lecture 4 roots of complex numbers characterization of · pdf file• roots of complex...
TRANSCRIPT
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