{"id":35160,"date":"2022-08-25T11:00:44","date_gmt":"2022-08-25T05:30:44","guid":{"rendered":"https:\/\/apboardsolutions.guru\/?p=35160"},"modified":"2022-08-25T11:00:44","modified_gmt":"2022-08-25T05:30:44","slug":"inter-2nd-year-maths-2a-complex-numbers-solutions-ex-1a","status":"publish","type":"post","link":"https:\/\/apboardsolutions.guru\/inter-2nd-year-maths-2a-complex-numbers-solutions-ex-1a\/","title":{"rendered":"Inter 2nd Year Maths 2A Complex Numbers Solutions Ex 1(a)"},"content":{"rendered":"

Practicing the Intermediate 2nd Year Maths 2A Textbook Solutions<\/a> Inter 2nd Year Maths 2A Complex Numbers Solutions Exercise 1(a) will help students to clear their doubts quickly.<\/p>\n

Intermediate 2nd Year Maths 2A Complex Numbers Solutions Exercise 1(a)<\/h2>\n

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

Question 1.
\nIf z1<\/sub> = (2, -1), z2<\/sub> = (6, 3), find z1<\/sub> – z2<\/sub>.
\nSolution:
\nz1<\/sub> = (2, -1), z2<\/sub> = (6, 3)
\n\u2234 z1<\/sub> – z2<\/sub> = (2 – 6, -1 – 3) = (-4, -4)<\/p>\n

Question 2.
\nIf z1<\/sub> = (3, 5) and z2<\/sub> = (2, 6), find z1<\/sub> . z2<\/sub>
\nSolution:
\nGiven z1<\/sub> = (3, 5) = 3 + 5i
\nand z2<\/sub> = (2, 6) = 2 + 6i
\nz1<\/sub> . z2<\/sub> = (3 + 5i) . (2 + 6i)
\n= 6 + 10i + 18i + 30i2<\/sup>
\n= 6 + 28i + 30(-1) [since i2<\/sup> = -1]
\n= -24 + 28i
\n= (-24, 28)<\/p>\n

\"Inter<\/p>\n

Question 3.
\nWrite the additive inverse of the following complex numbers.
\n(i) (\u221a3, 5)
\n(ii) (-6, 5) + (10, -4)
\n(iii) (2, 1) (-4, 6)
\nSolution:
\nThe additive inverse of (a, b) is (-a, -b)
\n(i) The additive inverse of (\u221a3, 5) is (-\u221a3, -5)
\n(ii) (-6, 5) + (10, -4)
\n= (-6 + 10, 5 + (-4))
\n= (4, 1)
\n\u2234 The additive inverse of (4, 1) is (-4, -1)
\n(iii) (2, 1) . (-4, 6)
\n= ((2 \u00d7 -4 – 1 \u00d7 6), (1 \u00d7 -4 + 2 \u00d7 6))
\n= (-8 – 6, -4 + 12)
\n= (-14, 8)
\n\u2234 The additive inverse of (-14, 8) is (14, -8)<\/p>\n

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

Question 1.
\nIf z1<\/sub> = (6, 3); z2<\/sub> = (2, -1), find z1<\/sub>\/z2<\/sub>.
\nSolution:
\nGiven z1<\/sub> = (6, 3) = 6 + 3i
\nand z2<\/sub> = (2, -1) = 2 – i<\/p>\n

Question 2.
\nIf z = (cos \u03b8, sin \u03b8), find (z – \\(\\frac{1}{z}\\))
\nSolution:
\nGiven z = (cos \u03b8, sin \u03b8) = cos \u03b8 + i sin \u03b8
\n\u21d2 \\(\\frac{1}{z}\\) = cos \u03b8 – i sin \u03b8
\n\u2234 z – \\(\\frac{1}{z}\\) = (cos \u03b8 + i sin \u03b8) – (cos \u03b8 – i sin \u03b8)
\n= 2 i sin \u03b8
\n= 0 + i (2 sin \u03b8)
\n= (0, 2 sin \u03b8)<\/p>\n

\"Inter<\/p>\n

Question 3.
\nWrite the multiplicative inverse of the following complex numbers.
\n(i) (3, 4)
\n(ii) (sin \u03b8, cos \u03b8)
\n(iii) (7, 24)
\n(iv) (-2, 1)
\nSolution:
\nThe multiplicative inverse of the complex number (a, b) is \\(\\left(\\frac{a}{a^{2}+b^{2}}, \\frac{-b}{a^{2}+b^{2}}\\right)\\)
\n(i) Multiplicative inverse of (3, 4) = \\(\\left(\\frac{3}{3^{2}+4^{2}} \\cdot \\frac{-4}{3^{2}+4^{2}}\\right)\\) = \\(\\left(\\frac{3}{25}, \\frac{-4}{25}\\right)\\)
\n(ii) Multiplicative inverse of (sin \u03b8, cos \u03b8) = \\(\\left(\\frac{\\sin \\theta}{\\sin ^{2} \\theta+\\cos ^{2} \\theta}, \\frac{-\\cos \\theta}{\\sin ^{2} \\theta+\\cos ^{2} \\theta}\\right)\\) = (sin \u03b8, -cos \u03b8)
\n(iii) Multiplicative inverse of (7, 24) = \\(\\left(\\frac{7}{7^{2}+24^{2}}, \\frac{-24}{7^{2}+24^{2}}\\right)\\) = \\(\\left(\\frac{7}{625}, \\frac{-24}{625}\\right)\\)
\n(iv) Multiplicative inverse of (-2, 1) = \\(\\left(\\frac{-2}{(-2)^{2}+(1)^{2}}, \\frac{-1}{(-2)^{2}+(1)^{2}}\\right)\\) = \\(\\left(-\\frac{2}{5},-\\frac{1}{5}\\right)\\)<\/p>\n","protected":false},"excerpt":{"rendered":"

Practicing the Intermediate 2nd Year Maths 2A Textbook Solutions Inter 2nd Year Maths 2A Complex Numbers Solutions Exercise 1(a) will help students to clear their doubts quickly. Intermediate 2nd Year Maths 2A Complex Numbers Solutions Exercise 1(a) I. Question 1. If z1 = (2, -1), z2 = (6, 3), find z1 – z2. Solution: z1 …<\/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\/35160"}],"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=35160"}],"version-history":[{"count":1,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/35160\/revisions"}],"predecessor-version":[{"id":35162,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/35160\/revisions\/35162"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/media?parent=35160"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/categories?post=35160"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/tags?post=35160"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}