Mathematics and Statistics Std XI (Part II) PDF

 The Constitution of India Chapter IV A Fundamental Duties ARTICLE 51A Fundamental Duties- It shall be the duty of every citizen of India- (a) to abide by the Constitution and respect its ideals and institutions, the National Flag and th...

 The Constitution of India Chapter IV A Fundamental Duties ARTICLE 51A Fundamental Duties- It shall be the duty of every citizen of India- (a) to abide by the Constitution and respect its ideals and institutions, the National Flag and the National Anthem; (b) to cherish and follow the noble ideals which inspired our national struggle for freedom; (c) to uphold and protect the sovereignty, unity and integrity of India; (d) to defend the country and render national service when called upon to do so; (e) to promote harmony and the spirit of common brotherhood amongst all the people of India transcending religious, linguistic and regional or sectional diversities, to renounce practices derogatory to the dignity of women; (f) to value and preserve the rich heritage of our composite culture; (g) to protect and improve the natural environment including forests, lakes, rivers and wild life and to have compassion for living creatures; (h) to develop the scientific temper, humanism and the spirit of inquiry and reform; (i) to safeguard public property and to abjure violence; (j) to strive towards excellence in all spheres of individual and collective activity so that the nation constantly rises to higher levels of endeavour and achievement; (k) who is a parent or guardian to provide opportunities for education to his child or, as the case may be, ward between the age of six and fourteen years.  The Coordination Committee formed by GR No. Abhyas - 2116/(Pra.Kra.43/16) SD - 4 Dated 25.4.2016 has given approval to prescribe this textbook in its meeting held on 20.06.2019 and it has been decided to implement it from the educational year 2019-20. Mathematics and Statistics (Arts and Science) Part - 2 STANDARD - XI Maharashtra State Bureau of Textbook Production and Curriculum Research, Pune - 411 004 Download DIKSHA App on your smartphone. If you scan the Q.R.Code on this page of your textbook, you will be able to access full text and the audio-visual tudy material relevant to each lesson provided as teaching and learning aid. First Edition : 2019 © Maharashtra State Bureau of Textbook Production Third Reprint : 2022 and Curriculum Research, Pune- 411 004. Maharashtra State Bureau of Textbook Production and Curriculum Research reserves all rights relating to the book. No part of this book should be reproduced without the written permission of the Director, Maharashtra State Bureau of Textbook Production and curriculum Research, Pune. Mathematics - II (Science and Arts) Cover, Illustrations and Computer Drawings Commitee Members Dr. Mangala Narlikar (Chairman) Shri. Sandip Koli, Dr. Sharad Gore (Member) Artist Shri. Prahallad Chippalagatti (Member) Shri. Prasad Kunte (Member) Co-ordinator Shri. Sujit Shinde (Member) Ujjwala Shrikant Godbole Smt. Prajakti Gokhale (Member) I/C Special Officer for Mathematics Shri. Pradipkumar Bhavsar (Member) Shri. Ramakant Sarode (Member) Shri. Milind Patil (Member) Production Smt. Ujjwala Godbole (Member-Secretary) Sachin Mehta Chief Production Officer Sanjay Kamble Production Officer Prashant Harne Asst. Production Officer Mathematics - II (Science and Arts) Study Group Members Paper Dr. Ishwar Patil Dr. Pradeep Mugale 70 GSM Cream wove Dr. Pradnyankumar Bhojankar Shri. Balkrishna Mapari Shri. Prafullchandra Pawar Shri. Uday Mahindrakar Print Order No. Smt. Nileema Khaladkar Shri. Devanand Bagul N/PB/2022-23/1,25,000 Smt. Swati Powar Shri. Sachin Batwal Printer Smt. Mahadevi Mane Shri. Swapnil Shinde S.P.BINDING WORKS,PUNE Shri. Dilip Panchamia Smt. Deepti Kharpas Shri. Pramod Deshpande Shri. Amit Totade Shri. Dhananjay Panhalkar Smt. Gauri Prachand Shri. Sharadchandra Walagade Smt. Supriya Abhyankar Publisher Shri. Vinayak Godbole Vivek Uttam Gosavi, Controller Maharashtra State Textbook Bureau, Prabhadevi Mumbai- 400 025  PREFACE Dear Students, Welcome to the eleventh standard! You have successfully completed your secondary education and have entered the higher secondary level. You will now need to learn certain mathematical concepts and acquire some statistical skills to add more applicability to your work. Maharashtra State Bureau of Textbook Production and Curriculum Research has modified and restructured the curriculum in Mathematics in accordance with changing needs. The curriculum of Mathematics is divided in two parts. Part 1 covers topics in Trignometry, Algebra, Co-ordinate Geometry and Statistics. Part 2 covers Complex Numbers, Sets and Relations, Calculus and Combinatorics. There is a special emphasis on applications. Activities are added in chapters for creative thinking. Some material will be made available on E-balbharati website (ebalbharati.in).It contains a list of specimen practical problems on each chapter. Students should complete the practical exercises under the guidance of their teachers. Maintain a journal and take teacher’s signature on every completed practical. You are encouraged to use modern technology in your studies. Explore the Internet for more recent information on topics in the curriculum. Get more examples and practice problems from the Internet. You will enjoy learning if you study the textbook thoroughly and manage to solve problems. On the title page Q.R. code is given. It will help you to get more knowledge and clarity about the contents. This textbook is prepared by Mathematics Subject Commitee and members of study group. This book has also been reviewed by senior teachers and subject experts. The Bureau is grateful to all of them and would like to thank for their contribution in the form of creative writing, constructive criticism, and valuable suggestions in making this book useful to you and helpful to your teachers. The Bureau hopes that the textbook will be received well by all stakeholders in the right spirit. You are now ready to study. Best wishes for a happy learning experience. Pune Date : 20 June 2019 (Dr. Sunil Magar) Indian Solar Date : 30 Jyeshtha 1941 Director Maharashtra State Bureau ofTextbook Production and Curriculum Research, Pune.  Mathematics and Statistics Std XI (Part II) Arts and Science Sr. Area Topic Competency Statement No The students will be able to - understand set of complex numbers and different ways of expressing complex Complex Complex numbers. 1 Numbers Numbers perform algebraic operations on complex numbers. simplify algebraic expressions involving complex numbers. Sequences Revise AP, learn GP and HP. Sequences and 2 Find the general term and the sum of the first n Series and Series terms of these sequences. count the number of arrangements of given Permutations Permutations, objects satisfying specific conditions. 3 and Combinations count the number of possible selections of combinations objects with certain conditions. Method of understand the method of induction and apply Method of Induction it to verify mathematical statements. Induction and 4 Binomial expand binomial expressions and find its theorem Binomial general term. Theorem simplify the binomial expression for negative index or fractional power. work with sets and operations on sets. construct sets from given conditions. 5 Sets solve problems on applications of set Sets and theory. Relations identify the types of relations. Relations study equivalence relations.  work with function defined on different domains. 6 Functions Functions identify different types of functions. carry out algebraic operations on functions. understand the concept of limit of a function. 7 Limits Limits determine the limits of functions if they exist. Define and study the continuity of a function 8 Continuity Continuity at a point and in an interval. understand and study the differentiability of a function. 9 Differentiation Differentiation understand and study differentiation of various functions.  INDEX Sr. No. Chapter Page No. 1 Complex Numbers 1 2 Sequences and Series 23 3 Permutations and Combination 43 4 Methods of Induction and Binomial Theorem 69 5 Sets and Relations 87 6 Functions 106 7 Limits 133 8 Continuity 160 9 Differentiation 179 Answers 196  1 COMPLEX NUMBERS 1.1 A Complex number : Let's Study 1.1(a) Imaginary Number : A complex number (C.N.) A number of the form bi, where b∈ R, b ≠ 0, Algebra of C.N. i= −1 is called an imaginary number. Geometrical Representation of C.N. 2 Polar & Exponential form of C.N. Ex : −25 = 5i, 2i, i, − 11i etc. 7 De Moivre's Theorem. Note: The number i satisfies following properties, Let's Recall i) i×0=0 ii) If a∈R, then −a 2 = 2 2 Algebra of real numbers. iii) If a, b∈R, and ai = bi then a = b Solution of linear and quadratic equations Representation of a real number on the number 1.1 (b) Complex Number : line Definition : A number of the form a+ib, where Representation of point in a plane a, b∈ R and i = −1 with i2 = −1 is called a Trigonometric ratios complex number and is usually denoted by z. That is z = a+ib, a, b∈R and i = −1 Introduction: Here 'a' is called the real part of z and is denoted Consider, the equation x2 + 1 = 0. This by Re(z) or R(z). 'b' is called imaginary part of z equation has no solution in the set of real numbers and is denoted by Im(z) or I(z) because there is no real number whose square is −1. We need to extend the set of real numbers to a The set of complex numbers is denoted by C larger set, which would include solutions of such ∴ C = {a+ib / a, b∈ R, and i = −1 } equations. Ex : We introduce a symbol i (greek letter iota) z a+ib Re(z) Im(z) such that i = −1 and i = −1. i is called as an 2 2+4i 2+4i 2 4 imaginary unit or an imaginary number. 5i 0+5i 0 5 Swiss mathematician Leonard Euler 3−4i 3−4i 3 −4 (1707-1783) was the first mathematician to 5+ −16 5+4i 5 4 introduce the symbol i with i = −1 and i2 = −1. 2+ 5 i 2+ 5 i 2 5 7+ 3 (7+ 3 )+0i (7+ 3 ) 0 1 Note : 1.2.2 Conjugate of a Complex Number: 1) A complex number whose real part is zero Definition : The conjugate of a complex is called a purely imaginary number. Such a number z = a + ib is defined as a − ib and is number is of the form z = 0 + ib = ib denoted by z 2) A complex number whose imaginary part is Ex : 1) zero is a real number. z z z = a + 0i = a, is a real number. 3 + 4i 3 − 4i 3) A complex number whose both real and 7i −2 −7i −2 imaginary parts are zero is the zero complex 3 3 number. 0 = 0 + 0i 5i −5i 4) The set R of real numbers is a subset of the 2+ 3 2+ 3 set C of complex numbers. 7+ 5i 7− 5i 5) The real part and imaginary part cannot be 2) Properties of z combined to form single term. e.g. 2 + 3i ≠ 5i 1) ( z) = z 1.2 Algebra of Complex Numbers : 2) If z = z , then z is purely real. 1.2.1 Equality of two Complex Numbers : 3) If z = −z , then z is purely imaginary. Definition : Two complex numbers z1 = a+ib Now we define the four fundamental and z2 = c + id are said to be equal if their operations of addition, subtraction, multiplication and division of complex numbers. corresponding real and imaginary parts are equal. 1.2.3 Addition of complex numbers : i.e. a + ib = c + id if and only if a = c and b = d Let z1 = a+ib and z2 = c+id Ex. : i) If x + iy = 4 + 3i then x = 4 and y = 3 then z1+ z2 = (a+ib) + (c+id) = (a+c) + (b+d) i Ex. : ii) If 7a + i (3a − b) = 21 − 3i then find a and b. In other words, Re(z1+ z2) = Re(z1) + Re(z2) and Im(z1+ z2) = Im(z1) + Im(z2) Solution : 7a + (3a − b) i = 21 − 3i By equality of complex numbers Ex. 1) (2 + 3i) + (4 + 3i) = (2+4) + (3+3)i 7a = 21 ∴a=3 = 6 + 6i and 3a − b = −3 ∴ 3(3) + 3 = b 2) (−2 + 5i) + (7 + 3i) + (6 − 4i) ∴ 12 = b = [(−2) +7+6] + [5 + 3 + (− 4)]i Note : The order relation (inequality) of complex = 11 + 4i number can not be defined. Hence, there does not Properties of addition : If z1, z2, z3 are complex exist a smaller or greater complex number than numbers then given complex number. We cannot say i < 4. i) z1+ z2 = z2+ z1 (commutative) ii) z1 + (z2+ z3) = (z1+ z2) + z3 (associative) 2  iii) z1 + 0 = 0 + z1 = z1 (identity) Properties of Subtraction : iv) z + z = 2Re(z) (Verify) 1) z − z = 2 Im(z) (Verify) v) (z1+ z2) = z1 + z2 1) (z1− z2) = z1 − z2 1.2.4 Scalar Multiplication : 1.2.6 Multiplication of complex numbers: If z = a+ib is any complex number, then for Let z1 = a+ib and z2 = c+id. We denote every real number k, define kz = ka + i (kb) multiplication of z1 and z2 as z1.z2 and is given by Ex. 1) If z = 7 + 3i then 5z = 5(7 + 3i) = 35 + 15i z1.z2 = (a+ib)(c+id) = a(c+id)+ib(c+id) 2) z1 = 3 − 4i and z2 = 10 − 9i then = ac + adi + bci + i2bd 2z1 + 5z2 = 2(3 − 4i) + 5(10 − 9i) = ac + (ad+bc)i − bd (∵ i2 = −1) = 6 − 8i + 50 − 45i z1.z2 = (ac−bd) + (ad+bc)i = 56 − 53i Note: 1) 0.z = 0(a + ib) = 0 + 0i = 0 Ex. 1) z1 = 2+3i, z2 = 3−2i ∴ z1.z2 = (2+3i)(3−2i) = 2(3−2i) + 3i(3−2i) 1.2.5 Subtraction of complex numbers : = 6 − 4i + 9i − 6i2 Let z1 = a+ib, z2 = c+id then define = 6 − 4i + 9i + 6 (∵ i2 = −1) z1− z2 = z1+ (−1)z2 = (a+ib) + (−1)(c+id) = 12 + 5i = (a+ib) + (−c − id) Ex. 2) z1 = 2−7i, z2 = 4−3i, z3 = 1+i then = (a−c) + i (b−d) (2 z1). (z2). (z3) = 2(2−7i). (4−3i). (1+i) Hence, Re(z1− z2) = Re(z1) − Re(z2) = (4−14i). [4+4i−3i−3i2] Im(z1− z2) = Im(z1) − Im(z2) = (4−14i). [7+i] Ex. 1) z1 = 4+3i, z2 = 2+i = 28 + 4i−98i−14i2 ∴ z1− z2 = (4+3i) − (2+i) = 42−94i = (4−2) + (3−1) i Properties of Multiplication : = 2 + 2i i) z1.z2 = z2.z1 (commutative) 2) z1 = 7+i, z2 = 4i, z3 = −3+2i ii) (z1.z2).z3 = z1.(z2..z3) (associative) then 2z1 − (5z2 + 2z3) iii) (z1.1) = 1.z1 = z1 (identity) = 2(7+i) − [5(4i) + 2(−3+2i)] = 14 + 2i − [20i − 6 + 4i] iv) (z1.z2) = z1. z2 (Verify) = 14 + 2i − [− 6 + 24i] v) If z = a+ib, z.z = a2 + b2 = 14 + 2i + 6 − 24i = 20 − 22i 3 1.2.7. Powers of i : We have −1 = i, i2 = −1, 1.2.8 Division of complex number : i3 = −i, i4 = 1. Let us consider in, where n is a Let z1 = a+ib and z2 = c+id be any two positive integer and n > 4. complex numbers such that z2 ≠ 0 We divide n by 4 and obtain the remainder 'r'. Now, Let m be the quotient. z1 a+ib z2 = c+id where z2 ≠ 0 i.e. c+id ≠ 0 Then, n = 4m + r, where 0 ≤ r < 4 ∴ in = i4m + r = i4m. ir = (i4)m. ir = 1.ir = ir Multiply and divide by conjugate of z2. z1 a+ib c−id Similarly, ∴ = × z2 c+id c−id 1 i4m = 1 = 4m = i−4m (a + ib) (c − id) i = (c + id) (c − id) i2 = −1, i3 = i × i2 = −i (ac+bd)+(bc − ad)i = 1 i i c 2 +d 2 i−1 = = 2 = = −i i i −1 ac+bd bc ad The above equations help us to find ik for any = i c 2 +d 2 c 2 +d 2 integer k.  ac+bd  bc ad Where  2 2   R and 2 2 R  c +d  c +d Illustration : If z1 = 3+2i, & z2 = 1+i, z1 3+2i then z = 1+i 2 By multiplying numerator and denominator with z2=1−i z1 3+2i 1−i z = 1+i × 1−i Fig. 1.1 2 3−3i+2i−2i2 e.g. (i) i50 = (i4)12. i2 = i2 = −1 = 1+1 (ii) i318 = (i4)79. i2 = −1 5−i = 2 (iii) i999 = (i4)249+3 = (i4)249. i3 = −i z1 5 1 ∴ z = 2 − i Remark : In general, 2 2 i4n = 1, i4n+1 = i, Properties of Division : i4n+2 = −1, i4n+3 = −i where n ∈ Z 1 1 i i 1) = i × = −1 = −i i i 1 a−ib 2) = 2 2 a+ib a +b 4  Ex. 4 : If x + 2i + 15i6y = 7x + i3 (y+4) SOLVED EXAMPLES find x + y, given that x, y ∈ R. Ex. 1 : Write (1+2i) (1+3i) (2+i)−1 in the form Solution : a+ib x + 2i + 15i6y = 7x + i3 (y+4) ∴ x + 2i − 15y = 7x − (y+4) i Solution : (1+2i)(1+3i) (∵ i6 = −1, i3 = −i) (1+2i) (1+3i) (2+i)−1 = 2+i ∴ x − 15y + 2i = 7x − (y+4) i 1+3i+2i+6i 2 −5+5i 2−i = = 2+i × Equating real and imaginary parts, we get 2+i 2−i −10+5i+10i−5i2 −5+15i x − 15y = 7x and 2 = −(y+4) = = 4+1 (∴ i2 = −1) 4−i2 ∴ −6x − 15y = 0.... (i) y+6 = 0....(ii) −5+15i ∴ y = −6, x = 15 [Solving (i) and (ii)] = = −1 + 3i 5 ∴ x + y = 15 − 6 = 9 1 2 3 5 Ex. 2 : (Activity) Express + i2 + 3 + i4 i i 3 in the form of (a + ib).  3 i Ex. 5 : Show that  +  = i. Solution : i2 = −1, i3 = −i, i4 = 1  2 2 1 1 1 1 Solution : = , i2 = , 3 = , i4 = i i 3 3  3 i   3 +i  1 2 3 5 L.H.S. =  +  =   ∴ i + i2 + 3 + i4 i  2 2  2  ( 3) ( 3) i + 3 3 2 = 1( )+2( )+3( )+5( ) +3 3i 2 + (i )3 = +i = ( 2) 3 ∴ a= , b= 3 3 + 9i − 3 3 − i = Ex. 3 : If a and b are real and 8 (i4+3i)a + (i−1)b + 5i3 = 0, find a and b. 8i = 8 Solution : (i +3i)a + (i−1)b + 5i = 0+0i 4 3 = i i.e. (1+3i)a + (i−1)b − 5i = 0+0i = R.H.S. ∴ a + 3ai + bi − b − 5i = 0+0i i.e. (a−b) + (3a+b−5)i = 0+0i By equality of complex numbers, we get a−b = 0 and 3a+b−5 = 0 EXERCISE 1.1 ∴ a=b and 3a+b = 5 1) Simplify : ∴ 3a+a = 5 ∴ 4a = 5 i) −16 + 3 −25 + −36 − −625 5 ∴a = 4 ii) 4 −4 + 5 −9 − 3 −16 5 ∴a=b = 4 5 2) Write the conjugates of the following 8) Show that 1 + i10 + i20 + i30 is a real number. complex numbers 9) Find the value of i) 3+i ii) 3− i iii) − 5 − 7i i) i49 + i68 + i89 + i110 iv) − −5 v) 5i vi) 5−i ii) i + i2 + i3 + i4 vii) 2 + 3i viii) cosq + i sinq i592 + i590 + i588 + i586+ i584 3) Find a and b if 10) Simplify : i582 + i580 + i578 + i576+ i574 i) a + 2b + 2ai = 4 + 6i 11) Find the value of 1 + i2 + i4 + i6+ i8 +...+i20 ii) (a-b) + (a+b)i = a + 5i 12) Show that 1 + i10 + i100 − i1000 = 0. iii) (a+b) (2 + i) = b + 1 + (10 + 2a)i iv) abi = 3a − b + 12i 13) Is (1 + i14 + i18 + i22) a real number? Justify your answer. 1 v) = 3 − 2i a + ib 14) Evaluate :  i 37  167  vi) (a + ib) (1 + i) = 2 + i  i  4 4) Express the following in the form of a+ib,  1 15) Prove that (1+i) × 1   = 16. 4 a, b∈R i = −1. State the values of a and b.  i i) (1+2i)(−2+i) ii) (1+i)(1−i)−1 i6+i7+i8+ i9 16) Find the value of i(4+3i) (2+i) i2+i3 iii) (1−i) iv) (3−i) (1+2i) 1  3i −1 − 3i 17) If a = , b= then 2 2 1 i  2 3+2i 3−2i v)   vi) + show that a2 = b and b2 = a.  1 i  2−5i 2+5i 2  3 x y vii) (1+i)−3 viii) 18) If x+iy = (a+ib)3, show that a + b = 4(a2−b2) 4  3 a+3i 19) If = 1−i, show that (5a−7b) = 0. 2+ib ix) ( 5  2 4 )  (1  9 )  (2  3i )(2  3i ) a + ib 4i 8  3i 9  3 20) If x+iy = , x) (2+3i)(2−3i) xi) 11 c + id 3i  4i10  2 a2+b2 prove that (x2+y2)2 = 5) Show that (1  3i )3 is a real number. c2+d2 1+i  2 21) If (a+ib) = , then prove that (a2+b2) = 1. 6) Find the value of  3   (i6−i7)(1+i11) 1−i  i 7) Evaluate the following :  7 i 3 7 i 3  22) Show that    is real.  7 i 3 7  i 3  i) i35 ii) i888 iii) i93 iv) i116 1 23) If (x+iy)3 = y+vi the show that v) i403 vi) 58 vii) i−888 i y v viii) i30 + i40 + i50 + i60    = 4(x −y ) 2 2  x y  6 24) Find the value of x and y which satisfy the put b2 = m following equations (x, y∈R) ∴ m2+6m−16 = 0 i) (x+2y) + (2x−3y) i + 4i = 5 ∴ (m+8)(m−2) = 0 x+1 y−1 ∴ m = −8 or m = 2 ii) + =i 1+i 1−i i.e. b2 = −8 or b2 = 2 (x+iy) 2+i 9 iii) + = 13 (1+i) but b is a real number ∴ b2 ≠ −8 2+3i 2−3i iv) If x(1+3i) + y(2−i) − 5 + i3 = 0, find x+y So, b2 = 2 ∴ b= ± 2 v) If x+2i+15i6y = 7x+i3 (y+4), find x+y For, b = 2 , a = 2 2 1.3 Square root of a complex number : ∴ 6 + 8i = 2 2 + 2i= 2 (2+ i) Consider z = x+iy be any complex number For, b = − 2 , a = −2 2 Let x + iy = a+ib, a, b∈R ∴ 6 + 8i = −2 2 − 2 i = − 2 (2+ i) On squaring both the sides, we get ∴ 6 + 8i = ± 2 (2+ i) x+iy = (a+ib)2 x+iy = (a2−b2) + (2ab) i Ex. 2 : Find the square root of 3 − 4i Equating real and imaginary parts, we get Solution : x = (a2−b2) and y = 2ab Let 3 − 4i = a+ib a, b∈R Solving these equations simultaneously, we can On squaring both the sides, we have get the values of a and b. 3 − 4i = (a+ib)2 ∴ 3 − 4i = (a2−b2) + (2ab) i Solved Examples: Equating real and imaginary parts, we have Ex.1 : Find the square root of 6+8i. a2−b2 = 3, 2ab = −4 Solution : As (a2+b2)2 = (a2−b2)2 + (2ab)2 Let, 6 + 8i = a+ib (a, b∈R) (a2+b2)2 = 32 + (−4)2 = 9 + 16 = 25 On squaring both the sides, we get (a2+b2)2 = 52 6+8i = (a+ib) 2 ∴ a2+b2 = 5 ∴ 6+8i = (a2−b2) + (2ab) i Solving a2+b2 = 5 and a2−b2 = 3 we get Equating real and imaginary parts, we have 2a2 = 8 6 = a2−b2.... (1) a2 = 4 ∴ a = ±2 8 = 2ab.... (2) 4 For a = 2; ∴ a = b −4 −4 4 2 b= = = −1 ∴ (1) becomes 6 =   − b2 2a 2(2) b −4 16 For, a = −2, b = =1 i.e. 6 = 2 − b2 2(−2) b ∴ 3 − 4i = 2 − i or −2 + i ∴ b4+6b2−16 = 0 i.e. (b2)2 + 6b2 − 16 = 0 7 1.4 Fundamental Theorem of Algebra : Ex. 2 : Solve x2−(2 3 + 3i) x + 6 3 i = 0 'A polynomial equation with real coefficients Solution : Given equation is has at least one root' in C. or 'A polynomial equation with complex x2−(2 3 + 3i) x + 6 3 i = 0 coefficients and of degree n has n complex roots'. The method of finding the roots of 1.4.1 Solution of a Quadratic Equation in ax2 + bx + c = 0, is applicable even if a, b, c complex number system : b  b 2  4ac Let the given equation be ax2 + bx + c = 0 are complex numbers. where x = 2a where a, b, c∈R and a ≠ 0 Here, a = 1, b = −(2 3 + 3i), c = 6 3 i ∴ The solution of this quadratic equation is given by b2 − 4ac = [−(2 3 + 3i)]2 − 4 × 1 × 6 3 i b  b 2  4ac = 12 − 9 + 12 3 i − 24 3 i x = 2a = 3 − 12 3 i Hence, the roots of the equation ax2 + bx + c = 0 = 3 (1 − 4 3 i) b  b 2  4ac −b − b 2 − 4ac are and So, the given equation has complex roots. These 2a 2a roots are given by The expression (b2−4ac) = D is called the discriminant. 2  3  3i  3(1  4 3 i ) If D < 0 then the roots of the given quadratic x = 2 equation are complex. Now, we shall find 1 − 4 3 i Note : If p + iq is the root of equation ax2 + bx + c = 0 where a, b, c∈R and a ≠ 0 then p − iq is also Let a + ib = 1 − 4 3 i a solution of the given equation. Thus, complex ∴ a2 − b2 + 2iab = 1− 4 3 i roots occur in conjugate pairs. ∴ a2 − b2 = 1 and 2ab = −4 3 Solved Examples : Ex. 1 : Solve x2 + x + 1 = 0 a2 − b2 = 1 and ab = −2 3 Solution : Given equation is x2 + x + 1 = 0 Consider (a2 + b2)2 = (a2 − b2)2 + 4 a2b2 Comparing with ax2 + bx + c = 0 we get = 1 + 4 (12) = 49 a = 1, b = 1, c = 1 ∴ a2 + b2 = 7...(1) These roots are given by and also a2 − b2 = 1...(2) b  b  4ac 2 x = Solving (1) and (2) 2a 2a2 = 8 ∴ a2 = 4 1  3 = ∴ a = ±2 and b = ± 3 2 1  3 i ∴ We have four choices = 2 a = 2, b= 3... (3) 1  3 i −1 − 3 i ∴ Roots are and a = 2, b=− 3... (4) 2 2 8  a = −2, b= 3... (5) Ex. 4 : If x = −5 + 2 −4 , find the value of a = −2, b=− 3... (6) x4 + 9x3 + 35x2 − x + 64. With this combination, the condition Solution : x = −5 + 2 −4 = −5 + 2(2i) = −5 + 4i ab = −2 3 should also be satisfied. Let P(x) = x4 + 9x3 + 35x2 − x + 64 We can verify the values of a and b given by Form the quadratic expression q(x) with real (4) and (5) satisfy the condition and that from (3) coefficients and roots α = −5+4i and α = −5−4i. and (6) do not satisfy the condition. α α = (−5+4i)(−5−4i) = 25 + 16 = 41 Hence, a = 2, b=− 3 α + α = (−5+4i) + (−5−4i) = −10 and a = −2, b= 3 Therefore q(x) = x2 + 10x + 41 ∴ 1 − 4 3 i = ± (2 − 3 i) Now q(α) = α2 + 10α + 41 = 0...(1) ∴ The roots are given by We divide the given polynomial x = (2 ) 3 + 3i ± 3 2 − 3 i( ) p(x) by q(x). x2−x+4 2 So, the roots are 2 3 and 3i. x2+10x+41 x4+ 9x3+35x2 −x+64 x4+10x3+41x2 Ex. 3 : Find the value of x3 − x2 + 2x + 10 − x3 + 6x2 − x when x = 1 + 3 i. − x3 −10x2 − 41x Solution : Since x = 1 + 3i 4x2 + 40x + 64 ∴ x−1 = 3i 4x2 + 40x +164 squaring both sides, we get − 100 (x−1)2 = ( 3 i)2 ∴ p(x) = (x2 + 10x + 41) (x2 − x + 4) − 100 ∴ x2 − 2x + 1 = 3i2 ∴ p(α) = q(α) (α2 − α + 4) − 100 i.e. x2 −2x +1 = −3 ∴ p(α) = 0 − 100 = − 100 (By I) ∴ x2 − 2x = −4........ (I) x3− x2 +2x + 10 = x3 − (x2 − 2x) + 10 (By I) EXERCISE 1.2 = x3 − (−4) + 10 = x3 + 14 = (1 + 3 i)3 + 14 1) Find the square root of the following complex numbers = 1 + 3 3 i − 9 − 3 3 i + 14 i) −8−6i ii) 7+24i iii) 1+4 3 i =6 iv) 3+2 10 i v) 2(1− 3 i) 9 2) Solve the following quadratic equations. is called the real axis. Similarly, y = Im(z) is represented on the Y-axis, so the Y-axis is called i) 8x2 + 2x + 1 = 0 the imaginary axis. ii) 2x2 − 3 x + 1 = 0 iii) 3x2 − 7x + 5 = 0 iv) x2− 4x + 13 = 0 3) Solve the following quadratic equations. i) x2 + 3ix + 10 = 0 ii) 2x2 + 3ix + 2 = 0 iii) x2 + 4ix − 4 = 0 iv) ix2− 4x − 4i = 0 4) Solve the following quadratic equations. Fig. 1.2 i) x2 − (2+i) x −(1−7i) = 0 e.g. (1) (1, 2) ≡ 1 + 2i (2) −4+3i ≡ (−4, 3) ii) x − (3 2 +2i) x + 6 2 i = 0 2 (3) (0, 0) ≡ 0 + 0i (4) 5+0i ≡ (5, 0) iii) x2 − (5−i) x + (18+i) = 0 (5) (0, −1) ≡ 0 − i (6) −2−2i ≡ (−2,−2) iv) (2 + i)x2− (5−i) x +2 (1−i) = 0 A diagram which represents complex 5) Find the value of numbers by points in a plane with reference to the real and imaginary axes is called Argand's i) x3 − x2+x+46, if x = 2+3i. diagram on complex plane. 25 ii) 2x3 − 11x2+44x+27, if x = 3−4i. 1.5.1 Modulus of z : 5 iii) x3 + x2−x+22, if x = 1−2i. If z = a+ib is a complex number then the modulus of z, denoted by | z | or r, is defined as iv) x4 +9 x3+35x2−x+4, if x = −5+ −4. | z | = a 2 + b 2. (From fig. 1.3), point P(a, b) v) 2x4 + 5x3+7x2−x+41, if x = −2− 3 i. represents the complex number z = a+ib. 1.5 Argand Diagram or Complex Plane : ∴r=|z|= a 2 + b 2 = OP A complex number z = x + iy, x, y∈R and Hence, modulus of z is the distance of i = −1 is expressed as a point in the plane whose point P from the origin where the point P co-ordinates are ordered pair (x, y). Jean Robert represents the complex number z in the plane. Argand used the one to one correspondence e.g. For, z = 4 + 3i, between a complex number and the points in a Modulus of z = | z | = 16 + 9 = 25 = 5 the plane. 1.5.2 Argument of z : Let z = x + iy be a complex number. Then the point P(x, y) represents the complex OP makes an angle q with positive direction number z = x + iy (fig.1.2) i.e. x + iy ≡ (x, y), of X-axis. q is called the argument or amplitude x = Re(z) is represented on the X-axis. So, X-axis of the complex number z = a+ib, denoted by arg(z). 10  b ∴ q = tan−1 =   = arg(z), a e.g. If z = 2+2i then 2 arg(z) = q = tan−1   2 p ∴ tan−1(1) = 4 Note : If tan x = y then its inverse funtion is given by x = tan−1 y or x = arc tan y Fig. 1.3 eg: p 1  1  p b a 1) As tan = then tan−1 =  = ∴ sinq = r , cosq = r , r ≠ 0 6 3  3 6 ∴ b = rsinq , a = rcosq   p 2) As tan    = − tan = −1 then b  4 4 and tanq = a , if a ≠ 0  tan−1 (−1) =  4 1.5.3 Argument of z in different quadrants/axes : θ = arg z b z = a + ib Example Quadrant/Axis = tan−1   , from Example a (0 ≤ θ < 2π) On positive real a > 0, b = 0 z=3 θ=0 θ=0 (X) axis b θ = tan−1   , a > 0, b > 0 z=1+i In quadrant I a p 1 p θ = tan−1   = (0 < θ < ) 1   4 2 On Positive p p a = 0, b < 0 z = 5i θ= θ= imaginary (Y) axis 2 2 b  1  θ = tan−1   + π θ = tan−1   +π a < 0, b > 0 z=− 3 +i In quadrant II p a  3 ( < θ < π)  5p 2 = +π=. 6 6 On negative real a < 0, b = 0 z = −6 θ=π θ=π (X) axis 11  b  3 θ = tan−1   + π θ = tan−1  +π a < 0, b < 0 In quadrant III a  1  z = −1 − 3i 3p   π 4p (π < θ < ) = +π= ) 2 3 3 On negative 3p 3p a = 0, b < 0 z = −2i θ= 2 θ= imaginary (Y) axis 2 b  1  θ = tan−1   + 2π θ = tan−1   p a  1  a > 0, b < 0 z=1−i In quadrant IV 3p  ( < θ < 2π) = + 2π = 4 2 4 v) | z 1 z 2 | = | z 1| | z 2| SOLVED EXAMPLES z1 |z | vi) = |z1| , z2 ≠ 0 Ex. 1 : If z = 1+3i, find the modulus and amplitude z2 2 of z. vii) | z1 + z2| ≤ | z1| + | z2| Solution : z = 1+3i here a=1, b=3 and a, b > 0 viii) | z1 + z2|2 = | z1|2 + | z2|2 + 2Re (z1 z2 ) ∴ | z | = 12 + 32 = 10 ix) | z1 − z2|2 = | z1|2 + | z2|2 − 2Re (z1 z2 ) b 3 amp z = tan−1   = tan−1   = tan−1 (3) x) | z1 + z2|2 + | z1− z2|2 = 2(| z1|2 + | z2|2) a 1 xi) | az1 − bz2|2 + | bz1+ az2|2 Ex. 2 : Find the modulus, argument of the complex number −7 + 24i. = (a2+b2) (| z1|2 + | z2|2) where a, b∈R Solution : let z = −7+24i a = −7, b = 24 Properties of arguments : ∴|z|= (7) + (24) = 2 2 625 = 25 1) arg (z1. z2) = arg z1 + arg z2 Here, z1 2) arg = arg z1 − arg z2 b z2 arg z = tan−1   = tan−1   24 a 3) arg z + arg z = 0, arg (z z ) = 0  7  1 4) arg z = − arg z = arg As a < 0, b > 0, q lies in 2nd quadrant. z 5) If arg z = 0 then z is real Properties of modulus of complex numbers : 1.5.4 Polar form of a complex number : If z1, z2, z3 are complex numbers, then Let the complex number z = a+ib be i) | z | = 0 ⇔ z = 0 i.e. Re (z) = Im (z) = 0 represented by the point P(a, b) (see fig 1.4) ii) | z | = | −z | = | z | = |− z | b Let m∠XOP = q = tan−1   and l(OP) = r iii) −| z | ≤ Re (z) ≤ | z |; −| z | ≤ Im (z) ≤ | z | a = a 2 + b 2 >0, then P(r, q) are called the Polar iv) z z = | z |2 Co-ordinates of P. 12  We call the origin as pole. (figure 1.4) −z z −z z Fig.1.4 Fig. 1.5 As a = rcosq, b = rsinq Thus, q together with r gives the position of ∴ z = a + ib becomes the point A in the Argand's diagram. Hence, from the figure 1.5, z = rcosq + irsinq p 3p ∴ z = r(cosq + isinq) arg z = 45° = , arg (− z ) = 135° = 4 4 This is called polar form of complex number z = a + ib 5p 7p arg (−z) = 225° = , arg z = 315° = 4 4 1.5.5 Exponential form : Ex. 2 : Represent the following complex numbers It is known and can be proved using special in the polar form and in the exponential form series that eiq = cosq + isinq i) 4+4 3 i ii) −2 iii) 3i iv) − 3 + i ∴ z = a + ib = r (cosq + isinq) = r eiq Solution : where r = | z | and q = arg z is called an exponential form of complex number. i) Let, z = 4 + 4 3 i a = 4, b = 4 3 Solved Example: Ex. 1 : Represent the complex numbers r = 42 + (4 3 ) 2 = 16 + 48 = 64 = 8 z = 1+i, z = 1−i, − z = −1+i, −z = −1−i in As q lies in quadrant I Argand's diagram and hence find their arguments from the figure. b 4 3 q = tan−1   = tan−1   Solution : a  4  arg z is the angle made by the segment OA p with the positive direction of the X-axis. (Fig.1.5) = tan−1( 3 ) = or 60° 3 13 ∴ The polar form of z = 4 + 4 3 i is  1  π = tan−1   +π= −6 +π z = r (cosθ + i sinθ) − 3 5π z = 8 (cos 60° + i sin 60°) = 6 π π ∴ The polar form of = 8 (cos + i sin ) 3 3 z = r (cosθ + i sinθ) π  i  ∴ The exponential form of z = 8 e 3 5π 5π z = 2 (cos 6 + i sin ) 6 ii) Let z = −2 ∴ The exponential form of  5π  i ∴ a = −2, b = 0 z = re = 2 e iθ  6   Hence, r = (−2) 2 + 02 = 4=2 3π i Ex. 3 : Express z =.e 4 2.e in the a + ib form. As point (−2, 0) lies on negative real axis 3π i θ = πc or 180° Solution: z =.e 4 2.e = r eiθ ∴ The polar form of z = 2(cos180°+i sin180°) 3π ∴r= ,θ= 4 2.e = 2 (cos π + isin π) As the polar form of z is ∴ The exponential form of z = 2eiπ z = r (cosθ + i sinθ) iii) Let z = 3i 3π 3π = 2.e(cos + i sin ) 4 4 a=0, b=3 By using allied angles results in trigonometry, we Hence, r = 02 + 32 = 3 get As point (0, 3) lies on positive imaginary axis 3π  π π 1 πc cos = cos  π −  = − cos =− 4  4 4 2 θ = 2 or 90° 3π  π π 1 sin = sin  π −  = sin = ∴ The polar form of z = 3(cos90° + i sin90°) 4  4 4 2 π π = 3 (cos 2 + i sin 2 )  −1  1   ∴ z = 2.e +i  i π  2  2  ∴ The exponential form of z = 3e 2 = −1 + i 5π i π i iv) Let, z = − 3 + i Ex. 4 : Express (i) 3.e 12 × 4.e 12 ∴a=− 3, b=1  π π  2  cos + i sin  r = (− 3 ) + 1 = 3 + 1 = 4 = 2 2 2  12 12  ii) in a + ib form  5π 5π  As (− 3 , 1) lies in quadrant II 2  cos + i sin   6 6  b q = tan−1   = π a 14  5π i π i Solution: (i) 3.e 12 × 4.e 12 3) If z = 3 + 5i then represent the z , z , − z , − z in Argand's diagram.  5π π   + i = (3 × 4) e 12 12  4) Express the following complex numbers in 6π i π i polar form and exponential form. = 12 e = 12 e 12 2 1 i) −1+ 3 i ii) −i iii) −1 iv) 1+i  π π = 12  cos + i sin  1+ 2i 1+ 7i  2 2 v) vi) 1−3i (2−i)2 = 12 (0 + i) = 12i 5) Express the following numbers in the form  π π  π x+iy 2  cos + i sin  i  12 12  e 2.e 12 π π (ii) = 5π i) 3 (cos 6 + isin 6 )  5π 5π  i 2  cos + i sin  2e 6  6 6  7π 7π ii) 2.e(cos 4 + isin 4 )  2   12π − 56π  i  2   − 34π  i =  e = e   5π   5π   2   2  iii) 7  cos  −  + i sin  −    6   6   2    3π   3π    cos  −  + i sin  −   -4π 5π i =  π i i  2   4   4  iv) e 3 v) e 3 vi) e 6  2  1  1  =    − +i−  6) Find the modulus and argument of the  2  2  2  1+ 2i complex number. 1 i 1−3i = − − 2 2 7) Convert the complex number i−1 EXERCISE 1.3 z= π π in the polar form. cos 3 + isin 3 1) Find the modulus and amplitude for each of 8) For z = 2+3i verify the following : the following complex numbers. i) ( z ) = z ii) z z = |z|2 i) 7 − 5i ii) i iii) −8 + 15i 3 + 2.e iii) (z+ z ) is real iv) z − z = 6i iv) −3(1−i) v) −4−4i vi) 3 −i 9) z1 = 1 + i, z2 = 2 − 3i. Verify the following : vii) 3 viii) 1 + i iv) 1 + i 3 i) z 1+ z 2 = z1 + z2 x) (1+2i)2 (1−i) 4 + 3i sin θ  ii) z1− z2 = z1 − z2 2) Find real values of q for which    1 − 2i sin θ  is purely real. iii) z1. z2 = z1. z2 z1 z1 iv) z2 = z 2 15 1.6 De Moivres Theorem:  π  π = cos  8 ×  + i sin  8 ×  If z1 = r1e iθ1 and z2 = r2e iθ2  3  3 Then z1. z2 = (r1eiθ1) (r2eiθ2)  2π   2π  = cos   − i sin   = r1.r2 ei(θ1 + θ2)  3   3  That is if two complex numbers are  8π 2π   − 2π =  multiplied then their modulii get multiplied and  3 3  arguments get added.  π  π = cos  π −  + i sin  π −   3  3 z1 r1eiθ1  r1  i (θ1 −θ2 ) = =  .e π  π  z2 r2 eiθ2  r2  = − cos   + i sin   3 3 That is, if one complex numbers is divided by the other, then their modulii get divided and 1 3 = − +i arguments get substracted. 2 2 In 1730, De Moivre propsoed the  π π  15 following theorem for finding the power of (ii)  cos − i sin   10 10  a complex number z = r (cosθ + i sinθ), as [r(cosθ + i sinθ)]n = rn (cos nθ + i sin nθ)]  15π   15π  = cos   − i sin   for any n ∈ z.  10   10  The proof of this theorem can be given using  3π   3π  = cos   − i sin   the Method of Induction (Chapter 4).  2   2   π  π For example: = cos  π +  − i sin  π +   2  2 i) (cosθ + i sinθ)5 = (cos5θ + i sin5θ) π  π ii) (cosθ + i sinθ)−1 = cos(−θ) + i sin (−θ) = − cos − i  − sin  2  2 2 2  2  = −0 − i (−1) = i iii) (cosθ + i sinθ) 3 = cos  θ  + i sin  θ 3  3  Solved Examples: (cos 5θ + i sin 5θ ) 2 (iii) (cos 4θ − i sin 4θ )3 Ex. 1 Use De Moivres Theorom and simplify. cos 10θ + i sin 10θ 8 =  π π cos 12θ − i sin 12θ i)  cos + i sin   3 3 cos 10θ + i sin 10θ = cos ( −12θ ) + i sin ( −12θ ) 15  π π  ii)  cos − i sin   10 10  = cos [10θ − (−12θ)] + i sin [10θ − (−12θ)] ( cos 5θ + i sin 5θ ) 2 iii) = cos 22θ + i sin 22θ ( cos 4θ − i sin 4θ ) 3 Ex. 2 Express (1 + i)4 in a + ib form. 8  π π Solution : Let z = 1 + i Solution: (i)  cos + i sin   3 3 ∴ x = 1, y = 1 16  r = 11  2 , 1 = [(−1)2+2×(−1)×i 3 +(i 3 )2] 4 x θ = tan−1 1 y = (1−2i 3 −3) π 4 = tan−1 1 = 1 4 = (−2−2i 3 )    4 ∴ z = r(cosθ + isinθ) = 2 cos  i sin   4 4 −1−i 3 = 2 4 2     1  i 3  ∴ z4 = (1 + i)4 = 2 cos  i sin   4 4 Similarly it can be verified that    2  1  i 3  4 4  = = 2 cos  i sin 2  4 4  Thus cube roots of unity are 1, = 2  cos   i sin   2 1  i 3  1  i 3  = 2  1  i(0)  ,  2  2   2  1   2 = − 2 + 0i 2 = 1  i 3  1  i 3  Let = w, then   = w2 2  2  1.7 Cube roots of unity : Hence, cube roots of unity are 1, w, w2 Number 1 is often called unity. Let x be the 2 cube root of unity i.e. 1  1  i 3   1  i 3  where w =   and w2 =   ∴ x3 = 1  2   2  ∴ x3 −1 = 0 2p i 4p i Also note that 1 = e2pi, w = e 3 , w2 = e 3 ∴ (x −1)(x2 + x +1) = 0 ∴ x −1 = 0 or x2 + x +1 = 0 Properties of 1, w, w2 i) w is complex cube root of 1. 1  (1) 2  4 11 ∴ x = 1 or x = ∴ w3 = 1 2 1 1  3 ii) w3 −1 = 0 ∴ x = 1 or x = i.e. (w −1) (w2 +w+1) = 0 2 ∴ w =1 or w2 +w+1 = 0 1  i 3 ∴ x = 1 or x = but w ≠1 2 ∴ Cube roots of unity are ∴ w2 +w+1 = 0 1 1 1  i 3 −1 − i 3 iii) w2 = and w2 = w 1, , w 2 2 Among the three cube roots of unity, one is iv) w = 1 so w3n = 1 3 real and other two roots are complex conjugates v) w4 = w3.w = w so w3n+1 = w of each other. vi) w5 = w2.w3 = w2.1 = w2 So w3n+2 = w2 2  1  i 3  vii) w = w2 =    2  viii) w 2 = w 17 1.8 Set of points in complex plane Illustration: If z = x + iy represents the variable point For z1 = 2 +3i , z2 = 1 +i and z = x + iy P(x,y) and z1 = x1 + iy1, represents the fixed point A (x1,y1) then (i) z−z1 represents the length of (i) z − z1 = ( x + iy ) − (2 + 3i ) = x + iy − 2 − 3i AP = ( x − 2) + i ( y − 3) = ( x − 2) + ( y − 3) 2 2 represents the distance between (x,y) and (2,3). (ii) If z − z1 = 5 , then Fig. 1.6 ( x + iy ) − (−1 + i ) = ( x + 1) + i ( y − 1) = 5 (2) z−z1= a represents the circle with centre A(x1,y1) and radius a. ∴ ( x + 1) 2 + ( y − 1) 2 = 5 ∴ ( x + 1) 2 + ( y − 1) 2 = 52 represents the circle with centre (−1,1) and = P(z) radius 5. (iii) If z − z1 = z − z2 then A(z1) ( x + iy ) − (2 + 3i ) = ( x + iy ) − (−1 + i ) ∴ ( x − 2) + i ( y − 3) = ( x + 1) + i ( y − 1) ∴ ( x − 2) 2 + ( y − 3) 2 = ( x + 1) 2 + ( y − 1) 2 Fig. 1.7 ∴ x2 − 4 x + 4 + y 2 − 6 y + 9 = x2 + 2x + 1 + y 2 − 2 y + 1 x2 − 4 x + 4 + y 2 − 6 y + 9 = x2 + 2x + 1 + y 2 − 2 y + 1 (3) z−z1=z−z2represents the perpendicular bisector of the line joining the points A and ∴ −6 x + 4 y + 11 = 0 i.e. 6 x + 4 y − 11 = 0 B. represents the perpendicular bisector of line joining points (2,3) and (−1,1). (z2) SOLVED EXAMPLES A(z1) Ex. 1 : If w is a complex cube root of unity, then prove that (z) 1 1 i) + w2 = −1 w ii) (1+ w2)3 = −1 iii) (1−w+w2)3 = −8 Fig. 1.8 Solution : Given, w is a complex cube root of unity. ∴ w3 = 1 Also w2 +w+1 = 0 18  ∴ w2 +1 = − w and w +1 = − w2 = [(1−w)(1−w2)]2 1 1 w+1 −w2 i) + w2 = = w2 = −1 = (1−w2−w+w3)2 w w2 ii) (1+ w2)3 = (−w)3 = −w3 = −1 = [1−(w2+w)+1]2 = [1−(−1)+1]2 = (1+1+1)2 = (3)2 = 9 iii) (1−w+w2)3 = (1+w2−w)3 = (−w−w)3 (∵ 1 +w2 = −w) Ex. 3 : If w is a complex cube root of unity such that x=a+b, y=aw+bw2 and z=aw2+bw, a, b∈R = (−2w)3 = −8 w3 = −8 × 1 = −8 prove that Ex. 2 : If w is a complex cube root of unity, then i) x+y+z=0 ii) x3+y3+z3=3(a3+b3) show that Solution : Since w is a complex cube root of unity i) (1−w+w2)5 + (1+w−w2)5 = 32 ∴ w3 = 1 and w2 +w+1 = 0 but w ≠ 1 given ii) (1−w)(1−w ) (1−w )(1−w ) = 9 2 4 5 ∴ w2 +1 = −w, w+1 = − w2, w3 = w6 = 1 Solution : i) x+y+z = a+b+aw+bw2+aw2+bw i) Since w = 1 3 = a(1+w+w2)+b(1+w+w2) and w ≠ 1 ∴ w +w+1 = 0 2 = a.0 + b.0 Also w2 +1 = − w, w2 +w = −1 = 0 and w +1 = − w2 Now, (1−w+w2)5 ii) x3 = a3+3a2b+3ab2+b3 = (−w−w)5 (∵ 1 +w2 = −w) y3 = a3w3+3a2bw4+3ab2w5+b3w6 = (−2w)5 z3 = a3w6+3a2bw5+3ab2w4+b3w3 = −32w5 adding the above three equations x3+y3+z3 = a3(1+w3+w6)+3a2b(1+w+w2) + (1+w−w2)5 = (−w2−w2)5 3ab2 (1+w2+w)+b3(1+w6+w3) = (−2w2)5 = 3a3 +3a2b(0) +3ab2(0)+3b3 = −32w 10 = 3(a3 + b3) ∴ (1−w+w2)5 + (1+w−w2)5 = −32w5−32w10 Ex. 4 : Prove that = −32w (1+w ) 5 2 1+wn+w2n = 3, if n is multiple of 3 = −32w5×(−w) = 32w6 = 32(w3)2 1+wn+w2n = 0, if n is not multiple of 3, n∈N = 32 × (1)2 = 32 Solution : for n, k∈N, if n is multiple of 3 then n=3k and if n is not multiple of 3 then n=3k+1 or ii) (1−w)(1−w2) (1−w4)(1−w5) 3k+2 = (1−w)(1−w2) (1−w3.w)(1−w3.w2) ∴ if n is multiple of 3 = (1−w)(1−w ) (1−w)(1−w ) 2 2 i.e. n=3k = (1−w)2(1−w2)2 then 1+wn+w2n = 1+w3k+w2×3k 19  = 1+(w3)k+(w3)2k 4) If α and β are the complex cube root of unity, show that = 1+(1)k+(1)2k (a) α2+β2+αβ = 0 =1+1+1 (b) α4+β4+α−1β−1 = 0 =3 If n = 3k + r , r = 1,2. 5) If x = a+b, y = αa+βb and z = aβ+bα where α and β are the complex cube-roots of unity, As w is complex root of Unity. show that xyz = a3+b3 wr, r = 1,2 is also complex root of Unity. 6) Find the equation in cartesian coordinates of ∴1+w +w =0 r 2r the locus of z if we have, 1+wn+w2n=0, if n is not a multiple of 3. (i) z = 10 (ii) z − 3 = 2 (iii) z − 5 + 6i = 5 (iv) z + 8 = z − 4 (v) z − 2 − 2i = z + 2 + 2i EXERCISE 1.4 (vi) z + 3i = 1 z − 6i 1) Find the value of i) w18 ii) w21 iii) w−30 iv) w−105 7) Use De Moivres theorem and simplify the following ( cos 2θ + i sin 2θ ) 7 2) If w is a complex cube root of unity, show i) that ( cos 4θ + i sin 4θ ) 3 i) (2−w)(2−w2) =7 cos 5θ + i sin 5θ ii) (1+w−w2)6 = 64 ii) ( cos 3θ − i sin 3θ ) 2 iii) (1+w)3−(1+w2)3 = 0 4  7π 7π  iv) (2+w+w ) −(1−3w+w ) = 65 2 3 2 3  cos + i sin  13 13  v) (3+3w+5w2)6 −(2+6w+2w2)3 = 0 iii)  6  4π 4π  a+bw+cw2  cos − i sin  vi) = w2  13 13  c+aw+bw2 vii) (a + b) + (aw + bw2) + (aw2 + bw) = 0 8) Express the following in the form a + ib, a, b∈ R, using De Moivre's theorem. viii) (a − b) (a − bw) (a − bw2) = a3 − b3 ix) (a + b)2 + (aw + bw2)2 + (aw2 + bw)2 = i) (1 − i)5 ii) (1 + i)6 iii) (1 − 3 i)4 6ab iv) (−2 3 − 2i)5 3) If w is a complex cube root of unity, find the value of Let's Remember 1 i) w+ ii) w2+w3+w4 iii) (1+w2)3 w ∙ A number a+ib, where a, b∈ R and i = −1 , iv) (1−w−w2)3 + (1−w+w2)3 is a complex number. v) (1+w)(1+w2)(1+w4)(1+w8) ∙ Let z1 = a+ib and z2 = c+id. Then z1 + z2 = (a+c) + (b+d)i 20  z1 − z2 = (a−c) + (b−d)i 4) If w is a complex cube root of unity, then the z1. z2 = (ac−bd) + (ad+bc)i value of w99+w100+w101 is : z1  ac  bd   bc  ad  A) −1 B) 1 C) 0 D) 3   i z z z2  c 2  d 2   c 2  d 2  5) If z=r(cosq+isinq), then the value of + z z ∙ For any non-zero complex number z = a+ib A) cos2q B) 2cos2q C) 2cosq D) 2sinq 1 a −b = +i 2 2 z a2+b2 a +b 6) If w(≠1) is a cube root of unity and (1+w)7 ∙ For any positive integer k, = A + Bw, then A and B are respectively the i4k = 1, i4k+1 = i, i4k+2 = −1, i4k+3 = − i numbers A) 0, 1 B) 1,1 C) 1, 0 D) −1,1 ∙ The conjugate of z = a+ib is z , is given by z = a−ib 7) The modulus and argument of (1+i 3 )8 are respectively ∙ The polar form of the complex number