{"id":22048,"date":"2024-01-10T15:50:14","date_gmt":"2024-01-10T10:20:14","guid":{"rendered":"https:\/\/apboardsolutions.in\/?p=22048"},"modified":"2024-01-11T12:27:43","modified_gmt":"2024-01-11T06:57:43","slug":"inter-2nd-year-maths-2a-complex-numbers-solutions-ex-1c","status":"publish","type":"post","link":"https:\/\/apboardsolutions.in\/inter-2nd-year-maths-2a-complex-numbers-solutions-ex-1c\/","title":{"rendered":"Inter 2nd Year Maths 2A Complex Numbers Solutions Ex 1(c)"},"content":{"rendered":"

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

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

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

Question 1.
\nExpress the following complex numbers in modulus-amplitude form.
\n(i) 1 – i
\nSolution:
\n1 – i
\nLet 1 – i = r(cos \u03b8 + i sin \u03b8)
\nEquating real and imaginary parts
\nr cos \u03b8 = 1, r sin \u03b8 = -1
\n\u21d2 \u03b8 lies in IV quadrant
\nSquaring and adding
\nr2<\/sup> (cos2<\/sup>\u03b8 + sin2<\/sup>\u03b8) = 1 + 1 = 2
\n\u21d2 r2<\/sup> = 2
\n\u21d2 r = \u221a2
\ntan \u03b8 = – 1
\n\u21d2 \u03b8 = \\(-\\frac{\\pi}{4}\\)
\n\u2234 1 – i = \u221a2(cos(\\(-\\frac{\\pi}{4}\\)) + i sin(\\(-\\frac{\\pi}{4}\\)))<\/p>\n

(ii) 1 + i\u221a3
\nSolution:
\n1 + i\u221a3 = r (cos \u03b8 + i sin \u03b8)
\nEquating real and imaginary parts
\nr cos \u03b8 = 1 ………(1)
\nr sin \u03b8 = \u221a3 ……..(2)
\n\u03b8 lies in the I quadrant
\nSquaring and adding (1) and (2)
\nr2<\/sup> (cos2<\/sup>\u03b8 + sin2<\/sup>\u03b8) = 1 + 3
\n\u21d2 r2<\/sup> = 4
\n\u21d2 r = 2
\nDividing (2) by (1)
\n\\(\\frac{r \\sin \\theta}{r \\cos \\theta}\\) = \u221a3
\n\u21d2 tan \u03b8 = \u221a3
\n\u21d2 \u03b8 = \\(\\frac{\\pi}{3}\\)
\n\u2234 1 + i\u221a3 = 2(cos \\(\\frac{\\pi}{3}\\) + i sin \\(\\frac{\\pi}{3}\\))<\/p>\n

\"Inter<\/p>\n

(iii) -\u221a3 + i
\nSolution:
\n-\u221a3 + i = r(cos \u03b8 + i sin \u03b8)
\nEquating real and imaginary parts
\nr cos \u03b8 = -\u221a3 …….(1)
\nr sin \u03b8 = 1 …….(2)
\n\u21d2 \u03b8 lies in II quadrant
\nSquaring and adding (1), (2)
\nr2<\/sup> (cos2<\/sup>\u03b8 + sin2<\/sup>\u03b8) = 3 + 1 = 4
\n\u21d2 r2<\/sup> = 4
\n\u21d2 r = 2
\nDividing (2) by (1)
\n\\(\\frac{r \\sin \\theta}{r \\cos \\theta}=-\\frac{1}{\\sqrt{3}}\\)
\n\u21d2 tan \u03b8 = \\(-\\frac{1}{\\sqrt{3}}\\) and \u03b8 lies in II quadrant
\n\u21d2 \u03b8 = 180\u00b0 – 30\u00b0 = 150\u00b0 = \\(\\frac{5 \\pi}{6}\\)
\n\u2234 -\u221a3 + i = 2(cos \\(\\frac{5 \\pi}{6}\\) + i sin \\(\\frac{5 \\pi}{6}\\))<\/p>\n

(iv) -1 – i\u221a3
\nSolution:
\n1 – i\u221a3 = r (cos \u03b8 + i sin \u03b8)
\nEquating real and imaginary parts
\nr cos \u03b8 = -1, r sin \u03b8 = -\u221a3
\nSquaring and adding
\nr2<\/sup> (cos2<\/sup>\u03b8 + sin2<\/sup>\u03b8) = 1 + 3
\n\u21d2 r2<\/sup> = 4
\n\u21d2 r = 2
\n\u2234 cos \u03b8 = \\(-\\frac{1}{2}\\), sin \u03b8 = \\(-\\frac{\\sqrt{3}}{2}\\)
\n\u21d2 \u03b8 = -120\u00b0 = \\(-\\frac{2 \\pi}{3}\\)
\n\u2234 -1 – i\u221a3 = 2[cos(\\(-\\frac{2 \\pi}{3}\\)) + i sin(\\(-\\frac{2 \\pi}{3}\\))]<\/p>\n

Question 2.
\nSimplify -2i(3 + i) (2 + 4i) (1 + i) and obtain the modulus of that complex number.
\nSolution:
\n-2i(3 + i) (2 + 4i) (1 + i)
\n= (-6i – 2i2<\/sup>) (2 + 2i + 4i + 4i2<\/sup>)
\n= (2 – 6i) (-2 + 6i)
\n= -4 + 12i – 36i2<\/sup> + 12i
\n= 32 + 24i
\n= 8(4 + 3i)
\nModulus = |8(4 + 3i)|
\n= 8\\(\\sqrt{(4)^{2}+(3)^{2}}\\)
\n= 8(5)
\n= 40<\/p>\n

Question 3.
\n(i) If z \u2260 0, find Arg z + Arg \\((\\bar{z})\\).
\nSolution:
\nIf z = x + iy, then Arg(z) = \\(\\tan ^{-1}\\left(\\frac{y}{x}\\right)\\)
\nand \\((\\bar{z})\\) = x – iy, then Arg (z) = \\(\\tan ^{-1}\\left(\\frac{-y}{x}\\right)\\) = –\\(\\tan ^{-1}\\left(\\frac{y}{x}\\right)\\)
\n\u2234 Arg (z) + Arg \\((\\bar{z})\\) = \\(\\tan ^{-1}\\left(\\frac{y}{x}\\right)\\) + (-\\(\\tan ^{-1}\\left(\\frac{y}{x}\\right)\\)) = 0<\/p>\n

(ii) If z1<\/sub> = -1 and z2<\/sub> = -i, then find Arg(z1<\/sub>z2<\/sub>)
\nSolution:
\nz1<\/sub> = -1
\n\u21d2 z1<\/sub> = cos \u03c0 + i sin \u03c0
\n\u21d2 Arg z1<\/sub> = \u03c0
\nz2<\/sub> = -i
\n\u21d2 z2<\/sub> = \\(\\cos \\left(-\\frac{\\pi}{2}\\right)+i \\sin \\left(-\\frac{\\pi}{2}\\right)\\)
\n\u21d2 Arg z2<\/sub> = \\(-\\frac{\\pi}{2}\\)
\n\u2234 Arg (z1<\/sub>z2<\/sub>) = Arg z1<\/sub> + Arg z2<\/sub>
\n= \u03c0 – \\(\\frac{\\pi}{2}\\)
\n= \\(\\frac{\\pi}{2}\\)<\/p>\n

(iii) If z1<\/sub> = -1, z2<\/sub> = i then find Arg\\(\\left(\\frac{\\mathbf{z}_{1}}{\\mathbf{z}_{2}}\\right)\\)
\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\u2234 Arg\\(\\left(\\frac{\\mathbf{z}_{1}}{\\mathbf{z}_{2}}\\right)\\) = Arg z1<\/sub> – Arg z2<\/sub>
\n= \u03c0 – \\(\\frac{\\pi}{2}\\)
\n= \\(\\frac{\\pi}{2}\\)<\/p>\n

\"Inter<\/p>\n

Question 4.
\n(i) If (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.
\nSolution:
\ncos \u03b8 + i sin \u03b8
\n= (cos 2\u03b1 + i sin 2\u03b1) (cos 2\u03b2 + i sin 2\u03b2)
\n= cos 2\u03b1 . cos 2\u03b2 + i sin 2\u03b1 . cos 2\u03b2 + i cos 2\u03b1 sin 2\u03b2 + i2<\/sup> sin 2\u03b1 . sin 2\u03b2
\n= (cos 2\u03b1 . cos 2\u03b2 – sin 2\u03b1 . sin 2\u03b2) + i(sin 2\u03b1 cos 2\u03b2 + cos 2\u03b1 sin 2\u03b2)
\n= cos 2(\u03b1 + \u03b2) + i sin 2(\u03b1 + \u03b2)
\n\u2234 \u03b8 = 2(\u03b1 + \u03b2)<\/p>\n

(ii) If \u221a3 + i = r (cos \u03b8 + i sin \u03b8), then find the value of \u03b8 in radian measure.
\nSolution:
\nGiven that
\n\u221a3 + i = r(cos \u03b8 + i sin \u03b8)
\n\u21d2 r cos \u03b8 = \u221a3, r sin \u03b8 = 1
\n\u21d2 r2<\/sup> (cos2<\/sup>\u03b8 + sin2<\/sup>\u03b8) = 3 + 1
\n\u21d2 r2<\/sup> = 4
\n\u21d2 r = 2
\n\u2234 cos \u03b8 = \\(\\frac{\\sqrt{3}}{2}\\) and sin \u03b8 = \\(\\frac{1}{2}\\)
\n\u21d2 \u03b8 = \\(\\frac{\\pi}{6}\\)<\/p>\n

(iii) If x + iy = cis \u03b1 . cis \u03b2 then find the value of x2<\/sup> + y2<\/sup>.
\nSolution:
\nGiven x + iy = cis \u03b1 . cis \u03b2
\n\u21d2 x + iy = (cos \u03b1 + i sin \u03b1) (cos \u03b2 + i sin \u03b2)
\n\u21d2 x + iy = cos(\u03b1 + \u03b2) + i sin(\u03b1 + \u03b2)
\nEquating real and imaginary parts
\nx = cos(\u03b1 + \u03b2) and y = sin(\u03b1 + \u03b2)
\n\u2234 x2<\/sup> + y2<\/sup> = cos2<\/sup>(\u03b1 + \u03b2) + sin2<\/sup>(\u03b1 + \u03b2) = 1<\/p>\n

(iv) 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|\\)
\nSolution:
\n\\(\\frac{z_{2}}{z_{1}}\\), (z1<\/sub> \u2260 0) is purely imaginary.
\nWe can suppose that \\(\\frac{z_{2}}{z_{1}}\\) = iy, where y \u2208 R – {0}
\n\"Inter<\/p>\n

(v) If (\u221a3 + i)100<\/sup> = 299<\/sup> (a + ib), then show that a2<\/sup> + b2<\/sup> = 4.
\nSolution:
\n\"Inter<\/p>\n

Question 5.
\n(i) If z = x + iy and |z| = 1, find the locus of z.
\nSolution:
\nGiven z = x + iy
\nAlso |z| = 1
\n\u21d2 |x + iy| = 1
\n\u21d2 \\(\\sqrt{x^{2}+y^{2}}=1\\)
\n\u21d2 x2<\/sup> + y2<\/sup> = 1
\n\u2234 Locus of z is x2<\/sup> + y2<\/sup> = 1<\/p>\n

(ii) If the amplitude of (z – 1) is \\(\\frac{\\pi}{2}\\), then find the locus of z.
\nSolution:
\nGiven z = x + iy
\nz – 1 = x + iy – 1 = (x – 1) + iy
\nSince Amplitude (z – 1) is \\(\\frac{\\pi}{2}\\)
\nReal part of z – 1 is zero.
\n\u2234 x – 1 = 0
\n\u2234 Locus of z is x – 1 = 0<\/p>\n

(iii) If the Arg \\(\\overline{\\mathbf{z}}_{1}\\) and Arg z2<\/sub> are \\(\\frac{\\pi}{5}\\) and \\(\\frac{\\pi}{3}\\) respectively, find (Arg z1<\/sub> + Arg z2<\/sub>)
\nSolution:
\nArg \\(\\overline{\\mathbf{z}}_{1}\\) = \\(\\frac{\\pi}{5}\\)
\n\u21d2 Arg \\(\\overline{\\mathbf{z}}_{1}\\) = -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}\\) = \\(\\frac{2 \\pi}{15}\\)<\/p>\n

\"Inter<\/p>\n

(iv) If z = \\(\\frac{1+2 i}{1-(1-i)^{2}}\\) then find Arg(z).
\nSolution:
\nGiven 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}{1+2 i}\\)
\n= 1
\n= cos 0\u00b0 + i sin 0\u00b0
\n\u2234 Arg (z) = 0\u00b0<\/p>\n

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

Question 1.
\nSimplify the following complex numbers and find their modulus.
\n(i) \\(\\frac{(2+4 i)(-1+2 i)}{(-1-i)(3-i)}\\)
\n(ii) \\(\\frac{(1+i)^{3}}{(2+i)(1+2 i)}\\)
\nSolution:
\n\"Inter
\n\"Inter<\/p>\n

Question 2.
\n(i) If (1 – i) (2 – i) ( 3 – i) …… (1 – ni) = x – iy then prove that 2 . 5 . 10 …….. (1 + n2<\/sup>) = x2<\/sup> + y2<\/sup>.
\nSolution:
\nGiven (1 – i) (2 – i) (3 – i) (1 – ni) = x – iy
\n|1 – i| |2 – i| |3 – i| …… |1 – ni| = |x – iy|
\n\\(\\sqrt{1+1} \\sqrt{4+1} \\sqrt{9+1} \\ldots \\sqrt{1+n^{2}}=\\sqrt{x^{2}+y^{2}}\\)
\n\u2234 2 . 5 . 10 ……. (1 + n2<\/sup>) = x2<\/sup> + y2<\/sup><\/p>\n

(ii) If the real part of \\(\\frac{z+1}{z+i}\\) is 1, then find the locus of z.
\nSolution:
\n\"Inter<\/p>\n

(iii) If |z – 3 + i | = 4 determine the locus of z.
\nSolution:
\nLet z = x + iy
\nGiven |z – 3 + i| = 4
\n\u21d2 |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 of z is x2<\/sup> + y2<\/sup> – 6x + 2y – 6 = 0<\/p>\n

(iv) If |z + ai| = |z – ai| then find the locus of z.
\nSolution:
\nLet z = x + iy
\nGiven |z + ai | = |z – ai|
\n\u21d2 |x + iy + ai| = |x + iy – ai|
\n\u21d2 |x + i(y + a)| = |x + i(y – a)|
\n\u21d2 \\(\\sqrt{x^{2}+(y+a)^{2}}=\\sqrt{x^{2}+(y-a)^{2}}\\)
\n\u21d2 x2<\/sup> + (y + a)2<\/sup> = x2<\/sup> + (y – a)2<\/sup>
\n\u21d2 (y + a)2<\/sup> – (y – a)2<\/sup> = 0
\n\u21d2 4ay = 0
\n\u21d2 y = 0
\n\u2234 The locus of z is y = 0<\/p>\n

\"Inter<\/p>\n

Question 3.
\nIf z = (x + iy) and if the point P in the Argand plane represents z, then describe geometrically the locus of P satisfying the equation.
\n(i) |2z – 3| = 7
\nSolution:
\n|2z – 3| = 7
\nLet z = x + iy
\nGiven |2z – 3| = 7
\n\u21d2 |2(x + iy) – 3| = 7
\n\u21d2 |(2x – 3) + i(2y)| = 7
\n\u21d2 \\(\\sqrt{(2 x-3)^{2}+(2 y)^{2}}\\) = 7
\n\u21d2 4x2<\/sup> – 12x + 9 + 4y2<\/sup> = 49
\n\u21d2 4x2<\/sup> + 4y2<\/sup> – 12x – 40 = 0
\n\u21d2 x2<\/sup> + y2<\/sup> – 3x – 10 = 0
\n\u2234 The locus of z is x2<\/sup> + y2<\/sup> – 3x – 10 = 0
\nThis equation represents a circle.
\nCentre = (\\(\\frac{3}{2}\\), 0)
\nRadius = \\(\\sqrt{\\frac{9}{4}-(-10)}\\) = \\(\\frac{7}{2}\\) units<\/p>\n

(ii) |z|2<\/sup> = 4 Re(z + 2)
\nSolution:
\n|z|2<\/sup> = 4 Re (z + 2)
\nGiven |z|2<\/sup> = 4 Re (z + 2)
\n\u21d2 |x + iy|2<\/sup> = 4 Re (x + iy + z)
\n\u21d2 (\\(\\sqrt{x^{2}+y^{2}}\\)) = 4(x + 2)
\n\u21d2 x2<\/sup> + y2<\/sup> = 4x + 8
\n\u2234 The locus of z is x2<\/sup> + y2<\/sup> – 4x – 8 = 0
\nThis equation represents a circle. Center = (2, 0)
\nRadius = \\(\\sqrt{4-(-8)}\\) = 2\u221a3 units.<\/p>\n

(iii) |z + i|2<\/sup> – |z – i|2<\/sup> = 2
\nSolution:
\nGiven |z + i|2<\/sup> – |z –\u00a0 i|2<\/sup> = 2
\n\u21d2 |x + iy + i|2<\/sup> – |x + iy – i|2<\/sup> = 2
\n\u21d2 |x + i(y + 1)|2<\/sup> – |x + i(y – 1)|2<\/sup> = 2
\n\u21d2 \\(\\left[\\sqrt{x^{2}+(y+1)^{2}}\\right]^{2}\\) – \\(\\left[\\sqrt{x^{2}+(y-1)^{2}}\\right]^{2}\\) = 2
\n\u21d2 x2<\/sup> + (y + 1 )2<\/sup> – [x2<\/sup> + (y – 1)2<\/sup>] = 2
\n\u21d2 x2<\/sup> + (y + 1)2<\/sup> – x2<\/sup> – (y – 1)2<\/sup> = 2
\n\u21d2 4y = 2
\n\u21d2 2y – 1 = 0
\nThe locus of P is 2y – 1 = 0
\nThis equation represents a straight line parallel to the x-axis.<\/p>\n

(iv) |z + 4i| + |z – 4i| = 10
\nSolution:
\nGiven |z + 4i| + |z – 4i| = 10
\n\u21d2 |x + iy + 4i| + |x + iy – 4i| = 10
\n\u21d2 |x + i(y + 4)| + |x + i(y – 4)| = 10
\n\u21d2 \\(\\sqrt{x^{2}+(y+4)^{2}}+\\sqrt{x^{2}+(y-4)^{2}}\\)
\n\u21d2 \\(\\sqrt{x^{2}+(y+4)^{2}}\\) = 10 – \\(\\sqrt{x^{2}+(y-4)^{2}}\\)
\n\u21d2 x2<\/sup> + (y + 4)2<\/sup> = 100 + x2<\/sup> + (y – 4)2<\/sup> – 20 \\(\\sqrt{x^{2}+(y-4)^{2}}\\)
\n\u21d2 16y – 100 = – 20\\(\\sqrt{x^{2}+(y-4)^{2}}\\)
\n\u21d2 4y – 25 = -5\\(\\sqrt{x^{2}+(y-4)^{2}}\\)
\n\u21d2 (4y – 25)2<\/sup> = 25[x2<\/sup> + (y – 4)2<\/sup>]
\n\u21d2 16y2<\/sup> – 200y + 625 = 25x2<\/sup> + 25y2<\/sup> – 200y + 400
\n\u21d2 25x2<\/sup> + 9y2<\/sup> = 225
\n\u2234 The locus of P is 25x2<\/sup> + 9y2<\/sup> = 225
\n\u21d2 \\(\\frac{x^{2}}{9}+\\frac{y^{2}}{25}\\) = 1
\nThis equation represents an ellipse.
\nCenter = (0, 0)
\na2<\/sup> = 9, b2<\/sup> = 25
\nWe know a2<\/sup> – b2<\/sup>(1 – e2<\/sup>)
\n\u21d2 9 = 25(1 – e2<\/sup>)
\n\u21d2 1 – e2<\/sup> = \\(\\frac{9}{25}\\)
\n\u21d2 e = 1 – \\(\\frac{9}{25}\\) = \\(\\frac{16}{25}\\)
\n\u2234 Eccentricity = \\(\\frac{4}{5}\\) and major axis parallel to y-axis.<\/p>\n

\"Inter<\/p>\n

Question 4.
\n(i) If z1<\/sub>, z2<\/sub> are two non-zero complex numbers satisfying |z1<\/sub> + z2<\/sub>| = |z1<\/sub>| + |z2<\/sub>|, show that Arg z1<\/sub> – Arg z2<\/sub> = 0.
\nSolution:
\n\"Inter
\n\"Inter
\n\"Inter<\/p>\n

(ii) z = 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.
\nSolution:
\nLet z = x + iy and a = \u03b1 + i\u03b2
\n\\(\\frac{z-a}{z+\\bar{a}}=\\frac{(x+i y)-(\\alpha+i \\beta)}{(x+i y)+(\\alpha-i \\beta)}\\)
\n\\(\\left|\\frac{z-a}{z+\\bar{a}}\\right|\\) = 1
\n\u21d2 \\(\\frac{|z-a|}{|z+\\bar{a}|}\\) = 1
\n\u21d2 |z – a| = |z + \\(\\overline{\\mathrm{a}}\\)|
\n\u21d2 |(x – \u03b1)| + i(y – \u03b2)| = |(x + \u03b1) + i(y – \u03b2)|
\n\u21d2 \\(\\sqrt{(x-\\alpha)^{2}+(y-\\beta)^{2}}=\\sqrt{(x+\\alpha)^{2}+(y-\\beta)^{2}}\\)
\n\u21d2 (x – \u03b1)2<\/sup> + (y – \u03b2)2<\/sup> = (x + \u03b1)2<\/sup> + (y – \u03b2)2<\/sup>
\n\u21d2 (x – \u03b1)2<\/sup> = (x + \u03b1)2<\/sup>
\n\u21d2 (x + \u03b1)2<\/sup> – (x – \u03b1)2<\/sup> = 0
\n\u21d2 4\u03b1x = 0
\nRe(a) \u2260 0
\n\u21d2 a \u2260 0
\n\u21d2 x = 0
\nLocus of P is x = 0 i.e., Y-axis<\/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(c) will help students to clear their doubts quickly. Intermediate 2nd Year Maths 2A Complex Numbers Solutions Exercise 1(c) I. Question 1. Express the following complex numbers in modulus-amplitude form. (i) 1 – i Solution: 1 – … 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.in\/wp-json\/wp\/v2\/posts\/22048"}],"collection":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/comments?post=22048"}],"version-history":[{"count":1,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts\/22048\/revisions"}],"predecessor-version":[{"id":22049,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts\/22048\/revisions\/22049"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/media?parent=22048"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/categories?post=22048"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/tags?post=22048"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}