{"id":20765,"date":"2022-08-16T12:20:14","date_gmt":"2022-08-16T06:50:14","guid":{"rendered":"https:\/\/apboardsolutions.com\/?p=20765"},"modified":"2022-08-16T12:35:54","modified_gmt":"2022-08-16T07:05:54","slug":"inter-2nd-year-maths-2a-complex-numbers-important-questions","status":"publish","type":"post","link":"https:\/\/apboardsolutions.com\/inter-2nd-year-maths-2a-complex-numbers-important-questions\/","title":{"rendered":"Inter 2nd Year Maths 2A Complex Numbers Important Questions"},"content":{"rendered":"

Students get through Maths 2A Important Questions<\/a> Inter 2nd Year Maths 2A Complex Numbers Important Questions which are most likely to be asked in the exam.<\/p>\n

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

Question 1.
\nIf z1<\/sub> = -1, z2<\/sub> = i then find Arg \\(\\left(\\frac{z_{1}}{z_{2}}\\right)\\) (AP Mar. 17) (TS Mar.\u2019 16; May \u201811)
\nSolution:
\nZ1<\/sub> = -1 = cos \u03c0 + i sin \u03c0
\n\u21d2 Arg z1<\/sub> = \u03c0
\nz2<\/sub> = i = cos \\(\\frac{\\pi}{2}\\) + i sin \\(\\frac{\\pi}{2}\\)
\n\u21d2 Arg z2<\/sub> = \\(\\frac{\\pi}{2}\\)
\n\u21d2 Arg \\(\\left(\\frac{z_{1}}{z_{2}}\\right)\\) = Arg z1<\/sub> – Arg z2<\/sub> = \u03c0 – \\(\\frac{\\pi}{2}\\).
\n= \\(\\frac{\\pi}{2}\\)<\/p>\n

Question 2.
\nIf z = 2 – 3i, show that z2<\/sup> – 4z + 13 = 0. (Mar. \u201808)
\nSolution:
\n\u2234 z = 2 – 3i
\n\u21d2 z – 2 = -3i
\n\u21d2 (z – 2)2<\/sup> = (-3i)2<\/sup>
\n\u21d2 z2<\/sup> – 4z + 4 = 9i2<\/sup>
\n\u21d2 z2<\/sup> – 4z + 4 = 9(-1)
\n\u21d2 z2<\/sup> – 4z + 13 = 0<\/p>\n

\"Inter<\/p>\n

Question 3.
\nFind the multiplicative inverse of 7 + 24i. (TS Mar. 16)
\nSolution:
\nSince (x + iy)\\(\\left[\\frac{x-i y}{x^{2}+y^{2}}\\right]\\) = 1, it follows that the multiplicative inverse of (x + iy) is \\(\\frac{x-i y}{x^{2}+y^{2}}\\)
\nHence the multiplicative inverse of 7 + 24i is
\n\"Inter<\/p>\n

Question 4.
\nWrite the following complex numbers in the form A + iB. (2 – 3i) (3 + 4i)\u00a0 (AP Mar. ’17)
\nSolution:
\n(2 – 3i) (3 + 4i) = 6 + 8i – 9i – 12i2<\/sup>
\n= 6 – i + 12
\n= 18 – i = 18 + i(-1)<\/p>\n

Question 5.
\nWrite the following complex numbers in the form A + iB. (1 + 2i)3<\/sup> (TS Mar. ’17)
\nSolution:
\n(1 + 2i)3<\/sup> = 1 + 3.i2<\/sup>.2i + 3.1. 4i2<\/sup> + 8i3<\/sup>
\n= 1 + 6i – 12 – 8i
\n= -11 – 2i = (-11) + i(-2)<\/p>\n

Question 6.
\nWrite the conjugate of the following complex number \\(\\frac{5 i}{7+i}\\) (AP Mar. ’15)
\nSolution:
\n\"Inter<\/p>\n

Question 7.
\nFind a square root for the complex number 7 + 24i. (Mar. \u201814)
\nSolution:
\n7 + 24i
\n\"Inter
\n\"Inter<\/p>\n

\"Inter<\/p>\n

Question 8.
\nFind a square root for the complex number 3 + 4i\u00a0 (Mar. ’13)
\nSolution:
\n\"Inter<\/p>\n

Question 9.
\nExpress the following complex numbers in modulus amplitude form. 1 – i (AP Mar. 15)
\nSolution:
\n1 – i
\nLet 1 – i = r (cos \u03b8 + i sin \u03b8)
\nEquating real and imaginary parts
\nr cos \u03b8 = 1
\nr sin \u03b8 = -1
\n\u21d2 \u03b8 lies in IV quadrant .
\nSquaring and adding
\nr2<\/sup> (cos2<\/sup> \u03b8 + sin2<\/sup> \u03b8) = 1 + 1 = 2
\nr2<\/sup> = 2 \u21d2 r = \\(\\sqrt{2}\\)
\ntan \u03b8 = -1
\n\u21d2 \u03b8 = -\u03c0\/4
\n\"Inter<\/p>\n

Question 10.
\nExpress the complex numbers in modulus \u2014 amplitude form 1 + i\\(\\sqrt{3}\\) (TS Mar. ’17)
\nSolution:
\n1 + i\\(\\sqrt{3}\\) = r (cos \u03b8 + i sin \u03b8)
\nEquating real and imaginary parts
\nr cos \u03b8 = 1 —– (1)
\nr sin \u03b8 = \\(\\sqrt{3}\\) —– (2)
\n\u03b8 lies in I quadrant
\nSquaring and adding (1) and (2)
\nr2<\/sup> (cos2<\/sup> \u03b8 – sin2<\/sup> \u03b8) = 1 + 3
\nr2<\/sup> = 4 \u21d2 r = 2
\nDividing (2) by (1)
\n\"Inter<\/p>\n

Question 11.
\nIf the Arg \\(\\overline{\\mathbf{z}}_{1}\\) and Arg \\(z_{2}\\) are \\(\\frac{\\pi}{5}\\) and \\(\\frac{\\pi}{3}\\) respectively, find (Arg z1<\/sub> + Arg z2<\/sub>) (AP Mar. ’16)
\nSolution:
\nArg \\(\\overline{\\mathbf{z}}_{1}\\) = \\(\\frac{\\pi}{5}\\) \u21d2 Arg z1<\/sub> = – Arg z1<\/sub> = – \\(\\frac{\\pi}{5}\\)
\nArg z2<\/sub> = \\(\\frac{\\pi}{3}\\)
\n\u2234 Arg z1<\/sub> + Arg z2<\/sub> = – \\(\\frac{\\pi}{5}\\) + \\(\\frac{\\pi}{3}\\)
\n= \\(\\frac{-3 \\pi+5 \\pi}{15}\\) = \\(\\frac{2 \\pi}{15}\\)<\/p>\n

Question 12.
\nIf |z – 3 + i| = 4 determine the locus of z. (May. ’14)
\nSolution:
\nLet z = x + iy
\nGiven |z – 3 + i| = 4
\n|x + iy – 3 + i| = 4
\n\u21d2 (x – 3) + i(y + 1) = 4
\n\u21d2 \\(\\sqrt{(x-3)^{2}+(y+1)^{2}}\\) = 4
\n\u21d2 (x – 3)2<\/sup> + (y + 1)2<\/sup> = 16
\n\u21d2 x2<\/sup> – 6x + 9 + y2<\/sup> + 2y + 1 = 16
\n\u21d2 x2<\/sup> + y2<\/sup> – 6x + 2y – 6 = 0
\n\u2234 The locus \u00f5f z is x2<\/sup> + y2<\/sup> – 6x + 2y – 6 = 0<\/p>\n

\"Inter<\/p>\n

Question 13.
\nThe points P, Q denote the complex numbers z1<\/sub>, z2<\/sub> in the Argand diagram. O is the origin. If z1<\/sub>z2<\/sub> + z2<\/sub>z1<\/sub> = 0, show that POQ = 90\u00b0. (Mar. \u201807)
\nSolution:
\nLet z1<\/sub> = x1<\/sub> + iy1<\/sub> and z2<\/sub> = x2<\/sub> + iy2<\/sub>
\n\"Inter<\/p>\n

Question 14.
\nFind the real and imaginary parts of the complex number \\(\\frac{a+i b}{a-i b}\\). (TS Mar. 15)
\nSolution:
\n\"Inter<\/p>\n

Question 15.
\nWrite z = –\\(\\sqrt{7}\\) + i\\(\\sqrt{21}\\) in the polar form. (Mar. ’11)
\nSolution:
\n\"Inter
\nSince the given point lies in the second quadrant we look for a solution of tan \u03b8 = –\\(\\sqrt{3}\\) that lies in \\(\\left[\\frac{\\pi}{2}, \\pi\\right]\\), we find that \u03b8 = \\(\\frac{2 \\pi}{3}\\) is such a solution.
\n\"Inter<\/p>\n

\"Inter<\/p>\n

Question 16.
\nz = x + iy and the point P represents z in the Argand plane and \\(\\left|\\frac{z-a}{z+\\bar{a}}\\right|\\) = 1, Re (a) \u2260 0, then find the locus of P. (TS Mar. ’17)
\nSolution:
\nLet z = x + iy and a = \u03b1 + i\u03b2
\n\"Inter
\nLocus of P is x = 0 i.e., Y – axis<\/p>\n

Question 17.
\nIf x + iy = \\(\\frac{1}{1+\\cos \\theta+i \\sin \\theta}\\), show that 4x2<\/sup> – 1 = 0 (AP Mar. ’16, TS Mar. ’17, ’15, ’06 )
\nSolution:
\n\"Inter
\nEquating real parts on both sides, we have
\nx = \\(\\frac{1}{2}\\)
\n2x = 1
\n\u21d2 4x2<\/sup> = 1
\n4x2<\/sup> – 1 = 0<\/p>\n

Question 18.
\nIf (\\(\\sqrt{3}\\) + 1)100<\/sup> = 299<\/sup> (a + ib), then show that a2<\/sup> + b2<\/sup> = 4. (AP Mar. \u201816)
\nSolution:
\n\"Inter
\n\"Inter<\/p>\n

Question 19.
\nShow that the points in the Argand diagram represented by the complex numbers 2 + 2i, -2 – 2i, 2\\(\\sqrt{3}\\) + 2\\(\\sqrt{3}\\)i are the vertices of an equilateral triangle. (Mar \u201807)
\nSolution:
\nA (2, 2), B (-2, -2), C (-2\\(\\sqrt{3}\\), 2\\(\\sqrt{3}\\)) represents the given complex number in the Argand diagram.
\n\"Inter<\/p>\n

Question 20.
\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. (June 04) (TS Mar. ’16; AP Mar.’15 ’05; May ’05)
\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\"Inter
\n\u2234 AB2<\/sup> = BC2<\/sup> = CD2<\/sup> = DA2<\/sup>
\n\u21d2 AB = BC = CD = DA
\nAC2<\/sup> = (-2 – 4)2<\/sup> + (7 + 3)2<\/sup>
\n= 36 +100 = 136
\n(BD)2<\/sup> = (-\\(\\frac{3}{2}\\) – \\(\\frac{7}{2}\\))2<\/sup> + (\\(\\frac{1}{2}\\) – \\(\\frac{7}{2}\\))2<\/sup>
\n= 25 + 9 = 34
\nAC \u2260 BD
\nA, B, C, D are the vertices of a Rhombus.<\/p>\n

Question 21.
\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 exists three real numbers p, q, r not all zero, satisfying pz1<\/sub> + qz2<\/sub> + rz3<\/sub> = 0 and p + q + r = 0. (Mar. \u201807)
\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> = \\(\\frac{p z_{1}+q z_{2}}{p+q}\\)
\n\u21d4 z3<\/sub> divides line segment joining z1<\/sub>, z2<\/sub> in the ratio q : p
\n\u21d4 z1<\/sub>, z2<\/sub>, z3<\/sub> are collinear<\/p>\n

Question 22.
\nIf the amplitude of \\(\\left(\\frac{z-2}{z-6 i}\\right) \\frac{\\pi}{2}\\), find its locus. (Mar. ’06)
\nSolution:
\nLet z = (x + iy)
\n\"Inter
\nHence a = 0 and b > 0
\n\u2234 x(x – 2) + y(y – 6) = 0
\nor x2<\/sup> + y2<\/sup> – 2x – 6y = 0.<\/p>\n

\"Inter<\/p>\n

Question 23.
\nIf x + iy = \\(\\frac{3}{2+\\cos \\theta+i \\sin \\theta}\\) then show that x2<\/sup> + y2<\/sup> = 4x – 3 (TS Mar ’17)
\nSolution:
\n\"Inter
\nEquating real and imaginary parts on both sides, we have
\n\"Inter
\n\"Inter
\n\"Inter<\/p>\n

Question 24.
\nExpress \\(\\frac{4+2 i}{1-2 i}\\) + \\(\\frac{3+4 i}{2+3 i}\\) in the form a + ib, a \u2208 R, b \u2208 R.
\nSolution:
\n\"Inter<\/p>\n

Question 25.
\nFind the real and imaginary parts of the complex number \\(\\frac{a+i b}{a-i b}\\) (TS Mar ’15)
\nSolution:
\n\"Inter<\/p>\n

Question 26.
\nExpress (1 – 3)3<\/sup> (1 + i) in the form of a + ib.
\nSolution:
\n(1 – i)3<\/sup> (1 + j) = (1 – j)2<\/sup> (1 – i) (1 + j)
\n= (1 + i2<\/sup> – 2i) (12<\/sup> – i2<\/sup>)
\n= (1 – 1 – 2i) (1 + 1)
\n= 0 – 4i = 0 + i (-4)<\/p>\n

Question 27.
\nFind the multiplicative inverse of 7 + 24i.\u00a0 (TS. Mar. ’16 )
\nSolution:
\nSince (x + iy)\\(\\left[\\frac{x-i y}{x^{2}+y^{2}}\\right]\\) = 1, it follows that the multiplicative inverse of \"Inter<\/p>\n

Question 28.
\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
\n\"Inter
\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, excluding the point (0, 2).<\/p>\n

Question 29.
\nIf 4x + i (3x – y) = 3 -6i where x and y are real numbers, then find the values of x and y.
\nSolution:
\n\u2235 4x + i(3x – y) = 3 – 6i
\nEquating real and imaginary parts, we get 4x = 3 and 3x – y = -6
\n4x = 3 and 3x – y = -6
\n\u21d2 x = 3\/4 and 3\\(\\left(\\frac{3}{4}\\right)\\) – y = -6
\n\\(\\frac{9}{4}\\) + 6 = y
\n\u21d2 y = \\(\\frac{33}{4}\\)
\n\u2234 x = \\(\\frac{3}{4}\\) and y = \\(\\frac{33}{4}\\)<\/p>\n

\"Inter<\/p>\n

Question 30.
\nIf z = 2 – 3i, show that z2<\/sup> – 4z + 13 = 0. (Mar. ’08)
\nSolution:
\n\u2234 z = 2 – 3i
\n\u21d2 z – 2 = – 3i
\n\u21d2 (z – 2)2<\/sup> = (-3i)2<\/sup>
\n\u21d2 z2<\/sup> – 4z + 4 = 9i2<\/sup>
\n\u21d2 z2<\/sup> – 4z + 13 = 0
\n\u21d2 z2<\/sup> – 4z + 4 = 9<\/p>\n

Question 31.
\nFind the complex conjugate of (3 + 4i) (2 – 3i).
\nSolution:
\nThe given complex number is
\n(3 + 4i) (2 – 3i) = 6 + 8i – 9i – 12i2<\/sup>
\n= 6 – i – 12(-1) = 18 + i
\nIts complex conjugate is 18 + i<\/p>\n

Question 32.
\nShow that z1<\/sub> = \\(\\frac{2+11 i}{25}\\), z2<\/sub> = \\(\\frac{-2+i}{(1-2 i)^{2}}\\), are conjugate to each other.
\nSolution:
\n\"Inter
\nSince, this complex number is the conjugate of \\(\\frac{2+11 i}{25}\\), the two given complex numbers
\nare conjugate to each other.<\/p>\n

Question 33.
\nFind the square root of (-5 + 12i).
\nSolution:
\nWe have \\(\\sqrt{a+i b}\\) =
\n\"Inter
\nIn this example a = -5, b = 12
\n\"Inter<\/p>\n

Question 34.
\nWrite z = –\\(\\sqrt{7}\\) + i\\(\\sqrt{21}\\) in the polar form. (Mar ’11)
\nSolution:
\n\"Inter
\nSince the given point lies in the second quadrant we look for a solution of
\ntan \u03b8 = – \\(\\sqrt{3}\\) that lies in [\\(\\frac{\\pi}{2}\\), \u03c0] we find that \u03b8 = \\(\\frac{2 \\pi}{3}\\) is such a solution.
\n\u2234 –\\(\\sqrt{7}\\) + i\\(\\sqrt{21}\\) = 2\\(\\sqrt{7}\\) cis \\(\\frac{2 \\pi}{3}\\)
\n(or) 2\\(\\sqrt{7}\\)(cos \\(\\frac{2 \\pi}{3}\\) + i sin \\(\\frac{2 \\pi}{3}\\))<\/p>\n

Question 35.
\nExpress -1 – i in polar form with principle value of the amplitude.
\nSolution:
\nLet -1 – i = r (cos \u03b8 + i sin \u03b8), then
\n-1 = r cos \u03b8, -1 = r sin \u03b8, tan \u03b8 = 1 \u2014\u2014\u2014 (1)
\n\u2234 r2<\/sup> = 2
\n\u21d2 r = \u00b1\\(\\sqrt{2}\\)
\nSince \u03b8 is positive, -\u03c0 < \u03b8 < \u03c0, the value \u03b8 satisfying the equation (1) is
\n\u03b8 = -135\u00b0 = \\(\\frac{-3 \\pi}{4}\\)
\n\"Inter<\/p>\n

Question 36.
\nIf the amplitude of \\(\\left(\\frac{z-2}{z-6 i}\\right) \\frac{\\pi}{2}\\), find its locus.\u00a0 (Mar. ’06)
\nSolution:
\n\"Inter
\nBy hypothesis, amplitude of a + ib = \\(\\frac{\\pi}{2}\\)
\nSo \\(\\frac{\\pi}{2}\\) = tan-1<\/sup> \\(\\frac{b}{a}\\)
\nHence a = 0 and b > 0
\n\u2234 x(x – 2) + y(y – 6) = 0
\nor x2<\/sup> + y2<\/sup> – 2x – 6y = 0.<\/p>\n

Question 37.
\nShow that the equation of any circle in the complex plane is of the form z\\(\\overline{\\mathbf{z}}\\) + b\\(\\overline{\\mathbf{z}}\\) + b\\(\\overline{\\mathbf{z}}\\) + c = 0, 1(b \u2208 C, c \u2208 R).
\nSolution:
\nAssume the general form of the equation of a circle in cartesian co-ordinates as
\nx2<\/sup> + y2<\/sup> + 2gx + 2fy + c = 0, (g, f \u2208 R) —— (1)
\nTo write this equation in the complex variable form, let (x, y) = z.
\nThen \\(\\frac{z+\\bar{z}}{2}\\) = x, \\(\\frac{z-\\bar{z}}{2 i}\\)
\n= y = \\(\\frac{-i(z-\\bar{z})}{2}\\)
\n\u2234 x2<\/sup> + y2<\/sup> = |z|2<\/sup> = z\\(\\overline{\\mathbf{z}}\\)
\nSubstituting these results in equation (1), we obtain
\nz\\(\\overline{\\mathbf{z}}\\) + g(z + \\(\\overline{\\mathbf{z}}\\)) + f(z – \\(\\overline{\\mathbf{z}}\\))(-i) + c = 0
\ni.e., z\\(\\overline{\\mathbf{z}}\\) + (g – if)z + (g + if)\\(\\overline{\\mathbf{z}}\\) + c = 0 ——-(2)
\nIf (g + if) = b, then equation (2) can be written as z\\(\\overline{\\mathbf{z}}\\) + \\(\\overline{\\mathbf{b}}\\)z + b\\(\\overline{\\mathbf{z}}\\) + c = 0<\/p>\n

\"Inter<\/p>\n

Question 38.
\nShow that the complex numbers z satisfying z2<\/sup> + \\((\\overline{\\mathbf{z}})^{2}\\) = 2 constitute a hyperbola.
\nSolution:
\nSubstituting z = x + iy in the given equation
\nz2<\/sup> + (\\(\\overline{\\mathbf{z}}\\))2<\/sup> = 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> + 2ixy + x2<\/sup> – y2<\/sup> – 2ixy = 2
\ni.e., x2<\/sup> – y2<\/sup> = 1.
\nSince, this equation denotes a hyperbola, all the complex numbers satisfying
\n\"Inter
\nlie on the hyperbola x2<\/sup> – y2<\/sup> = 1.<\/p>\n

Question 39.
\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 P, Q, R respectively. Then P = (1, 3), Q = (4, -3), R = (5, -5). The 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.
\nSince the slope of PQ is the slope of QR, the points P, Q and R are collinear.<\/p>\n

Question 40.
\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
\nA = -4 + 3i = (-4, 3)
\nB = 2 – 3i = (2, -3)
\nThen equation of the straight line \\(\\overleftrightarrow{\\mathrm{AB}}\\) is
\ny – 3 = \\(\\frac{3+3}{-4-2}\\)(x + 4)
\ni.e., x + y + 1 = 0.<\/p>\n

Question 41.
\nz = x + iy represents a point in the Argand plane, find the locus of z. Such that |z| = 2.
\nSolution:
\n|z| = 2, z = x + iy
\nif \\(\\sqrt{x^{2}+y^{2}}\\) = 2
\nif \\(\\sqrt{x^{2}+y^{2}}\\) = 2
\nif and only if x2<\/sup> + y2<\/sup> = 4
\nThe equation x2<\/sup> + y2<\/sup> = 4 represents the circle with centre at the origin (0, 0) and radius 2 units.
\n\u2234 The locus of |z| = 2 is the circle
\nx2<\/sup> + y2<\/sup> = 4<\/p>\n

Question 42.
\nThe point P represents a complex number z in the Argand plane. If the amplitude of z is \\(\\frac{\\pi}{4}\\), determine the locus of P.
\nSolution:
\nLet z = x + iy.
\nBy hypothesis, amplitude of z = \\(\\frac{\\pi}{4}\\)
\nHence tan-1<\/sup> \\(\\left(\\frac{y}{x}\\right)\\) = \\(\\frac{\\pi}{4}\\) and \\(\\frac{y}{x}\\) = tan \\(\\frac{\\pi}{4}\\)
\nHence x = y
\n\u2234 The locus of P is x = y.<\/p>\n

\"Inter<\/p>\n

Question 43.
\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 the quotient \\(\\frac{z-i}{z-1}\\) is not defined if z = 1.
\n\"Inter
\n\u2234 The locus of P is the circle
\nx2<\/sup> + y2<\/sup> – x – y = 0
\nexcluding the point (1, 0).<\/p>\n

Question 44.
\nDescribe geometrically the following subsets of C.
\ni) {z \u2208 C| |z – 1 + i| = 1}
\nii) {z \u2208 C| |z + i| \u2264 3|
\nSolution:
\ni) Let S = {z \u2208 C| z – 1 + i| = 1}
\nIf we write z = (x, y), then
\nS = {(x, y) \u2208 R2<\/sup>||x + iy – 1 + i| = 1}
\n= {x, y) \u2208 R2<\/sup> || x + i(y – 1)| \u2264 3}
\n= {(x, y) \u2208 R2<\/sup> || (x – 1)2<\/sup> + (y + 1)2<\/sup> = i}
\nHence S is a circle with centre (1, -1) and radius 1 unit.<\/p>\n

ii) Let S’ = {z \u2208 C || z + i| \u2264 3}
\nThen S = {(x, y \u2208 R2<\/sup> || x + iy + i| \u2264 3}
\n= {(x, y) \u2208 R2<\/sup> || x2<\/sup> + i(y + 1) \u2264 3}
\n= {(x, y) \u2208 R2<\/sup> || x2<\/sup> + (y + 1)2<\/sup> \u2264 9}
\nHence S’ is the closed circular disc with centre at (0, -1) and radius 3 units.<\/p>\n","protected":false},"excerpt":{"rendered":"

Students get through Maths 2A Important Questions Inter 2nd Year Maths 2A Complex Numbers Important Questions which are most likely to be asked in the exam. Intermediate 2nd Year Maths 2A Complex Numbers Important Questions Question 1. If z1 = -1, z2 = i then find Arg (AP Mar. 17) (TS Mar.\u2019 16; May \u201811) … Read more<\/a><\/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.com\/wp-json\/wp\/v2\/posts\/20765"}],"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=20765"}],"version-history":[{"count":16,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/posts\/20765\/revisions"}],"predecessor-version":[{"id":20930,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/posts\/20765\/revisions\/20930"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/media?parent=20765"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/categories?post=20765"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/tags?post=20765"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}