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  • Lecture 4

    Roots of complex numbers

    Characterization of a polynomial by its roots

    Techniques for solving polynomial equations

  • ROOTS OF COMPLEX NUMBERS

    Def.:

    A number u is said to be an n-th root of complex number zif un = z, and we write u = z1/n.

    Th.:

    Every complex number has exactly n distinct n-th roots.

    Let z = r(cos + i sin ); u = (cos+ i sin). Then

    r(cos + i sin ) = n(cos+ i sin)n = n(cosn+ i sinn)

    n = r , n = + 2k (k integer)Thus = r1/n , = /n+ 2k/n .

    n distinct values for k from 0 to n 1. (z 6= 0)

    So u = z1/n = r1/n[

    cos

    (

    n+

    2k

    n

    )

    + i sin

    (

    n+

    2k

    n

    )]

    , k = 0, 1, . . . , n1

  • Note. f(z) = z1/n is a multi-valued function.

    1 ( . . 1 0)n nx i e x= ! =

    1= e2m! " 11/n = e2m! /n

    =cos(2m!

    n

    )+sin(2m!

    n

    )

    !"#$%&'()(*(+,-(.//,0(/1(2+3,4(*(

    11/5 = cos(2m!

    5) + sin(

    2m!

    5) (m = 0,1,2,3,4).

    11n

    x/

    ! =

  • Example 2. Find all cubic roots of z = 1 + i:u = (1 + i)1/3

    u = (2)1/3

    [

    cos

    (

    3

    4

    1

    3+

    2k

    3

    )

    + i sin

    (

    3

    4

    1

    3+

    2k

    3

    )]

    , k = 0, 1, 2

    that is,

    k = 0 : 21/6(

    cos 4 + i sin4

    )

    k = 1 : 21/6(

    cos 1112 + i sin1112

    )

    k = 2 : 21/6(

    cos 1912 + i sin1912

    )

    k=2

    k=0

    k=1

    Equivalently:

    u = (1 + i)1/3 = e(1/3) ln(1+i) = e(1/3)[ln

    2+i(3/4+2k)]

    = (2)1/3ei(/4+2k/3)

  • !""#$%"&%'"()*"+,-($%

    1

    1 0( )n n

    n nP z a z a z a

    !

    !" + + + .!

    ( ) 0izP z = = ! z

    i is a root

    [Gauss, 1799]

    proof of fundamental theorem of algebra is given in the course

    Functions of a complex variable, Short Option S1

  • 1

    1 20

    1

    1

    1

    1

    ( )( ) ( )

    ( ( 1) )

    n n

    n n n

    nn

    n

    j

    n n n

    n

    j

    j

    j

    z z za z a z a a z z z

    a z z z z

    !

    !

    !

    = =

    + + + = ! ! !

    = ! + + ! ." #

    ! !

    !

    !"#$#%&'$()(*+,#,-./0*.1(#/,20,(&),$..&),

    3..&),.4,-./0*.1(#/),

    1

    1 0( )n n

    n nP z a z a z a

    !

    !" + + + .!

    ( ) 0izP z = = ! z

    i is a root

    1 0 ( 1)

    ! !

    nn

    j jn n

    a a

    a az z! = ! ; = !" #

    !.1-#$(*+,%.'5%('*&),.4,6*78,#*9,6:,

  • !"#"$%&'()'*+$!%&'*,-.$/$2

    2 1 0a x a x a+ +

    11 2

    2

    ( )a

    x xa

    = ! +0

    1 2

    2

    .a

    x xa

    =0&1$,2$),,3.$ 4),(&+3$,2$),,3.$

    ( )21 1 2 01,2

    2

    4

    2

    a a a a

    xa

    ! !=

    52$+,167!89$),,3.$+,1!$:-$+,167!8$+,-;'3!$6':).$

    !-!)'7$.,7&*,-.$-,3$'?':7'@7!$2,)$A:#A!)$,)(!)$6,7B-,1:'7.$C%&')*+.$'-($'@,?!D$

    E'-$F-($.,7&*,-.$:-$.6!+:'7$+'.!.G"$

    2,)$),,3.$

    H$

  • Example 1:

    z5 + 32 = 0

    The solutions of the given equation are the fifth roots of 32:

    (32)1/5 = 321/5[

    cos

    (

    5+

    2k

    5

    )

    + i sin

    (

    5+

    2k

    5

    )]

    , k = 0, 1, 2, 3, 4

    that is,

    k = 0 : 2(

    cos 5 + i sin5

    )

    k = 1 : 2(

    cos 35 + i sin35

    )

    k = 2 : 2k = 3 : 2

    (

    cos 75 + i sin75

    )

    k = 4 : 2(

    cos 95 + i sin95

    )

    Re z

    k=0

    k=1

    k=2

    k=3

    k=4

    Im z

  • (z + )7+ (z ! )7 = 0

    (z +

    z ! )7 = !1= e(2m+1)"

    !z +

    z " = e(2m+1)# /7

    ! z(1" e(2m+1)# /7 ) = "(1+ e(2m+1)# /7 )

    ! z =

    e(2m+1)" /7 +1

    e(2m+1)" /7 #1

    = e(2m+1)! /14 + e"(2m+1)! /14

    e(2m+1)! /14 " e"(2m+1)! /14= 2cos( 2m+1

    14! )

    2sin( 2m+114

    ! )= cot( 2m+1

    14! )

    !"#$%$&$'$($)$*+

    ,-.!/01+&+2+34456+47+/40894!:.06+

  • (z + )7 + (z ! )7 = 0

    r n rx y

    !!"#$%&'(")#)""*#+,"#-'".-%")+#'/## ( )

    nx y+%)#

    0

    1

    2

    3

    4

    5

    1( )

    1 1( )

    1 2 1( )

    1 3 3 1( )

    1 4 6 4 1( )

    1 5 10 10 5 1( )

    x y

    x y

    x y

    x y

    x y

    x y

    +

    +

    +

    +

    +

    +

    0,"1"#23"#-')4")%")+&5#'6+2%)"*#/3'7#821-2&91#+3%2):&"#;#

    (z + )7 + (z ! )7 = 0 " z7 ! 21z5 + 35z3 ! 7z = 0

    1 7 21 35 35 21 7 1

  • 7 5 3

    6 4 2

    3 2 2

    21 35 7 0

    21 35 7 0 0

    21 35 7 0 ( )

    z z z z

    z z z or z

    w w w w z

    ! + ! =

    " ! + ! = =

    " ! + ! = #

    !"#$%&'(')*+$#,-*.%)$

    /*)$0#$1&'2#)$')$*)%3"#&$4%&5$6$

    3 221 35 7 0w w w! + ! =7#)/#$3"#$&%%38$%4$ *$

    2 2 114

    cot ( )mw !+= ( 0 1 2)m = , ,

    2 2 2 1140

    cot ( ) 21mm

    !+=

    " =#9-5$%4$&%%38$

    (z + )7 + (z ! )7 = 0 " z7 ! 21z5 + 35z3 ! 7z = 0

    , 3mz m =

    :;*5#3$*)%3"#&$4%&5$

  • Pascals triangle = table of binomial coefficients(

    n

    r

    )

    = n!r!(nr)!

    i.e., coefficients of xrynr in (x+ y)n

    [Pascal, 1654]

    1

    1 1

    1 2 1

    1 3 3 1

    1 4 6 4 1

    1 5 10 10 5 1

    1 6 15 20 15 6 1

    1 7 21 35 35 21 7 1

    r-th element of n-th row given by sum of two elements above it in (n 1)-th row:

    (

    n

    r

    )

    =

    (

    n 1

    r 1

    )

    +

    (

    n 1

    r

    )

    [use (x+ y)n = (x+ y)n1x+ (x+ y)n1y ]

  • Historical note

    binomial coefficients already known in the middle ages:

    Pascals triangle first discovered by Chinese mathematicians of 13thcentury to find coefficients of (x+ y)n

    (

    n

    r

    )

    = n!r!(nr)! in Hebrew writings of 14th century is

    number of combinations of n objects taken r at a time

    Pascal (1654) rediscovers triangle and most importantly unitesalgebraic and combinatorial viewpoints

    theory of probability; proof by induction

  • 3 27 7 1 0z z z+ + + = .

    !"#$# %&'()*+#*",-./*#0)*+*#()*#1&2*+/34&5#*61,7'0#&'(#'9:4'18#;#

    1 8 28 56 70 56 28 8 1,8?,/@8#(+4,&5/*#48#

    8 8 7 5 312

    3 2 2

    [( 1) ( 1) ] 8 56 56 8

    8 [ 7 7 1] ( )

    z z z z z z

    z w w w w z

    + ! ! = + + +

    = + + + " .

    8'#

    A'0##8 8( 1) ( 1) 0z z+ ! ! = #0)*#

    2 81

    1em iz

    z

    ! /+

    "=

    #4B*B#0)*#

    z =

    em! /4 +1

    em! /4 "1= "cot(m! / 8) (m = 1,2,,7),

    8'#()*#+''(8#'=#()*#54:**61,7',+*##

    2cot ( 8)z m!= " /

    C##

    1 2 3m = , ,

    B##

    0

    1

    2

    3

    4

    5

    1( )

    1 1( )

    1 2 1( )

    1 3 3 1( )

    1 4 6 4 1( )

    1 5 10 10 5 1( )

    y

    x y

    x y

    x y

    x y

    x y

    +

    +

    +

    +

    +

    +

  • Example

    Question from 2008 Paper

    Find all the solutions of the equation(

    z + i

    z i

    )n

    = 1 ,

    and solve

    z4 10z2 + 5 = 0 .

    (

    z + i

    z i

    )n

    = 1 = ei(+2N) , N integer z + i

    z i= ei(/n+2N/n) , N = 0, 1, . . . , n1

    Then : z = iei(/n+2N/n) + 1

    ei(/n+2N/n) 1= i

    cos[(1 + 2N)/(2n)]

    i sin[(1 + 2N)/(2n)]= cotg

    (1 + 2N)

    2n.

    Forn = 5 : (z+i)5 = (zi)5 z(z410z2+5) = 0 . Then the 4 roots of z410z2+5 = 0 are

    cotg

    10, cotg

    3

    10, cotg

    7

    10, cotg

    9

    10.

  • 2 22 2 2 2 2 2

    2 2

    2 ( 1)( 2 cos )( 2 cos ) ( 2 cos )

    m mz a m

    z az a z az a z az az a m m m

    ! ! !" "= " + " + " + .

    "!

    !"#$####%&'(#)&*)##

    zr= acos

    r!

    m a

    2 cos2r!

    m" a

    2

    =a(cosr!

    m

    1" cos2r!

    m)=ae r! /m (r=0,1,,m).

    +''),#

    21

    ma =-.*/012#3'.430.1)# 5678#*1/#9678#0/.1:3*;#

    2 2 2 2 2 2 2 22 ( 1)( ) ( )( 2 cos )( 2 cos ) ( 2 cos )m

    Q z z a z az a z az a z az am m m

    ! ! !"# " " + " + " + .!

    P(z) ! z2m

    " a2m

    z

    r= aer! i /m )6+''),#

  • This concludes part A of the course.

    A. Complex numbers

    1 Introduction to complex numbers

    2 Fundamental operations with complex numbers

    3 Elementary functions of complex variable

    4 De Moivres theorem and applications

    5 Curves in the complex plane

    6 Roots of complex numbers and polynomials