Mathematics Exam > Mathematics Notes > Topic-wise Tests & Solved Examples for Mathematics > Properties Of Entire Function
Properties Of Entire Function | Topic-wise Tests & Solved Examples for Mathematics PDF Download
| Download, print and study this document offline |
Please wait while the PDF view is loading
Page 1
Free coaching of B.Sc (h) maths & JAM
For more 8130648819
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
< ??
Page 2
Free coaching of B.Sc (h) maths & JAM
For more 8130648819
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
< ??
Free coaching of B.Sc (h) maths & JAM
For more 8130648819
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:
Page 3
Free coaching of B.Sc (h) maths & JAM
For more 8130648819
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
< ??
Free coaching of B.Sc (h) maths & JAM
For more 8130648819
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:
Free coaching of B.Sc (h) maths & JAM
For more 8130648819
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
Page 4
Free coaching of B.Sc (h) maths & JAM
For more 8130648819
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
< ??
Free coaching of B.Sc (h) maths & JAM
For more 8130648819
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:
Free coaching of B.Sc (h) maths & JAM
For more 8130648819
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
Free coaching of B.Sc (h) maths & JAM
For more 8130648819
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
Page 5
Free coaching of B.Sc (h) maths & JAM
For more 8130648819
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
< ??
Free coaching of B.Sc (h) maths & JAM
For more 8130648819
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:
Free coaching of B.Sc (h) maths & JAM
For more 8130648819
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
Free coaching of B.Sc (h) maths & JAM
For more 8130648819
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
Free coaching of B.Sc (h) maths & JAM
For more 8130648819
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 ?
Read More
|
27 docs|150 tests
|
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.
Related Exams
About this Document
|
4.71/5 Rating |
|
Dec 22, 2024 Last updated |
Document Description: Properties Of Entire Function for Mathematics 2024 is part of Topic-wise Tests & Solved Examples for Mathematics preparation.
The notes and questions for Properties Of Entire Function have been prepared according to the Mathematics exam syllabus. Information about Properties Of Entire Function covers topics
like and Properties Of Entire Function Example, for Mathematics 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for Properties Of Entire Function.
Introduction of Properties Of Entire Function in English is available as part of our Topic-wise Tests & Solved Examples for Mathematics
for Mathematics & Properties Of Entire Function in Hindi for Topic-wise Tests & Solved Examples for Mathematics course.
Download more important topics related with notes, lectures and mock test series for Mathematics
Exam by signing up for free. Mathematics: Properties Of Entire Function | Topic-wise Tests & Solved Examples for Mathematics
Description
Full syllabus notes, lecture & questions for Properties Of Entire Function | Topic-wise Tests & Solved Examples for Mathematics - Mathematics | Plus excerises question with solution to help you revise complete syllabus for Topic-wise Tests & Solved Examples for Mathematics | Best notes, free PDF download
Information about Properties Of Entire Function
In this doc you can find the meaning of Properties Of Entire Function defined & explained in the simplest way possible. Besides explaining types of
Properties Of Entire Function theory, EduRev gives you an ample number of questions to practice Properties Of Entire Function tests, examples and also practice Mathematics
tests
|
27 docs|150 tests
|
Download as PDF
|
Explore Courses for Mathematics exam
|
|
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
Related Searches
Objective type Questions
,Important questions
,video lectures
,Previous Year Questions with Solutions
,past year papers
,Viva Questions
,study material
,Sample Paper
,MCQs
,Properties Of Entire Function | Topic-wise Tests & Solved Examples for Mathematics
,Summary
,ppt
,Extra Questions
,shortcuts and tricks
,Free
,Properties Of Entire Function | Topic-wise Tests & Solved Examples for Mathematics
,Exam
,Semester Notes
,practice quizzes
,Properties Of Entire Function | Topic-wise Tests & Solved Examples for Mathematics
,mock tests for examination
;
Additional Information about Properties Of Entire Function for Mathematics Preparation
Properties Of Entire Function Free PDF Download
The Properties Of Entire Function is an invaluable resource that delves deep into the core of the Mathematics exam.
These study notes are curated by experts and cover all the essential topics and concepts, making your preparation more efficient and effective.
With the help of these notes, you can grasp complex subjects quickly, revise important points easily,
and reinforce your understanding of key concepts. The study notes are presented in a concise and easy-to-understand manner,
allowing you to optimize your learning process. Whether you're looking for best-recommended books, sample papers, study material,
or toppers' notes, this PDF has got you covered. Download the Properties Of Entire Function now and kickstart your journey towards success in the Mathematics exam.
Importance of Properties Of Entire Function
The importance of Properties Of Entire Function cannot be overstated, especially for Mathematics aspirants.
This document holds the key to success in the Mathematics exam.
It offers a detailed understanding of the concept, providing invaluable insights into the topic.
By knowing the concepts well in advance, students can plan their preparation effectively.
Utilize this indispensable guide for a well-rounded preparation and achieve your desired results.
Properties Of Entire Function Notes
Properties Of Entire Function Notes offer in-depth insights into the specific topic to help you master it with ease.
This comprehensive document covers all aspects related to Properties Of Entire Function.
It includes detailed information about the exam syllabus, recommended books, and study materials for a well-rounded preparation.
Practice papers and question papers enable you to assess your progress effectively.
Additionally, the paper analysis provides valuable tips for tackling the exam strategically.
Access to Toppers' notes gives you an edge in understanding complex concepts.
Whether you're a beginner or aiming for advanced proficiency, Properties Of Entire Function Notes on EduRev are your ultimate resource for success.
Properties Of Entire Function Mathematics Questions
The "Properties Of Entire Function Mathematics Questions" guide is a valuable resource for all aspiring students preparing for the
Mathematics exam. It focuses on providing a wide range of practice questions to help students gauge
their understanding of the exam topics. These questions cover the entire syllabus, ensuring comprehensive preparation.
The guide includes previous years' question papers for students to familiarize themselves with the exam's format and difficulty level.
Additionally, it offers subject-specific question banks, allowing students to focus on weak areas and improve their performance.
Study Properties Of Entire Function on the App
Students of Mathematics can study Properties Of Entire Function alongwith tests & analysis from the EduRev app,
which will help them while preparing for their exam. Apart from the Properties Of Entire Function,
students can also utilize the EduRev App for other study materials such as previous year question papers, syllabus, important questions, etc.
The EduRev App will make your learning easier as you can access it from anywhere you want.
The content of Properties Of Entire Function is prepared as per the latest Mathematics syllabus.
|
© EduRev
|
Education Revolution
|
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup