maximum modulus principle:
DESCRIPTION
Maximum Modulus Principle: If f is analytic and not constant in a given domain D, then |f(z)| has no maximum value in D. That is, there is no z 0 in the domain such that |f(z)| |f(z 0 )| for all points z in D. Proof: Assume that |f(z)| does have a maximum value in D. R. z 0. C R. - PowerPoint PPT PresentationTRANSCRIPT
Maximum Modulus Principle:
If f is analytic and not constant in a given domain D, then |f(z)| has no maximum value in D.
That is, there is no z0 in the domain such that |f(z)||f(z0)| for all points z in D.
Proof: Assume that |f(z)| does have a maximum value in D.
2
0
00 )(2
1)( drezfzf i
2
0
00 )(2
1)( drezfzf i
z0
R
CR
Alternatively
Theorem: If f is analytic, continuous and not constant in a closed bounded region D, then the maximum value of |f(z)| is achieved only on the boundary of D.
Some other aspects of the maximum modulus theorem:
Assume that f(z) is not 0 in a region R. Then if f(z) is analytic in R, then so is 1/f(z).Result: minimum of |f(z)| also occurs on the boundary.
. then ),(),()( If )( uivuivuzg eeeeeyxivyxuzg
Since the max and min of |f(z)| are on the boundary, so is the max and min of u(x,y).
Same applies to v(x,y).
Indented Contour
• The complex functions f(z) = P(z)/Q(z) of the improper integrals (2) and (3) did not have poles on the real axis. When f(z) has a pole at z = c, where c is a real number, we must use the indented contour as in Fig 19.13.
Fig 19.13
Suppose f has a simple pole z = c on the real axis. If
Cr is the contour defined by
THEOREM 19.17Behavior of Integral as r →
,irecz then,0
rCr
czfidzzf ),)((Res)(lim0
THEOREM 19.17 proofProofSince f has a simple pole at z = c, its Laurent series is f(z) = a-1/(z – c) + g(z) where a-1 = Res(f(z), c) and g is analytic at c. Using the Laurent series and the parameterization of Cr, we have
(12)
21
001 )(
)(
II
derecgirdre
irea
dzzf
iii
i
Cr
THEOREM 19.17 proofFirst we see
Next, g is analytic at c and so it is continuous at c and is bounded in a neighborhood of the point; that is, there exists an M > 0 for which |g(c + rei)| M.
Hence
It follows that limr0|I2| = 0 and limr0I2 = 0.We complete the proof.
) ),((Res9
1
01011
czfiia
idadre
ireaI i
i
rMMdrderecgirI ii 002 )(
Example 5Evaluate the Cauchy principal value of
Solution Since the integral is of form (3), we consider the contour integral
dx
xxx
x
)22(
sin2
)22(
1)( ,
)22( 22
zzzzf
zzz
dzeC
iz
Fig 19.14
f(z) has simple poles at z = 0 and z = 1 + i in the upper half-plane. See Fig 19.14.
Example 5 (2)
• Now we have
(13)
Taking the limits in (13) as R and r 0, from Theorem 19.16 and 19.17, we have
)1,)((iRes2
)0,)((Res)22(
P.V. 2
iezf
ezfidxxxx
e
iz
izix
C C
r
R C
R
r
iz
R riezfi )1,)((Res2
Example 5 (3)
Now
Therefore,
)1(4
)1 ,)((Res
21
)0 ,)((Res
1
ie
iezf
ezf
iiz
iz
)1(4
i221
)22(P.V.
1
2 ie
idxxxx
e iix
Example 5 (4)
Using e-1+i = e-1(cos 1 + i sin 1), then
)]1cos1(sin1[2)22(
sinP.V.
)1cos1(sin2)22(
cosP.V.
12
12
edxxxx
x
edxxxx
x
Indented Paths
iBdzzf
RxzC
B
Rxz
xzf(z)
C
00
20
0
20
0
)(lim Then,
taken.isdirection clockwise the
and where,|| circle a of halfupper thedenotes (ii)
; residue with and
96) (Fig. ||0disk punctured ain tion representa seriesLaurent a
with axis, real on the point aat pole simple a has function a (i)
thatSuppose :Theorem
x0
C
2
sinlim
0
20
dx
x
xI
dzz
edx
x
x iz
Imsin
C
CR
I1I2
021 CCII R
applies) Lemma s(Jordan' 011
Rx
2
0
2
0
sin2
0
sincos2
0
)sin(cos
2
0
)sin(cos2
0
)sin(cos
zero? togoes over integral theHope
dededeediedzz
e
diedeie
edz
z
e
C
iii
C
iz
iiii
ii
C
iz
21
sin
)(
)sin(sinIdu
u
udu
u
udx
x
xI
C
CR
I1I2
021 CCII R
origin. at the pole simple a has z
eiz
1
!3!2!11
z
1
exists.tion representa seriesLaurent
0
32
B
iziziz
z
eiz
iiBdzz
e
C
iz
0
1
)(
find ely toAlternativ
00
0
eB
ez
Biz
dzz
edx
x
x
idzz
e
dzz
edz
z
edz
z
e
iz
iz
C
iz
C
iziz
R
00
limImsin
0
Contour Integration Example
The graphical interpretation
)2( .2/
)sin(
2
1)0( Find
dp
x
xxf
)sin()(
>> x=[-10*pi:0.1:10*pi];>> plot(x,sin(x)./x)>> grid on>> axis([-10*pi 10*pi -0.4 1])
axis Real on the is )sin()sin(
)( iyxzz
z
x
xzf
>> x=[-10*pi:0.1:10*pi];>> plot3(x,zeros(size(x)),sin(x)./x)>> grid on>> axis([-10*pi 10*pi -1 1 -0.4 1])
z
ezg
iyxzz
ezf
iz
iz
)(
axis Real on the is Im)(
z
zzg
)cos()(Re
z
zzg
)sin()(Im
>> mesh(x,y,sin(z)./z)
axis Real on the is )sin()sin(
)( iyxzz
z
x
xzf
>> x=[-10*pi:0.1:10*pi];>> y = [-3:0.1:3].';>> z=ones(size(y))*x+i.*(y*ones(size(x)));>> mesh(x,y,cos(z)./z)
z
zzg
)cos()(Re
z
zzg
)sin()(Im
Cauchy’s Inequality: If f is analytic inside and on CR and M is the maximum value of f on CR, then
z0
R
CR
),2,1(!
)( 0)( n
R
Mnzf
nn
Proof:
2
0)1(1
01
0
00
)( )(
2
!
)(
)(
2
!)( dRie
eR
eRzf
i
ndz
zz
eRzf
i
nzf i
nin
i
Cn
in
R
2
0)1(1
02
0)1(1
0)(
2
!)(
2
!dRie
eR
eRzfndRie
eR
eRzf
i
n i
nin
i
inin
i
nn
i
nin
i
R
MndR
R
MndRie
eR
eRzfn !
2
!)(
2
! 2
01
2
0)1(1
0
iR eRzzC 0 :as written becan on point Any :Note
),2,1(!
)( 0)( n
R
Mnzf
nn
R
Mzf )(' 0
As R goes to infinity, then f’(z) must go to zero, everywhere. Then f(z) must be constant.
Liouville’s Theorem: If f is entire and bounded in the complex plane, then f(z) is constant throughout the plane.
Proof:
2
0
0
2
0
0
00 )(
2
1)(
2
1)(
2
1)( drezfdire
re
rezf
idz
zz
zf
izf ii
i
i
CR
Gauss’s Mean Value Theorem: If f is analytic within and on a given circle, its value at the center is the arithmetic mean of its values on the circle.
0
11
plane halfleft On the
22
3
dzzz
ei
i
z
C1
dzz
zdy
iy
iydy
iy
iydx
x
x i
i
22222222 1
3cosh
1
3cosh
1
3cosh
1
3cos
dz
zz
eedz
zz
zdz
z
z i
i
zzi
i
i
i
22
33
2222 112
1
11
3cosh
1
3cosh
The integral does not go to zero on the circle, the integral can’t be solved this way.
RRiRz
RRiRz
RiRz
eee
eee
ee
3sin3cos33
3sin3cos33
sin3cos33
,For
,For
dxx
exdx
x
exdx
x
axx iaxiax
4Im
4Im
4
sin4
3
4
3
4
3
4/34/4/4/3
334
4
2,2,2,2
4,4,4,4
04
iiii
iiii
eeeex
eeeex
x
044 4
3
4
3
R
R
R
x
x
4Res2
4Res2
44 4
3
)4/3exp(24
3
)4/exp(24
3
4
3
z
zi
z
zi
z
ezdz
z
eziziz
C
iaziaz
RC
0R
Rx
y
;|| circle a oexterior t
are that 0 plane halfupper in the points allat analytic is )(function a (i)
that Suppose
0Rz
yzzf
; where)0( semicircle a denotes (ii) 0RReRzC iR
0lim where
,|)(|such that constant positive a is there,on points allfor (iii)
RR
RRR
M
MzfMCz
0)(lim ,constant positiveevery for Then, RC
iaz
Rdzezfa
Jordan’s Lemma
dzz
edx
x
x zi
22
3
22 1Re
1
3cos
dz
iziz
edz
z
e zizi
22
3
22
3
1
01
1
1
12222
RRM
Rz
C1
C2
dz
iziz
edz
z
e zizi
22
3
22
3
1 01
1
1
12222
RRM
Rz
C2
4323
2
3
23)('
)(
iz
eizizeiz
iz
ez
iz
zizi
zi
333
22
3
16
412Res
ie
ieei
iziz
e zi
iz
33
22
3 2)(2
1 eieidz
z
e zi