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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
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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).
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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.
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Fig 19.13
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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
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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
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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 )(
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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
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Fig 19.14
f(z) has simple poles at z = 0 and z = 1 + i in the upper half-plane. See Fig 19.14.
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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
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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
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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
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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
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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
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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
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Contour Integration Example
The graphical interpretation
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)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])
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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])
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z
ezg
iyxzz
ezf
iz
iz
)(
axis Real on the is Im)(
z
zzg
)cos()(Re
z
zzg
)sin()(Im
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>> 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
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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
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),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.
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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
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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
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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
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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
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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