COMPLEX NUMBERS EXAMPLES & SOLUTIONS

1 Producedby The Open University of Sri Lanka20150 COMPLEXNUMBERSEXAMPLES& SOLUTIONSD epartment of Mathematics & PhilosophyThe Open University of Sri LankaProducedby The Open University of Sri Lanka20151 EXAMPLES for COMPLEX numbersQuestion(01)(i) Find the real values ofxandysuch that(1)2(2 3 )33i xii y iiii ++++= +(ii)Find the real values ofxandyare the COMPLEX numbers23ix y and24xyi conjugate of each other.(iii)Find the square roots of4 4i+(iv)Find the COMPLEX numberZsatisfying the equation12583 ZZi = and418ZZ = (v) Find real such that3 21 2 sini sini + is(a) real(b) imaginarySolution(i)(1)2(2 3 )33i xii y iiii ++++= +{}{}(1)2 (3)(2 3 )(3)(3)(3)i xiii y iiiii ++ +++ = +222(3)(1)62(2 3 )(3)39ii xiiii yi iii+ ++++ + = 222(33)6(6 923 )39 1ii i xiiii yi ii+ +++ ++= +(4 2 )(9 7 )9110i xi yii + + = COMPLEX NUMBERS .

2 EXAMPLES & SOLUTIONSP roducedby The Open University of Sri Lanka20152[]491(19 27 ) 0xyixy+ + = 491 (1)xy+ =19 (2)xy =By solving equations (1) and (2)825x =375y=(ii)2234ix yxyi = 223() 4ix yxyi = ++2() 3xy +=24x y =24xy = 43yy =234 0yy+ =(4)(1)0yy+ =4y= or1y=when4y= 2414x == 1x = 1y=2441x == (Not real)1x = and4y= (iii)Let24 4zi= +and()zx iy=+4 4zi =+2()()zx iy x iy=++2222xixy i y=++222()2zxyi xy= +Producedby The Open University of Sri Lanka2015322()24 4xyi xyi += +224xy =24xy=2yx={}2224xx =4244 0xx + =22(2)0x =22x =2x= 222y == 22zi=+or22zi= (iv) (1)83zzi = (2)8zz = Letzx iy= +2212 (12)12(12)zxiyzxy = + = +228(8)8(8)zix i yzixy = + =+ 224 (4)4(4)zxiyzxy = + = +228 (8)8(8)zxiyzxy = + = +From (1);22221253 (12)5(8)83zxyxyzi = +=+ {}22229 (12)25(8).

3 (1)xyxy +=+ Producedby The Open University of Sri Lanka20154 From (2)418zz = 2222(4)(8)xyxy += +2222(4)(8)xyxy += +22228161664xxyxxy ++= ++848x=6x = form (1);22229 (6 12)25 6(8)yy +=+ 229 (36)25 36 (8)yy +=+ 216400(2500 324) 0yy + =2164002176 0yy +=225136 0yy +=(17)(8)0yy =17y=or8y=6 17zi= +or6 8zi= +(v)3 2 sin(3 2 sin )(1 2 sin )1 2 sin(1 2 sin )(1 2 sin )iiiziii +++== +222(3 4 sin)8 sin1 4 sin1 4 sini =+++Ifzis real28 sin01 4 sin =+2 =( 1)2nn =+ n Ifzis imaginary223 4 sin01 4 sin =+Producedby The Open University of Sri Lanka2015523sin4 =3sin2 = 3 = 3n = n Question(02)(i)Express the following COMPLEX NUMBERS in the polar form(a)232ii + (b)(1)(1)(13 )iii ++(c)21 7(2)ii+ (d)

4 0 0i+(ii)Find the modulus and the principal value of the argumentof the followingcomplex NUMBERS (a)1 31 2ii (b)1111iiii + + (c)1 cossini ++(d)21 (1)1 2ii +Producedby The Open University of Sri Lanka20156(iii)Find the modulus and the principal value of the argument of each of the followingcomplex NUMBERS (a)(13 )(1)ii (b)( 33 )(4 4 )ii + (c)( 33 )(4 4 )ii + (iii)Find the square roots of(a)5 12i+(b)15 8i+(c)24 10i+Solution(2)(a)()222223(3)(2)6512(2)( 2)5iiiiiziiii + ==== ++ 211222i = {}22 (cos(4))sin(4))i = + (){}{}2( 2 ) cos(4)sin4cos(4)sin(4)ii = + + []2 cos(44)sin(44)i = + []2 cos(2)sin(2)i = + (b)1(1)(1)(13 )(1)(13 )(1)(1)(13 )(13 )iiiiziiiiii ==+++ + 22212 (13 )2 (13 )(1)(1 3 ) + == Producedby The Open University of Sri Lanka201571(3)4i= 231422i = {}1cos( 56)sin( 56)2i = + (c).

5 2221 71 71 7(1 7 )(3 4 )(2)24(3 4 )(3 4 )(3 4 )iiiiiziiiiii+++++==== + +23 282525ii++=25 25( 1)25ii +== +212 12i = + []2 cos 34sin 3 4i =+(d)0 0zi= +is not possible to write in the form{}cossinzri =+Where0r> Since there is no value such thatcos0 =andsin0 = There is no argument for Zero COMPLEX number .(iii)(a)221 3(1 3 ) (1 2 )1 655 51 2(1 2 ) (1 2 )1 45iiiiiiziiii + ====++ {}1212 ( 12 )ii= = + []2 cos( 34)sin( 34)i = + 2z =andarg( )34z = (b)[][]22211111(1)(1)(1)1(1)(1)(1)1iiiii iiiziiiii + + + += ==+ + []2( 2 )22 cos(2)sin(2)2iii == = + Producedby The Open University of Sri Lanka201582z=arg( )2z = (c)1 cossinzi = ++22 cos (2) 2 sin (2) cos (2)i =+(){}2 cos (2) cos(2)sin2zi =+2 cos2z =arg( )2z =(d).

6 221 (1)1 (1 2)1 211 21 21 2ii iiziii ++====+++{}1 cos 0sin 0i=+1z=arg( )0z=(iii)(a)(13 )(1)zii= 13112. 22222ii = 2 cos()sin() . 2 cos()sin()3344ii = + + []2 2 cos( (34)) sin( (34)) = ++ +[]2 2 cos( 7 12)sin( 7 12)i = + ( )7 12 Arg z = 2 2z=Producedby The Open University of Sri Lanka20159 Question(03)(a)Given that the COMPLEX numberZand its conjugateZsatisfy theequation212 6Z ZiZi+=+find the possible values ofZ.(b)IfZx iy= +and2Za ib= +where, , ,x y a bare real,prove that2222xaba=++By solving the equation42625 0ZZ++=for2Z,or otherwise express each of the fourroots of the equation in the formx iy+.

7 Solution(a).212 6z zizi+=+Letzx iy= +,x y zx iy= 22()z zxy=+222 () 12 6xyi x iyi+++=+22(2 ) 212 6xyyxii + +=+22212xyy + =and26x=3x =223212yy+ =223 0yy =(3)(1) 0yy +=3y =or1y= 3 3zi= +or3zi= (b)zx iy= +2222()()2zx iyxyi xy=+= +2za ib= +Producedby The Open University of Sri Lanka20151022axy = and2bxy=2byx=222baxx = 2224baxx= 422440xaxb =222416168aabx +=2222aabx +=Sincex 20x 2222xaab= ++42625 0zz++=22(3)16 0z++=()2223(4 )zi+= 23 4zi= +or23 4zi= Letzx iy= +and23 4za ibi= += +2222( 3)43 2x = + =21x=1x = and4222byx === 1 2zi= or( 1 2 )zi= + The roots of the equation42625 0zz++=are1 2i+,1 2i +,1 2i ,1 2i Produced by The Open University of Sri Lanka 2015 11 Course Team Authors Web Content Developer Mr.

8 S. M. Lal Chulawansa Miss. J. I. Y. Jayaweera Miss. M. S. S. Fernando The Open University of Sri Lanka Nawala, Nugegoda, Sri Lanka OER Transformation 2015 2014, Open University of Sri Lanka (OUSL). OUSL OER is developed by the Centre for Educational Technology and Media. Except where otherwise noted, content on this site is licensed under a Creative Commons Attribution-NonCommercial- Share Alike License