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Lecture 4 • Roots of complex numbers • Characterization of a
17
Lecture 4 • Roots of complex numbers • Characterization of a polynomial by its roots • Techniques for solving polynomial equations
Transcript of Lecture 4 • Roots of complex numbers • Characterization of a
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 θ + i sin θ); u = ρ(cosα+ i sinα). Then
r(cos θ + i sin θ) = ρn(cosα+ i sinα)n = ρn(cosnα+ i sinnα)
⇒ ρn = r , nα = θ + 2πk (k integer)
Thus ρ = r1/n , α = θ/n+ 2πk/n .
n distinct values for k from 0 to n− 1. (z 6= 0)
So u = z1/n = r1/n[
cos
(
θ
n+
2πk
n
)
+ i sin
(
θ
n+
2πk
n
)]
, k = 0, 1, . . . , n−1
Note. f(z) = z1/n is a “multi-valued” function.
1 ( . . 1 0)n nx i e x= ! =
1= e2m! ı " 11/n = e2m! ı /n
=cos(2m!
n
)+ısin(2m!
n
)
!"#$%&'()(*(+,-(.//,0(/1(2+3,4(*(
11/5 = cos(2m!
5) + ısin(
2m!
5) (m = 0,1,2,3,4).
11n
x/
! =
Example 2. Find all cubic roots of z = −1 + i:u = (−1 + i)1/3
u = (√2)1/3
[
cos
(
3π
4
1
3+
2πk
3
)
+ i sin
(
3π
4
1
3+
2πk
3
)]
, k = 0, 1, 2
that is,
k = 0 : 21/6(
cos π4 + i sin π
4
)
k = 1 : 21/6(
cos 11π12 + i sin 11π
12
)
k = 2 : 21/6(
cos 19π12 + i sin 19π
12
)
k=2
k=0
k=1
• Equivalently:
u = (−1 + i)1/3 = e(1/3) ln(−1+i) = e(1/3)[ln√
2+i(3π/4+2kπ)]
= (√2)1/3ei(π/4+2kπ/3)
!""#$%"&%'"()*"+,-($%
1
1 0( ) n n
n nP z a z a z a
!
!" + + + .!
( ) 0izP z = = ! z
i is a root
[Gauss, 1799]
• proof of fundamental theorem of algebra is given in the course
“Functions of a complex variable”, Short Option S1
1
1 20
1
1
1
1
( )( ) ( )
( ( 1) )
n n
n n n
nn
n
j
n n n
n
j
j
j
z z za z a z a a z z z
a z z z z
!
!
!
= =
+ + + = ! ! !
= ! + + ! ." #
! !
!
!"#$#%&'$()(*+,#,-./0*.1(#/,20,(&),$..&),
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.$
( )2
1 1 2 0
1,2
2
4
2
a a a a
xa
! ± !
=
52$+,167!89$),,3.$+,1!$:-$+,167!8$+,-;&#'3!$6':).$
<$
8$
8$
=,,3./$
>!-!)'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
(
π
5+
2πk
5
)
+ i sin
(
π
5+
2πk
5
)]
, k = 0, 1, 2, 3, 4
that is,
k = 0 : 2(
cos π5 + i sin π
5
)
k = 1 : 2(
cos 3π5 + i sin 3π
5
)
k = 2 : −2
k = 3 : 2(
cos 7π5 + i sin 7π
5
)
k = 4 : 2(
cos 9π5 + i sin 9π
5
)
Re z
k=0
k=1
k=2
k=3
k=4
Im z
(z + ı)7+ (z ! ı)7
= 0
(z + ı
z ! ı)7 = !1= e(2m+1)" ı
!z + ı
z " ı= e(2m+1)# ı /7
! z(1" e(2m+1)# ı /7 ) = "ı(1+ e(2m+1)# ı /7 )
! z = ı
e(2m+1)" ı /7+1
e(2m+1)" ı /7#1
= ıe(2m+1)! ı /14 + e"(2m+1)! ı /14
e(2m+1)! ı /14 " e"(2m+1)! ı /14= ı2cos( 2m+1
14! )
2ısin( 2m+114
! )= cot( 2m+1
14! )
!"#$%$&$'$($)$*+
,-.!/01+&+2+34456+47+/40894!:.06+
(z + ı)7 + (z ! ı)7 = 0
r n rx y
!!"#$%&&#'(")#)""*#+,"#-'".-%")+#'/## ( )nx y+%)#
0
1
2
3
4
5
1( )
1 1( )
1 2 1( )
1 3 3 1( )
1 4 6 4 1( )
1 5 10 10 5 1( )
x y
x y
x y
x y
x y
x y
+
+
+
+
+
+
0,"1"#23"#-')4")%")+&5#'6+2%)"*#/3'7#821-2&91#+3%2):&"#;#
(z + ı)7 + (z ! ı)7 = 0 " z7! 21z5 + 35z3 ! 7z = 0
1 7 21 35 35 21 7 1<+,#3'$#'/#821-2&91#+3%2):&"#%1# 1'#
=>27?&"#@#;#2)#2&+"3)2A4"#/'37#
Solution:
z = cot( 2m+114
! )
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 ( ) 21m
m!+
=" =#9-5$%4$&%%38$
(z + ı)7 + (z ! ı)7 = 0 " z7! 21z5 + 35z3 ! 7z = 0
, 3mz m =
:;*5<+#$=$6$>#3$*)%3"#&$4%&5$
Pascal’s triangle = table of binomial coefficients(
n
r
)
= n!r!(n−r)!
i.e., coefficients of xryn−r 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)n−1x+ (x+ y)n−1y ]
Historical note
♦ binomial coefficients already known in the middle ages:
• “Pascal’s triangle” first discovered by Chinese mathematicians of 13th
century to find coefficients of (x+ y)n
•(
n
r
)
= n!r!(n−r)! 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+ + + = .
!"#$# %&'()*+#*",-./*#0)*+*#()*#1&2*+/34&5#*61,7'0#&'(#'9:4'18#;#
1 8 28 56 70 56 28 8 1<()#+'0#'=#>,8?,/@8#(+4,&5/*#48#
8 8 7 5 312
3 2 2
[( 1) ( 1) ] 8 56 56 8
8 [ 7 7 1] ( )
z z z z z z
z w w w w z
+ ! ! = + + +
= + + + " .
8'#
A'0##8 8( 1) ( 1) 0z z+ ! ! = #0)*&##
2 81
1em iz
z
! /+
"=
#4B*B#0)*&##
z =
em! ı /4+1
em! ı /4"1
= "ıcot(m! / 8) (m = 1,2,…,7),
8'#()*#+''(8#'=#()*#54:*&#*61,7'&#,+*##
2cot ( 8)z m!= " /
C##
1 2 3m = , ,
B##
0
1
2
3
4
5
1( )
1 1( )
1 2 1( )
1 3 3 1( )
1 4 6 4 1( )
1 5 10 10 5 1( )
� y
x y
x y
x y
x y
x y
+
+
+
+
+
+
Example
Question from 2008 Paper
Find all the solutions of the equation(
z + i
z − i
)n
= −1 ,
and solve
z4 − 10z2 + 5 = 0 .
(
z + i
z − i
)n
= −1 = ei(π+2Nπ) , N integer ⇒z + i
z − i= ei(π/n+2Nπ/n) , N = 0, 1, . . . , n−1
Then : z = iei(π/n+2Nπ/n) + 1
ei(π/n+2Nπ/n)− 1
= icos[π(1 + 2N)/(2n)]
i sin[π(1 + 2N)/(2n)]= cotg
π(1 + 2N)
2n.
Forn = 5 : (z+i)5 = −(z−i)5 ⇒ z(z4−10z2+5) = 0 . Then the 4 roots of z4−10z2+5 = 0 are
cotgπ
10, cotg
3π
10, cotg
7π
10, cotg
9π
10.
2 22 2 2 2 2 2
2 2
2 ( 1)( 2 cos )( 2 cos ) ( 2 cos )
m mz a m
z az a z az a z az az a m m m
! ! !" "= " + " + " + .
"!
!"#$####%&'(#)&*)##
zr= acos
r!
m± a
2 cos2r!
m" a
2
=a(cosr!
m
±ı 1" cos2r!
m)=ae± ır! /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 )
mQ z z a z az a z az a z az a
m m m
! ! !"# " " + " + " + .!
P(z) ! z2m
" a2m
z
r= aer! i /m )6+''),#<#
0=.=#%&'(#)&*)## P(z) = Q(z) where
• 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 Moivre’s theorem and applications
5 Curves in the complex plane
6 Roots of complex numbers and polynomials
