{"id":11344,"date":"2024-03-08T01:07:44","date_gmt":"2024-03-07T19:37:44","guid":{"rendered":"https:\/\/tsboardsolutions.in\/?p=11344"},"modified":"2024-03-11T17:53:56","modified_gmt":"2024-03-11T12:23:56","slug":"ts-inter-2nd-year-maths-2a-complex-numbers-important-questions","status":"publish","type":"post","link":"https:\/\/tsboardsolutions.in\/ts-inter-2nd-year-maths-2a-complex-numbers-important-questions\/","title":{"rendered":"TS Inter 2nd Year Maths 2A Complex Numbers Important Questions"},"content":{"rendered":"

Students must practice these\u00a0TS Inter 2nd Year Maths 2A Important Questions<\/a> Chapter 1 Complex Numbers to help strengthen their preparations for exams.<\/p>\n

TS Inter 2nd Year Maths 2A Complex Numbers Important Questions<\/h2>\n

Question 1.
\nExpress \\(\\frac{4+2 i}{1-2 i}+\\frac{3 r 4 i}{2+3 i}\\) in the form a + bi, a \u2208 R, b \u2208 R.
\nSolution:
\n\"TS<\/p>\n

\"TS<\/p>\n

Question 2.
\nFind the real and Imaginary parts of the complex number \\(\\frac{a+i b}{a-i b}\\)
\nSolution:
\n\"TS<\/p>\n

Question 3.
\nExpress (1 – i)3<\/sup>(1+i) in the of on a+ib.
\nSolution:
\n(1 – i)3 <\/sup>(1 + i) = (1 -i)2<\/sup> (1 – 1) (1 + 1)
\n– (1 +i2 <\/sup>– 2i)(12<\/sup>– i2<\/sup>)
\n(1 – 1 – 2i) (1 +1) 2(0 – 2i)
\n= 0\u00a0 – 4i = 0 + (i – 4)<\/p>\n

\"TS<\/p>\n

Question 4.
\nFind the multiplicative Inverse of 7 + 24i.
\nSolution:
\nSince \\((x+i y)\\left(\\frac{x-i y}{x^2+y^2}\\right)=1\\) it follows that the multiplicative inverse of
\n(x+iy) is \\(\\frac{x-i y}{x^2+y^2}\\)
\nHence the multiplicative inverse of 7 + 24i is
\n\\(\\frac{7-24 i}{(7)^2+(24)^2}=\\frac{7-24 i}{49+576}=\\frac{7-241}{625}\\)<\/p>\n

Question 5.
\nDetermine the locus of z, z \u2260 2i, such that Re \\(\\left(\\frac{z-4}{z-2 i}\\right)=0\\)
\nSolution:
\nLet z = x + iy, then
\n\"TS
\n\"TS
\nThe ratio on the R.H.S is zero
\ni.e., x2<\/sup> – 4x + y2<\/sup> – = 0 if and only if (x – 2)2<\/sup> (y – 1)2<\/sup> =5.
\n\u21d4 x,y\u2260(0, 2) and (x – 2)2<\/sup>+(y – 1)2<\/sup>=5
\nHence the locus of the given point representing the complex number is the circle with (2, 1) as centre and \\(\\sqrt{5}\\) units as radius except for the point (0, 2).<\/p>\n

\"TS<\/p>\n

Question 6.
\nIf 4x+i(3x – y) = 3 – 6i where x and y are real numbers, then find the values of x and y.
\nSolution:
\nWe have 4x+i(3x-y)=3+i(-6).
\nEquating the real and imaginary parts in the above equation, we get
\n4x = 3, 3x\u00a0 – y = – 6. Upon solving the simultaneous equations, we get
\nx = 3\/4 and y = 33\/4.<\/p>\n

Question 7.
\nIf z=2 – 3i, then show that z2 <\/sup>– 4z+ 13=0.
\nSolution:
\nz = 2 – 3i \u21d2 z – 2= – 3i = (z -2)2<\/sup>=(-3i)2<\/sup>
\n\u21d2 z2<\/sup> + 4 – 4z = – 9
\n\u21d2 z2<\/sup>– 4z+ 13=0<\/p>\n

Question 8.
\nFind the complex conjugate of (3+4i) (2-3i).
\nSolution:
\nThe given complex number
\n(3+4i) (2-3i) = 6 – 9i + 8i + 12 = 18 – i
\nIts complex conjugate = 18 + i.<\/p>\n

Question 9.
\nShow that \\(z_1=\\frac{2+11 i}{25}, \\quad z_2=\\frac{-2+i}{(1-2 i)^2}\\)
\nSolution:
\n\"TS
\nSince this complex number is the conjugate of \\(\\frac{2+11 i}{25}\\) the given complex numbers z1<\/sub>, z2<\/sub> are conjugate to each other.<\/p>\n

Question 10.
\nFind the square roots of (-5+ 12f).
\nSolution:
\nFrom 1.2.8, we have
\n\"TS<\/p>\n

\"TS<\/p>\n

Question 11.
\nWrite \\(z=-\\sqrt{7}+i \\sqrt{21}\\) in the polar form.
\nSolution:
\n\"TS<\/p>\n

Question 12.
\nExpress – 1 – i in polar form with principal value of the amplitude.
\nSolution:
\nLet\u00a0 – 1\u00a0 – 1 = r (cos \u03b8 + i sin \u03b8).
\nThen – i = rcos\u03b8,- 1 = r sin\u03b8 and tan\u03b8 = 1 …………….. (1)
\n\u2234 r2<\/sup> = 2, i.e., r = \u00b1 \\(\\sqrt{2}\\)
\nSince r is positive, r = \\(\\sqrt{2}\\)
\nSince \u2018\u03b8\u2019 satisfies – \u03c0 \u2264 0 < \u03c0, the value of \u03b8 satisfying the equation (1) is \u03b8\u00a0\\(=\\frac{-3 \\pi}{4}\\)
\n\u2234 \\(-1-i=\\sqrt{2}\\left[\\cos \\left(-\\frac{3 \\pi}{4}\\right)+i \\sin \\left(\\frac{-3 \\pi}{4}\\right)\\right]\\)<\/p>\n

\"TS<\/p>\n

Question 13.
\nIf the amplitude of \\(\\left(\\frac{z-2}{z-6 i}\\right)=\\frac{\\pi}{2}\\),find its locus.
\nSolution:
\n\"TS
\nThe points satisfying (1) and (2) constitute the arc of the circle x2<\/sup> + y2<\/sup> – 2x – Gy = 0 intercepted by the diameter
\n3x + y – 6 = 0 not containing the origin and excluding the points (0, 6) and (2, 0). Hence this arc is the required locus.<\/p>\n

Question 14.
\nShow that the equation of any circle in the complex plane is of the form
\n\\(\\mathbf{z} \\overline{\\mathbf{z}}+\\mathbf{b} \\overline{\\mathbf{z}}+\\overline{\\mathbf{b}} \\mathbf{z}+\\mathrm{c}=\\mathbf{0},(\\mathbf{b} \\in \\mathrm{C} ; c \\in R)\\)
\nSolution:
\nAssume the general form of the equation of a circle in Cartesian coordinates as
\nx2<\/sup>+y2<\/sup>+2gx+2fy+c=0, (g,f \u2208 R) …………………. (1)
\nTo write this equation in the complex variable form.
\nLet (x, y) = z. Then
\n\"TS<\/p>\n

\"TS<\/p>\n

Question 15.
\nShow that the complex numbers z satisfying \\(z^2+\\bar{z}^2=2\\) constitute a hyperbola.
\nSolution:
\nSubstituting z = x + \u00a1y in the given equation \\(z^2+\\bar{z}^2=2\\) we obtain the Cartesian form of the given equation.
\n\u2234 (x+iy)2<\/sup>+(x-iy)2<\/sup>=2
\ni.e., x2<\/sup> – y2<\/sup> + 2 ixy + x2<\/sup> – y2<\/sup> – 2ixy = 2
\nor 2x2<\/sup> + 2(iy)2<\/sup> = 2
\ni.e., x2<\/sup> – y2<\/sup> = 1
\nSince this equation denotes a hyperbola, all the complex numbers satisfying \\(z^2+\\bar{z}^2=2\\)\u00a0constitute the hyperbola x2<\/sup> – y2<\/sup>\u00a0= 1.<\/p>\n

Question 16.
\nShow that the points In the Argand diagram represented by the complex numbers 1 + 3i, 4 – 3i, 5 – 5i are collinear.
\nSolution:
\nLet the three complex numbers be represented in the Argand plane by the points
\nP, Q, R respectively. Then P = (1, 3),Q = (4, – 3) and R = (5, – 5).
\nThe slope of the line segment joining P, Q is \\(\\frac{3+3}{1-4}=\\frac{6}{-3}=-2\\)
\nSimilarly, the slope of the line segment joining Q, R is \\(\\frac{-3+5}{4-5}=\\frac{2}{-1}=-2\\). Since the slope of PQ is the slope of QR, the points P, Q, R are collinear.<\/p>\n

Question 17.
\nFind the equation of the straight line joining the points represented by (-4 + 3i), (2-3i) in the Argand plane.
\nSolution:
\nTake the given points as A= – 4 + 3i = (-4,3)\u00a0 B=2-3i = (2,-3).
\nThen the equation of the straight line\u00a0\\(\\overline{\\mathrm{AB}}\\) is
\ny – 3 = \\(\\frac{3+3}{-4-2}\\) (x +4)
\ni.e, x + y +1 = 0<\/p>\n

Question 18.
\nz=x+iy represents a point in the Argand plane. Find the locus of z such that lzl = 2.
\nSolution:
\nLet z = x + \u00a1y, Then |z|=2 if and only if
\n|x + iy| = 2 if and only if 4x2<\/sup> + y2<\/sup> = 2 if and only if x2<\/sup> + y2<\/sup> = 4.
\nx2<\/sup> + y2<\/sup> = 4 represents a circle with centre at (0,0) and radius 2.
\n\u2234 The locus of |z|=2 is the circle x2<\/sup>+ y2<\/sup> = 4.<\/p>\n

\"TS<\/p>\n

Question 19.
\nThe point P represents a complex number z in the Argand plane. If the amplitude of \\(\\mathrm{z} is \\frac{\\pi}{4}\\), determine the locus of P.
\nSolution:
\nLet z=x+i y
\n\"TS<\/p>\n

Question 20.
\nIf the point P denotes the complex number z=x+iy in the Argand plane and if \\(\\frac{z-i}{z-1}\\) is a purely imaginary number, find the locus of P.
\nSolution:
\nWe note that \\(\\frac{z-1}{z-1}\\) is not defined If z = 1.
\n\"TS
\ni.e., x2<\/sup> + y2<\/sup>– x- y = 0 and (x, y) \u2260 (1, 0).
\n\u2234 The locus of P is the circle
\nx2<\/sup> + y2<\/sup>– x – y = 0 excluding the point (1, 0).<\/p>\n

\"TS<\/p>\n

Question 21.
\nDescribe geometrically the following subsets of C:
\n(i) { z \u2208 C| | z – 1+i | = 1
\n(ii) { z \u2208 C| | z + 1+i| \u2264 3
\nSolution:
\n(i) Let S = {z \u2208 C| | z – 1+i | = 1)
\nIf we write z (x, y), then
\n\"TS
\nHence S is a circle with centre (1, – 1) and radius 1 unit.
\n\"TS
\nHence s\u2019 is the closed circular disc with centre at (0, – 1) and radius 3 units.<\/p>\n","protected":false},"excerpt":{"rendered":"

Students must practice these\u00a0TS Inter 2nd Year Maths 2A Important Questions Chapter 1 Complex Numbers to help strengthen their preparations for exams. TS Inter 2nd Year Maths 2A Complex Numbers Important Questions Question 1. Express in the form a + bi, a \u2208 R, b \u2208 R. Solution: Question 2. Find the real and Imaginary … Read more<\/a><\/p>\n","protected":false},"author":2,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[3],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/11344"}],"collection":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/users\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/comments?post=11344"}],"version-history":[{"count":5,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/11344\/revisions"}],"predecessor-version":[{"id":11514,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/11344\/revisions\/11514"}],"wp:attachment":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/media?parent=11344"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/categories?post=11344"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/tags?post=11344"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}