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 z
if un = z, and we write u = z 1/n .
Th.:
• Every complex number has exactly n distinct n-th roots.
Let z = r(cos θ + i sin θ); u = ρ(cos α + i sin α). Then
r(cos θ + i sin θ) = ρn (cos α + i sin α)n = ρn (cos nα + i sin nα)
⇒ ρn = r ,
nα = θ + 2πk
Thus ρ = r 1/n ,
(k integer)
α = θ/n + 2πk/n .
n distinct values for k from 0 to n − 1. (z 6= 0)
So u = z 1/n = r 1/n cos
θ
2πk
+
n
n
+ i sin
θ
2πk
+
n
n
,
k = 0, 1, . . . , n−1
Note. f (z) = z 1/n is a “multi-valued” function.
!"#$%&'()(*(+,-(.//,0(/1(2+3,4(*(
x n = 1 (i.e. x n ! 1 = 0)
! x = 11/n
1 = e2m! ı
" 11/ n = e2m! ı / n
2m!
2m!
=cos(
)+ısin(
)
n
n
2m!
2m!
1 = cos(
) + ısin(
) (m = 0,1,2,3,4).
5
5
1/5
Example 2. Find all cubic roots of z = −1 + i:
u = (−1 + i)1/3
√ 1/3
3π 1 2πk
3π 1 2πk
u = ( 2)
cos
+
+
+ i sin
,
4 3
3
4 3
3
that is,
π
π
1/6
k=0: 2
cos 4 + i sin 4
11π
11π
1/6
k=1
k=1: 2
cos 12 + i sin 12
19π
19π
1/6
k=2: 2
cos 12 + i sin 12
k = 0, 1, 2
k=0
k=2
• Equivalently:
u = (−1 + i)
1/3
(1/3) ln(−1+i)
(1/3)[ln
=e
=e
√ 1/3 i(π/4+2kπ/3)
= ( 2) e
√
2+i(3π/4+2kπ)]
!""#$%"&%'"()*"+,-($%
P ( z ) " an z n + an !1 z n !1 + ! + a0 .
P( z = zi ) = 0
! zi is a root
[Gauss, 1799]
• proof of fundamental theorem of algebra is given in the course
“Functions of a complex variable”, Short Option S1
3..&),.4,-./0*.1(#/),
P ( z ) " an z n + an !1 z n !1 + ! + a0 .
! zi is a root
P( z = zi ) = 0
!"#$#%&'$()(*+,#,-./0*.1(#/,20,(&),$..&),
an z n + an !1 z n !1 + ! + a0 = an ( z ! z1 )( z ! z2 )! ( z ! zn )
n
n
n
= an ( z ! z
n !1
"z
j =1
;
!
!
a0
= (!1) n # z
an
j
+ ! + (!1)
# z ).
j
j =1
!.1-#$(*+,%.'5%('*&),.4,6*78,#*9,6:,
an !1
= !" z
an
j
j
n
a2 x 2 + a1 x + a0
!"#"$%&'()'*+$!%&'*,-.$/$
!a ±
(
=
1
=,,3./$
x1,2
a12 ! 4a2 a0
)
<$
2a2
8$
52$+,167!89$),,3.$+,1!$:-$+,167!8$+,-;&#'3!$6':).$
0&1$,2$),,3.$
•
a1
= !( x1 + x2 )
a2
4),(&+3$,2$),,3.$
8$
a0
= x1.x2
a2
2,)$),,3.$
>!-!)'7$.,7&*,-.$-,3$'?':7'@7!$2,)$A:#A!)$,)(!)$6,7B-,1:'7.$C%&')*+.$'-($'@,?!D$
H$
E'-$F-($.,7&*,-.$:-$.6!+:'7$+'.!.G"$
Example 1:
z 5 + 32 = 0
• The solutions of the given equation are the fifth roots of −32:
(−32)1/5 = 321/5 cos
π 2πk
+
5
5
+ i sin
π 2πk
+
5
5
,
that is,
π
+ i sin 5
3π
+ i sin 5
k = 0 : 2 cos π5
k = 1 : 2 cos 3π
5
k = 2 : −2
7π
7π
k = 3 : 2 cos 5 + i sin 5
9π
9π
k = 4 : 2 cos 5 + i sin 5
k=1
k = 0, 1, 2, 3, 4
Im z
k=0
k=2
Re
k=4
k=3
,-.!/01+&+2+34456+47+/40894!:.06+
7
7
(z + ı) + (z ! ı) = 0
z+ı 7
(
) = !1 = e(2m+1)" ı
z!ı
!
z+ı
= e(2m+1)# ı /7
z"ı
! z(1 " e(2m+1)# ı /7 ) = "ı(1 + e(2m+1)# ı /7 )
e(2m+1)" ı /7 + 1
! z = ı (2m+1)" ı /7
e
#1
2cos( 2m+1
!)
e(2m+1)! ı /14 + e "(2m+1)! ı /14
2m+1
14
= ı (2m+1)! ı /14
=
ı
=
cot(
!)
14
"(2m+1) ! ı /14
2m+1
2ısin( 14 ! )
e
"e
!"#$%$&$'$($)$*+
=>27?&"#@#;#2)#2&+"3)2A4"#/'37#
7
7
(z + ı) + (z ! ı) = 0
x r y n!r
!"#$%&&#'(")#)""*#+,"#-'".-%")+#'/##
%)#
( x + y)n
0,"1"#23"#-')4")%")+&5#'6+2%)"*#/3'7#821-2&91#+3%2):&"#;#
( x + y )0
( x+ y )1
( x + y )2
( x+ y )3
( x + y )4
( x+ y )5
1
1
1
1
1
1
2
3
4
5
<+,#3'$#'/#821-2&91#+3%2):&"#%1#
z = cot( 2m+1
!)
14
1
3
6
10
Solution:
1
1
4
10
1
5
1
1 7 21 35 35 21 7 1
1'#
(z + ı)7 + (z ! ı)7 = 0 " z 7 ! 21z 5 + 35z 3 ! 7z = 0
:;*5<+#$=$6$>#3$*)%3"#&$4%&5$
!"#$%&'(')*+$#,-*.%)$
(z + ı)7 + (z ! ı)7 = 0 " z 7 ! 21z 5 + 35z 3 ! 7z = 0
/*)$0#$1&'2#)$')$*)%3"#&$4%&5$6$
z 7 ! 21z 5 + 35 z 3 ! 7 z = 0
"
zm , m = 3
z 6 ! 21z 4 + 35 z 2 ! 7 = 0 or
z=0
" w3 ! 21w2 + 35w ! 7 = 0 ( w # z 2 )
7#)/#$3"#$&%%38$%4$
w3 ! 21w2 + 35w ! 7 = 0
w = cot 2 ( 2 m14+1 ! )
2
9-5$%4$&%%38$
*&#$
(m = 0,1, 2)
" # m =0 cot 2 ( 2 m14+1 ! ) = 21
Pascal’s triangle = table of binomial coefficients
n
r
i.e., coefficients of xr y n−r in (x + y)n
=
n!
r!(n−r)!
[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−1 x + (x + y)n−1 y ]
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
n!
in Hebrew writings of 14th century is
•
= r!(n−r)!
r
number of combinations of n objects taken r at a time
♦ Pascal (1654) rediscovers triangle and most importantly unites
algebraic and combinatorial viewpoints
−→ theory of probability; proof by induction
!"#$#
%&'()*+#*",-./*#0)*+*#()*#1&2*+/34&5#*61,7'&#48#&'(#'9:4'18#;#
3
2
z + 7 z + 7 z + 1 = 0.
<()#+'0#'=#>,8?,/@8#(+4,&5/*#48#
1
2
( + y )0
( x+ y )1
( x + y )2
( x+ y )3
( x + y )4
( x+ y )5
1
1
1
1
1
1
1 8 28 56 70 56 28 8 1 8'#
[( z + 1)8 ! ( z ! 1)8 ] = 8 z 7 + 56 z 5 + 56 z 3 + 8 z
= 8 z[ w3 + 7 w2 + 7 w + 1] ( w " z 2 ).
A'0##
( z + 1)8 ! ( zC##! 1)8 = 0
#0)*&##
z +1
z "1
= e 2 m! i/ 8
e m! ıB#/4# + 1
= "ıcot(m! / 8) (m = 1,2,…,7),
#4B*B#0)*&## z =
m! ı /4
e
"1
8'#()*#+''(8#'=#()*#54:*&#*61,7'&#,+*##
z = " cot 2 (m!/8)
m = 1, 2, 3
2
3
4
5
1
1
3
6
10
1
4
10
1
5
1
Example
Question from 2008 Paper
Find all the solutions of the equation
n
z+i
= −1 ,
z−i
and solve
z 4 − 10z 2 + 5 = 0 .
z+i
z−i
n
= −1 = ei(π+2N π) , N integer ⇒
z+i
= ei(π/n+2N π/n) , N = 0, 1, . . . , n−1
z−i
ei(π/n+2N π/n) + 1
π(1 + 2N )
cos[π(1 + 2N )/(2n)]
Then : z = i i(π/n+2N π/n)
= cotg
.
=i
i sin[π(1 + 2N )/(2n)]
2n
e
−1
For n = 5 : (z+i)5 = −(z−i)5 ⇒ z(z 4 −10z 2 +5) = 0 . Then the 4 roots of z 4 −10z 2 +5 = 0 are
3π
7π
9π
π
, cotg
, cotg
, cotg
.
cotg
10
10
10
10
!"#$####%&'(#)&*)##
z 2m " a 2m
!
2!
(m " 1)!
= ( z 2 " 2az cos + a 2 )( z 2 " 2az cos
+ a 2 ) ! ( z 2 " 2az cos
+ a 2 ).
2
2
z "a
m
m
m
0=.=#%&'(#)&*)## P(z) = Q(z) where
P(z) ! z 2m " a 2m
6+''),#<#
Q( z ) # ( z 2 " a 2 )( z 2 " 2az cos
+''),#
zr = a cos
=a(cos
-.*/012#3'.430.1)#
zr = ae r! i / m )
!
2!
(m " 1)!
+ a 2 )( z 2 " 2az cos
+ a 2 ) ! ( z 2 " 2az cos
+ a 2 ).
m
m
m
r!
r!
± a 2 cos 2
" a2
m
m
r!
r!
±ı 1 " cos 2
)=ae±ır! / m
m
m
a2 m = 1
(r=0,1,…,m).
5678#*1/#9678#0/.1:3*;#
• 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