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CH 5 PROPERTIES OF ENTIRE FUNCTION
5.1 The Cauchy Integral Formula and Taylor Expansion for Entire Functions
We now show that if f is entire and if
g ( z ) {
f ( z ) f ( a )
z a
, if z a
f
( a ) , if z a
.
then the I nt egr al T heor em ( . ) and C losed C ur eve T heor em ( . ) ap p ly to g as w ell
as to f . (Note that since f is entire, g is continuous; however, it is not obvious that g is entire). We begin by
showing that the Rectangle Theorem applies to g.
5.1 rectangle thm (II):
If f is entire and if g(z) = 8
( ) ( )
, z a
f
( a ) , z a
then g ( z ) dz ,
Wh ere is t he b ound ary of a rectan g le R.
Proof : Since, f is entire implies g is continuous on R
i.e. g must be boundary on R
i. e. ? s. that | g ( z ) | ? z R
Case 1: if a ex t( R)
? g is analytic within and on R.
i.e. g ( z ) dz by rectangle thm
case 2 I f a b ound a r y ( R) i. e . a
then, we subdivide R into six sub-rectangle as shown below(sand note pg 1)
a 1 say, also, the orientation of R 2 & R 5 are shown.
lso, let k : boundary of the k
th
sub- r ect an g le ( k , , …. )
1 : that sub- r ect an g le w hich conta in ‘a’
W. L . O.G ( 1) < ?? /A for any arbitrary choice.
? g is a na l. Wit hin o n R 2, R 3 …. . R 6
ote, S in c e, 2 , 3 , …. . 6 d oes not conta in ‘a’
g ( z ) dz for k = 2, 3, 4, 5, 6 [rectangle thm I]
? g ( z ) dz ? … .
But , g ( z ) dz . g ( z ) dz /
= g ( z ) dz only due to the cancellation involved
i.e. g ( z ) dz « . l ( r
)
b y L f or mu la
i.e. g ( z ) dz «
i.e. g ( z ) dz ( ?? was arbitrary)
case 3: I f a I nt erior ( R)
this time we divide R into nine sub-rectangle, as in the previous case (i.e. ?? ( 1)< ?? etc.)
note along the eight sub- r ect an g le n ot cont ai ni ng ‘ a’
g ( z ) dz k , , …. . ( Rec. t hm I )
? g ( z ) dz
g ( z ) dz
= g ( z ) dz
fig sand pg 2
i.e. g ( z ) dz
< ??
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CH 5 PROPERTIES OF ENTIRE FUNCTION
5.1 The Cauchy Integral Formula and Taylor Expansion for Entire Functions
We now show that if f is entire and if
g ( z ) {
f ( z ) f ( a )
z a
, if z a
f
( a ) , if z a
.
then the I nt egr al T heor em ( . ) and C losed C ur eve T heor em ( . ) ap p ly to g as w ell
as to f . (Note that since f is entire, g is continuous; however, it is not obvious that g is entire). We begin by
showing that the Rectangle Theorem applies to g.
5.1 rectangle thm (II):
If f is entire and if g(z) = 8
( ) ( )
, z a
f
( a ) , z a
then g ( z ) dz ,
Wh ere is t he b ound ary of a rectan g le R.
Proof : Since, f is entire implies g is continuous on R
i.e. g must be boundary on R
i. e. ? s. that | g ( z ) | ? z R
Case 1: if a ex t( R)
? g is analytic within and on R.
i.e. g ( z ) dz by rectangle thm
case 2 I f a b ound a r y ( R) i. e . a
then, we subdivide R into six sub-rectangle as shown below(sand note pg 1)
a 1 say, also, the orientation of R 2 & R 5 are shown.
lso, let k : boundary of the k
th
sub- r ect an g le ( k , , …. )
1 : that sub- r ect an g le w hich conta in ‘a’
W. L . O.G ( 1) < ?? /A for any arbitrary choice.
? g is a na l. Wit hin o n R 2, R 3 …. . R 6
ote, S in c e, 2 , 3 , …. . 6 d oes not conta in ‘a’
g ( z ) dz for k = 2, 3, 4, 5, 6 [rectangle thm I]
? g ( z ) dz ? … .
But , g ( z ) dz . g ( z ) dz /
= g ( z ) dz only due to the cancellation involved
i.e. g ( z ) dz « . l ( r
)
b y L f or mu la
i.e. g ( z ) dz «
i.e. g ( z ) dz ( ?? was arbitrary)
case 3: I f a I nt erior ( R)
this time we divide R into nine sub-rectangle, as in the previous case (i.e. ?? ( 1)< ?? etc.)
note along the eight sub- r ect an g le n ot cont ai ni ng ‘ a’
g ( z ) dz k , , …. . ( Rec. t hm I )
? g ( z ) dz
g ( z ) dz
= g ( z ) dz
fig sand pg 2
i.e. g ( z ) dz
< ??
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i.e. g ( z ) dz
= 0 ( ?? > 0 was arbi.)
corollary 5.2 vivek: Suppose g is as before, then the integral theorem & the closed curve theorem , both apply
to g.
Sol
n
Since, g is continuous , the proof of thm 4.15 & that of 4.16 apply, without any modification of g.
For instance, while proceeding for integral thm.
Define G(z) = g ( t ) dt
& then = |
( ) ( )
g ( z ) |
( using the conti nuit y of g L f or mu la)
Prob : ( )
dz
( , )
= 2
i
if n if n
? ( z a )
dz 2
, n i , n ( , )
Lemma f or C auch y ’s I nt egr al fo r m ula,
Let C
C ( , ) i. e . cir cle w it h cer te r rad ius .
i.e. C *z | z | +. S upp ose ‘a’ i s conta i ne d in C then
i
Proof : consider, dz
L et z e
i ,
d z ie
i d
?
( ) = d
fig sand pg 3
= i d
= 2 i
Now ,
Consider
( )
k , k +
?
( )
= ( )
d
=
e
d
=
.
/
= [ e
e
] = 0
( )
2
i
if k if k , k
Consider,
( ) ( )
=
0 1
( ) ( )
where w =
Now , |w| = |
| | |
| |
| |
<1
( | a| | a | )
Dista nce, if any po in t wi thin cir cle f r om cen te r alw ays l ess t han or eq ual to ( r ad ius)
?
( w )
=
( w w
w
)
=
( )
?
= 0
( )
1
dz the right-hand side series is uniformly convergent throughout C ,
and therefore using term by term integration
=
( )
=2 i …. = 2 i
5.2 Corollary
Suppose g is as above. Then the Integral Theorem and the Closed Curve Theorem apply to g.
Proof:
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CH 5 PROPERTIES OF ENTIRE FUNCTION
5.1 The Cauchy Integral Formula and Taylor Expansion for Entire Functions
We now show that if f is entire and if
g ( z ) {
f ( z ) f ( a )
z a
, if z a
f
( a ) , if z a
.
then the I nt egr al T heor em ( . ) and C losed C ur eve T heor em ( . ) ap p ly to g as w ell
as to f . (Note that since f is entire, g is continuous; however, it is not obvious that g is entire). We begin by
showing that the Rectangle Theorem applies to g.
5.1 rectangle thm (II):
If f is entire and if g(z) = 8
( ) ( )
, z a
f
( a ) , z a
then g ( z ) dz ,
Wh ere is t he b ound ary of a rectan g le R.
Proof : Since, f is entire implies g is continuous on R
i.e. g must be boundary on R
i. e. ? s. that | g ( z ) | ? z R
Case 1: if a ex t( R)
? g is analytic within and on R.
i.e. g ( z ) dz by rectangle thm
case 2 I f a b ound a r y ( R) i. e . a
then, we subdivide R into six sub-rectangle as shown below(sand note pg 1)
a 1 say, also, the orientation of R 2 & R 5 are shown.
lso, let k : boundary of the k
th
sub- r ect an g le ( k , , …. )
1 : that sub- r ect an g le w hich conta in ‘a’
W. L . O.G ( 1) < ?? /A for any arbitrary choice.
? g is a na l. Wit hin o n R 2, R 3 …. . R 6
ote, S in c e, 2 , 3 , …. . 6 d oes not conta in ‘a’
g ( z ) dz for k = 2, 3, 4, 5, 6 [rectangle thm I]
? g ( z ) dz ? … .
But , g ( z ) dz . g ( z ) dz /
= g ( z ) dz only due to the cancellation involved
i.e. g ( z ) dz « . l ( r
)
b y L f or mu la
i.e. g ( z ) dz «
i.e. g ( z ) dz ( ?? was arbitrary)
case 3: I f a I nt erior ( R)
this time we divide R into nine sub-rectangle, as in the previous case (i.e. ?? ( 1)< ?? etc.)
note along the eight sub- r ect an g le n ot cont ai ni ng ‘ a’
g ( z ) dz k , , …. . ( Rec. t hm I )
? g ( z ) dz
g ( z ) dz
= g ( z ) dz
fig sand pg 2
i.e. g ( z ) dz
< ??
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i.e. g ( z ) dz
= 0 ( ?? > 0 was arbi.)
corollary 5.2 vivek: Suppose g is as before, then the integral theorem & the closed curve theorem , both apply
to g.
Sol
n
Since, g is continuous , the proof of thm 4.15 & that of 4.16 apply, without any modification of g.
For instance, while proceeding for integral thm.
Define G(z) = g ( t ) dt
& then = |
( ) ( )
g ( z ) |
( using the conti nuit y of g L f or mu la)
Prob : ( )
dz
( , )
= 2
i
if n if n
? ( z a )
dz 2
, n i , n ( , )
Lemma f or C auch y ’s I nt egr al fo r m ula,
Let C
C ( , ) i. e . cir cle w it h cer te r rad ius .
i.e. C *z | z | +. S upp ose ‘a’ i s conta i ne d in C then
i
Proof : consider, dz
L et z e
i ,
d z ie
i d
?
( ) = d
fig sand pg 3
= i d
= 2 i
Now ,
Consider
( )
k , k +
?
( )
= ( )
d
=
e
d
=
.
/
= [ e
e
] = 0
( )
2
i
if k if k , k
Consider,
( ) ( )
=
0 1
( ) ( )
where w =
Now , |w| = |
| | |
| |
| |
<1
( | a| | a | )
Dista nce, if any po in t wi thin cir cle f r om cen te r alw ays l ess t han or eq ual to ( r ad ius)
?
( w )
=
( w w
w
)
=
( )
?
= 0
( )
1
dz the right-hand side series is uniformly convergent throughout C ,
and therefore using term by term integration
=
( )
=2 i …. = 2 i
5.2 Corollary
Suppose g is as above. Then the Integral Theorem and the Closed Curve Theorem apply to g.
Proof:
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We observe that since g is continuous, the proofs of Theorem 4.15 and 4.16 apply, without any modification,
to g.
CLOSED CURVE THEOREM FOR g
let f be an en ti r e g d efine d as g ( z ) {
f ( z ) f ( a )
z a
, if z a
f
( a ) , if z a
.
Then g ( z ) dz ? smooth closed curve C.
Proof y in te g r al t hm , ? a n e n ti r e fu n
c
G s.that
? g ( z ) dz G
( z ) dz G ( z ( b ) ) G ( z ( a ) )
p r op . ( C is cl osed )
5.3 Cauchy Integral Formula: Suppose that f is entire , that ‘a’ i s som e comple x no., an d that C i s the cur v e,
C : Re
i
, , with R > |a|
T hen ,
i
?
f ( z )
z a
dz f ( a ) ? a I nt ( C ( , R ) ) Re
Re
C ( , R ) R | a |
Proof: Define
g ( z ) {
f ( z ) f ( a )
z a
f ( a )
if z a
if z a
Since, f is entire, the closed curve theorem apply to g
? g ( z ) dz
i . e . ?
f ( z ) f ( a )
z a
dz
i . e . ?
f ( z )
z a
dz f ( a ) ?
dz
z a
( , )
i . e . f ( a ) ?
z a
dz
?
f ( z )
z a
dz
i . e . f ( a ) . i ?
f ( z )
z a
dz as ?
z a
dz
i
i . e . f ( a ) i
?
f ( z )
z a
dz
Pro b . ?
sin z
z
dz
| |
by C auch y
s I nt egr al f or mu la i f ( a ) ?
f ( z )
z a
dz
H ere f(z) sin z , a C C( , )
then ?
sin z
z | | dz ?
f ( z )
z a
dz
| |
=2 i f(0) = 2 i sin 0 = 0
Pro b ?
cos z
z ( z )
dz
| |
?
cos z
z z
dz | | ta k e f ( z ) cos z
z a
By above mentioned result
?
cos z
z ( z )
dz i .
cos i
i
| |
T ayl or ’s expansion of an entire fun
c
: let f is entire , it has a power series representation. In fact f
k
(0) exits
f or k , , , 3 , … . . and ( ) ?
f
( )
k z
let z , w e find a s uc h that z «a i . e. | z | | a| C be the circl e
Problem based on Cauchy integral formula (Generalized)/n
th
derivative formula on last page
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CH 5 PROPERTIES OF ENTIRE FUNCTION
5.1 The Cauchy Integral Formula and Taylor Expansion for Entire Functions
We now show that if f is entire and if
g ( z ) {
f ( z ) f ( a )
z a
, if z a
f
( a ) , if z a
.
then the I nt egr al T heor em ( . ) and C losed C ur eve T heor em ( . ) ap p ly to g as w ell
as to f . (Note that since f is entire, g is continuous; however, it is not obvious that g is entire). We begin by
showing that the Rectangle Theorem applies to g.
5.1 rectangle thm (II):
If f is entire and if g(z) = 8
( ) ( )
, z a
f
( a ) , z a
then g ( z ) dz ,
Wh ere is t he b ound ary of a rectan g le R.
Proof : Since, f is entire implies g is continuous on R
i.e. g must be boundary on R
i. e. ? s. that | g ( z ) | ? z R
Case 1: if a ex t( R)
? g is analytic within and on R.
i.e. g ( z ) dz by rectangle thm
case 2 I f a b ound a r y ( R) i. e . a
then, we subdivide R into six sub-rectangle as shown below(sand note pg 1)
a 1 say, also, the orientation of R 2 & R 5 are shown.
lso, let k : boundary of the k
th
sub- r ect an g le ( k , , …. )
1 : that sub- r ect an g le w hich conta in ‘a’
W. L . O.G ( 1) < ?? /A for any arbitrary choice.
? g is a na l. Wit hin o n R 2, R 3 …. . R 6
ote, S in c e, 2 , 3 , …. . 6 d oes not conta in ‘a’
g ( z ) dz for k = 2, 3, 4, 5, 6 [rectangle thm I]
? g ( z ) dz ? … .
But , g ( z ) dz . g ( z ) dz /
= g ( z ) dz only due to the cancellation involved
i.e. g ( z ) dz « . l ( r
)
b y L f or mu la
i.e. g ( z ) dz «
i.e. g ( z ) dz ( ?? was arbitrary)
case 3: I f a I nt erior ( R)
this time we divide R into nine sub-rectangle, as in the previous case (i.e. ?? ( 1)< ?? etc.)
note along the eight sub- r ect an g le n ot cont ai ni ng ‘ a’
g ( z ) dz k , , …. . ( Rec. t hm I )
? g ( z ) dz
g ( z ) dz
= g ( z ) dz
fig sand pg 2
i.e. g ( z ) dz
< ??
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i.e. g ( z ) dz
= 0 ( ?? > 0 was arbi.)
corollary 5.2 vivek: Suppose g is as before, then the integral theorem & the closed curve theorem , both apply
to g.
Sol
n
Since, g is continuous , the proof of thm 4.15 & that of 4.16 apply, without any modification of g.
For instance, while proceeding for integral thm.
Define G(z) = g ( t ) dt
& then = |
( ) ( )
g ( z ) |
( using the conti nuit y of g L f or mu la)
Prob : ( )
dz
( , )
= 2
i
if n if n
? ( z a )
dz 2
, n i , n ( , )
Lemma f or C auch y ’s I nt egr al fo r m ula,
Let C
C ( , ) i. e . cir cle w it h cer te r rad ius .
i.e. C *z | z | +. S upp ose ‘a’ i s conta i ne d in C then
i
Proof : consider, dz
L et z e
i ,
d z ie
i d
?
( ) = d
fig sand pg 3
= i d
= 2 i
Now ,
Consider
( )
k , k +
?
( )
= ( )
d
=
e
d
=
.
/
= [ e
e
] = 0
( )
2
i
if k if k , k
Consider,
( ) ( )
=
0 1
( ) ( )
where w =
Now , |w| = |
| | |
| |
| |
<1
( | a| | a | )
Dista nce, if any po in t wi thin cir cle f r om cen te r alw ays l ess t han or eq ual to ( r ad ius)
?
( w )
=
( w w
w
)
=
( )
?
= 0
( )
1
dz the right-hand side series is uniformly convergent throughout C ,
and therefore using term by term integration
=
( )
=2 i …. = 2 i
5.2 Corollary
Suppose g is as above. Then the Integral Theorem and the Closed Curve Theorem apply to g.
Proof:
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We observe that since g is continuous, the proofs of Theorem 4.15 and 4.16 apply, without any modification,
to g.
CLOSED CURVE THEOREM FOR g
let f be an en ti r e g d efine d as g ( z ) {
f ( z ) f ( a )
z a
, if z a
f
( a ) , if z a
.
Then g ( z ) dz ? smooth closed curve C.
Proof y in te g r al t hm , ? a n e n ti r e fu n
c
G s.that
? g ( z ) dz G
( z ) dz G ( z ( b ) ) G ( z ( a ) )
p r op . ( C is cl osed )
5.3 Cauchy Integral Formula: Suppose that f is entire , that ‘a’ i s som e comple x no., an d that C i s the cur v e,
C : Re
i
, , with R > |a|
T hen ,
i
?
f ( z )
z a
dz f ( a ) ? a I nt ( C ( , R ) ) Re
Re
C ( , R ) R | a |
Proof: Define
g ( z ) {
f ( z ) f ( a )
z a
f ( a )
if z a
if z a
Since, f is entire, the closed curve theorem apply to g
? g ( z ) dz
i . e . ?
f ( z ) f ( a )
z a
dz
i . e . ?
f ( z )
z a
dz f ( a ) ?
dz
z a
( , )
i . e . f ( a ) ?
z a
dz
?
f ( z )
z a
dz
i . e . f ( a ) . i ?
f ( z )
z a
dz as ?
z a
dz
i
i . e . f ( a ) i
?
f ( z )
z a
dz
Pro b . ?
sin z
z
dz
| |
by C auch y
s I nt egr al f or mu la i f ( a ) ?
f ( z )
z a
dz
H ere f(z) sin z , a C C( , )
then ?
sin z
z | | dz ?
f ( z )
z a
dz
| |
=2 i f(0) = 2 i sin 0 = 0
Pro b ?
cos z
z ( z )
dz
| |
?
cos z
z z
dz | | ta k e f ( z ) cos z
z a
By above mentioned result
?
cos z
z ( z )
dz i .
cos i
i
| |
T ayl or ’s expansion of an entire fun
c
: let f is entire , it has a power series representation. In fact f
k
(0) exits
f or k , , , 3 , … . . and ( ) ?
f
( )
k z
let z , w e find a s uc h that z «a i . e. | z | | a| C be the circl e
Problem based on Cauchy integral formula (Generalized)/n
th
derivative formula on last page
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C : { w | w R e
, }
C ( , R )
? C auch y ’s I nt egr al fo r mu la
f ( z ) i
?
f ( w )
w z
dw
? | z | | a | , act ually | z | | a |
not e w z
w . z
w
/
w
. z
w
/
w
[ z
w
.
z
w
/
] ( | z | | w |
i . e . f ( w )
w z
f ( w )
w
z .
f ( w )
w
z
.
f ( w )
w
i . e .
i
?
f ( w )
w z
i
?
f ( w )
w
dw
z . 4
i
?
f ( w )
w
dw
5 z
. 4
i
?
f ( w )
w
5 i . e .
f ( z ) ? C
z
w her e C
i
?
f ( w )
w
dw
… … … . . ( i )
This is a power series representation of f(z). Here C k depends upon the Curve C.
We have to prove that C k is independent of a, we have
f(z) = C 0+C 1z+C 2z
2
….
f ( z ) ? C
z
is unifor ml y conv erg en t
? I t ca n be differentiated term by term
H en ce , f ( z ) ? k C
z
f ( z ) ? k ( k ) C
z
i. e. f ( ) C 1 , f ( ) C 2 , f ( ) C 3 ………
follows, f
k
(0)= k!C k ? k
or C k =
( )
…… …… . . ( ii )
by (i) & (ii)
f ( z ) ? C
z
?
f
( ) z
k z hence p r ov ed
…… …… …… … …… …… … … …… …… …… … …… …… … … …… …… …… … …… …… … … …. .
f ( ) C
C
z C
z
C
f ( )
f
( ) C
C
z 3 C
z
C
f
( )
f ( ) C
C
z C
f
( )
Corollary 5.6 An entire fun
c
is infinitely differentiable .
Sol
n
let f be an entire fun
c
i.e. f has a valid Tay lor ’ s exp en sion
i . e . f ( z ) ?
f
( )
k z
i.e. f
k
exists ? k
i.e. f is infinitely differentiable
alternatively,
let f be an entire fun
c
f ( z ) can b e d efine d as ever y w he r e cgt p ow er series.
ever y p ow er ser ie s can b e d iff infinitely many times in its region of convergence.
f ( z ) can b e d iff in f in it ely many times i n i ts reg ion of conv erg en ce i n w hol e comple x plan e
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CH 5 PROPERTIES OF ENTIRE FUNCTION
5.1 The Cauchy Integral Formula and Taylor Expansion for Entire Functions
We now show that if f is entire and if
g ( z ) {
f ( z ) f ( a )
z a
, if z a
f
( a ) , if z a
.
then the I nt egr al T heor em ( . ) and C losed C ur eve T heor em ( . ) ap p ly to g as w ell
as to f . (Note that since f is entire, g is continuous; however, it is not obvious that g is entire). We begin by
showing that the Rectangle Theorem applies to g.
5.1 rectangle thm (II):
If f is entire and if g(z) = 8
( ) ( )
, z a
f
( a ) , z a
then g ( z ) dz ,
Wh ere is t he b ound ary of a rectan g le R.
Proof : Since, f is entire implies g is continuous on R
i.e. g must be boundary on R
i. e. ? s. that | g ( z ) | ? z R
Case 1: if a ex t( R)
? g is analytic within and on R.
i.e. g ( z ) dz by rectangle thm
case 2 I f a b ound a r y ( R) i. e . a
then, we subdivide R into six sub-rectangle as shown below(sand note pg 1)
a 1 say, also, the orientation of R 2 & R 5 are shown.
lso, let k : boundary of the k
th
sub- r ect an g le ( k , , …. )
1 : that sub- r ect an g le w hich conta in ‘a’
W. L . O.G ( 1) < ?? /A for any arbitrary choice.
? g is a na l. Wit hin o n R 2, R 3 …. . R 6
ote, S in c e, 2 , 3 , …. . 6 d oes not conta in ‘a’
g ( z ) dz for k = 2, 3, 4, 5, 6 [rectangle thm I]
? g ( z ) dz ? … .
But , g ( z ) dz . g ( z ) dz /
= g ( z ) dz only due to the cancellation involved
i.e. g ( z ) dz « . l ( r
)
b y L f or mu la
i.e. g ( z ) dz «
i.e. g ( z ) dz ( ?? was arbitrary)
case 3: I f a I nt erior ( R)
this time we divide R into nine sub-rectangle, as in the previous case (i.e. ?? ( 1)< ?? etc.)
note along the eight sub- r ect an g le n ot cont ai ni ng ‘ a’
g ( z ) dz k , , …. . ( Rec. t hm I )
? g ( z ) dz
g ( z ) dz
= g ( z ) dz
fig sand pg 2
i.e. g ( z ) dz
< ??
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i.e. g ( z ) dz
= 0 ( ?? > 0 was arbi.)
corollary 5.2 vivek: Suppose g is as before, then the integral theorem & the closed curve theorem , both apply
to g.
Sol
n
Since, g is continuous , the proof of thm 4.15 & that of 4.16 apply, without any modification of g.
For instance, while proceeding for integral thm.
Define G(z) = g ( t ) dt
& then = |
( ) ( )
g ( z ) |
( using the conti nuit y of g L f or mu la)
Prob : ( )
dz
( , )
= 2
i
if n if n
? ( z a )
dz 2
, n i , n ( , )
Lemma f or C auch y ’s I nt egr al fo r m ula,
Let C
C ( , ) i. e . cir cle w it h cer te r rad ius .
i.e. C *z | z | +. S upp ose ‘a’ i s conta i ne d in C then
i
Proof : consider, dz
L et z e
i ,
d z ie
i d
?
( ) = d
fig sand pg 3
= i d
= 2 i
Now ,
Consider
( )
k , k +
?
( )
= ( )
d
=
e
d
=
.
/
= [ e
e
] = 0
( )
2
i
if k if k , k
Consider,
( ) ( )
=
0 1
( ) ( )
where w =
Now , |w| = |
| | |
| |
| |
<1
( | a| | a | )
Dista nce, if any po in t wi thin cir cle f r om cen te r alw ays l ess t han or eq ual to ( r ad ius)
?
( w )
=
( w w
w
)
=
( )
?
= 0
( )
1
dz the right-hand side series is uniformly convergent throughout C ,
and therefore using term by term integration
=
( )
=2 i …. = 2 i
5.2 Corollary
Suppose g is as above. Then the Integral Theorem and the Closed Curve Theorem apply to g.
Proof:
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We observe that since g is continuous, the proofs of Theorem 4.15 and 4.16 apply, without any modification,
to g.
CLOSED CURVE THEOREM FOR g
let f be an en ti r e g d efine d as g ( z ) {
f ( z ) f ( a )
z a
, if z a
f
( a ) , if z a
.
Then g ( z ) dz ? smooth closed curve C.
Proof y in te g r al t hm , ? a n e n ti r e fu n
c
G s.that
? g ( z ) dz G
( z ) dz G ( z ( b ) ) G ( z ( a ) )
p r op . ( C is cl osed )
5.3 Cauchy Integral Formula: Suppose that f is entire , that ‘a’ i s som e comple x no., an d that C i s the cur v e,
C : Re
i
, , with R > |a|
T hen ,
i
?
f ( z )
z a
dz f ( a ) ? a I nt ( C ( , R ) ) Re
Re
C ( , R ) R | a |
Proof: Define
g ( z ) {
f ( z ) f ( a )
z a
f ( a )
if z a
if z a
Since, f is entire, the closed curve theorem apply to g
? g ( z ) dz
i . e . ?
f ( z ) f ( a )
z a
dz
i . e . ?
f ( z )
z a
dz f ( a ) ?
dz
z a
( , )
i . e . f ( a ) ?
z a
dz
?
f ( z )
z a
dz
i . e . f ( a ) . i ?
f ( z )
z a
dz as ?
z a
dz
i
i . e . f ( a ) i
?
f ( z )
z a
dz
Pro b . ?
sin z
z
dz
| |
by C auch y
s I nt egr al f or mu la i f ( a ) ?
f ( z )
z a
dz
H ere f(z) sin z , a C C( , )
then ?
sin z
z | | dz ?
f ( z )
z a
dz
| |
=2 i f(0) = 2 i sin 0 = 0
Pro b ?
cos z
z ( z )
dz
| |
?
cos z
z z
dz | | ta k e f ( z ) cos z
z a
By above mentioned result
?
cos z
z ( z )
dz i .
cos i
i
| |
T ayl or ’s expansion of an entire fun
c
: let f is entire , it has a power series representation. In fact f
k
(0) exits
f or k , , , 3 , … . . and ( ) ?
f
( )
k z
let z , w e find a s uc h that z «a i . e. | z | | a| C be the circl e
Problem based on Cauchy integral formula (Generalized)/n
th
derivative formula on last page
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C : { w | w R e
, }
C ( , R )
? C auch y ’s I nt egr al fo r mu la
f ( z ) i
?
f ( w )
w z
dw
? | z | | a | , act ually | z | | a |
not e w z
w . z
w
/
w
. z
w
/
w
[ z
w
.
z
w
/
] ( | z | | w |
i . e . f ( w )
w z
f ( w )
w
z .
f ( w )
w
z
.
f ( w )
w
i . e .
i
?
f ( w )
w z
i
?
f ( w )
w
dw
z . 4
i
?
f ( w )
w
dw
5 z
. 4
i
?
f ( w )
w
5 i . e .
f ( z ) ? C
z
w her e C
i
?
f ( w )
w
dw
… … … . . ( i )
This is a power series representation of f(z). Here C k depends upon the Curve C.
We have to prove that C k is independent of a, we have
f(z) = C 0+C 1z+C 2z
2
….
f ( z ) ? C
z
is unifor ml y conv erg en t
? I t ca n be differentiated term by term
H en ce , f ( z ) ? k C
z
f ( z ) ? k ( k ) C
z
i. e. f ( ) C 1 , f ( ) C 2 , f ( ) C 3 ………
follows, f
k
(0)= k!C k ? k
or C k =
( )
…… …… . . ( ii )
by (i) & (ii)
f ( z ) ? C
z
?
f
( ) z
k z hence p r ov ed
…… …… …… … …… …… … … …… …… …… … …… …… … … …… …… …… … …… …… … … …. .
f ( ) C
C
z C
z
C
f ( )
f
( ) C
C
z 3 C
z
C
f
( )
f ( ) C
C
z C
f
( )
Corollary 5.6 An entire fun
c
is infinitely differentiable .
Sol
n
let f be an entire fun
c
i.e. f has a valid Tay lor ’ s exp en sion
i . e . f ( z ) ?
f
( )
k z
i.e. f
k
exists ? k
i.e. f is infinitely differentiable
alternatively,
let f be an entire fun
c
f ( z ) can b e d efine d as ever y w he r e cgt p ow er series.
ever y p ow er ser ie s can b e d iff infinitely many times in its region of convergence.
f ( z ) can b e d iff in f in it ely many times i n i ts reg ion of conv erg en ce i n w hol e comple x plan e
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C or oll ar y . I f f is e nt ire an d if a b e a ny c om p lex no. then ,
f ( z ) f ( a ) f
( a ) ( z a )
f
( a ) ( z a )
Sol
n
define g(t) = f(t+a) ? t
? g is a n e nt ire f un
c
hence has a Tay lor ’s ser ie s exp an sion
i . e . g ( t ) ?
g
( )
k t
g ( ) g
( )
t g
( )
t
t rep lace b y z a i . e . t z a
g ( z a ) g ( ) g
( )
( z a ) g
( )
( z a )
f ( z ) f ( a ) f
( a )
( z a ) f
( a )
( z a )
Pro p ositi o n . if f be en ti r e f un c g ( z ) {
f ( z ) f ( a )
z a
, if z a then g is also en ti r e fu n
.
f
( a ) , if z a
Poof: Since f is entire fun
c
? f ( z ) has T ayl or ’s expa nsi on
? Tay lor ’s expa nsi on a b out the p oint ‘a’ is
f(z) f ( a) f ( a)(z a)
( )
( z a )
an d f or z a
( ) ( )
f
( a )
( )
( z a )
g(z) = f
( a )
( )
( z a ) …… …( i)
by def
n
of g , the eq
n
(i) is also valid for z=a
? g ( z ) is a n e nt ire f un
c
because g is a power series exp. which is convergent everywhere
g ( z ) mu st b e a na ly ti c on w hol e of
en ti r e
F in d the p ow er series r ep r ese nt at ion of
Que/pg 64 =
( i ) f ( z ) z
aro und z ( ii ) f ( z ) e
aro und z }
Sol
n
Since f is entire
y Tay lo r ’s ser ie s exp r essi on
f ( z ) f ( ) f
( )
( z ) f
( )
( z )
?
f
( )
k ( z )
i . e . z
( z ) ( z )
only , hig her d eriv at ive g r ea te r than g oes to z ero .
(ii) Since f(z)=e
z
is entire
y Tay lo r ’s ser ie s exp en sion
f ( z ) f ( ) f
( )
( z ) f
( )
( z )
?
f
( ) ( z )
k
e
e
e
( z ) e
( z )
?
( )
e
6 ( z ) ( z )
( z )
3 7 e
?
( z )
k
Que 3(a)/pg 64 Show that an odd entire fun
c
has only odd terms in its power series expansion about z=0
(b) What if odd replace by even ?
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FAQs on Properties Of Entire Function - Topic-wise Tests & Solved Examples for Mathematics
| 1. What is an entire function in mathematics? |  |
An entire function in mathematics is a complex-valued function that is defined and holomorphic on the entire complex plane. This means that the function is differentiable at every point in the complex plane.
| 2. What are the properties of entire functions? | |
Some properties of entire functions include:
- An entire function is infinitely differentiable at every point in the complex plane.
- Entire functions are holomorphic on the entire complex plane, which means they are analytic and have a power series representation.
- Entire functions are bounded on compact subsets of the complex plane.
- Entire functions have no poles or essential singularities.
- The zeros of an entire function can only be isolated points or accumulate at infinity.
| 3. Can an entire function have an essential singularity? | |
No, an entire function cannot have an essential singularity. An essential singularity is a type of singularity that cannot be removed by any analytic function. Since an entire function is defined and holomorphic on the entire complex plane, it cannot have any singularities, including essential singularities.
| 4. Are all polynomials entire functions? | |
Yes, all polynomials are entire functions. A polynomial is a function of the form f(z) = a_n z^n + a_{n-1} z^{n-1} + ... + a_1 z + a_0, where n is a non-negative integer and a_0, a_1, ..., a_n are complex constants. Polynomials are holomorphic on the entire complex plane, making them entire functions.
| 5. Are entire functions limited to the complex plane? | |
Yes, entire functions are defined and holomorphic on the complex plane. The complex plane consists of all complex numbers, which can be represented as a combination of a real part and an imaginary part. The behavior and properties of entire functions are specifically studied within the context of the complex plane.
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