/sup>\u2264\u03b8\u2264180<sup>0<\/sup>.<\/p>\n<p>\u03b8 = 2\u03c0\/3<\/p>\n<p>Z = 8 (cos (2\u03c0\/3) + i sin (2\u03c0\/3))<\/p>\n<p>\u2234 Polar form of -16 \/ (1 + i\u221a3) is 8 (cos (2\u03c0\/3) + i sin (2\u03c0\/3))<\/p>\n<p><strong>2. Write (i<sup>25<\/sup>)<sup>3<\/sup>&nbsp;in polar form.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given: Z = (i<sup>25<\/sup>)<sup>3<\/sup><\/p>\n<p>= i<sup>75<\/sup><\/p>\n<p>= i<sup>74<\/sup>. i<\/p>\n<p>= (i<sup>2<\/sup>)<sup>37<\/sup>. i<\/p>\n<p>= (-1)<sup>37<\/sup>. i<\/p>\n<p>= (-1). i<\/p>\n<p>= \u2013 i<\/p>\n<p>= 0 \u2013 i<\/p>\n<p>So now,<\/p>\n<p>|Z| =&nbsp;<strong>\u221a<\/strong>(x<sup>2<\/sup>&nbsp;+ y<sup>2<\/sup>)<\/p>\n<p>=&nbsp;<strong>\u221a<\/strong>(0<sup>2<\/sup>&nbsp;+ (-1)<sup>2<\/sup>)<\/p>\n<p>=&nbsp;<strong>\u221a<\/strong>(0 + 1)<\/p>\n<p>=&nbsp;<strong>\u221a<\/strong>1<\/p>\n<p>\u03b8 = tan<sup>-1<\/sup>&nbsp;(|y| \/ |x|)<\/p>\n<p>= tan<sup>-1<\/sup>&nbsp;(1 \/ 0)<\/p>\n<p>= tan<sup>-1<\/sup>&nbsp;\u221e<\/p>\n<p>Since x \u2265 0, y &lt; 0 complex number lies in 4<sup>th<\/sup>&nbsp;quadrant and the value of \u03b8 is -90<sup>0<\/sup>\u2264\u03b8\u22640<sup>0<\/sup>.<\/p>\n<p>\u03b8 = -\u03c0\/2<\/p>\n<p>Z = 1 (cos (-\u03c0\/2) + i sin (-\u03c0\/2))<\/p>\n<p>= 1 (cos (\u03c0\/2) \u2013 i sin (\u03c0\/2))<\/p>\n<p>\u2234 Polar form of (i<sup>25<\/sup>)<sup>3<\/sup>&nbsp;is 1 (cos (\u03c0\/2) \u2013 i sin (\u03c0\/2))<\/p>\n<p><strong>3.<\/strong>&nbsp;<strong>Express the following complex numbers in the form&nbsp;r (cos \u03b8 + i sin \u03b8):<\/strong><\/p>\n<p><strong>(i) 1 + i tan \u03b1<\/strong><\/p>\n<p><strong>(ii) tan \u03b1 \u2013 i<\/strong><\/p>\n<p><strong>(iii) 1 \u2013 sin \u03b1 + i cos \u03b1<\/strong><\/p>\n<p><strong>(iv) (1 \u2013 i) \/ (cos \u03c0\/3 + i sin \u03c0\/3)<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p><strong>(i)&nbsp;<\/strong>1 + i tan \u03b1<\/p>\n<p>Given: Z = 1 + i tan \u03b1<\/p>\n<p>We know that the polar form of a complex number Z = x + iy is given by Z = |Z| (cos \u03b8 + i sin \u03b8)<\/p>\n<p>Where,<\/p>\n<p>|Z| = modulus of complex number =&nbsp;<strong>\u221a<\/strong>(x<sup>2<\/sup>&nbsp;+ y<sup>2<\/sup>)<\/p>\n<p>\u03b8 = arg (z) = argument of complex number = tan<sup>-1<\/sup>&nbsp;(|y| \/ |x|)<\/p>\n<p>We also know that tan \u03b1 is a periodic function with period \u03c0.<\/p>\n<p>So \u03b1 is lying in the interval [0, \u03c0\/2) \u222a (\u03c0\/2, \u03c0].<\/p>\n<p>Let us consider case 1:<\/p>\n<p>\u03b1 \u2208 [0, \u03c0\/2)<\/p>\n<p>So now,<\/p>\n<p>|Z| = r =&nbsp;<strong>\u221a<\/strong>(x<sup>2<\/sup>&nbsp;+ y<sup>2<\/sup>)<\/p>\n<p>=&nbsp;<strong>\u221a<\/strong>(1<sup>2<\/sup>&nbsp;+ tan<sup>2<\/sup>&nbsp;\u03b1)<\/p>\n<p>=&nbsp;<strong>\u221a<\/strong>( sec<sup>2<\/sup>&nbsp;\u03b1)<\/p>\n<p>= |sec \u03b1| since, sec \u03b1 is positive in the interval [0, \u03c0\/2)<\/p>\n<p>\u03b8 = tan<sup>-1<\/sup>&nbsp;(|y| \/ |x|)<\/p>\n<p>= tan<sup>-1<\/sup>&nbsp;(tan \u03b1 \/ 1)<\/p>\n<p>= tan<sup>-1<\/sup>&nbsp;(tan \u03b1)<\/p>\n<p>= \u03b1 since, tan \u03b1 is positive in the interval [0, \u03c0\/2)<\/p>\n<p>\u2234 Polar form is Z = sec \u03b1 (cos \u03b1 + i sin \u03b1)<\/p>\n<p>Let us consider case 2:<\/p>\n<p>\u03b1 \u2208 (\u03c0\/2, \u03c0]<\/p>\n<p>So now,<\/p>\n<p>|Z| = r =&nbsp;<strong>\u221a<\/strong>(x<sup>2<\/sup>&nbsp;+ y<sup>2<\/sup>)<\/p>\n<p>=&nbsp;<strong>\u221a<\/strong>(1<sup>2<\/sup>&nbsp;+ tan<sup>2<\/sup>&nbsp;\u03b1)<\/p>\n<p>=&nbsp;<strong>\u221a<\/strong>( sec<sup>2<\/sup>&nbsp;\u03b1)<\/p>\n<p>= |sec \u03b1|<\/p>\n<p>= \u2013 sec \u03b1 since, sec \u03b1 is negative in the interval (\u03c0\/2, \u03c0]<\/p>\n<p>\u03b8 = tan<sup>-1<\/sup>&nbsp;(|y| \/ |x|)<\/p>\n<p>= tan<sup>-1<\/sup>&nbsp;(tan \u03b1 \/ 1)<\/p>\n<p>= tan<sup>-1<\/sup>&nbsp;(tan \u03b1)<\/p>\n<p>= -\u03c0 + \u03b1 since, tan \u03b1 is negative in the interval (\u03c0\/2, \u03c0]<\/p>\n<p>\u03b8 = -\u03c0 + \u03b1 [since, \u03b8 lies in 4<sup>th<\/sup>&nbsp;quadrant]<\/p>\n<p>Z = -sec \u03b1 (cos (\u03b1 \u2013 \u03c0) + i sin (\u03b1 \u2013 \u03c0))<\/p>\n<p>\u2234 Polar form is Z = -sec \u03b1 (cos (\u03b1 \u2013 \u03c0) + i sin (\u03b1 \u2013 \u03c0))<\/p>\n<p><strong>(ii)&nbsp;<\/strong>tan \u03b1 \u2013 i<\/p>\n<p>Given: Z = tan \u03b1 \u2013 i<\/p>\n<p>We know that the polar form of a complex number Z = x + iy is given by Z = |Z| (cos \u03b8 + i sin \u03b8)<\/p>\n<p>Where,<\/p>\n<p>|Z| = modulus of complex number =&nbsp;<strong>\u221a<\/strong>(x<sup>2<\/sup>&nbsp;+ y<sup>2<\/sup>)<\/p>\n<p>\u03b8 = arg (z) = argument of complex number = tan<sup>-1<\/sup>&nbsp;(|y| \/ |x|)<\/p>\n<p>We also know that tan \u03b1 is a periodic function with period \u03c0.<\/p>\n<p>So \u03b1 is lying in the interval [0, \u03c0\/2) \u222a (\u03c0\/2, \u03c0].<\/p>\n<p>Let us consider case 1:<\/p>\n<p>\u03b1 \u2208 [0, \u03c0\/2)<\/p>\n<p>So now,<\/p>\n<p>|Z| = r =&nbsp;<strong>\u221a<\/strong>(x<sup>2<\/sup>&nbsp;+ y<sup>2<\/sup>)<\/p>\n<p>=&nbsp;<strong>\u221a<\/strong>(tan<sup>2<\/sup>&nbsp;\u03b1 + 1<sup>2<\/sup>)<\/p>\n<p>=&nbsp;<strong>\u221a<\/strong>( sec<sup>2<\/sup>&nbsp;\u03b1)<\/p>\n<p>= |sec \u03b1| since, sec \u03b1 is positive in the interval [0, \u03c0\/2)<\/p>\n<p>= sec \u03b1<\/p>\n<p>\u03b8 = tan<sup>-1<\/sup>&nbsp;(|y| \/ |x|)<\/p>\n<p>= tan<sup>-1<\/sup>&nbsp;(1\/tan \u03b1)<\/p>\n<p>= tan<sup>-1<\/sup>&nbsp;(cot \u03b1) since, cot \u03b1 is positive in the interval [0, \u03c0\/2)<\/p>\n<p>= \u03b1 \u2013 \u03c0\/2 [since, \u03b8 lies in 4<sup>th<\/sup>&nbsp;quadrant]<\/p>\n<p>Z = sec \u03b1 (cos (\u03b1 \u2013 \u03c0\/2) + i sin (\u03b1 \u2013 \u03c0\/2))<\/p>\n<p>\u2234 Polar form is Z = sec \u03b1 (cos (\u03b1 \u2013 \u03c0\/2) + i sin (\u03b1 \u2013 \u03c0\/2))<\/p>\n<p>Let us consider case 2:<\/p>\n<p>\u03b1 \u2208 (\u03c0\/2, \u03c0]<\/p>\n<p>So now,<\/p>\n<p>|Z| = r =&nbsp;<strong>\u221a<\/strong>(x<sup>2<\/sup>&nbsp;+ y<sup>2<\/sup>)<\/p>\n<p>=&nbsp;<strong>\u221a<\/strong>(tan<sup>2<\/sup>&nbsp;\u03b1 + 1<sup>2<\/sup>)<\/p>\n<p>=&nbsp;<strong>\u221a<\/strong>( sec<sup>2<\/sup>&nbsp;\u03b1)<\/p>\n<p>= |sec \u03b1|<\/p>\n<p>= \u2013 sec \u03b1 since, sec \u03b1 is negative in the interval (\u03c0\/2, \u03c0]<\/p>\n<p>\u03b8 = tan<sup>-1<\/sup>&nbsp;(|y| \/ |x|)<\/p>\n<p>= tan<sup>-1<\/sup>&nbsp;(1\/tan \u03b1)<\/p>\n<p>= tan<sup>-1<\/sup>&nbsp;(cot \u03b1)<\/p>\n<p>= \u03c0\/2 + \u03b1 since, cot \u03b1 is negative in the interval (\u03c0\/2, \u03c0]<\/p>\n<p>\u03b8 = \u03c0\/2 + \u03b1 [since, \u03b8 lies in 3<sup>th<\/sup>&nbsp;quadrant]<\/p>\n<p>Z = -sec \u03b1 (cos (\u03c0\/2 + \u03b1) + i sin (\u03c0\/2 + \u03b1))<\/p>\n<p>\u2234 Polar form is Z = -sec \u03b1 (cos (\u03c0\/2 + \u03b1) + i sin (\u03c0\/2 + \u03b1))<\/p>\n<p><strong>(iii)<\/strong>&nbsp;1 \u2013 sin \u03b1 + i cos \u03b1<\/p>\n<p>Given: Z = 1 \u2013 sin \u03b1 + i cos \u03b1<\/p>\n<p>By using the formulas,<\/p>\n<p>Sin<sup>2<\/sup>&nbsp;\u03b8 + cos<sup>2<\/sup>&nbsp;\u03b8 = 1<\/p>\n<p>Sin 2\u03b8 = 2 sin \u03b8 cos \u03b8<\/p>\n<p>Cos 2\u03b8 = cos<sup>2<\/sup>&nbsp;\u03b8 \u2013 sin<sup>2<\/sup>&nbsp;\u03b8<\/p>\n<p>So,<\/p>\n<p>z= (sin<sup>2<\/sup>(\u03b1\/2) + cos<sup>2<\/sup>(\u03b1\/2) \u2013 2 sin(\u03b1\/2) cos(\u03b1\/2)) +&nbsp;<em>i<\/em>&nbsp;(cos<sup>2<\/sup>(\u03b1\/2) \u2013 sin<sup>2<\/sup>(\u03b1\/2))<\/p>\n<p>=&nbsp;(cos(\u03b1\/2) \u2013 sin(\u03b1\/2))<sup>2<\/sup>&nbsp;+&nbsp;<em>i<\/em>&nbsp;(cos<sup>2<\/sup>(\u03b1\/2) \u2013 sin<sup>2<\/sup>(\u03b1\/2))<\/p>\n<p>We know that the polar form of a complex number Z = x + iy is given by Z = |Z| (cos \u03b8 + i sin \u03b8)<\/p>\n<p>Where,<\/p>\n<p>|Z| = modulus of complex number =&nbsp;<strong>\u221a<\/strong>(x<sup>2<\/sup>&nbsp;+ y<sup>2<\/sup>)<\/p>\n<p>\u03b8 = arg (z) = argument of complex number = tan<sup>-1<\/sup>&nbsp;(|y| \/ |x|)<\/p>\n<p>Now,<\/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-46.png\" alt=\"RD Sharma Solutions for Class 11 Maths Chapter 13 \u2013 Complex Numbers image - 46\"><\/p>\n<p>We know that sine and cosine functions are periodic with period 2\u03c0<\/p>\n<p>Here we have 3 intervals:<\/p>\n<p>0 \u2264 \u03b1 \u2264 \u03c0\/2<\/p>\n<p>\u03c0\/2 \u2264 \u03b1 \u2264 3\u03c0\/2<\/p>\n<p>3\u03c0\/2 \u2264 \u03b1 \u2264 2\u03c0<\/p>\n<p>Let us consider case 1:<\/p>\n<p>In the interval 0 \u2264 \u03b1 \u2264 \u03c0\/2<\/p>\n<p>Cos (\u03b1\/2) &gt; sin (\u03b1\/2) and also 0 &lt; \u03c0\/4 + \u03b1\/2 &lt; \u03c0\/2<\/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-47.png\" alt=\"RD Sharma Solutions for Class 11 Maths Chapter 13 \u2013 Complex Numbers image - 47\"><\/p>\n<p>\u2234 Polar form is Z =&nbsp;<strong>\u221a<\/strong>2 (cos (\u03b1\/2) \u2013 sin (\u03b1\/2)) (cos (\u03c0\/4 + \u03b1\/2) + i sin (\u03c0\/4 + \u03b1\/2))<\/p>\n<p>Let us consider case 2:<\/p>\n<p>In the interval \u03c0\/2 \u2264 \u03b1 \u2264 3\u03c0\/2<\/p>\n<p>Cos (\u03b1\/2) &lt; sin (\u03b1\/2) and also \u03c0\/2 &lt; \u03c0\/4 + \u03b1\/2 &lt; \u03c0<\/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-48.png\" alt=\"RD Sharma Solutions for Class 11 Maths Chapter 13 \u2013 Complex Numbers image - 48\"><\/p>\n<p>Since, (1 \u2013 sin \u03b1) &gt; 0 and cos \u03b1 &lt; 0 [Z lies in 4<sup>th<\/sup>&nbsp;quadrant]<\/p>\n<p>= \u03b1\/2 \u2013 3\u03c0\/4<\/p>\n<p>\u2234 Polar form is Z = \u2013<strong>\u221a<\/strong>2 (cos (\u03b1\/2) \u2013 sin (\u03b1\/2)) (cos (\u03b1\/2 \u2013 3\u03c0\/4) + i sin (\u03b1\/2 \u2013 3\u03c0\/4))<\/p>\n<p>Let us consider case 3:<\/p>\n<p>In the interval 3\u03c0\/2 \u2264 \u03b1 \u2264 2\u03c0<\/p>\n<p>Cos (\u03b1\/2) &lt; sin (\u03b1\/2) and also \u03c0 &lt; \u03c0\/4 + \u03b1\/2 &lt; 5\u03c0\/4<\/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-49.png\" alt=\"RD Sharma Solutions for Class 11 Maths Chapter 13 \u2013 Complex Numbers image - 49\"><\/p>\n<p>\u03b8 = tan<sup>-1<\/sup>&nbsp;(tan (\u03c0\/4 + \u03b1\/2))<\/p>\n<p>= \u03c0 \u2013 (\u03c0\/4 + \u03b1\/2) [since, \u03b8 lies in 1st quadrant and tan\u2019s period is \u03c0]<\/p>\n<p>= \u03b1\/2 \u2013 3\u03c0\/4<\/p>\n<p>\u2234 Polar form is Z = \u2013<strong>\u221a<\/strong>2 (cos (\u03b1\/2) \u2013 sin (\u03b1\/2)) (cos (\u03b1\/2 \u2013 3\u03c0\/4) + i sin (\u03b1\/2 \u2013 3\u03c0\/4))<\/p>\n<p><strong>(iv)&nbsp;<\/strong>(1 \u2013 i) \/ (cos \u03c0\/3 + i sin \u03c0\/3)<\/p>\n<p>Given: Z = (1 \u2013 i) \/ (cos \u03c0\/3 + i sin \u03c0\/3)<\/p>\n<p>Let us multiply and divide by (1 \u2013 i<strong>\u221a<\/strong>3), 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-50.png\" alt=\"RD Sharma Solutions for Class 11 Maths Chapter 13 \u2013 Complex Numbers image - 50\"><\/p>\n<p>We know that the polar form of a complex number Z = x + iy is given by Z = |Z| (cos \u03b8 + i sin \u03b8)<\/p>\n<p>Where,<\/p>\n<p>|Z| = modulus of complex number =&nbsp;<strong>\u221a<\/strong>(x<sup>2<\/sup>&nbsp;+ y<sup>2<\/sup>)<\/p>\n<p>\u03b8 = arg (z) = argument of complex number = tan<sup>-1<\/sup>&nbsp;(|y| \/ |x|)<\/p>\n<p>Now,<\/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-51.png\" alt=\"RD Sharma Solutions for Class 11 Maths Chapter 13 \u2013 Complex Numbers image - 51\"><\/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-52.png\" alt=\"RD Sharma Solutions for Class 11 Maths Chapter 13 \u2013 Complex Numbers image - 52\"><\/p>\n<p>Since x &lt; 0, y &lt; 0 complex number lies in 3<sup>rd<\/sup>&nbsp;quadrant and the value of \u03b8 is 180<sup>0<\/sup>\u2264\u03b8\u2264-90<sup>0<\/sup>.<\/p>\n<p>= tan<sup>-1<\/sup>&nbsp;(2 +&nbsp;<strong>\u221a<\/strong>3)<\/p>\n<p>= -7\u03c0\/12<\/p>\n<p>Z =&nbsp;<strong>\u221a<\/strong>2 (cos (-7\u03c0\/12) + i sin (-7\u03c0\/12))<\/p>\n<p>=&nbsp;<strong>\u221a<\/strong>2 (cos (7\u03c0\/12) \u2013 i sin (7\u03c0\/12))<\/p>\n<p>\u2234 Polar form of (1 \u2013 i) \/ (cos \u03c0\/3 + i sin \u03c0\/3) is&nbsp;<strong>\u221a<\/strong>2 (cos (7\u03c0\/12) \u2013 i sin (7\u03c0\/12))<\/p>\n<p>To Know more about the <a href=\"https:\/\/cbse.nic.in\/\" target=\"_blank\" rel=\"noopener\">CBSE<\/a>&nbsp;Class 11 Maths exam, ask questions in the comments.<\/p>\n<h2><span class=\"ez-toc-section\" id=\"faqs-on-rd-sharma-solutions-for-class-11-maths-chapter-13-exercise-134\"><\/span>FAQs on RD Sharma Solutions for Class 11 Maths Chapter 13 Exercise 13.4<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-1630497567024\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"how-many-questions-exist-in-rd-sharma-solutions-for-class-11-maths-chapter-13-exercise-134\"><\/span>How many questions exist in RD Sharma Solutions for Class 11 Maths Chapter 13 Exercise 13.4?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>There are 6 questions in\u00a0RD Sharma Solutions Class 11 Maths Chapter 13 Exercise 13.4.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1630497635980\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"from-where-can-i-download-the-pdf-of-rd-sharma-solutions-class-11-maths-chapter-13-ex-134\"><\/span>From where can I download the PDF of RD Sharma Solutions Class 11 Maths Chapter 13 Ex 13.4?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>You can find the download link from the above blog.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1630497661682\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"how-much-does-downloading-the-pdf-of-rd-sharma-solutions-for-class-11-maths-chapter-13-exercise-134-cost\"><\/span>How much does downloading the PDF of RD Sharma Solutions for Class 11 Maths Chapter 13 Exercise 13.4 cost?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>You can download it 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 13.4: You can download the Free PDF of RD Sharma Solutions Class 11 Maths without any hassle from this blog. Subject matter experts have made the solutions easy to understand. To know more about the RD Sharma Solutions Class 11 Maths Chapter 13 Exercise 13.4, read &#8230; <a title=\"RD Sharma Class 11 Solutions Chapter 13 Exercise 13.4 (Updated for 2024)\" class=\"read-more\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-13-exercise-13-4\/\" aria-label=\"More on RD Sharma Class 11 Solutions Chapter 13 Exercise 13.4 (Updated for 2024)\">Read more<\/a><\/p>\n","protected":false},"author":243,"featured_media":122803,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"fifu_image_url":"","fifu_image_alt":""},"categories":[73411,2985,73410],"tags":[3428,73334,73564],"amp_enabled":true,"_links":{"self":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/67951"}],"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=67951"}],"version-history":[{"count":5,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/67951\/revisions"}],"predecessor-version":[{"id":505762,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/67951\/revisions\/505762"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media\/122803"}],"wp:attachment":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media?parent=67951"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/categories?post=67951"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/tags?post=67951"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}