Looking for a Tutor Near You?

Post Learning Requirement »
x

Choose Country Code

x

Direction

x

Ask a Question

x

x
x
For better view click here
x
Hire a Tutor
DISCOVER

Mathematics

Loading...

Published in: Mathematics
2,440 Views

01_Straight_Line 02_Circle 03_Permutation_and_combination 04_Complex_numbers 05_Application_of_Derivative 06_Binomial_theorem 07_Progression__Series.pdf 08_Definite_Integral.pdf 09_Indefinite_Integral.pdf 10_Probability.pdf

Score G / Mumbai

year of teaching experience

Qualification:

Teaches: Chemistry, English, Mathematics, Physics, Bank Clerical, Bank PO, SSC Exams, UPSC Exam, GATE Exam, IIT JEE Advanced, IIT JEE Mains, NATA Exam, Civil, Computer, Electronics, IT, IELTS, Spoken English, TOEFL, French, Spanish, Biology, Medical Entrance Exams, NEET, GMAT, GRE, PTE, SAT

Contact this Institute
  1. COMPLEX NUMBERS INTRODUCTION A number in the form of a + ib, where a, b are real numbers and i = NCI is called a complex number. A complex number can also be defined as an ordered pair of real numbers a and b and may be written as (a, b), where the first number denotes the real part and the second number denotes the imaginary part. If z = a + ib, then the real part of z is denoted by Re (z) and the imaginary part by 1m (z). A complex number is said to be purely real if Im(z) = 0, and is said to be purely imaginary if Re(z) = 0. The complex number 0 = 0 + i0 is both purely real and purely imaginary. 1 as imaginary unit, denoted by i. Thus, i NCI Symbol i : We define positive square root of 12 = Properties of i (i) (ii) For any integer n, 14 n = 1, 1 For example : PO 4 4 x 501 1 Also i X 124 1 For any integer n, 14 n + i 411+1 That is, the sum of four consecutive powers of i is zero. For example : 193 + 194 + 195 + 196 = 0 Complex number : A number of the form x + iy, where x and y are real numbers, is called a complex number, denoted by z. Thus z = x + iy, xe R, y e Ris a complex number. We define x = Real part of z, denoted by Re(z) y = Imaginary part of z, denoted by Im(z) + y 2 = Modulus or absolute value of z, denoted by Izl Properties of z : (ii) (iii) (iv) (v) (vi) (vii) If Re(z) = 0, then z = iy is called a purely imaginary number. If Im(z) = 0, then z = x is called a purely real number. z = 0 = 0 + i0 is both purely real as well as purely imaginary. Order relation (> or
  2. Complex Number Geometrical representation of complex numbers A complex number z = x + iycan be represented by Y (imaginary axis) a point P, whose Cartesian coordinates are (x, y) referred to axes OX and OY, usually called real and imaginary axes respectively. Point P is called the image of the complex number z and the 0 z is called the affix of the point P. The conjugate i of the number z is the affix of image Q of the point P in the real axis. Now, the modulus of z, 01 —Tt 11—01 2 X (real axis) The angle XOP is called the argument or amplitude of z, denoted by arg(z) or amp(z). Thus arg(z) 0 = tan x If we take OP= R, then x = R cos0, and y = R sine. Then z = x + R(cos0+ isinØ. This is known as trigonometric or polar form of the complex number z. Also z = R (cos0 + i sin0) = re10. This is known as Euler's formula. Again if Zl and represent two points P and Qin the Argand plane, then IZI - 41 represents the distance PQ. Principal value of Argument : In general the arg(z) of a complex number z is the solution of the simultaneous equation x cos0 2 and sino 2 Clearly the argument (z), i.e., 0 cannot be unique. 2n1T+ 0, n is an integer, is also an argument of z. The value of 0 such that -IT < 0 < is called the principal value of the argument. The argument of the complex number 0 is not defined. The principal value of argument (0) of the complex number z = x + iyfor different combinations of x and y are shown in following figures: z(x, y) z(x, y) 0 x In each case = tan and x 0 2 x 0 X O 0 EXAMPLE 2 : SOLUTION : z(x, y) + Sin — 3 Represent the given complex numbers in polar form: (i) (1 + ié)2 / 4i(1 — ix/ä) (ii) sin u — icos (u acute) (1 (1 -2 +2iN6 4 (l iUä) 2 1 and — cos + i sin— 2 2 (iii) 1 + cos —
  3. 3 cos + i sin Hence 4 (1 iv6) 2 2 (ii) Real part > 0; Imaginary part < 0 COMPLEX NUMBER 1 iTt/2 2 argument of sin — i cos is in the nature of a negative acute angle. . sin u — icos u = (iii) 1 + cos —+i sin— 3 COS 2 2 cos + i sin (I 2 2 —+i.2 sin — cos — 6 6 It in-/6 2cos—e 6 cos — + i sin — 2 cos — 6 6 6 Properties of conjugate of a complex number 6 (iii) (vi) z —i = 2ilm(z) 2 (iv) ü = Izl (ii) z +2 = 2Re(z) 2 Note: The properties (v) and (vi) can be extended to any number of complex number. (vii) (x) (xii) (xiii) —z 2 2 (viii) (xi) +214 2Re(i14) (ix) 2Re(zli2) z z EXAMPLE 3 : SOLUTION : z is purely real. —i z is purely imaginary. If Iz- 2 + il < 2 then find the greatest and least value of I z I. Given that Iz-2+il>IIzI-12-iII Iz-2+i121 Izl- From (i) and (ii) Hence greatest value of Izl is 6+2 and least value of Izl is 2. Properties of Modulus of a Complex Number Izl = 0 —z = 0 Izl > 0 for any complex number z, (iii) (iv) (vi) (ix) (xii) 2 2 IZI +41 +14 — , can be extended to any number of complex numbers. IZI z z (v) - 1, i.e. —is a unimodular complex number. z 2 2 1 1 1 21 I I - 21 - Il Il Il ± 41 < + Z Izl Re(z) Izl (viii) (x) Izl élzl —Izl < Im(z) < Izl +ilZ2) = +1412 +2Re(zli2) 2{IZ112 + z 212} +lz
  4. Complex Number Properties of Argument of Complex Numbers arg(z14) = arg(Zl) +arg(z2) , can be extended to any number of complex numbers. 4 (ii) (iv) (vi) arg arg(zn) arg(Zl) — arg(z2) n arg(z) arg(i) = —arg(z) (v) (vii) z arg 2arg(z) arg(z) = 0 iff z is purely real. arg(Z) = ± — iff z is purely imaginary. 2 EXAMPLE 4 : SOLUTION : Find out the principal arguments of the following complex numbers. (ii) 3 -4i (iii) -3 + 4i (i) tan-14/3 4 (ii) tan-I 3 (iii) It + tan-I (-4/3) 4 (iv) -Tt + tan 3 Concept of Rotation in Complex Plane Let z z z represent points A, B, C respectively on plane. Then AB=lz2 41, AC the complex 41. Let 0 be the counter clockwise and BC Iza angle ZBAC, then 0 = arg z We may write C(Z3) 0 A(ZI) B(Z2) x o z (ii) z (cos0 +isin0) = —(cos0 +isin0) AB AB Multiplying a complex number by i represents a rotation of angle — counter-clockwise about origin. 2 27t Multiplying a complex number by CD represents a rotation of angle about origin clockwise or 3 anticlockwise. EXAMPLE 5 : SOLUTION : ABCD is a rhombus. Its diagonals AC and BD intersect at M such that BD = 2AC. If the points D and M represent the complex number 1 + i and 2 — i respectively, find the complex number(s) representing A. Let A be z. The position MA can be obtained by rotating MD anticlockwise through an angle , simultaneously length gets halved. 2 1 iTt/2 . z (2—1) 2 1 2 1 1 2 z 2 2 Another position of A corresponds getting interchanged and in that 1 number of Ais 1+—i+2—i=3 2 2 . The complex number of A is either to A and C the complex 1 2 1 2
  5. 5 De Moivre theorem If ne l, then (cos0+ isin0)n = cos no + isin no COMPLEX NUMBER —, q then (cos0+ i Sino) n will have Q values one of which is given by cos no + i If n e Q, say n q sin no. (P and Q are integers) EXAMPLE 6 : SOLUTION : If n be a positive integer, prove that (1 (1 i)2n 0 2+1 2+1 — Sin— 4 (1 + i)2n i)2n (1 2 2 cos 4 cos — 4 i sin 4 if n be odd if — be even 2 if — be odd 2 2n 2 cos E +i sin — nit 2 2 nit 2 cos — i sin — 2 2 nit (l + i)2n + (l i)2n 2 cos E + i sin + cos — i sin — 2 2 2 2+1 cos 2 If n be odd = 2m + 1, then RHS = 2 cos (2m + 1) — 2 also even so that n - - 4k, then RHS = 2 2 If n be even and — else RHS = 2+1 cos 2 2+1 Cube Roots of Unity Let 2=1 > z3 1=0 (z -1) (E + z + 1) = O z = lor z COS (2kTt) = 2 +1 2 z Also, o and are called imaginary cube roots of unity and one the roots of z2 + z + 1 2 2 , we generally represent CD 2 27T 1 + 2m 2m and 2 2 2 2 cos — + i sin — 3 3 411 — cos—+isin— 3 3 3 . 47t 3
  6. Complex Number -2 -1 -2 6 If u, p, Y are roots off 3k + 3x + 7 = 0 (and c,) is cube roots of unity), then find the value of EXAMPLE 7 : SOLUTION : (1-1 B-l —1 We have 01—1 (x- 1)3+8 0 3 = (-2)3 2 (cube roots of unity) 2 2 20) Here = Then 1, 1 -20, 1 20), —1 — (1)1/3 = 1, (D, CD 2 20) 2 20) . 20), Y -2 -20 1 1 -202 -20 -202 2 CD CD CD + + CD 302 2 Properties of o and o 1 + = 0, in general 1 n + 22n = 3 or 0 according as n is a multiple of 3 or not (n e 0. 311+2 2 = o and CD (ii) (iv) (v) (vi) (vii) a/ = 1; in general = 1, CD 2 2 6 and CD The cube roots of unity represent the vertices of an equilateral triangle inscribed in a unit circle with centre at origin on the complex plane. One vertex is always on positive real axis. 2 If is a real cube root of a real number then its other roots are and . If a complex number z is such that IRe(z)I : = 1 : Uä ordä:l, then z can be expressed in terms of i, (Dor (D . For any real a, b, c; a + bcc) + CCD The nth Roots of Unity Let 1 = cos2k1T+ isin2k1T, ke I cos Z = (COS2k1t + isin 2k1t)l/n n n 27t 27t If we represent cos—+isin— by u, then the roots of unity are 1, u, u , n n Properties of nth Roots of Unity 2+...+ an-I = 0 n—l 0 n '0-1 = (-IY+I + i Sin n 2kTC and 0 n n The points represented by the roots of unity are located at the vertices of a regular polygon of n sides inscribed in a unit circle with centre at the origin. One vertex being on the positive real axis. - Ada) 27T n 27T n x 27T Al (1) n All (an)
  7. 7 EXAMPLE 8 : SOLUTION : COMPLEX NUMBER Find the cube roots of 4 - 443 i. 2 . Cube roots of 4 —46 i are given by 2k1t + U 1/3 k = 0, 1, 2 and pi/ 3 = 81/3 = 2 (positive real cube root of 8) z = p cis 3 27t + U + 01 Thus z = 2 cis , 2 cis 3 3 , 2 cis are the required roots. 3 1 Here u is given by cos u = — and sin u = 2 3 ALITER Let or, or, since CD = Therefore, z 2e Geometrical Applications 2 (4 44301/3 -h/3 1/3 (1)1/3 —W9 ,2e .03 and2e .03 i2t/3 2 i4Tt/3 CD = e —W9 , 24 w 9 and 2e ill Tt/9 (ii) (iii) (iv) (v) (vi) (vii) Distance between two points A and B represented by complex numbers Zl and 4 is 14-41. Affix of a point P dividing the join of point A and B with affices Zl and 4 in the ratio m: n, internally is mz2 + nzl mz2 — nzl m+n ; externally is m—n Affix of mid point of A(ZI) and B(Z2) is 2 Affix of centroid of AABC, with vertices A(ZI), B(Z2) and c(z3) is 3 3 z Equation of straight line passing through two points A(ZI) and B(Z2) in complex form is Zl z or 21 1 1 1 0 General equation of a straight line in complex plane is äz+ai +b number and b is a constant real number. Slope of this line = i(a - ä) Re(a) Im(a) O, where a is a constant complex Iäzo +aio +bl Distance of a given point P(zo) from the line äz+ai +b POD O is given by 21aI
  8. Complex Number (viii) Equation of a circle of radius R and —ZIZ2 —Z2Z3 -1 centre at point C(zo) is Iz- = R. Iz - > R represents the points lying outside the circle. Iz - < R represents the points lying inside the circle. 8 PG) 0 C(z) (ix) (x) (xi) (xii) Any point on the circle Iz- = R can be given by z = zo + re10 General equation of a circle in complex plane is given by zi+ai +äz +b = 0, where b e R. Its center is at the Ia12-b> 0. Equation of is (z —4) (z —22 Let and 4 Then arg point a a and radius Ia12 —b. The circle is C with affix circle described on a line segment AB, as z2) (i —21) , where Zl and are affices of points A and B. be two given complex numbers. PG) 1 2 u, 0 < u < It represents all points z lying on the arc of a circle. If 01 O, — z lies on the major arc (excluding 2 points A and B). BOQ If 01 2' , z lies on the minor arc (excluding points A and B). (xiii) Four points A(ZI), B(Z2), c(z3) and D(Z4) taken in order are concyclic if real iff diameter is purely (xiv) Iz- + Iz ellipse. (xv) Iz- Zil + Iz 41 = a, a e represents an ellipse if IZI - 41 < a. Points Zl and 4 represent the foci of 41 = a, a e R- {0} represents an hyperbola if IZI - 41 > Ial. Points Zl and represents the foci of hyperbola. (xvi) The triangle whose vertices are equilateral if and only if the points represented by the complex numbers 4, 4, is 1 1 1 —Z3Z1 — O. z 3 EXAMPLE 9 : SOLUTION : Interpret Geometrically the complex number 'z' which satisfied the following inequality logl/2 Iz—11+4 In order the log is to be defined, Iz- Il -2 > 0 Iz—11+4 1 Also, Iz-11-2 Iz- 10 which is always true. Hence the inequality will hold for all 'z' satisfying the condition that Iz- Il > 2. Geometrically, it represents the exterior of a circle with center (1 + 01) and radius '2'. 3
  9. 9 EXAMPLE 10 : SOLUTION : COMPLEX NUMBER If Ilz + 21 - Iz- 211 = a2, ze C representing a hyperbola for a e R, then find the values of a. Here foci are at -2 and 2 at a distance at 4. Hence the given equation represents a hyperbola if 4 i.e. a e (-2, 2).

Need a Tutor or Coaching Class?

Post an enquiry and get instant responses from qualified and experienced tutors.

Post Requirement

Related Study Notes

Upload Study Notes

If you have your own Study Notes which you think can benefit others, please upload on LearnPick. For each approved Study Note you will get 100 Credit Points and 100 Activity Score which will increase your profile visibility.

Upload Now
Home > Study Notes > Mathematics