{"id":4037,"date":"2023-12-14T04:52:49","date_gmt":"2023-12-13T23:22:49","guid":{"rendered":"https:\/\/wordpress-505192-1602719.cloudwaysapps.com\/?p=4037"},"modified":"2023-12-15T17:29:02","modified_gmt":"2023-12-15T11:59:02","slug":"samacheer-kalvi-12th-maths-solutions-chapter-2-ex-2-3","status":"publish","type":"post","link":"https:\/\/samacheerkalviguru.com\/samacheer-kalvi-12th-maths-solutions-chapter-2-ex-2-3\/","title":{"rendered":"Samacheer Kalvi 12th Maths Solutions Chapter 2 Complex Numbers Ex 2.3"},"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 2.3<\/h2>\n

Question 1.
\nIf z1<\/sub> = 1 – 3i, z2<\/sub> = -4i, and z3<\/sub> = 5, show that
\n(i) (z1<\/sub> + z2<\/sub>) + z3<\/sub> = z1<\/sub> + (z2<\/sub> + z3<\/sub>)
\n(ii) (z1 <\/sub>z2<\/sub>) z3<\/sub> = z1 <\/sub>(z2<\/sub> z3<\/sub>)
\nSolution:
\n(i) z1<\/sub> = 1 – 3i, z2<\/sub> = -4i, z3<\/sub> = 5
\n(z1<\/sub> + z2<\/sub>) + z3<\/sub>
\n= (1 – 3i – 4i) + 5
\n= (1 – 7i) + 5
\n= 6 – 7i …… (1)
\nz1<\/sub> + (z2<\/sub> + z3<\/sub>)
\n= (1 – 3i) + (-4i + 5)
\n= 6 – 7i ……. (2)
\nfrom (1) & (2) we get
\n\u2234 (z1<\/sub> + z2<\/sub>) + z3<\/sub> = z1<\/sub> + (z2<\/sub> + z3<\/sub>)<\/p>\n

\"Samacheer<\/p>\n

(ii) (z1<\/sub> z2<\/sub>) z3<\/sub> = [(1 – 3i) (-4i]5
\n= [-4i + 12 i2<\/sup>]5
\n= [-12 – 4i] 5
\n= -60 – 20i ……. (1)
\nz1<\/sub> (z2<\/sub> z3<\/sub>) = (1 – 3i)[(-4i)(5)]
\n= (-20i + 60 i2<\/sup>)
\n= -20i – 60 ……. (2)
\nfrom (1) & (2) we get
\n\u2234 (z1<\/sub> z2<\/sub>)(z3<\/sub>) = z1 <\/sub>(z2<\/sub> z3<\/sub>)<\/p>\n

Question 2.
\nIf z1<\/sub> = 3, z2<\/sub> = -7i, and z3<\/sub> = 5 + 4i, show that
\n(i) z1<\/sub> (z2<\/sub> + z3<\/sub>) = z1<\/sub> z2<\/sub> + z1<\/sub> z3<\/sub>
\n(ii) (z1<\/sub> z2<\/sub>) z3<\/sub> = z1<\/sub> z3<\/sub> + z2<\/sub> z3<\/sub>
\nSolution:
\n(i) z1<\/sub> = 3, z2<\/sub> = -7i, z3<\/sub> = 5 + 4i
\nz1<\/sub> (z2<\/sub> + z3<\/sub>) = 3 (-7i + 5 + 4i)
\n= 3 (5 – 3i)
\n= 15 – 9i …… (1)
\nz1<\/sub> z2<\/sub> + z1<\/sub> z3<\/sub> = 3 (-7i) + 3 (5 + 4i)
\n= -21i + 15 + 12i
\n= 15 – 9i …… (2)
\nfrom (1) & (2), we get
\n\u2234 z1<\/sub> (z2<\/sub> + z3<\/sub>) = z1<\/sub> z2<\/sub> + z1<\/sub> z3<\/sub><\/p>\n

(ii) (z1<\/sub> + z2<\/sub>) z3<\/sub> = (3 – 7i) (5 + 4i)
\n= 15 + 12i – 35i – 28 i2<\/sup>
\n= 15 – 23i + 28
\n= 43 – 23i ……. (1)
\nz1<\/sub> z3<\/sub> + z2<\/sub> z3<\/sub> = 3(5 + 4i) – 7i(5 + 4i)
\n= 15 + 12i – 35i – 28 i2<\/sup>
\n= 15 – 23i + 28
\n= 43 – 23i ….. (2)
\nfrom (1) & (2), we get
\n\u2234 (z1<\/sub> + z2<\/sub>) z3<\/sub> = z1<\/sub> z3<\/sub> + z2<\/sub> z3<\/sub><\/p>\n

Question 3.
\nIf z1<\/sub> = 2 + 5i, z2<\/sub> = -3 – 4i, and z3<\/sub> = 1 + i, find the additive and multiplicative inverse of z1<\/sub>, z2<\/sub>, and z3<\/sub>.
\nSolution:
\nz1<\/sub> = 2 + 5i
\n(\u0430) Additive inverse of z1<\/sub> = -z1<\/sub> = -(2 + 5i) = -2 – 5i
\n(b) Multiplicative inverse of
\n\\(z_{1}=\\frac{1}{z_{1}}=\\frac{1}{2+5 i} \\times \\frac{2-5 i}{2-5 i}=\\frac{(2-5 i)}{4+25}=\\frac{2-5 i}{29}\\)<\/p>\n

z2<\/sub> = -3 – 4i
\n(a) Additive inverse of z2<\/sub> = -z2<\/sub> = -(-3 – 4i) = 3 + 4i
\n(b) Multiplicative inverse of
\n\\(z_{2}=\\frac{1}{z_{2}} \\frac{1}{-3-4 i} \\times \\frac{-3+4 i}{-3+4 i}=\\frac{-3+4 i}{9+16}=\\frac{-3+4 i}{25}\\)<\/p>\n

z3<\/sub> = 1 + i
\n(\u0430) Additive inverse of z3<\/sub> = -z3<\/sub> = -(1 + i) = -1 – i
\n(b) Multiplicative inverse of
\n\\(z_{3}=\\frac{1}{z_{3}} \\frac{1}{1+i} \\times \\frac{1-i}{1-i}=\\frac{1-i}{2}\\)<\/p>\n

Samacheer Kalvi 12th Maths Solutions Chapter 2 Complex Numbers Ex 2.3 Additional Problems<\/h3>\n

Question 1.
\nIf z1<\/sub> = 4 – 7i, z2<\/sub> = 2 + 3i and z3<\/sub> = 1 + i show that.
\n(i) z1<\/sub> + (z2<\/sub> + z3<\/sub>) = (Z1<\/sub> + z2<\/sub>) + Z3<\/sub>
\n(ii) (Z1<\/sub> z2<\/sub>) z3<\/sub> = Z1<\/sub> (z2<\/sub> z3<\/sub>)
\nSolution:
\n(i) Z1<\/sub> + (z2<\/sub> + z13<\/sub>) = 4 – 7i + (2 + 3i + 1 + i) = 4 – 7i + (3 + 4i) = 7 – 3i
\n(z1<\/sub> + z2<\/sub>) + z3<\/sub> = (4 – 7i + 2 + 3i) + (1 + i) = (6 – 4i) + (1 + i)
\n= 7 – 3i …(2)
\nFrom (1) and (2) we get, z1<\/sub> + (z2<\/sub> + z3<\/sub>) = (z1<\/sub> + z2<\/sub>) + z3<\/sub><\/p>\n

(ii) (z1<\/sub>z2<\/sub>) z3<\/sub> = (4 – 7i) + (2 + 3i) (1 + i) = (8 + 12i – 14 i + 21) (1 + i)
\n= (29 – 2i)(1 + i) = 29 + 29i – 2i + 2
\n= 31 + 27i ….(1)
\nZ1<\/sub> (z2<\/sub>z3<\/sub>) = (4 – 7i) [(2 + 3i) (1 + i)] = 4 – 7i [2 + 2i + 3i – 3]
\n= 4 – 7i[5i – 1] = 20i – 4 + 35 + 7i
\n= 31 + 27i …(2)
\nFrom (1) and (2) we get, (Z1<\/sub>Z2<\/sub>) z3<\/sub> = z1<\/sub> (z2<\/sub>z3<\/sub>)<\/p>\n

\"Samacheer<\/p>\n

Question 2.
\nGiven z1<\/sub> = 1 + i, z2<\/sub> = 4 – 3i and z3<\/sub> = 2 + 5i verify that.
\nZ1<\/sub>(Z2<\/sub> Z3<\/sub>) = Z1<\/sub> Z2<\/sub> – z1<\/sub>z3<\/sub>
\nSolution:
\nZ1<\/sub> = 1 + i,
\nz2<\/sub> = 4 – 3i,
\nz3<\/sub> = 2 + 5i
\nZ1<\/sub> (z2<\/sub> – z3<\/sub>) = 1 + i[(4 – 3i) – (2 + 5i)] = 1 + i[2 – 8i] = 2 – 8i + 2i + 8
\n= 10 – 6i …(1)
\nZ1<\/sub> z2<\/sub> = (1 + i) (4 – 3i) = 4 – 3i + 4i + 3 = 7 + i
\nZ1<\/sub>Z3<\/sub>= (1 + i) (2 + 5i) = 2 + 5i + 2i – 5 = 7i – 3
\nZ1<\/sub>Z2<\/sub> – Z1<\/sub> z3<\/sub> = 7 + i – (7i – 3) = 7 + i – 7i + 3
\n= 10 – 6i …(2)
\nFrom (1) and (2) we get, Z1<\/sub> (z2<\/sub> – z3<\/sub>) = z1<\/sub> z2<\/sub> – z1<\/sub> z3<\/sub><\/p>\n

Question 3.
\nGiven z1<\/sub> = 4 – 7i and z2<\/sub> = 5 + 6i find the additive and multiplicative inverse of z1<\/sub> + z2<\/sub> and Z1<\/sub> – Z2<\/sub>.
\nSolution:
\nZ1<\/sub> = 4 – 7i, z2<\/sub> = 5 + 6i
\nZ1<\/sub> + z2<\/sub> = 4 – 7i + 5 + 6i = 9 + i
\nAdditive inverse of z1<\/sub> + z2<\/sub> = – (z1<\/sub> + z2<\/sub>) = – (9 – i) = i – 9
\n\"Samacheer<\/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 2.3 Question 1. If z1 = 1 – 3i, z2 = -4i, and z3 = 5, …<\/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":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/posts\/4037"}],"collection":[{"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/users\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/comments?post=4037"}],"version-history":[{"count":1,"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/posts\/4037\/revisions"}],"predecessor-version":[{"id":60145,"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/posts\/4037\/revisions\/60145"}],"wp:attachment":[{"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/media?parent=4037"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/categories?post=4037"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/tags?post=4037"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}