Lecture 4
• Characterization of a polynomial by its roots
• Techniques for solving polynomial equations
• Final comments and examples on Part A (complex nos)
!""#$%"&%'"()*"+,-($%
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 .
P( z = zi ) = 0
! zi is a root
!"#$#%&'$()(*+,#,-./0*.1(#/,20,(&),$..&),
an z n + an !1 z n !1 + ! + a0 = an ( z ! z1 )( z ! z2 )! ( z ! zn )
n
n
n
= an ( z ! z
n !1
"z
j =1
!.1-#$(*+,%.'5%('*&),.4,6*78,#*9,6:,
an !1
= !" z
an
j
;
a0
= (!1) n # z
an
j
j
+ ! + (!1)
n
# z ).
j
j =1
a2 x 2 + a1 x + a0
!"#"$%&'()'*+$!%&'*,-.$/$
(
=
!a1 ± a12 ! 4a2 a0
=,,3./$
x1,2
)
<$
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 z
k=4
k=3
,-.!/01+&+2+34456+47+/40894!:.06+
(z + ı)7 + (z ! ı)7 = 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#
(z + ı)7 + (z ! ı)7 = 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
[Zhu Shije]
n
n!
•
= r!(n−r)!
in Hebrew writings of 14th century [Levi ben Gershon]:
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
!"#$#
%&'()*+#*",-./*#0)*+*#()*#1&2*+/34&5#*61,7'&#48#&'(#'9:4'18#;#
1
( + y )0
( x+ y )1
( x + y )2
( x+ y )3
( x + y )4
( x+ y )5
3
2
z + 7 z + 7 z + 1 = 0.
<()#+'0#'=#>,8?,/@8#(+4,&5/*#48#
1
2
1
1
1
1
1
1 8 28 56 70 56 28 8 1 8'#
[( z + 1)8 ! ( z ! 1)8 ] = 8 z 7 + 56 z 5 + 56 z 3 + 8 z
= 8 z[ w3 + 7 w2 + 7 w + 1] ( w " z 2 ).
A'0##
( z + 1)8 ! ( zC##! 1)8 = 0
#0)*&##
z +1
z "1
= e 2 m! i/ 8
e m! ıB#/4# + 1
= "ıcot(m! / 8) (m = 1,2,…,7),
#4B*B#0)*&## z =
e m! ı /4 " 1
8'#()*#+''(8#'=#()*#54:*&#*61,7'&#,+*##
z = " cot 2 (m!/8)
m = 1, 2, 3
2
3
4
5
1
1
3
6
10
1
4
10
1
5
1
!"#$####%&'(#)&*)##
z 2m " a 2m
!
2!
(m " 1)!
2
2
2
2
2
=
(
z
"
2
az
cos
+
a
)(
z
"
2
az
cos
+
a
)
!
(
z
"
2
az
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 the main topics of 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
• We next give final comments and examples.
An application of complex algebra to plane geometry
p
0
2a
q
2d
s
2b
2c
Arbitrary quadrilateral.
Construct squares on each side.
Then
segments joining the centres of
opposite squares are perpendicular
and of equal length.
r
2a+2b+2c+2d=0
iπ/2
Center p: a+a e
= a(1+i) ; likewise q=2a+b(1+i)
Thus A
r=2a+2b+c(1+i); s=2a+2b+2c+d(1+i)
s−q = b(1−i)+2c+d(1+i); B r−p = a(1−i)+2b+c(1+i) .
i π/2
A and B perpendicular and of equal length means B=A e
,
i.e. B=iA, i.e. A+ i B=0. Verify that indeed A+iB=0.
COMPLEX FUNCTIONS AS MAPPINGS
f : z 7→ w = f (z)
• Complex functions can be viewed as mapping sets of points in C (e.g. curves,
regions) into other sets of points in C.
Example:
w = f (z) = ez .
Set z = x + iy ;
w = ρeiφ . Then ρ = ex ; φ = y .
Im z
f
b2
f
b1
lines x = a −→ circles ρ = ea
lines y = b −→ rays φ = b
Im w
1111111111
0000000000
0000000000
1111111111
0000000000
1111111111
0000000000
1111111111
0000000000
1111111111
0000000000
1111111111
a
1
a2
b
2
b
1
f
Re z
Re w
• For any subset A of C, the “image of A through f ” is the set of points w
such that w = f (z) for some z belonging to A, and is denoted f (A).
• f maps A on to f (A)
Examples
• Let f (z) = 1/z. What does f do to the interior and exterior of the
unit circle |z| = 1? What does f do to points on the unit circle?
1
w = f (z) =
;
z
Im z
1 −iθ
z = re ⇒ w = e
r
Im w
iθ
f
1
Re z
1
Re w
⊲ f maps the interior of the unit circle on to the exterior, and viceversa.
Unit circle is mapped on to itself.
• Let f (z) = (z − 1)/(z + 1). What is the image through f of the
imaginary axis?
[Answ.: unit circle]
And of the right and left half planes? [Answ.: interior and exterior of
unit circle]
• Let f (z) = ln z (principal branch). What is the image through f
of the upper half plane?
[Answ.: horizontal strip between 0 and iπ]
From Maths Collection Paper 2009
Sketch the locus of points in the complex plane for
(a) Im(z 2 + i) = −1
(b) arg(z 2 + i) = π/2
,
2
2 2
z=x+iy;z =x −y +2ixy
(a)
1+2xy=−1=
Im z
1 Re z
xy=−1
y=−1/x
2
2
θ = arctg [(1 + 2 x y) / (x − y )] = π / 2
Im z
Re z
(b)
=
2
x =y
2
=
x=y
2
x=−y,1−2x >0
Maths Paper 2007
Sketch the curve in the complex plane
z = at eibt ,
−6 ≤ t ≤ 6
where a and b are real constants given by a = 1.1, b = π/3.
Im z
1.1
1
−6
1.1
6
Re z
Maths Collection Paper 2006
♦ Find Re and Im of 1/(1 − z), where z = reiθ , and hence
P∞ n
sum the series n=0 r cos(nθ) for r < 1.
1
r sin θ
1 − r cos θ
1 − z∗
+
i
=
=
1−z
|1 − z|2
1 + r2 − 2r cos θ
1 + r2 − 2r cos θ
|
|
{z
}
{z
}
Re
∞
X
rn cos(nθ) =
∞
X
Im
rn Re einθ = Re
= Re
♦ Show that 1 +
cos θ
2
+
(reiθ )n
n=0
n=0
n=0
∞
X
1 − r cos θ
1
=
1 − reiθ
1 + r2 − 2r cos θ
cos 2θ
4
+
cos 3θ
8
+ ... =
4−2 cos θ
5−4 cos θ
.
∞
∞
∞ „ iθ «n
X
X
X
cos(nθ)
1
e
inθ
=
Re
e
=
Re
2n
2n
2
n=0
n=0
n=0
= Re
1 − cos θ/2
4 − 2 cos θ
1
=
=
1 − eiθ /2
1 + 1/4 − cos θ
5 − 4 cos θ
Maths Paper 2008
Prove the identity
N
−1
X
sin(N x/2)
cos(nx) =
cos((N − 1)x/2) .
sin(x/2)
n=0
N
−1
X
cos(nx) = Re
n=0
N
−1
X
einx = Re (
n=0
= Re
"
∞
X
einx −
n=0
iN x
1−e
∞
X
n=0
inx
e
#
∞
X
einx
)
n=N
| P{z }
inx
eiN x ∞
n=0 e
1 − eiN x
= Re
1 − eix
sin(N x/2)
eiN x/2 [−2i sin(N x/2)]
i(N −1)x/2
=
= Re
Re
e
sin(x/2)
eix/2 [−2i sin(x/2)]
sin(N x/2)
cos((N − 1)x/2)
=
sin(x/2)