Chapter 4 Integrals Complex integral is extremely important, mathematically elegant.
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Transcript of Chapter 4 Integrals Complex integral is extremely important, mathematically elegant.
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Chapter 4 Integrals Complex integral is extremely important, mathematically elegant.
30. Complex-Valued Functions w(t) First consider derivatives and definite integrals of complex-valued functions w of a real variable t.
(1) )()()( titutwLet
Real function of t
(2) )(')(')()(' titudt
tdwtw
Provided they exist.
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0 0 0
0 0 0
0 0 0 0
0 0 0 0
0 0
For each complex constant
( ) ( ) ( )
( ) ( )
( ) ( )
( )(
i
x iy
x yzd dz w t u idt dt
d x u y i y xdtd dx u y i y u xdt dtx iy u
0 0
' ')
. . ( ) '( )
idi e z w t z w tdt
Various other rules for real-valued functions of t apply here.
tzeZtzedtdge 000 ..
However, not every rule carries over.
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Example: Suppose w(t) is continuous on .bta
. . or ( ( ) and ( ) are continuous)i e u t t
The “mean value theorem” for derivatives no longer applies.
There is a number c in a<t<b such that
abawbwcw
)()()('
a c b
0)0()2(
1 )('
20 )(
wwbut
Zeroneveritietw
tonitetwconsider
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Definite Integral of w(t) over a t b
is defined as ) ( ) ( ) (5)b b bw(t dt u t dt i t dta a a
when exists
Re ( ) Re ( ) Im Im ( ) b b b bw t dt w t dt w(t)dt w t dta a a a
Can verify that
1 2 1 2
( ) ( )
( ( ) ( )) ( ) ( )
( ) ( ) ( )
b bcw t dt c w t dta ab b bw t w t dt w t dt w t dta a ab c bw t dt w t dt w t dta a c
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Anti derivative (Fundamental theorem of calculus)
( )|
( ) ( ) ( )
,
( ) ( ) ( )
( )|
( ) ( ) ( ) ( )
( )
b ba a
iV t
W t U t iV tif W'(t) w(t)then U'(t) u(t) V'(t) υ(t)b b bw t dt u t dt i t dta a a
U t
U b iV b U a iV a
W b W
( )a
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00 0
00
0
0
0 0 0
0
To Prove ( )
Let
Re
Re
Re
b bw(t) dt w(t) dt a ba a
iθb b w dt r e w dt ra a
iθbr e w dta
-iθbe w dtaiθb (e w) dta
iθ iθ iθBut (e w) e w e w w
r
b w dtab bor wdt w dta a
realmust be real
Real part of real number is itself
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31. Contours
Integrals of complex-valued functions of a complex variable are defined on curves in the complex plane, rather than on just intervals of the real line.
A set of points z=(x, y) in the complex plane is said to be an arc if
x x(t) , y y(t) (a t b)
where x(t) and y(t) are continuous functions of real t.
非任意的組合
This definition establishes a continuous mapping of interval
into the xy, or z, plane; and the image points are ordered according to increasing values of t.
bta
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It is convenient to describe the points of arc C by
( ) ( )
where ( ) ( ) ( )
z z t a t b
z t x t i y t
The arc C is a simple arc, or a Jordan arc, if it does not cross itself.
1 2 1 2( ) when i.e. z(t ) z t t t
When the arc C is simple except that z(b)=z(a),
we say that C is a simple closed curve, or a Jordan Curve.
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0
2
0 2Simple closed curve
0 2
Ex :iθ z e ( θ π)
iθ also z z R e ( θ π)
3
0 2
Ex :iθ z e ( θ π)
10 11 2
Ex :x ix x
zx i x
0 1 2
1
y
x
2
40 2i
Ex : z e ( θ π)
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arc C : z(t) x(t) i y (t)
Suppose that x’(t) and y’(t) exist and are continuous throughout bta
C is called a differentiable arc
z'( ) '( ) '( )Since t x t iy t
2 2'( ) '( ) '( ) integrable over
real valuedz t x t y t is a t b
function
The length of the arc is defined as
0
ba
z'(t) dtb
L z'(t) Δta
Δt
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L is invariant under certain changes in the parametric
representation for C
To be specific, Suppose that
t (τ) (α τ β)
where is a real valued function mapping the interval
β)τ(α onto the interval .
bta We assume that
is continuous with a continuous derivative
We also assume that 0 for each'(τ) τ ( increases with )t
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if arc is
the fact that
allows us to write
β L z' (τ) '(τ) dταSo., C z Z(τ) z (τ) (α τ β)
Z'(τ) z' (τ) '(τ)
β L Z'(τ) dτα
Exercise 6(b)
Exercise 10
If is a differentiable arcand 0
z z(t) (a t b) z'(t) in a t b
Then the unit tangent vector
0z'(t)( 不代表水平,而是在此處停頓 長度不增加)
z'(t)T z'(t)
is well defined for all t in that open interval.
Such an arc is said to be smooth.
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For a smooth arc z z(t) (a t b)
is continuous on (we agree)and non zero on
z'(t) a t b a t b
A contour, or piecewise smooth arc, is an arc consisting of a finite number of smooth arcs joined end to end.
If z=z(t) is a contour, z(t) is continuous , Whereas z’(t) is piecewise continuous.
When only initial and final values of z(t) are the same, a contour is called a simple closed contour
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32. Contour Integrals
Integrals of complex valued functions f of the complex variable z: Such an integral is defined in terms of the values f(z) along a given contour C, extending from a point z=z1 to a point z=z2 in the complex plane. (a line integral)
Its value depends on contour C as well as the functions f.
Written as 2
1( ) or ( ) zf z dz f z dzc z
When value of integral is independent of the choice of the contour.
Choose to define it in terms of
not as lim .w(t) dt, Σ
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Suppose that z z(t) (a t b)
represents a contour C, extending from z1=z(a) to z2=z(b).
Let f(z) be piecewise continuous on C.
Or f [z(t)] is piecewise continuous on bta
The contour integral of f along C is defined a
(( ) ( 2)) ( ) bf z dz f z t z' t dt c a
Since C is a contour, z’(t) is piecewise continuous on bta t 的變化 define contour C
Section 31
So the existence of integral (2) is ensured.
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From section 30
0 0
( ) ( )
and ( ) ( ) ( ) ( )
z f z dz z f z dzc c f z g z dz f z dz g z dzc c c
0: complex constantz
Associated with contour C is the contour –C From z2 to z1
Parametric representation of -C
( )z z t (-b t -a)
z2=z(b)
z1=z(a)
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order of –C must also follow increasing parameter value
order of C follows (t increasing)
bta
a b -t 大 小
atb
( ) ( )
( ) w ( )
( ) ( ) ( ) ( )( )( )
( )
z t w ta f w t ' t dtb
dw t dz t dz t d tw' t dt dt d t dt
z' t
Thus
( ) ( ) ( )af z dz f z t z' t dtc b
where z’(-t) denotes the derivative of z(t) with respect
to t, evaluated at –t. |( ) ( )or ( ) t t
dz t dz t d t dt
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After a change of variable,
) ) ) )
τ ta τ b
a bf(z) dz f z(τ z'(τ dτ f z(τ z'(τ dτc b a f(z) dzc
dtd
1 2
1 2 1 0 0 2
1 2
if C C C z z z z z z
f(z) dz f(z) dz f(z) dzc c c
Definite integrals in calculus can be interpreted as areas, and they have other interpretations as well.
Except in special cases, no corresponding helpful interpretation, geometric or physical, is available for integrals in the complex plane.
z1
C1
z0z2
CC2
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33. Examples
Ex. 1
By def.
Note:
on the circle
for z
y
C2
0
-2
x
when is 2 ( )2 2
i
CI zdz C z e
2
2
2 2
2 2
2 (2 ) '
2 (2 ) 4 4
i i
i i
I e e d
e ie d i d i
42, 4, or
Hence .C
z zz z zdz
iz
22
2
2 2 2
2 2
2 (2 ) '
2 (2 ) 4 0.
i i
C
i i i
I zdz e e d
e ie d i e d
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x
y
A B
C1 C2
0
1+i1C1
Ex2. 1 2: contour , : segment C OAB C OB
2( ) 3 ( ).f z y x i x z x iy
1 1
0 0( ) .
2OA
if z dz yidy i ydy
1 1 12 2
0 0 0
1( ) (1 3 )1 (1 ) 3 ( ) - .
2ABf z dz x i x dx x dx i x dx i
1
1( ) .
2C
if z dz
2
1 12 2
0 0( ) 3 (1 ) 3(1 ) 1- .
Cf z dz i x i dx i x dx i
1 1 22
1( ) ( ) ( ) .
2C C C C
if z dz f z dz f z dz
:
(0 1)
OB
z x ix
x
:
0
(0 1)
OA
z iy
y
:
(0 1)
AB
z x i
x
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Ex3
Want to evaluate
Note that
dep. on end points only. indep. of the arc.
Let C denote an arbitrary smooth arc z=z(t), bta
2 2 2
2 22 1
2 2
2 2
2
2 1221
( ) ( ) ( )
zz
z
zzdz
z t z b z abI
az
z
b
C aI z dz z(t) z'(t) dt
2
2z(t)d z(t) z' (t)
dt
Integral of z around a closed contour in the plane is zero
c z dz 0
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Ex4. Semicircular path 3 0iθC : z e ( θ π)
起點 終點Although the branch (sec. 26) p.77. /2 0 0 2iθf(z) r e ( r , θ π)
of the multiple-valued function z1/2 is not defined at the initial point z=3 of the contour C, the integral
12zI dzc of that branch nevertheless exists.
For the integrand is piecewise continuous on C.
C
30-3
323 3 3 sin 0
2 2right hand limit at 0 3 0.
iθz(θ) eθ θiθ /f z(θ) e cos i ( θ π)
θ is i
/ 20
3 3
Hence ( ) is continuous on 0 when its value at 0
is defined as 3 Consequently 2 3 1 .iθ iθ e ie
f z θ π θ
. , I dθ ( i)
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34. Antiderivatives
- There are certain functions whose integrals from z1 to z2 are independent of path.
The theorem below is useful in determining when integration is independent of path and, moreover, when an integral around a closed path has value zero.
- Antiderivative of a continuous function f : a function F such that F’(z)=f(z) for all z in a domain D.
- note that F is an analytic function.
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Theorem: Suppose f is continuous on a domain D.
The following three statements are equivalent.
(a) f has an antiderivative F in D.
(b) The integrals of f(z) along contours lying entirely in D
and extending from any fixed point z1 to any fixed point
z2 all have the same value.
(c) The integrals of f(z) around closed contours lying
entirely in D all have value zero.
Note: The theorem does not claim that any of these statements
is true for a given f in a given domain D.
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Pf:
has am antiderivative
z
A:(a) (b) (a) f F.
d F z(t) F' z(t) z'(t) f z(t) z'(t)dt
b b f(z) dz f z(t) '(t)dt F z(t) F z(b) F z(a)ac a
2 1 F(z ) F(z )
(b)
c f(z)dz
c c c f(z)dzf(z)dzc f(z)dzf(z)dz
c f(z)dzc f(z) dz
CC C(c)B:(b)
01 1 2
02
021
21
(c)z1
C2
C1
z2
1 2 1 20 0
C :(c) (b) f(z) dz f(z) dz f(z) dzc c -c c c
(b)
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0
0 0
Define ( ) ( )
( ) ( ) ( ) ( ) ( )
(choose integration path from to as
z F z f s dszz z z z z F z z F z f s ds f s ds f s dsz z z
z z ds z z z z z
1 1( )
a line segment)
( )
( ) ( ) 1( ) ( ) ( )
f zz z
z z z z f(z) ds f z dsz zz zF z z -F z f z f s - f z dszz z
( ) ( )( ) ( ) ( )
Since is continuouswhenever
1 1
when is close to so that
lim
z zz
F z z -F zf z f s f z
z
f , f(s)- f(z) ε s- z δ
ds ε z ε z z
z z z, z δ
z
( ) ( ) or0
F z z f z F'(z) f(z) z
(c) -->(a)
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35. Examples2
3
131+i 2 3
00
Ex1. ( ) continuous
1 has an antiderivative ( ) throughout the plane.3
1 2 Hence (1 ) ( 1 )
3 3 3
for every contour from 0 to 1 .
i
f z z
F z z
zz dz i i
z z i
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For any contour from z1 to z2 that does not pass through the origin .
20 for : 2 (- )
zdz iC z ec
Note that: zzf 1)( can not be evaluated in a similar way
though derivative of any branch F(z) of log z is , 1
z
F(z) is not differentiable, or even defined, along its branch cut. (p.77)
which
t throughoudomaina in lienot does C C. circle theintersect ray theepoint where at theexist tofails cut, branch the
from the origin is used to form raya if ,particular (InF'(z)
1'( ) .F z
z
2
22
2 1 21 1 1 2
1
z1
1 1 1
Ex2. is continuous every where except at the origin.
It has an antiderivative in the domain 0
consequently 0 0zz
zz
dz
z z z z
- .
z , (z , z )
z
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.Ex3
The principal branch Log z of the logarithmic function serves as an antiderivative of the continuous function 1/z throughout D.
Hence
when the path is the arc 22 2π πiθz e (- θ )
excluded
Let : 0, Arg .D z z
222 2
Log Log(2 ) Log( 2 )
(ln2 ) (ln2 )2 2 2 2
iii i
dzz i i
z
i i i π π (- θ )
(compare with p.98)
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1
12
12
12
4
Evaluate
where is the branch
2 0 0 2
Cz
Ex .
dz
z θi
z r e (r , θ π)
C1
-3
C2
3
(2)
C1 is any contour from z=-3 to z=3, that lies above the x-axis. (Except end points)
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1. The integrand is piecewise continuous on C1, and the integral therefore exists.
2. The branch (2) of z 1/2 is not defined on the ray in particular at the point z=3. F(z) 不可積
3. But another branch.
is defined and continuous everywhere on C1.
4. The values of F1(z) at all points on C1 except z=3 coincide with those of our integrand (2); so the integrand can be replaced by F1(z).
)πθπ, - (riθ
er(z)F2
32
021
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Since an antiderivative of F1(z) is
12
1 11
333 0 22 33 3
2 3 1
zi π
i dz f (z) dz F (z) (e e )c -
( i)
We can write
32
32 2 32 01 3 3 2 2
i θ π πF (z) z r r e (r , - θ )
(cf. p. 100, Ex4)
-3 3
12
2For z dzc
Replace the integrand by the branch
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2
2
2
2 22
12
2 1
52 02 2
32 52Then 03 2 2
is an antiderivative of31 33 32 22 33 3
2 3 1
4 3
z
z
iθ π πf (z) r e (r , θ )
i θ π π F (z) r r e (r , θ )
f (z).i π
i πdz f (z) dz F (z) (e e )c
(- i)
dz -c c
1 2c c two contours can not be on the some domain
0
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36. Cauchy-Goursat Theorem
We present a theorem giving other conditions on a function f ensuring that the value of the integral of f(z) around a simple closed contour is zero.
Let C denote a simple closed contour z=z(t) described in the positive sense (counter clockwise). Assume f is analytic at each point interior to and on C.
),( bta
are continuous th, , ,
( ) ( ) ( )
( ) ( ) ( ) and ( ) ( )
( ) ( ) ( )
Green's theorem: If ( , ), ( , ),x y x yP Q Q
bf z dz f z t z' t dtc aif f z u x,y iv x,y , z x t i y t .
b bf z dz ux' vy' dt i v x' uy' dt.c a a
P x y Q x y P
roughout
( )the closed region consisting of all points interior to and on , xR yQ dAP
R CPdx Qdyc
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( ) ( ) )x y x yC R Rf z dz v u (u v dAdA i then
But(Cauchy-Riemann eg.) u v u v x y y x
when is analytic in and is continuous there
( ) 0
f R f'
f z dz Cauchy 's finding.c
Goursat was the first to prove that the condition of continuity on f’ can be omitted.
Cauchy-Goursat Theorem:
If f is analytic at all points interior to and on a simple closed contour C, then
( ) 0f z dzc
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tch-prob 36
37. Proof: Omit
38. Simply and Multiply Connected Domains
A simply connected domain D is a domain such that every simple closed contour within it encloses only points of D.
Multiply connected domain : not simply connected.
Can extend Cauchy-Goursat theorem to:
Thm 1: If a function f is analytic throughout a simply connected domain D, then
( ) 0f z dzc
for every closed contour C lying in D.
Not just simple closed contour as before.
n分成simpleclosed
contours
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tch-prob 37
Corollary 1. A function f which is analytic throughout a simply connected domain D must have an antiderivative in D.
section 34 for continuousgiven
analytic ( ) 0co
pf : f f f z dzc f
ntinuous has antiderivative f
Extend cauchy-goursat theorem to boundary of multiply connected domain
Theorem 2. If f is analytic within C and on C except for points
interior to any Ck, ( which is interior to C, ) then
( ) ( ) 01 k
nf z dz f z dzc c
k
C: simple closed contour, counter clockwise
Ck: Simple closed contour, clockwise
C1 C2C
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tch-prob 38
C2
C1Corollary 2.
Let C1 and C2 denote positively oriented simple closed contours, where C2 is interior to C1.
If f is analytic in the closed region consisting of those contours and all points between them, then
1 2
1 2
( ) ( )
from thm 2 ( ) ( ) 0
f z dz f z dzc cpf : , f z dz f z dzc -c
Principle of
deformation of paths.
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tch-prob 39
Example:
C is any positively oriented simple closed contour surrounding the origin.
0
0
0
2
0
Want to show that 2 .
Construct a positively oriented circle with center at the
origin and radius so small that lies entirely inside C.
Since 2
and since 1/ is anal
C
i
iC
dzi
zC
C
dz i ed i
z e
z
ytic everywhere except at =0,
we have 2 from corollary 2.C
z
dzi
z
C0
C
0
2 22 2
0 0
2
0
?
i i i
C C
C
zdz zdz e i e d i e d
z dz
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tch-prob 40
39. Cauchy Integral FormulaThm. Let f be analytic everywhere within and on a simple closed contour C, taken in the positive sense. If z0 is any point interior to C then,
00
00
)
)
( )1 (2
or( ) 2 (
f z dzf z c z- zπ i
f z dz π i f zc z z
(1) Cauchy integral formula
(Values of f interior to C are completely determined by the values of f on C)
2
2
2
9
9 ))(
. Let be 2
since ( ) is analytic within and on ,
and since is interior to .
2 (510( )(9 )
z
z
z
z
z
Ex C
f z C
-i C
z dz -i π dz π i c c z i- z z i
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tch-prob 41
C
C0
z0
Pf. of theorem:
0
00
00
-
-
is continuous at
( ) ( ) when
choose a positive :
z z
z z
f z
f z f z
C
since
0
( )f zz z is analytic in the closed region consisting of C
and C0 and all points between them, from corollary 2, section 38,
00 0
00
0 00 0 0
))
( ) ( )
( ) (( ) ( dz
f z dz f z dzc cz z z z
f z f z f z dz dzf z c c cz z z z z z
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tch-prob 42
0 0
0 00
00 0
0 0
0 00 0
0
0 0
) ))
) )
)
but 2
( ) ( ( ) (2 (
( ) ( ( ) (also
( ) (2
dz ic z zf z - f z f z f z
- π i f z dzc cz z z z
f z f z f z f z dz dzc cz z z z
f z f z dz πρc z z
2 π
00 0
00
)
)
( ) 2 ( 2
( ) 2 ( 0
f z dz π i f z πc z z
f z dz π i f zc z z
Non-negative constant arbitrary
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tch-prob 43
40. Derivatives of Analytic FunctionsTo prove : f analytic at a point
its derivatives of all orders exist at that point and are themselves analytic there.
0 interior to
in Cauchy integral formula on
z z C
z s C
( )1( )2
f s dsf z c s zπ i
( ) ( ) ( )1 1 12
( )12 ( )( )
f z z f z f s ( ) dsc s zz s z z zπ if s ds cπ i s z z s z
S C
zd
z
when 0 ,
where is theshortest distamefrom to on .
z d
d
z s C
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tch-prob 44
2
2
1 1but ( )( )( ) ( )
( )( )( )
f s dsc s z z s z s zf s ds z c s z z s z
2 2
-
Let denote the maximum value of ( ) on .
Let be the length of .Note that
( ) 0 as 0( )( ) ( )
lim ( ) ( ) (120
s z
s z z z
zz
z
M f s C
L Cd
s z z d -
MLf s ds z c s z z s z d - d
f z z - f z f s z π iz
2
2
3
( )
)
( )1'( )2 ( )
( )1 similarly for '' ( )( )
s z ds c #
f s ds f z cπ i s zf s ds f z cπ i s z
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tch-prob 45
Thm1. If f is analytic at a point, then its derivatives of all orders are also analytic functions at that pint.
In particular, when
( ( ( (
( ) ( , ) ( , ) is analytic at a point z ( )
the analyticity of ' eusures the continuity of ' there '( ) ) ) ) )
are conti
f z u x y iv x y x,y
f ff z u x,y i v x,y v x,y i u x,yx x y y
u , u , v , v x y x y
( ( (
nuous at ( )
similarly ''( ) ) ) )
x,y
f z u (x,y) i v x,y v x,y i u x,yxx xx yx yx
Corollary : If ( ) ( ) ( ) is analytic at a point , then have continuous partial derivatives of all orders at that point.
f z u x,y i v x,yz x iy u, v
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tch-prob 46
(0)
( )( 1)
( )0( 1)
0
( )
( )( )
Let ( ) denote ( ) and 0! 1, can use mathematicalinduction to verify that
( )!( ) ( 01 2 )2
( ) 2or!
nn
nn
s z
zz z
f z f z ,
f s dsnf z n , , ....cπ i
f z dz π i fc n
2
2
4 3 1( 0)
1: 1 ( )
( ) 2 8(0)3! 3
z
z
z
z
Ex C : f z e
f z dze π i π i dz f'''c cz
0
10 0
in , ( ) 1,
)
2:
2 0 ( 1 2 )(
Thm 2: If continuous in domain and ( ) 0 for every closed contour in ,c
n
f zEx z Cdz dz π i n , , ....c cz- z z- z
f Df z dz C D
then is analytic throughout .f D
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tch-prob 47
41. Liouville’s Theorem and the Fundamental Theorem of Algebra
Let z0 be a fixed complex number.
If f is analytic within and on a circle 0z z R
( )0 1
0
( )( )
( )!then ( 1 2 ).2
nn
zz zf z dzn f n , ,.......cπi
Let MR denote the Maximum value of )(zf
( )0 1
( )0
0
( )
( )
)
! 2 ( 1 2 )2!
( 1 2 ) Cauchy's inequality.
for 1 '(
n Rn
n Rn
R
MzR
z
nf πR n , ,....π
n Mf , n , ,.....
RM
n f zR
![Page 48: Chapter 4 Integrals Complex integral is extremely important, mathematically elegant.](https://reader035.fdocuments.net/reader035/viewer/2022062217/56814fb6550346895dbd6fc1/html5/thumbnails/48.jpg)
tch-prob 48
Thm 1 (Liouville’s theorem): If f is entire and bounded in the complex plane, then f(z) is constant throughout the plane.
0
0
0
)
)
: '(
for large, '( must be zero since is arbitrary, '( ) 0 everywhere is a constant
Mpf f zR
R f zz
f zf
finite
可以 Arbitrarily large
Thm 2 (Fundamental theorem of algebra):
Any polynomial2
0 1 2
0 0)
( ) ( 0)of degree ( 1) has at least one zero.i.e., there exists at least one point s.t. ( 0
nn n
P z a a z a z ......... a z an n
z P z
Pf. by contradiction
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tch-prob 49
Suppose that P(z) is not zero for any value of z.
Then is clearly entire and it is
also bounded in the complex plane.
1( )( )
f zP z
To show that it is bounded, first write
0 11 21 2
So that ( ) ( )
nn n n
nn
zz z za aa a
w ....
P z a w z
Can find a sufficiently large positive R such that
0 1 0 11 2 11 2 1
20 1 1, when ni
n i
n nn n n n n
az
z n
w z zz z z z z
a , i , , ........n- R
a a a aa a a.... .....
when2n
a z R
Generalized triangle inequality
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tch-prob 50
2
( )2 2
1 2Then ( )( )
nn n
n nnn nn
nn
aa w a w
a aP z a w z z R z R
f z z Ra RP z
So, is bounded in the region exterior to .
But is continuous in is bounded in .
is bounded in the entire plane.
From Liouville's theorem, ( ) is constant.
f z R
f z R f z R
f
f z
( ) is constant.
But ( ) is not constant We have a contradiction.
P z
P z
![Page 51: Chapter 4 Integrals Complex integral is extremely important, mathematically elegant.](https://reader035.fdocuments.net/reader035/viewer/2022062217/56814fb6550346895dbd6fc1/html5/thumbnails/51.jpg)
tch-prob 51
From the (F. T. 0. A) theorem
any polynomial P(z) of degree n can be expressed as
1 1) ( )( ) ( 10 sec 42
zP z z z Q (Ex. , )
Polynomial of degree n-1
1 2 2( ) ( )but ) ( z Q z z z Q
Polynomial of degree n-2
1 2 2
1 2
) ) ( )
) ) )
( ) ( (
( ( (n
zP z z z z z Q : : c z z z z ....... z z
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tch-prob 52
42. Maximum Moduli of FunctionsLemma. Suppose that f(z) is analytic throughout a neighborhood
0 0 0) of a point . If ( ) (z z ε z f z f z
for each point z in that neighborhood, then f(z) has the constant value f(z0) throughout the neighborhood .
00
2 00
200
)
( )
( )
( )1(2
121
2
iθ
iθ
iθ
f z ρe
ρe
f z ρe dθ
f z dzf z c z zπ i
iθ i ρ e dθπ i
π
0: 0 2iθCρ z z ρe ( θ π)
z1
z0
C
1 0
1 0
0
: Let be any point other than in the neighborhood.
Let be the distance between and .
: circle .
Pf z z
z z
C z z
f’s value at the center is the arithmetic mean of its values on the circle. ~ Gauss’s mean value theorem.
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tch-prob 53
200 0
0 0
2 20 0 00 0
200 0
) ( )
) )
( ) ( ) )
) ( )
1( (3)2
On the other hand, since ( ( (4)
2 (
1(2
i
i
i
i
f z ρe d
f z ρe d f z
f z ρe d
f z ..........π
f z ρe f z ..........
dθ π f z
f z π
(5) ......
From (3) and (5)
200 0
20 00
0 from (4), and continuous in
) ( )
( ) ( )
1(2
or 0
i
i
f z ρe
f z f z ρe
f z dθπ
dθ
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tch-prob 54
0 0
0 0
0 0
) )
)
)
( ( 0 2
( ) ( for all points in
( ) ( for all points in
From Ex 7(c), sec 22, (when the modulus of an analytic function is
if z f z ρe ( θ π)
f z f z z z- z ρ
f z f z z z- z ε
0)
constant in a domain, the function itself is constant there.)( ) ( for all points in the neighborhood.f z f z z
Thm. (maximum modulus principle)
If a function f is analytic and not constant in a given domain D, then has no maximum value in D. That is, there is no point z0 in the domain such that for all points z in it.
)(zf)0()( zfzf
![Page 55: Chapter 4 Integrals Complex integral is extremely important, mathematically elegant.](https://reader035.fdocuments.net/reader035/viewer/2022062217/56814fb6550346895dbd6fc1/html5/thumbnails/55.jpg)
tch-prob 55
d
N0 N1N2
z0 z1 z2
N n-1
Nnzn-1 zn
p
DPf:
-1's is in neighborhood
k kz z
Construct : k k
N z z d
0
is a connected open set.
There is a polygonal line lying entirely in that extends from , to any
other point in .
Let be the shortest distance from points on to the boundary of .
We then
D
L D z
P D
d L D
0 1 2
1
form a finite sequence of points , , , , (= ) along , where each point
is sufficiently close to the adjacent ones that
( 1, 2, , )
n
k k
z z z z P L
z z d k n
![Page 56: Chapter 4 Integrals Complex integral is extremely important, mathematically elegant.](https://reader035.fdocuments.net/reader035/viewer/2022062217/56814fb6550346895dbd6fc1/html5/thumbnails/56.jpg)
tch-prob 56
Assume f(z) has a max value in D at z0.
f(z) also has a max value in N0 at z0.
1 0 1 0
1 1
1 0
) ) )
)
from Lemma ) )
( ( ( in
( ) ( for all in
( ) ( (
f z f z z N
f z f z z N
f z f z f z
From Lemma, f(z) has constant value f(z0) throughout N0.
2 0 2 1
0
) ) ( in )
) )
( ( :
( ( for every in ( ) is constant a contradiction( ) has no maximum value in
n
z Nf z f z
f z f z z Df z f z D
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tch-prob 57
If a function f that is analytic at each point in the interior of a closed bounded region R is also continuous throughout R, then the modulus has a maximum value somewhere in R. (sec 14) p.41
)(zf
. ..( ) for all points in .
i ef z M z R
If is a constant fumction,then ( ) for all in
ff z M z R.
not continuouson boundary
Corollary: Suppose f is continuous in a closed bounded region R.
and that it is analytic and not constant in the interior of R.
Then Maximum value of in R occurs somewhere on the boundary R and never in the interior.
)(zf
Maximum at the boundary.