{"id":6902,"date":"2024-04-09T05:15:45","date_gmt":"2024-04-08T23:45:45","guid":{"rendered":"https:\/\/tsboardsolutions.in\/?p=6902"},"modified":"2024-04-10T17:26:56","modified_gmt":"2024-04-10T11:56:56","slug":"ts-inter-2nd-year-maths-2a-solutions-chapter-1s-ex-1c","status":"publish","type":"post","link":"https:\/\/tsboardsolutions.in\/ts-inter-2nd-year-maths-2a-solutions-chapter-1s-ex-1c\/","title":{"rendered":"TS Inter 2nd Year Maths 2A Solutions Chapter 1 Complex Numbers Ex 1(c)"},"content":{"rendered":"

Students must practice this TS Intermediate Maths 2A Solutions<\/a> Chapter 1 Complex Numbers Ex 1(c) to find a better approach to solving the problems.<\/p>\n

TS Inter 2nd Year Maths 2A Solutions Chapter 1 Complex Numbers Exercise 1(c)<\/h2>\n

I.
\nQuestion 1.
\nExpress the following complex numbers in modulus – amplitude form
\ni) 1 – i
\nii) 1 + i\u221a3
\niii) – \u221a3 + i
\niv) – 1 – i\u221a3
\nSolution:
\ni) (1 – i) = r cos \u03b8 + i r sin \u03b8
\nr cos \u03b8 = 1, r sin \u03b8 = – 1
\nr2<\/sup> (cos2<\/sup> + sin2<\/sup>) = 2
\nr2<\/sup> = 2
\nr = \u00b1 \u221a2
\ntan \u03b8 = – 1
\n\u03b8 = \\(-\\frac{\\pi}{4}\\)
\n\u221a2 [cos (\\(-\\frac{\\pi}{4}\\) ) + i sin (\\(-\\frac{\\pi}{4}\\) )]<\/p>\n

ii) 1 + i\u221a3 = r cos \u03b8 + r i sin \u03b8
\nr cos \u03b8 = 1 r sin \u03b8 = \u221a3
\nr2<\/sup> (cos2<\/sup> \u03b8 + sin2<\/sup> \u03b8) = 1 + 3
\nr2<\/sup> = 4
\nr = \u00b1 2
\ntan \u03b8 = \u221a3
\n\u03b8 = \\(\\frac{\\pi}{3}\\)
\n2 (cos \\(\\frac{\\pi}{3}\\) + i sin \\(\\frac{\\pi}{3}\\)).<\/p>\n

iii) – \u221a3 + i = r cos \u03b8 + r i sin \u03b8
\nr cos \u03b8 = – \u221a3
\nr sin \u03b8 = 1
\nr2<\/sup> (cos2<\/sup> \u03b8 + sin2<\/sup> \u03b8) = 3 + 1
\nr = \u00b1 2
\ntan \u03b8 = \\(\\frac{-1}{\\sqrt{3}}\\)
\n\\(2\\left(\\cos \\left(\\frac{5 \\pi}{6}\\right)+i \\sin \\left(\\frac{5 \\pi}{6}\\right)\\right)\\)
\n\\(2\\left(\\cos \\frac{5 \\pi}{6}+i \\sin \\frac{5 \\pi}{6}\\right)\\)<\/p>\n

iv) – 1 – \u221a3i = r cos \u03b8 + r i sin \u03b8
\nr cos \u03b8 = – 1
\nr sin \u03b8 = – \u221a3
\nr2<\/sup> (cos2<\/sup> \u03b8 + sin2<\/sup> \u03b8) = 4
\nr = \u00b1 2
\ntan \u03b8 = \u221a3
\n\u03b8 = \\(\\frac{2 \\pi}{3}\\)
\nHence 2 (cos \\(\\frac{2 \\pi}{3}\\) + i sin \\(\\frac{2 \\pi}{3}\\)).<\/p>\n

\"TS<\/p>\n

Question 2.
\nSimplify – 2i (3 + i) (2 + 4i) (1 + i) and obtain the modulus of that complex number.
\nSolution:
\nz = – 2i (6 + 12i + 2i – 4) (1 + i)
\n= – 2i (2 + 14i) (1 + i)
\n= – 2i (2 + 2i + 14i – 14)
\n= – 2i (- 12 + 16i)
\n= 24i + 32 = 8 (4 + 3i)
\n| z |2<\/sup> = 64.25
\n| z | = 8 \u00d7 5 = 40.<\/p>\n

Question 3.
\ni) If z \u2260 0 find Arg z + Arg \\(\\overline{\\mathbf{Z}}\\).
\nii) If z1<\/sub> = – 1 and z2<\/sub> = – i then find Arg(z1<\/sub>z2<\/sub>)
\niii) If z1<\/sub> = – 1 and z2<\/sub> = i then find Arg \\(\\left(\\frac{z_1}{z_2}\\right)\\).
\nSolution:
\ni) z = x + iy;
\n\\(\\overline{\\mathbf{Z}}\\) = x – iy
\nArg z = tan-1<\/sup> \\(\\frac{y}{x}\\)
\nArg \\(\\overline{\\mathbf{Z}}\\) = tan-1<\/sup> \\(\\frac{-y}{x}\\)
\nArg z + Arg \\(\\overline{\\mathbf{Z}}\\)
\ntan-1<\/sup> \\(\\frac{y}{x}\\) – tan-1<\/sup> \\(\\frac{-y}{x}\\)
\n0 when Arg z \u2260 n
\n2n when Arg z = n<\/p>\n

ii) z1<\/sub> = – 1 + 0i; z2<\/sub> = – i
\nArg (z1<\/sub>z2<\/sub>) = Arg z1<\/sub> + Arg z2<\/sub>
\n= tan-1<\/sup> \\(\\frac{0}{-1}\\) + tan-1<\/sup> \\(\\frac{(-1)}{0}\\)
\n= \u03c0 – \\(\\frac{\\pi}{2}\\) = \\(\\frac{\\pi}{2}\\)<\/p>\n

iii) z1<\/sub> = – 1; z2<\/sub> = i
\nArg \\(\\left(\\frac{z_1}{z_2}\\right)\\) = Arg z1<\/sub> + Arg z2<\/sub>
\ntan-1<\/sup> \\(\\frac{0}{-1}\\) – tan-1<\/sup> \\(\\frac{1}{0}\\)
\n= \u03c0 – \\(\\frac{\\pi}{2}\\) = \\(\\frac{\\pi}{2}\\).<\/p>\n

\"TS<\/p>\n

Question 4.
\ni) (cos 2\u03b1 + i sin 2\u03b1) (cos 2\u03b2 + i sin 2\u03b2) = cos \u03b8 + i sin \u03b8 then find the value of \u03b8.
\nii) If \u221a3 + i = r (cos \u03b8 + i sin \u03b8) then find the value of \u03b8 in radian measure.
\niu) If x + iy = cis \u03b1 . cis \u03b2 then find the value of x2<\/sup> + y2<\/sup>.
\niv) If \\(\\frac{z_2}{z_1}\\); z1<\/sub> \u2260 0 is an imaginary number then find the value of \\(\\left|\\frac{2 z_1+z_2}{2 z_1-z_2}\\right|\\).
\nv) If (\u221a3 + i)100<\/sup> = 299<\/sup> (a + ib) then show that a2<\/sup> + b2<\/sup> = 4.
\nSolution:
\ni) (cos 2\u03b1 + i sin 2\u03b1) (cos 2\u03b2 + i sin 2\u03b2) = cos \u03b8 + i sin \u03b8
\n(cos 2\u03b1 cos 2\u03b2 – sin 2\u03b1 sin 2\u03b2) + i(sin 2\u03b1 cos 2\u03b2 + sin 2\u03b2 cos 2\u03b1) = cos \u03b8 + i sin \u03b8
\ncos 2(\u03b1 + \u03b2) + i sin 2(\u03b1 + \u03b2) = cos \u03b8 + i sin \u03b8
\n\u03b8 = 2 (\u03b1 + \u03b2)<\/p>\n

ii) \u221a3 + i = r(cos \u03b8 + i sin \u03b8)
\nr cos \u03b8 = \u221a3
\nr sin \u03b8 = 1
\nr2<\/sup> (sin2<\/sup> \u03b8 + cos2<\/sup> \u03b8) = 4
\nr = \u00b1 2
\ntan \u03b8 = \\(\\frac{1}{\\sqrt{3}}\\)
\n\u03b8 = \\(\\frac{\\pi}{6}\\)<\/p>\n

iii) If x + iy = (cos \u03b1 + i sin \u03b1) (cos \u03b2 + i sin \u03b2)
\n(cos \u03b1 cos \u03b2 – sin \u03b1 sin \u03b2) + i(cos \u03b1 sin \u03b2 . sin\u03b1 cos \u03b2)
\nx + iy = cos(\u03b1 + \u03b2) + i sin (\u03b1 + \u03b2)
\nx = cos (\u03b1 + \u03b2)
\ny = sin (\u03b1 + \u03b2)
\nx2<\/sup> + y2<\/sup> = 1.<\/p>\n

iv) \\(\\frac{z_2}{z_1}=k i\\left|\\frac{2+\\frac{z_2}{z_1}}{2-\\frac{z_2}{z_1}}\\right|\\)
\n\\(\\left|\\frac{2+k i}{2-k i}\\right|=\\frac{\\sqrt{4+k^2}}{\\sqrt{4+k^2}}\\) = 1<\/p>\n

v) (\u221a3 + i)100<\/sup> = 299<\/sup> (a + ib)
\n|\u221a3 + i|100<\/sup> = 299<\/sup> |a + ib|
\n(\u221a4)100<\/sup> = 299<\/sup> \\(\\sqrt{a^2+b^2}\\)
\n2100<\/sup> = 299<\/sup> \\(\\sqrt{a^2+b^2}\\)
\n4 = a2<\/sup>+ b2<\/sup><\/p>\n

\"TS<\/p>\n

Question 5.
\ni) If z = x + iy and |z| = 1, then find the locus of z.
\nii) If Ihe amplitude of (z – 1) is \\(\\frac{\\pi}{2}\\) then find the locus of z.
\niii) If the Arg \\(\\overline{\\mathbf{z}}_1\\) and Arg \\(\\overline{\\mathbf{z}}_2\\) are \\(\\frac{\\pi}{5}\\) and \\(\\frac{\\pi}{3}\\) respectively then find Arg z1<\/sub> + Arg z2<\/sub>.
\niv) If z = \\(\\frac{1+2 i}{1-(1-i)^2}\\) then find Arg (z).
\nSolution:
\ni) z = x + iy
\n|z| = \\(\\sqrt{x^2+y^2}\\)
\n1 = x2<\/sup> + y2<\/sup>
\nLocus is circle.<\/p>\n

ii) z – 1 = (x – 1) + iy
\n\\(\\tan ^{-1} \\frac{y}{x-1}=\\frac{\\pi}{2}\\)
\nx – 1 = 0, y \u2260 0 also y > 0.<\/p>\n

iii) Arg \\(\\overline{\\mathrm{z}}_1\\) = \\(\\frac{\\pi}{5}\\)
\nArg z2<\/sub> = \\(\\frac{\\pi}{3}\\)
\nArg \\(\\overline{\\mathrm{z}}_1\\) = – Arg z1<\/sub> = \\(\\frac{-\\pi}{5}\\)
\nArg \\(\\overline{\\mathrm{z}}_1\\) + Arg z2<\/sub> = \\(\\frac{\\pi}{3}-\\frac{\\pi}{5}=\\frac{2 \\pi}{15}\\).<\/p>\n

iv) z = \\(\\frac{1+2 i}{1-(1-i)^2}\\)
\n= \\(\\frac{1+2 i}{1-1+2 i-i^2}\\)
\n= \\(\\frac{1+2 i}{2 i+1}\\) = 1
\nArg z = 0.<\/p>\n

\"TS<\/p>\n

II.
\nQuestion 1.
\nSimplify the following complex nunibers and find their modulus.
\ni) \\(\\frac{(2+4 i)(-1+2 i)}{(-1-i)(3-i)}\\)
\nii) \\(\\frac{(1+i)^3}{(2+i)(1+2 i)}\\)
\nSolution:
\ni) \\(\\frac{(2+4 i)(-1+2 i)}{(-1-i)(3-i)}\\)<\/p>\n

\"TS<\/p>\n

ii) z = \\(\\frac{(1+i)^3}{(2+i)(1+2 i)}\\)<\/p>\n

\"TS<\/p>\n

\"TS<\/p>\n

Question 2.
\ni) If(1 – i) (2 – i) (3 – i) …………. (1 – ni) = x – iy then prove that 2 . 5 . 10 ….. (1 + n2<\/sup>) = x2<\/sup> + y2<\/sup>.
\nii) If the real part of \\(\\frac{z+1}{z+i}\\) is 1,then find the locus of z.
\nlocus of z.
\niii) If |z – 3 + i| = 4 determine the locus of z.
\niv) If |z + ai| = |z – ai| then find the locus of z.
\nSolution:
\ni) (1 – i) (2 – i) (3 – i) ……….. (1 – ni) = x – iy
\nTaking modulus both sides
\n|(1 – i)| |(2 – i)| …………… |1 – ni| = |x – iy|
\n\u221a2 . \u221a5 ………….. \\(\\sqrt{1+n^2}=\\sqrt{x^2+y^2}\\)
\n2 . 5 . ………………… . (1 + n2<\/sup>) = x2<\/sup> + y2<\/sup><\/p>\n

ii) \\(\\frac{z+1}{z+i}\\)<\/p>\n

\"TS<\/p>\n

= k1<\/sub> + k2<\/sub>i
\nHere k1<\/sub> = 1
\nx2<\/sup> + y2<\/sup> + x + y = x2<\/sup> + (y + 1)2<\/sup>
\nx2<\/sup> + y2<\/sup> + x + y – x2<\/sup> + y2<\/sup> + 2y + 1
\nx – y = 1<\/p>\n

iii) |z – 3 + i| = 4
\n|(x – 3) + 1(y + 1)| = 4
\n(x – 3)2<\/sup> +(y + 1)2<\/sup> = 16
\nx2<\/sup> + y2<\/sup> – 6x + 2y + 10 = 16
\nx2<\/sup> + y2<\/sup> – 6x + 2y – 6 = 0<\/p>\n

iv) If |z+ ai| = |z – ai|
\n|x + (y – a)i| = |x + (y – a)i|
\nx2<\/sup> + (y + a)2<\/sup> = x2<\/sup>.(y – a)2<\/sup>
\ny = 0.<\/p>\n

\"TS<\/p>\n

Question 3.
\nlf z = x + iy and if the point P in the Argand plane represents z, then describe geometrically the locus of P satisfyIng the equations
\ni) |2z – 3| = 7
\nii) |z|2<\/sup> = 4 Re (z + 2)
\niii) |z + i|2<\/sup> – |z – i|2<\/sup> = 2
\niv) |z + 4i| + |z – 4| = 10
\nSolution:
\ni) |2z – 1| = 7
\n|2(x) – 3 + 2yi| = 7
\n\\(\\sqrt{(2 x-3)^2+4 y^2}\\) = 7
\n4x2<\/sup> – 12x + 9 + 4y = 49
\n4x2<\/sup> + 4y2<\/sup> – 12x – 40 = 0
\nx2<\/sup> + y2<\/sup> – 3x – 10 = 0.
\nCentre (\\(\\frac{3}{2}\\), 0) radius = \\(\\frac{7}{2}\\).<\/p>\n

ii) |z|2<\/sup> = 4 Re (z + 2)
\nx2<\/sup> + y2<\/sup> = 4 (x + 2)
\nx2<\/sup> + y2<\/sup> – 4x – 8 = 0
\nCircle centre (2, 0),
\nRadius = \u221a12 = 2\u221a3.<\/p>\n

iii) |z + i|2<\/sup> – |z – i|2<\/sup> = 2
\n(x)2<\/sup> + (y + 1)2<\/sup> – x2<\/sup> – (y – 1)2<\/sup> = 2
\n4y = 2
\n2y = 1
\n\u21d2 2y – 1 = 0.
\nLine parallel to x-axis.<\/p>\n

iv) |z + 4i| + |z – 4i| = 10
\n|(x + (y + 4)i)| + |(x + (y – 4)i| = 10<\/p>\n

\"TS<\/p>\n

\\(\\sqrt{x^2+(y+4)^2}+\\sqrt{x^2+(y-4)^2}\\) = 10
\nx2<\/sup> + (y + 4)2<\/sup> = (10 – \\(\\sqrt{x^2+(y-4)^2}\\))2<\/sup>
\nx2<\/sup> + (y + 4)2<\/sup> = 1oo + x2<\/sup> + (y – 4)2<\/sup> – 20\\(\\sqrt{x^2+(y-4)^2}\\)
\nSolving we get
\n25x2<\/sup> + 9y2<\/sup> = 225 is ellipse
\ncentre (0,0),
\neccentricity = e = \\(\\sqrt{\\frac{a^2-b^2}{a^2}}\\)
\n= \\(\\sqrt{\\frac{25-9}{25}}\\)
\ne = \\(\\frac{4}{5}\\).<\/p>\n

\"TS<\/p>\n

Question 4.
\nIf z1<\/sub>, z2<\/sub> are two non-zero complex numbers satisfying
\ni) |z1<\/sub> + z2<\/sub>| = |z1<\/sub>| + |z2<\/sub>| then show that Arg z1<\/sub> – Arg z2<\/sub> = 0.
\nii) If z = x + iy and the point P represents z in the Argand plane and \\(\\left|\\frac{\\mathbf{z}-\\mathbf{a}}{\\mathbf{z}+\\mathbf{a}}\\right|\\) = 1. Re(a) \u2260 0 then find the locus of P.
\nSolution:
\ni) |z1<\/sub> + z2<\/sub>| = |z1<\/sub>| + |z2<\/sub>|
\nSquaring both sides
\n|z1<\/sub> + z2<\/sub>|2<\/sup> = (|z1<\/sub>| + |z2<\/sub>|)2<\/sup>
\n(z1<\/sub> + z2<\/sub>) \\(\\left(\\bar{z}_1+\\bar{z}_2\\right)\\) = |z1<\/sub>|2<\/sup> + |z2<\/sub>|2<\/sup> + 2|z1<\/sub>| |z2<\/sub>|
\nz1<\/sub>\\(\\bar{z}_1\\) + z2<\/sub>\\(\\bar{z}_2\\) + z1<\/sub>\\(\\bar{z}_2\\) + z2<\/sub>\\(\\bar{z}_1\\) = |z1<\/sub>| + |z2<\/sub>| + 2|z1<\/sub>| |z2<\/sub>|
\n(x1<\/sub> + iy1<\/sub>) (x2<\/sub> – iy2<\/sub>) + (x2<\/sub> + iy2<\/sub>) (x1<\/sub> – iy1<\/sub>) = 2
\n2 (x1<\/sub>x2<\/sub> + y1<\/sub>y2<\/sub>) + i (y1<\/sub>x2<\/sub> – x1<\/sub>y2<\/sub> + x1<\/sub>y2<\/sub> – y1<\/sub>x2<\/sub>) = 2 \\(\\sqrt{\\mathrm{x}_1^2+\\mathrm{y}_1^2} \\sqrt{\\mathrm{x}_2^2+\\mathrm{y}_2^2}\\)
\nSquaring on both sides we get
\n(x1<\/sub>x2<\/sub> + y1<\/sub>y2<\/sub>)2<\/sup> = (x1<\/sub>2<\/sup> + y1<\/sub>2<\/sup>) (x2<\/sub>2<\/sup> + y2<\/sub>2<\/sup>)
\n(x1<\/sub>y2<\/sub> – y1<\/sub>x2<\/sub>)2<\/sup> = 0
\n\\(\\frac{\\mathrm{y}_2}{\\mathrm{x}_2}=\\frac{\\mathrm{y}_1}{\\mathrm{x}_1}\\)
\n\u2234 Arg z1<\/sub> – Arg z2<\/sub> = 0.<\/p>\n

ii) |z – a| = |z + a|
\nSquaring on both sides (x – a)2<\/sup> + y2<\/sup> = (x + a)2<\/sup> + y2<\/sup>
\n4xa = 0
\nx = 0
\nParallel to y – axis.<\/p>\n","protected":false},"excerpt":{"rendered":"

Students must practice this TS Intermediate Maths 2A Solutions Chapter 1 Complex Numbers Ex 1(c) to find a better approach to solving the problems. TS Inter 2nd Year Maths 2A Solutions Chapter 1 Complex Numbers Exercise 1(c) I. Question 1. Express the following complex numbers in modulus – amplitude form i) 1 – i ii) … 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\/6902"}],"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=6902"}],"version-history":[{"count":4,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/6902\/revisions"}],"predecessor-version":[{"id":7449,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/6902\/revisions\/7449"}],"wp:attachment":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/media?parent=6902"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/categories?post=6902"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/tags?post=6902"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}