{"id":21398,"date":"2022-08-18T16:06:44","date_gmt":"2022-08-18T10:36:44","guid":{"rendered":"https:\/\/apboardsolutions.com\/?p=21398"},"modified":"2023-03-30T14:11:59","modified_gmt":"2023-03-30T08:41:59","slug":"inter-2nd-year-maths-2a-complex-numbers-solutions-ex-1d","status":"publish","type":"post","link":"https:\/\/apboardsolutions.com\/inter-2nd-year-maths-2a-complex-numbers-solutions-ex-1d\/","title":{"rendered":"Inter 2nd Year Maths 2A Complex Numbers Solutions Ex 1(d)"},"content":{"rendered":"

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

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

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

Question 1.
\n(i) Find the equation of the perpendicular bisector of the line segment joining the points 7 + 7i, 7 – 7i in the Argand diagram.
\nSolution:
\nA(7, 7); B(7, -7) represents the given complex numbers in the Argand diagram.
\n\"Inter
\nO is the mid-point of AB.
\nCo-ordinates of O are \\(\\left(\\frac{7+7}{2}, \\frac{7-7}{2}\\right)\\) = (7, 0)
\nSlope of \\(\\overleftrightarrow{\\mathrm{AB}}=\\frac{7+7}{7-7}=\\frac{14}{0}\\) = \u221e
\nAB is parallel to Y-axis
\nPQ is perpendicular to AB
\nPQ is parallel to X-axis
\nSlope of PQ = 0
\nEquation of PQ is y – 0 = 0(x – 7)
\ni.e., y = 0<\/p>\n

(ii) Find the equation of the straight line joining the point -9 + 6i, 11 – 4i in the Argand plane.
\nSolution:
\nGiven points are -9 + 6i, 11 – 4i
\nLet A = (-9, 6); B = (11, -4)
\nEquation of the straight line AB is y – 6 = \\(\\frac{-4-6}{11+9}\\) (x + 9)
\n\u21d2 y – 6 = \\(\\frac{-1}{2}\\) (x + 9)
\n\u21d2 2y – 12 = -x – 9
\n\u21d2 x + 2y – 3 = 0<\/p>\n

\"Inter<\/p>\n

Question 2.
\nIf Z = x + iy and if the point P in the Argand plane represents Z, then describe geometrically the locus of z satisfying the equation.
\n(i) |z – 2 – 3i| = 5
\n(ii) 2|z – 2| = |z – 1|
\n(iii) Img z2<\/sup> = 4
\n(iv) \\({Arg}\\left(\\frac{z-1}{z+1}\\right)=\\frac{\\pi}{4}\\)
\nSolution:
\n(i) z = x + iy and |z – 2 – 3i| = 5
\n|z – 2 – 3i| = 5
\n\u21d2 |x + iy – 2 – 3i| = 5
\n\u21d2 |(x – 2) + i(y – 3)| = 5
\n\u21d2 \\(\\sqrt{(x-2)^{2}+(y-3)^{2}}\\) = 5
\n\u21d2 (x – 2)2<\/sup> + (y – 3)2<\/sup> = 25
\n\u21d2 x2<\/sup> – 4x + 4 + y2<\/sup> – 6y + 9 = 25
\n\u2234 Locus of P is x2<\/sup> + y2<\/sup> – 4x – 6y – 12 = 0<\/p>\n

(ii) 2|z – 2| = |z – 1|
\n2|x + iy – 2| = |x + iy – 1|
\n\u21d2 2|(x – 2) + iy| = |(x – 1) + iy|
\n\u21d2 \\(2 \\sqrt{(x-2)^{2}+y^{2}}=\\sqrt{(x-1)^{2}+y^{2}}\\)
\nSquaring both sides
\n\u21d2 4[(x – 2)2<\/sup> + y2<\/sup>] = (x – 1)2<\/sup> + y2<\/sup>
\n\u21d2 4(x2<\/sup> – 4x + 4 + y2<\/sup>) = x2<\/sup> – 2x + 1 + y2<\/sup>
\n\u21d2 4x2<\/sup> + 4y2<\/sup> – 16x + 16 = x2<\/sup> + y2<\/sup> – 2x + 1
\n\u2234 Locus of P is 3x2<\/sup> + 3y2<\/sup> – 14x + 15 = 0<\/p>\n

(iii) Img z2<\/sup> = 4
\n\u2235 z = x + iy
\n\u21d2 z2<\/sup> = (x + iy)2<\/sup>
\n\u21d2 z2<\/sup> = x2<\/sup> + i2<\/sup>y2<\/sup> + 2ixy
\n\u21d2 z2<\/sup> = (x2<\/sup> – y2<\/sup>) + i(2xy)
\n\u2234 Img (z2<\/sup>) = 2xy = 4
\n\u2234 The locus of P is xy = 2<\/p>\n

(iv) \\({Arg}\\left(\\frac{z-1}{z+1}\\right)=\\frac{\\pi}{4}\\)
\n\"Inter
\nSince \\({Arg}\\left(\\frac{z-1}{z+1}\\right)=\\frac{\\pi}{4}\\)
\n\u2234 \\(\\frac{2 y}{x^{2}+y^{2}-1}=\\tan \\frac{\\pi}{4}\\)
\n\u21d2 2y = x2<\/sup> + y2<\/sup> – 1
\n\u21d2 x2<\/sup> + y2<\/sup> – 2y – 1 = 0
\n\u2234 The locus of z is x2<\/sup> + y2<\/sup> – 2y – 1 = 0.<\/p>\n

\"Inter<\/p>\n

Question 3.
\nShow that the points in the Argand diagram represented by the complex numbers 2 + 2i, -2 – 2i, 2\u221a3 + 2\u221a3 i are the vertices of an equilateral triangle.
\nSolution:
\nA(2, 2), B(-2, -2), C(-2\u221a3, 2\u221a3) represent the given complex number in the Argand diagram.
\n\"Inter
\nAB2<\/sup> = (2 + 2)2<\/sup> + (2 + 2)2<\/sup>
\n= 16 + 16
\n= 32
\nBC2<\/sup> = (-2 + 2\u221a3)2<\/sup> + (-2 – 2\u221a3)2<\/sup>
\n= 4 + 12 – 8\u221a3 + 4 + 12 + 8\u221a3
\n= 32
\nAC2<\/sup> = (-2\u221a3 – 2)2<\/sup> + (2\u221a3 – 2)2<\/sup>
\n= 12 + 4 + 8\u221a3 + 12 + 4 – 8\u221a3
\n= 32
\nAB2<\/sup> = BC2<\/sup> = AC2<\/sup>
\n\u21d2 AB = BC = CA
\n\u2234 \u2206ABC is an Equilateral triangle.<\/p>\n

\"Inter<\/p>\n

Question 4.
\nFind the eccentricity of the ellipse whose equation is |z – 4| + |z – \\(\\frac{12}{5}\\)| = 10
\nSolution:
\nSP + S’P = 2a
\nS = (4, 0), S’ = (\\(\\frac{12}{5}\\), 0)
\n2a = 10 \u21d2 a = 5
\nSS’ = 2ae
\n\u21d2 4 – \\(\\frac{12}{5}\\) = 2 \u00d7 5e
\n\u21d2 \\(\\frac{8}{5}\\) = 10e
\n\u21d2 e = \\(\\frac{4}{25}\\)<\/p>\n

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

Question 1.
\nIf \\(\\frac{z_{3}-z_{1}}{z_{2}-z_{1}}\\) is a real number, show that the points represented by the complex numbers z1<\/sub>, z2<\/sub>, z3<\/sub> are collinear.
\nSolution:
\nLet z1<\/sub> = x1<\/sub> + iy1<\/sub>; z2<\/sub> = x2<\/sub> + iy2<\/sub>; z3<\/sub> = x3<\/sub> + iy3<\/sub>
\n\"Inter
\nGiven \\(\\frac{z_{3}-z_{1}}{z_{2}-z_{1}}\\) is a real number.
\nImaginary part = 0
\n\u21d2 (y3<\/sub> – y1<\/sub>) (x2<\/sub> – x1<\/sub>) – (x3<\/sub> – x1<\/sub>) (y2<\/sub> – y1<\/sub>) = 0
\n\u21d2 (y3<\/sub> – y1<\/sub>) (x2<\/sub> – x1<\/sub>) = (x3<\/sub> – x1<\/sub>) (y2<\/sub> – y1<\/sub>)
\n\u21d2 \\(\\frac{y_{3}-y_{1}}{x_{3}-x_{1}}=\\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\\)
\nThe points A(x1<\/sub>, y1<\/sub>), B(x2<\/sub>, y2<\/sub>), C(x3<\/sub>, y3<\/sub>) represents the complex numbers z1<\/sub>, z2<\/sub>, z3<\/sub> respectively.
\nSlope of \\(\\stackrel{\\leftrightarrow}{\\mathrm{AC}}\\) = Slope of \\(\\stackrel{\\leftrightarrow}{\\mathrm{AB}}\\)
\n\u2234 A, B, C are collinear.<\/p>\n

Question 2.
\nShow that the points in the Argand plane represented by the complex numbers 2 + i, 4 + 3i, 2 + 5i, 3i are the vertices of a square.
\nSolution:
\nA(2, 1), B(4, 3) C(2, 5), D(0, 3) represent the given complex number in the Argand plane
\n\"Inter
\nAB2<\/sup> = (2 – 4)2<\/sup> + (1 – 3)2<\/sup> = 4 + 4 = 8
\nBC2<\/sup> = (4 – 2)2<\/sup> + (3 – 5)2<\/sup> = 4 + 4 = 8
\nCD2<\/sup> = (2 – 0)2<\/sup> + (5 – 3)2<\/sup> = 4 + 4 = 8
\nDA2<\/sup> = (0 – 2)2<\/sup> + (3 – 1)2<\/sup> = 4 + 4 = 8
\nAB2<\/sup> = BC2<\/sup> = CD2<\/sup> = DA2<\/sup>
\n\u21d2 AB = BC = CD = DA ………(1)
\nAC2<\/sup> = (2 – 2)2<\/sup> + (1 – 5)2<\/sup> = 0 + 16 = 16
\nBD2<\/sup> = (4 – 0)2<\/sup> + (3 – 3)2<\/sup> = 16 + 0 = 16
\nAC2<\/sup> = BD2<\/sup>
\n\u21d2 AC = BD …….(2)
\nBy (1), (2)
\nA, B, C, D are the vertices of a square.<\/p>\n

\"Inter<\/p>\n

Question 3.
\nShow that the points in the Argand plane represented by the complex numbers -2 + 7i, \\(-\\frac{3}{2}+\\frac{1}{2} i\\), 4 – 3i, \\(\\frac{7}{2}\\)(1 + i) are the vertices of a rhombus.
\nSolution:
\nA(-2, 7), B(\\(-\\frac{3}{2}\\), \\(\\frac{1}{2}\\)), C(4, -3), D(\\(\\frac{7}{2}\\), \\(\\frac{7}{2}\\)) represents the given complex numbers in the Argand diagram.
\n\"Inter
\n\u2234 AB2<\/sup> = BC2<\/sup> = CD2<\/sup> = DA2<\/sup>
\n\u21d2 AB = BC = CD = DA ……….(1)
\nAC2<\/sup> = (-2 – 4)2<\/sup> + (7 + 3)2<\/sup>
\n= 36 + 100
\n= 136
\nBD2<\/sup> = \\(\\left(-\\frac{3}{2}-\\frac{7}{2}\\right)^{2}+\\left(\\frac{1}{2}-\\frac{7}{2}\\right)^{2}\\)
\n= 25 + 9
\n= 34
\nAC \u2260 BD ……..(2)
\n\u2234 A, B, C, D are the vertices of a Rhombus.<\/p>\n

Question 4.
\nShow that the points in the Argand diagram represented by the complex numbers z1<\/sub>, z2<\/sub>, z3<\/sub> are collinear, if and only if there exist three real numbers p, q, r not all zero, satisfying pz1<\/sub> + qz2<\/sub> + rz3<\/sub> = 0 and p + q + r = 0.
\nSolution:
\npz1<\/sub> + qz2<\/sub> + rz3<\/sub> = 0
\n\u21d4 rz3<\/sub> = -pz1<\/sub> – qz2<\/sub>
\n\u21d4 z3<\/sub> = \\(\\frac{-p z_{1}-q z_{2}}{r}\\) [\u2235 r \u2260 0]
\n\u2235 p + q + r = 0
\n\u21d4 r = -p – q
\n\u21d4 z3<\/sub> = \\(-\\frac{\\left(p z_{1}+q z_{2}\\right)}{-(p+q)}\\)
\n\u21d4 z3<\/sub> = \\(\\frac{p z_{1}+q z_{2}}{p+q}\\)
\n\u21d4 z3<\/sub> divides the line segment joining z1<\/sub>, z2<\/sub> in the ratio q : p
\n\u21d4 z1<\/sub>, z2<\/sub>, z3<\/sub> are collinear<\/p>\n

\"Inter<\/p>\n

Question 5.
\nThe points P, Q denotes the complex numbers z1<\/sub>, z2<\/sub> in the Argand diagram. O is the origin. If \\(\\bar{z}_{1} \\bar{z}_{2}+\\bar{z}_{2} \\bar{z}_{1}=0\\), show that POQ = 90\u00b0.
\nSolution:
\nLet z1<\/sub> = x1<\/sub> + iy1<\/sub> and z2<\/sub> = x2<\/sub> + iy2<\/sub>
\nThen P(x1<\/sub>, y1<\/sub>), Q(x2<\/sub>, y2<\/sub>), O(0, 0)
\n\\(\\overline{\\mathbf{z}}_{1}\\) = x1 – iy1, \\(\\overline{\\mathbf{z}}_{2}\\) = x2<\/sub> – iy2<\/sub>
\n\\(\\bar{z}_{1} \\bar{z}_{2}+\\bar{z}_{2} \\bar{z}_{1}\\) = (x1<\/sub> + iy1<\/sub>) (x2<\/sub> – iy2<\/sub>) + (x2<\/sub> + iy2<\/sub>) (x1<\/sub> – iy1<\/sub>) = 0
\n\u21d2 x1<\/sub>x2<\/sub> + y1<\/sub>y2<\/sub> – ix1<\/sub>y2<\/sub> + ix2<\/sub>y1<\/sub> + x1<\/sub>x2<\/sub> + y1<\/sub>y2<\/sub> – ix2<\/sub>y1<\/sub> + ix1<\/sub>y2<\/sub> = 0
\n\u21d2 2(x1<\/sub>x2<\/sub> + y1<\/sub>y2<\/sub>) = 0
\n\u21d2 x1<\/sub>x2<\/sub> + y1<\/sub>y2<\/sub> = 0
\n\u21d2 x1<\/sub>x2<\/sub> = -y1<\/sub>y2<\/sub>
\n\u21d2 \\(\\left(\\frac{-x_{1}}{y_{1}}\\right)\\left(-\\frac{x_{2}}{y_{2}}\\right)=-1\\)
\nSlope of OP \u00d7 Slope of OQ = -1
\n\u2234 OP, OQ are perpendicular
\n\u21d2 \u2220POQ = 90\u00b0<\/p>\n

Question 6.
\nThe complex number z has argument \u03b8, 0 < \u03b8 < \\(\\frac{\\pi}{2}\\) and satisfy the equation |z – 3i| = 3. Then prove that (cot \u03b8 – \\(\\frac{6}{z}\\)) = i
\nSolution:
\nLet z = cos \u03b8 + i sin \u03b8
\nGiven |z – 3i| = 3.
\n\u21d2 |(cos \u03b8 + i sin \u03b8) – 3i| = 3
\n\u21d2 |cos \u03b8 + i(sin \u03b8 – 3)| = 3
\n\u21d2 \\(\\sqrt{\\cos ^{2} \\theta+(\\sin \\theta-3)^{2}}\\) = 3
\n\u21d2 cos2<\/sup>\u03b8 + sin2<\/sup>\u03b8 – 6 sin \u03b8 + 9 = 9
\n\u21d2 1 – 6 sin \u03b8 = 0
\n\u21d2 6 sin \u03b8 = 1
\n\u21d2 sin \u03b8 = \\(\\frac{1}{6}\\)
\nsince 0 < \u03b8 < \\(\\frac{\\pi}{2}\\)
\n\"Inter
\n\u2234 cos \u03b8 = \\(\\frac{\\sqrt{35}}{6}\\)
\ncot \u03b8 = \u221a35
\n\u2234 cot \u03b8 – \\(\\frac{6}{z}\\) = cot \u03b8 – \\(\\frac{6}{\\cos \\theta+i \\sin \\theta}\\)
\n= cot \u03b8 – 6(cos \u03b8 – i sin \u03b8)
\n= \u221a35 – 6\\(\\left[\\frac{\\sqrt{35}}{6}-i \\cdot \\frac{1}{6}\\right]\\)
\n= \u221a35 – \u221a35 + i
\n= i
\nHence cot \u03b8 – \\(\\frac{6}{z}\\) = i<\/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(d) will help students to clear their doubts quickly. Intermediate 2nd Year Maths 2A Complex Numbers Solutions Exercise 1(d) I. Question 1. (i) Find the equation of the perpendicular bisector of the line segment joining the points … Read more<\/a><\/p>\n","protected":false},"author":5,"featured_media":39584,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[14],"tags":[],"jetpack_featured_media_url":"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/08\/Complex-Numbers-Solutions-Ex-1d.png","_links":{"self":[{"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/posts\/21398"}],"collection":[{"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/comments?post=21398"}],"version-history":[{"count":2,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/posts\/21398\/revisions"}],"predecessor-version":[{"id":21411,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/posts\/21398\/revisions\/21411"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/media\/39584"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/media?parent=21398"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/categories?post=21398"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/tags?post=21398"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}