Mathematics Exam > Mathematics Notes > Topic-wise Tests & Solved Examples for Mathematics > Complex Analysis (Part - 2) Solved Example
Complex Analysis (Part - 2) Solved Example | 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
Exercise 19: Suppose T . S how that the cor r esp ond in g set S ? is
a . a cir cle if T is a cir cle.
b . a cir cle minus ( , , ) if T is a line .
so l
let T be cir cle in the set of com p lex .
let it is d en ote d by
i . e . T * ( x , y ) ( x
y
) x Cy D + …. . ( i )
not e S is the ste r eog hr ap hic p r oje cti on of T ( on )
? ( x , y ) T ( , , ) S
using ste r eog r ap hic p r oje cti on f or mu lae ( 3 )
x , y , | z |
? f r om eq uation (i)
(
* (
* C (
* D
C D ( )
or ( D ) C D ( ii )
w hich is the eq uati on of a p lane in sp ace w hen a g en eral p oi nt has coo r d in at es ( , , ) .
ote that a p lane a sp her e in te r sect s in a cir cle .
S upp ose , D then ( i ) is a straigh t line in . ( x Cy D )
T hu s , the cor r esp ond in g set in the sp her e in g iven by the in te r sect ion of
w it h a p lane C D D by ( ii )
p assi ng thr oug ht ( , , ) S
w hich is a cir cle minus ( , , ) , sin ce ta k en .
if , D
T is ( x
y
) x Cy
e q
( ii ) b ecomes C
w hich is again a p lan e p assi ng ( , , )
b ut on the other hand ( , , ) S
i . e . this p lane in te r sect s the sp he r e
i . e . this in te r sect ion has to be a cir cle
i . e . S is a cir cle .
E xc erc ise . L et S * ( , , ) = + , w her e ,
w her e * ( , , ) + is the Rie mann sp her e ,
and let T be thecor r esp ond in g set ( x , y ) in . S how that T is the exte r ior of a cir cle
cen te r ed at .
So l
T ake D { ( x , y ) x
y
=
} , exte r ior of cir cle x
y
cen te r ed
at .
C lai m T D
to p r ov e T D . L et be a p oint in T cor r esp ond in g to a p oint in S , so that using
ste r eog r ap hic p r oje cti on , x
y
(
*
(
*
( )
( )
( )
now ,
=
( ) = ( ) = ( ever y thin g p ositi v e )
w hich is tru e as S
? T D . T hu s T D .
To p r ov e D T
if D , say ( x , y ) , then let ( , , ) be the cor r esp ond in g p oint on the sp her e
g iven by x
x
y
, y
x
y
, x
y
x
y
Page 2
Exercise 19: Suppose T . S how that the cor r esp ond in g set S ? is
a . a cir cle if T is a cir cle.
b . a cir cle minus ( , , ) if T is a line .
so l
let T be cir cle in the set of com p lex .
let it is d en ote d by
i . e . T * ( x , y ) ( x
y
) x Cy D + …. . ( i )
not e S is the ste r eog hr ap hic p r oje cti on of T ( on )
? ( x , y ) T ( , , ) S
using ste r eog r ap hic p r oje cti on f or mu lae ( 3 )
x , y , | z |
? f r om eq uation (i)
(
* (
* C (
* D
C D ( )
or ( D ) C D ( ii )
w hich is the eq uati on of a p lane in sp ace w hen a g en eral p oi nt has coo r d in at es ( , , ) .
ote that a p lane a sp her e in te r sect s in a cir cle .
S upp ose , D then ( i ) is a straigh t line in . ( x Cy D )
T hu s , the cor r esp ond in g set in the sp her e in g iven by the in te r sect ion of
w it h a p lane C D D by ( ii )
p assi ng thr oug ht ( , , ) S
w hich is a cir cle minus ( , , ) , sin ce ta k en .
if , D
T is ( x
y
) x Cy
e q
( ii ) b ecomes C
w hich is again a p lan e p assi ng ( , , )
b ut on the other hand ( , , ) S
i . e . this p lane in te r sect s the sp he r e
i . e . this in te r sect ion has to be a cir cle
i . e . S is a cir cle .
E xc erc ise . L et S * ( , , ) = + , w her e ,
w her e * ( , , ) + is the Rie mann sp her e ,
and let T be thecor r esp ond in g set ( x , y ) in . S how that T is the exte r ior of a cir cle
cen te r ed at .
So l
T ake D { ( x , y ) x
y
=
} , exte r ior of cir cle x
y
cen te r ed
at .
C lai m T D
to p r ov e T D . L et be a p oint in T cor r esp ond in g to a p oint in S , so that using
ste r eog r ap hic p r oje cti on , x
y
(
*
(
*
( )
( )
( )
now ,
=
( ) = ( ) = ( ever y thin g p ositi v e )
w hich is tru e as S
? T D . T hu s T D .
To p r ov e D T
if D , say ( x , y ) , then let ( , , ) be the cor r esp ond in g p oint on the sp her e
g iven by x
x
y
, y
x
y
, x
y
x
y
Free coaching of B.Sc (h) maths & JAM
For more 8130648819
w her e , = x
y
= ( x
y
) ( x
y
) ( ) =
( x
y
) =
( ) tru e b eca use D
p oint is in S and so in T , ? D T .
E xer cise S upp ose that z is the ste r og r ap hic of ( , , ) and
z
is the p r oje cti on of
( , , ) .
a . S how that ( , , )
b . S how that the f uncti on
z
, z , is r ep r ese nt at e d on b y r ota ti on ab out the
d ia mete r w it h en d p oint s ( , ,
* and (
, ,
* .
So l
we use the ste r eog r ap hic p r oje cti on f or mu lae
x
x
y
, y
x
y
, x
y
x
y
w her e z x iy is the ste r eog r ap hic p r oje cti on of ( , , )
g iven that
z
is the ste r eog r ap hic p r oje cti on of ( , , ) , we hav e us ing
z
x iy
x iy
x
y
x
x
y
i ( y
x
y
*
x
i y
, w her e x
x
x
y
, y
y
x
y
sin ce , ( x
)
( y
)
x
y
( x
y
)
x
y
, we oba in using
x
( x
)
( y
)
x
x
y
(
x
x
y
*
(
y
x
y
*
x
x
y
y
( x
)
( y
)
y
x
y
(
x
x
y
*
(
y
x
y
*
y
x
y
( x
)
( y
)
( x
)
( y
)
x
y
( , , ) ( , , ) not e ap p ly this f or z i
ANALYTIC POLYNOMIALS
First note that
x
y
ixy is a DIRECT FUNCTION of x+iy (its (x+iy)
2
& x
y
ixy is not a DIRECT FUNCTION of x+iy
Def
n
A polynomial p(x, y)will be called an na ly ti c p oly nomia l
? if ? ? k , s. t .
(means complete)
p ( x, y) ? 0 ? 1 ( x iy) ? 2(x+iy)
2
…. . ? n(x+iy)
n
Then, we will say that p(x, y) is a polynomial in z &
p(z) ? 0 ? 1 z ? 2z
2
…. . ? nz
n
S how that x
y
ixy is an an aly ti c p oly nomia l .
So l
p ( x , y ) x
y
ixy
? p ( x , y ) ( x iy )
p ( z ) z
? p ( x , y ) is an aly ti c p oly nomia l .
Page 3
Exercise 19: Suppose T . S how that the cor r esp ond in g set S ? is
a . a cir cle if T is a cir cle.
b . a cir cle minus ( , , ) if T is a line .
so l
let T be cir cle in the set of com p lex .
let it is d en ote d by
i . e . T * ( x , y ) ( x
y
) x Cy D + …. . ( i )
not e S is the ste r eog hr ap hic p r oje cti on of T ( on )
? ( x , y ) T ( , , ) S
using ste r eog r ap hic p r oje cti on f or mu lae ( 3 )
x , y , | z |
? f r om eq uation (i)
(
* (
* C (
* D
C D ( )
or ( D ) C D ( ii )
w hich is the eq uati on of a p lane in sp ace w hen a g en eral p oi nt has coo r d in at es ( , , ) .
ote that a p lane a sp her e in te r sect s in a cir cle .
S upp ose , D then ( i ) is a straigh t line in . ( x Cy D )
T hu s , the cor r esp ond in g set in the sp her e in g iven by the in te r sect ion of
w it h a p lane C D D by ( ii )
p assi ng thr oug ht ( , , ) S
w hich is a cir cle minus ( , , ) , sin ce ta k en .
if , D
T is ( x
y
) x Cy
e q
( ii ) b ecomes C
w hich is again a p lan e p assi ng ( , , )
b ut on the other hand ( , , ) S
i . e . this p lane in te r sect s the sp he r e
i . e . this in te r sect ion has to be a cir cle
i . e . S is a cir cle .
E xc erc ise . L et S * ( , , ) = + , w her e ,
w her e * ( , , ) + is the Rie mann sp her e ,
and let T be thecor r esp ond in g set ( x , y ) in . S how that T is the exte r ior of a cir cle
cen te r ed at .
So l
T ake D { ( x , y ) x
y
=
} , exte r ior of cir cle x
y
cen te r ed
at .
C lai m T D
to p r ov e T D . L et be a p oint in T cor r esp ond in g to a p oint in S , so that using
ste r eog r ap hic p r oje cti on , x
y
(
*
(
*
( )
( )
( )
now ,
=
( ) = ( ) = ( ever y thin g p ositi v e )
w hich is tru e as S
? T D . T hu s T D .
To p r ov e D T
if D , say ( x , y ) , then let ( , , ) be the cor r esp ond in g p oint on the sp her e
g iven by x
x
y
, y
x
y
, x
y
x
y
Free coaching of B.Sc (h) maths & JAM
For more 8130648819
w her e , = x
y
= ( x
y
) ( x
y
) ( ) =
( x
y
) =
( ) tru e b eca use D
p oint is in S and so in T , ? D T .
E xer cise S upp ose that z is the ste r og r ap hic of ( , , ) and
z
is the p r oje cti on of
( , , ) .
a . S how that ( , , )
b . S how that the f uncti on
z
, z , is r ep r ese nt at e d on b y r ota ti on ab out the
d ia mete r w it h en d p oint s ( , ,
* and (
, ,
* .
So l
we use the ste r eog r ap hic p r oje cti on f or mu lae
x
x
y
, y
x
y
, x
y
x
y
w her e z x iy is the ste r eog r ap hic p r oje cti on of ( , , )
g iven that
z
is the ste r eog r ap hic p r oje cti on of ( , , ) , we hav e us ing
z
x iy
x iy
x
y
x
x
y
i ( y
x
y
*
x
i y
, w her e x
x
x
y
, y
y
x
y
sin ce , ( x
)
( y
)
x
y
( x
y
)
x
y
, we oba in using
x
( x
)
( y
)
x
x
y
(
x
x
y
*
(
y
x
y
*
x
x
y
y
( x
)
( y
)
y
x
y
(
x
x
y
*
(
y
x
y
*
y
x
y
( x
)
( y
)
( x
)
( y
)
x
y
( , , ) ( , , ) not e ap p ly this f or z i
ANALYTIC POLYNOMIALS
First note that
x
y
ixy is a DIRECT FUNCTION of x+iy (its (x+iy)
2
& x
y
ixy is not a DIRECT FUNCTION of x+iy
Def
n
A polynomial p(x, y)will be called an na ly ti c p oly nomia l
? if ? ? k , s. t .
(means complete)
p ( x, y) ? 0 ? 1 ( x iy) ? 2(x+iy)
2
…. . ? n(x+iy)
n
Then, we will say that p(x, y) is a polynomial in z &
p(z) ? 0 ? 1 z ? 2z
2
…. . ? nz
n
S how that x
y
ixy is an an aly ti c p oly nomia l .
So l
p ( x , y ) x
y
ixy
? p ( x , y ) ( x iy )
p ( z ) z
? p ( x , y ) is an aly ti c p oly nomia l .
Free coaching of B.Sc (h) maths & JAM
For more 8130648819
S how that x
y
ixy is not an aly ti c p oly .
So l
Let if possible suppose, it is an analytic poly.
i.e. x
y
ixy ? ( x iy )
? x, y …… … …… . . ( *)
x
? x
taking y = 0
i.e. x
x x
x
x
i.e. b ut 2 = 1
i.e. x
y
iy ( x iy )
by (*)
a contradiction
E xample p ( x , y ) x
iv ( x , y ) is not an aly ti c f or any ch oice of the r ea l p oly nomia l
v ( x , y ) .
So l
f or a p oly nomia l in z can hav e a r ea l p art of d egr ee in x only if it is of the
f or m a z
bz c w it he a . In this case , how ever , the r ea l p art mu st conta in a y
,
i . e . y
b ut y
is not in p ( x , y )
? p ( x , y ) is not an aly ti c f or any v ( x , y )
def
n
let u(x, y) and v(x, y) be real valued function and
f(x, y) = u(x, y )+iv (x, y)
then the partial derivative of f (if exists) are
f x(x, y) = u x(x, y)+i v x(x, y)
f x(x, y) = u y(x, y)+i v y(x, y)
Proposition 2.3 A polynomial p(x, y) is analytic iff p y= ip x
Proof
forward part : let p(x, y) be analytic
? p ( x, y) ? 0 ? 1 ( x iy ) ? 2(x+iy)
2
…. . ? n(x+iy)
n
i.e. p x(x, y) = ( x i y ) 3
( x iy )
n ( x iy )
& p y(x, y) = i i ( x iy ) 3i ( x iy )
ni ( x iy )
Follows
P y(x, y)= ip x(x, y) done
Converse part
let p y = ip x [TS : p(x, y) is an analytic poly.]
let p(x, y) = q 0+q 1 +q 2 +......+q n
where, q n : sum of all n
th
degree term
WLOG q n ß 0x
n
ß 1x
n y ß 2x
n y
2
…. ß ny
n
We shall make use of the fact that
P y =ip x
i.e. (q n) y = i(q n) x
so let us 1
st
consider,
q n ( x, y) ß 0x
n
ß 1x
n y ß 2x
n y
2
…… ß n-2x
2
y
n ß n xy
n ß ny
n
i.e. (q n) y ß 1x
n ß 2x
n y 3 ß 3x
n 3
y
2
…. . ( n ) ß n x
2
y
n 3
( n ) ß n xy
n nß ny
n
& (q n) x n ß 0x
n ( n ) ß 1x
n y ( n ) ß 2x
n 3
y
2
…. . ß n xy
n ß n y
n +0
{
using q
i q
we g et ß
i n ß
. i . e . ß
.
n
/ ß
i / ß
i ( n ) ß
4 i . e . ß
i
( n )
.
n
/ ß
i i
.
n
/ ß
5
3 ß
i ( n ) ß
(
i . e . ß
i
( n )
3
ß
i n 3
. i
.
n
/ ß
i
.
n
3
/ ß
) ? in g en eral ß
i
.
n
k
/ ß
Page 4
Exercise 19: Suppose T . S how that the cor r esp ond in g set S ? is
a . a cir cle if T is a cir cle.
b . a cir cle minus ( , , ) if T is a line .
so l
let T be cir cle in the set of com p lex .
let it is d en ote d by
i . e . T * ( x , y ) ( x
y
) x Cy D + …. . ( i )
not e S is the ste r eog hr ap hic p r oje cti on of T ( on )
? ( x , y ) T ( , , ) S
using ste r eog r ap hic p r oje cti on f or mu lae ( 3 )
x , y , | z |
? f r om eq uation (i)
(
* (
* C (
* D
C D ( )
or ( D ) C D ( ii )
w hich is the eq uati on of a p lane in sp ace w hen a g en eral p oi nt has coo r d in at es ( , , ) .
ote that a p lane a sp her e in te r sect s in a cir cle .
S upp ose , D then ( i ) is a straigh t line in . ( x Cy D )
T hu s , the cor r esp ond in g set in the sp her e in g iven by the in te r sect ion of
w it h a p lane C D D by ( ii )
p assi ng thr oug ht ( , , ) S
w hich is a cir cle minus ( , , ) , sin ce ta k en .
if , D
T is ( x
y
) x Cy
e q
( ii ) b ecomes C
w hich is again a p lan e p assi ng ( , , )
b ut on the other hand ( , , ) S
i . e . this p lane in te r sect s the sp he r e
i . e . this in te r sect ion has to be a cir cle
i . e . S is a cir cle .
E xc erc ise . L et S * ( , , ) = + , w her e ,
w her e * ( , , ) + is the Rie mann sp her e ,
and let T be thecor r esp ond in g set ( x , y ) in . S how that T is the exte r ior of a cir cle
cen te r ed at .
So l
T ake D { ( x , y ) x
y
=
} , exte r ior of cir cle x
y
cen te r ed
at .
C lai m T D
to p r ov e T D . L et be a p oint in T cor r esp ond in g to a p oint in S , so that using
ste r eog r ap hic p r oje cti on , x
y
(
*
(
*
( )
( )
( )
now ,
=
( ) = ( ) = ( ever y thin g p ositi v e )
w hich is tru e as S
? T D . T hu s T D .
To p r ov e D T
if D , say ( x , y ) , then let ( , , ) be the cor r esp ond in g p oint on the sp her e
g iven by x
x
y
, y
x
y
, x
y
x
y
Free coaching of B.Sc (h) maths & JAM
For more 8130648819
w her e , = x
y
= ( x
y
) ( x
y
) ( ) =
( x
y
) =
( ) tru e b eca use D
p oint is in S and so in T , ? D T .
E xer cise S upp ose that z is the ste r og r ap hic of ( , , ) and
z
is the p r oje cti on of
( , , ) .
a . S how that ( , , )
b . S how that the f uncti on
z
, z , is r ep r ese nt at e d on b y r ota ti on ab out the
d ia mete r w it h en d p oint s ( , ,
* and (
, ,
* .
So l
we use the ste r eog r ap hic p r oje cti on f or mu lae
x
x
y
, y
x
y
, x
y
x
y
w her e z x iy is the ste r eog r ap hic p r oje cti on of ( , , )
g iven that
z
is the ste r eog r ap hic p r oje cti on of ( , , ) , we hav e us ing
z
x iy
x iy
x
y
x
x
y
i ( y
x
y
*
x
i y
, w her e x
x
x
y
, y
y
x
y
sin ce , ( x
)
( y
)
x
y
( x
y
)
x
y
, we oba in using
x
( x
)
( y
)
x
x
y
(
x
x
y
*
(
y
x
y
*
x
x
y
y
( x
)
( y
)
y
x
y
(
x
x
y
*
(
y
x
y
*
y
x
y
( x
)
( y
)
( x
)
( y
)
x
y
( , , ) ( , , ) not e ap p ly this f or z i
ANALYTIC POLYNOMIALS
First note that
x
y
ixy is a DIRECT FUNCTION of x+iy (its (x+iy)
2
& x
y
ixy is not a DIRECT FUNCTION of x+iy
Def
n
A polynomial p(x, y)will be called an na ly ti c p oly nomia l
? if ? ? k , s. t .
(means complete)
p ( x, y) ? 0 ? 1 ( x iy) ? 2(x+iy)
2
…. . ? n(x+iy)
n
Then, we will say that p(x, y) is a polynomial in z &
p(z) ? 0 ? 1 z ? 2z
2
…. . ? nz
n
S how that x
y
ixy is an an aly ti c p oly nomia l .
So l
p ( x , y ) x
y
ixy
? p ( x , y ) ( x iy )
p ( z ) z
? p ( x , y ) is an aly ti c p oly nomia l .
Free coaching of B.Sc (h) maths & JAM
For more 8130648819
S how that x
y
ixy is not an aly ti c p oly .
So l
Let if possible suppose, it is an analytic poly.
i.e. x
y
ixy ? ( x iy )
? x, y …… … …… . . ( *)
x
? x
taking y = 0
i.e. x
x x
x
x
i.e. b ut 2 = 1
i.e. x
y
iy ( x iy )
by (*)
a contradiction
E xample p ( x , y ) x
iv ( x , y ) is not an aly ti c f or any ch oice of the r ea l p oly nomia l
v ( x , y ) .
So l
f or a p oly nomia l in z can hav e a r ea l p art of d egr ee in x only if it is of the
f or m a z
bz c w it he a . In this case , how ever , the r ea l p art mu st conta in a y
,
i . e . y
b ut y
is not in p ( x , y )
? p ( x , y ) is not an aly ti c f or any v ( x , y )
def
n
let u(x, y) and v(x, y) be real valued function and
f(x, y) = u(x, y )+iv (x, y)
then the partial derivative of f (if exists) are
f x(x, y) = u x(x, y)+i v x(x, y)
f x(x, y) = u y(x, y)+i v y(x, y)
Proposition 2.3 A polynomial p(x, y) is analytic iff p y= ip x
Proof
forward part : let p(x, y) be analytic
? p ( x, y) ? 0 ? 1 ( x iy ) ? 2(x+iy)
2
…. . ? n(x+iy)
n
i.e. p x(x, y) = ( x i y ) 3
( x iy )
n ( x iy )
& p y(x, y) = i i ( x iy ) 3i ( x iy )
ni ( x iy )
Follows
P y(x, y)= ip x(x, y) done
Converse part
let p y = ip x [TS : p(x, y) is an analytic poly.]
let p(x, y) = q 0+q 1 +q 2 +......+q n
where, q n : sum of all n
th
degree term
WLOG q n ß 0x
n
ß 1x
n y ß 2x
n y
2
…. ß ny
n
We shall make use of the fact that
P y =ip x
i.e. (q n) y = i(q n) x
so let us 1
st
consider,
q n ( x, y) ß 0x
n
ß 1x
n y ß 2x
n y
2
…… ß n-2x
2
y
n ß n xy
n ß ny
n
i.e. (q n) y ß 1x
n ß 2x
n y 3 ß 3x
n 3
y
2
…. . ( n ) ß n x
2
y
n 3
( n ) ß n xy
n nß ny
n
& (q n) x n ß 0x
n ( n ) ß 1x
n y ( n ) ß 2x
n 3
y
2
…. . ß n xy
n ß n y
n +0
{
using q
i q
we g et ß
i n ß
. i . e . ß
.
n
/ ß
i / ß
i ( n ) ß
4 i . e . ß
i
( n )
.
n
/ ß
i i
.
n
/ ß
5
3 ß
i ( n ) ß
(
i . e . ß
i
( n )
3
ß
i n 3
. i
.
n
/ ß
i
.
n
3
/ ß
) ? in g en eral ß
i
.
n
k
/ ß
Free coaching of B.Sc (h) maths & JAM
For more 8130648819
{
r econsi d er q
( x , y ) ? ß
x
y
? i
.
n
k
/ ß
x
y
ß
? .
n
k
/ x
( iy )
ß
( x iy )
? p ( x , y ) q
q
( x , y ) q
( x , y ) q
( x , y ) ß
ß
( x iy ) ß
( x iy )
ß
( x iy )
i . e . p ( x , y ) is an an aly ti c p oly nomia l .
CAUCHY RIEMANN EQUATION (POLYNOMIAL CASE)
Let p(x, y) be a polynomial
? p ( x, y) u( x, y) i v( x , y )
NOTE: using previous theorem we can say that p(x, y) is an analytic polynomial.
iff p y(x, y)=i p x(x, y)
iff u y(x, y)+i v y(x, y) = i {u x(x, y)+ v x(x, y)}
iff u x = v y and u y v x called CAUCHY RIEMANN EQ.
E xample p g 3 non consta nt an aly ti c p oly nomia l can’t b e real v alued .
Reason : let p(x, y) = u+iv say
if p ossib le p ( x, y) u iv be non consta n t a na l. p oly nomia l s. t. p ( x, y) is real valu ed .
? v ( hence v x = v y = 0)
But, p is analytic
u x = v y and u y v x [C-R EQ
n
]
i.e. u x = u y = 0
i.e. u is constant, contradiction
E xample pg 3 Usi n g the C auch y Rie mann eq uati ons , one can v erif y that
x
y
ixy
is an aly ti c w hile x
y
ixy is not .
So l
x
y
ixy is analytic
let p ( x, y) = x
2
y
2
+2ixy
= u(x, y)+i v(x, y) say
Also, u x = 2x and v y = 2x
u y y an d v y = 2y
i.e. u x = v y and u y v x
i. e. p ( x, y) s at isfies the C R eq
n
i.e. p(x, y) is an analytic poly.
x
y
ixy is not an analytic poly.
let p ( x, y) x
y
i ( xy )
u(x, y)+iv(x, y) say
also, u x = 2x & v y x
u y = 2y & v y y
i.e. u x v y & u y v x
i.e. p ( x, y) d oes’t sa ti sf i es t he C R eq
n
? p(x, y) is not analytic.
Is x
2
+iy
2
an analytic poly.
Let p(x, y) = x
2
+iy
2
u(x, y) +iv(x, y) say
u = x
2
, v = y
2
u x=2x , v y=2y
? u x v y
i. e. C R eq
n
are not analytic
i.e. p(x, y) is not analytic.
I s x y ix
2
+iy
2
an analytic
Let p(x, y) = 2xy+i(y
2
x
2
)
u = 2xy , v = y
2
x
2-
Page 5
Exercise 19: Suppose T . S how that the cor r esp ond in g set S ? is
a . a cir cle if T is a cir cle.
b . a cir cle minus ( , , ) if T is a line .
so l
let T be cir cle in the set of com p lex .
let it is d en ote d by
i . e . T * ( x , y ) ( x
y
) x Cy D + …. . ( i )
not e S is the ste r eog hr ap hic p r oje cti on of T ( on )
? ( x , y ) T ( , , ) S
using ste r eog r ap hic p r oje cti on f or mu lae ( 3 )
x , y , | z |
? f r om eq uation (i)
(
* (
* C (
* D
C D ( )
or ( D ) C D ( ii )
w hich is the eq uati on of a p lane in sp ace w hen a g en eral p oi nt has coo r d in at es ( , , ) .
ote that a p lane a sp her e in te r sect s in a cir cle .
S upp ose , D then ( i ) is a straigh t line in . ( x Cy D )
T hu s , the cor r esp ond in g set in the sp her e in g iven by the in te r sect ion of
w it h a p lane C D D by ( ii )
p assi ng thr oug ht ( , , ) S
w hich is a cir cle minus ( , , ) , sin ce ta k en .
if , D
T is ( x
y
) x Cy
e q
( ii ) b ecomes C
w hich is again a p lan e p assi ng ( , , )
b ut on the other hand ( , , ) S
i . e . this p lane in te r sect s the sp he r e
i . e . this in te r sect ion has to be a cir cle
i . e . S is a cir cle .
E xc erc ise . L et S * ( , , ) = + , w her e ,
w her e * ( , , ) + is the Rie mann sp her e ,
and let T be thecor r esp ond in g set ( x , y ) in . S how that T is the exte r ior of a cir cle
cen te r ed at .
So l
T ake D { ( x , y ) x
y
=
} , exte r ior of cir cle x
y
cen te r ed
at .
C lai m T D
to p r ov e T D . L et be a p oint in T cor r esp ond in g to a p oint in S , so that using
ste r eog r ap hic p r oje cti on , x
y
(
*
(
*
( )
( )
( )
now ,
=
( ) = ( ) = ( ever y thin g p ositi v e )
w hich is tru e as S
? T D . T hu s T D .
To p r ov e D T
if D , say ( x , y ) , then let ( , , ) be the cor r esp ond in g p oint on the sp her e
g iven by x
x
y
, y
x
y
, x
y
x
y
Free coaching of B.Sc (h) maths & JAM
For more 8130648819
w her e , = x
y
= ( x
y
) ( x
y
) ( ) =
( x
y
) =
( ) tru e b eca use D
p oint is in S and so in T , ? D T .
E xer cise S upp ose that z is the ste r og r ap hic of ( , , ) and
z
is the p r oje cti on of
( , , ) .
a . S how that ( , , )
b . S how that the f uncti on
z
, z , is r ep r ese nt at e d on b y r ota ti on ab out the
d ia mete r w it h en d p oint s ( , ,
* and (
, ,
* .
So l
we use the ste r eog r ap hic p r oje cti on f or mu lae
x
x
y
, y
x
y
, x
y
x
y
w her e z x iy is the ste r eog r ap hic p r oje cti on of ( , , )
g iven that
z
is the ste r eog r ap hic p r oje cti on of ( , , ) , we hav e us ing
z
x iy
x iy
x
y
x
x
y
i ( y
x
y
*
x
i y
, w her e x
x
x
y
, y
y
x
y
sin ce , ( x
)
( y
)
x
y
( x
y
)
x
y
, we oba in using
x
( x
)
( y
)
x
x
y
(
x
x
y
*
(
y
x
y
*
x
x
y
y
( x
)
( y
)
y
x
y
(
x
x
y
*
(
y
x
y
*
y
x
y
( x
)
( y
)
( x
)
( y
)
x
y
( , , ) ( , , ) not e ap p ly this f or z i
ANALYTIC POLYNOMIALS
First note that
x
y
ixy is a DIRECT FUNCTION of x+iy (its (x+iy)
2
& x
y
ixy is not a DIRECT FUNCTION of x+iy
Def
n
A polynomial p(x, y)will be called an na ly ti c p oly nomia l
? if ? ? k , s. t .
(means complete)
p ( x, y) ? 0 ? 1 ( x iy) ? 2(x+iy)
2
…. . ? n(x+iy)
n
Then, we will say that p(x, y) is a polynomial in z &
p(z) ? 0 ? 1 z ? 2z
2
…. . ? nz
n
S how that x
y
ixy is an an aly ti c p oly nomia l .
So l
p ( x , y ) x
y
ixy
? p ( x , y ) ( x iy )
p ( z ) z
? p ( x , y ) is an aly ti c p oly nomia l .
Free coaching of B.Sc (h) maths & JAM
For more 8130648819
S how that x
y
ixy is not an aly ti c p oly .
So l
Let if possible suppose, it is an analytic poly.
i.e. x
y
ixy ? ( x iy )
? x, y …… … …… . . ( *)
x
? x
taking y = 0
i.e. x
x x
x
x
i.e. b ut 2 = 1
i.e. x
y
iy ( x iy )
by (*)
a contradiction
E xample p ( x , y ) x
iv ( x , y ) is not an aly ti c f or any ch oice of the r ea l p oly nomia l
v ( x , y ) .
So l
f or a p oly nomia l in z can hav e a r ea l p art of d egr ee in x only if it is of the
f or m a z
bz c w it he a . In this case , how ever , the r ea l p art mu st conta in a y
,
i . e . y
b ut y
is not in p ( x , y )
? p ( x , y ) is not an aly ti c f or any v ( x , y )
def
n
let u(x, y) and v(x, y) be real valued function and
f(x, y) = u(x, y )+iv (x, y)
then the partial derivative of f (if exists) are
f x(x, y) = u x(x, y)+i v x(x, y)
f x(x, y) = u y(x, y)+i v y(x, y)
Proposition 2.3 A polynomial p(x, y) is analytic iff p y= ip x
Proof
forward part : let p(x, y) be analytic
? p ( x, y) ? 0 ? 1 ( x iy ) ? 2(x+iy)
2
…. . ? n(x+iy)
n
i.e. p x(x, y) = ( x i y ) 3
( x iy )
n ( x iy )
& p y(x, y) = i i ( x iy ) 3i ( x iy )
ni ( x iy )
Follows
P y(x, y)= ip x(x, y) done
Converse part
let p y = ip x [TS : p(x, y) is an analytic poly.]
let p(x, y) = q 0+q 1 +q 2 +......+q n
where, q n : sum of all n
th
degree term
WLOG q n ß 0x
n
ß 1x
n y ß 2x
n y
2
…. ß ny
n
We shall make use of the fact that
P y =ip x
i.e. (q n) y = i(q n) x
so let us 1
st
consider,
q n ( x, y) ß 0x
n
ß 1x
n y ß 2x
n y
2
…… ß n-2x
2
y
n ß n xy
n ß ny
n
i.e. (q n) y ß 1x
n ß 2x
n y 3 ß 3x
n 3
y
2
…. . ( n ) ß n x
2
y
n 3
( n ) ß n xy
n nß ny
n
& (q n) x n ß 0x
n ( n ) ß 1x
n y ( n ) ß 2x
n 3
y
2
…. . ß n xy
n ß n y
n +0
{
using q
i q
we g et ß
i n ß
. i . e . ß
.
n
/ ß
i / ß
i ( n ) ß
4 i . e . ß
i
( n )
.
n
/ ß
i i
.
n
/ ß
5
3 ß
i ( n ) ß
(
i . e . ß
i
( n )
3
ß
i n 3
. i
.
n
/ ß
i
.
n
3
/ ß
) ? in g en eral ß
i
.
n
k
/ ß
Free coaching of B.Sc (h) maths & JAM
For more 8130648819
{
r econsi d er q
( x , y ) ? ß
x
y
? i
.
n
k
/ ß
x
y
ß
? .
n
k
/ x
( iy )
ß
( x iy )
? p ( x , y ) q
q
( x , y ) q
( x , y ) q
( x , y ) ß
ß
( x iy ) ß
( x iy )
ß
( x iy )
i . e . p ( x , y ) is an an aly ti c p oly nomia l .
CAUCHY RIEMANN EQUATION (POLYNOMIAL CASE)
Let p(x, y) be a polynomial
? p ( x, y) u( x, y) i v( x , y )
NOTE: using previous theorem we can say that p(x, y) is an analytic polynomial.
iff p y(x, y)=i p x(x, y)
iff u y(x, y)+i v y(x, y) = i {u x(x, y)+ v x(x, y)}
iff u x = v y and u y v x called CAUCHY RIEMANN EQ.
E xample p g 3 non consta nt an aly ti c p oly nomia l can’t b e real v alued .
Reason : let p(x, y) = u+iv say
if p ossib le p ( x, y) u iv be non consta n t a na l. p oly nomia l s. t. p ( x, y) is real valu ed .
? v ( hence v x = v y = 0)
But, p is analytic
u x = v y and u y v x [C-R EQ
n
]
i.e. u x = u y = 0
i.e. u is constant, contradiction
E xample pg 3 Usi n g the C auch y Rie mann eq uati ons , one can v erif y that
x
y
ixy
is an aly ti c w hile x
y
ixy is not .
So l
x
y
ixy is analytic
let p ( x, y) = x
2
y
2
+2ixy
= u(x, y)+i v(x, y) say
Also, u x = 2x and v y = 2x
u y y an d v y = 2y
i.e. u x = v y and u y v x
i. e. p ( x, y) s at isfies the C R eq
n
i.e. p(x, y) is an analytic poly.
x
y
ixy is not an analytic poly.
let p ( x, y) x
y
i ( xy )
u(x, y)+iv(x, y) say
also, u x = 2x & v y x
u y = 2y & v y y
i.e. u x v y & u y v x
i.e. p ( x, y) d oes’t sa ti sf i es t he C R eq
n
? p(x, y) is not analytic.
Is x
2
+iy
2
an analytic poly.
Let p(x, y) = x
2
+iy
2
u(x, y) +iv(x, y) say
u = x
2
, v = y
2
u x=2x , v y=2y
? u x v y
i. e. C R eq
n
are not analytic
i.e. p(x, y) is not analytic.
I s x y ix
2
+iy
2
an analytic
Let p(x, y) = 2xy+i(y
2
x
2
)
u = 2xy , v = y
2
x
2-
Free coaching of B.Sc (h) maths & JAM
For more 8130648819
u x=2y , v y=2y
u y=2x , v x x
i.e. u x=v y & u y v x
i. e. C R eq
n
satisfied
p ( x, y) is a na l .
Exercise 2: By comparing coefficients or by using the Cauchy Riemann equation, determine which are of the
following poly. is/ are analytic ?
(a) p(x+iy) = x
3x y
x i ( 3 x
y y
y )
=u(x, y)+iv(x, y) say
? u x = 3 x
3 y
v y = 3 x
y
u y xy v x = 6xy
i.e. u y v x and u x = v y
i. e. p ( x, y) s at ist i es C R eq
n
i.e. p(x+iy) is analytic (poly.)
(b) p(x+iy) = x
2
+iy
2
=u(x, y)+iv(x, y) say
? u x = 2x v y = 2y
i.e. u x v y
i. e. p ( x i y) d oes not s at ist ie s the C R e q
n
i.e. p(x+iy) is not an analytic poly.
(c) p(x+iy) = 2xy+i(y
2
x
2
)
= u(x, y)+iv(x, y) say
? u x = 2y v y =2y
u y = 2x v x x
i.e. u x=v y and u y v x etc.
E xer cise 3 S how that no non consta nt a na ly ti c p oly nomia l can t ake imagina r y v alue only .
Sol
n
let p(x, y) = u(x, y) +iv( x, y) b e a non con sta nt an al . Poly .
And if poss. sup. That p(x, y) takes imag. values only.
i.e. u(x, y) = 0
i.e. u x = u y = 0
i.e. v x = v y = 0 , using C R eq
n
]
i.e. v(x, y) is constant.
i.e. p(x, y) = 0+i. constant
= a constant poly.
Contradiction.
E xer cise . F in d the d eriv at ive p
( z ) of the an aly ti c p oly nomia ls in ( ) . S how that in
ea ch case p
( z ) p
. E xplai n .
( a ) p(x+iy) = x
3x y
x i ( 3 x
y y
y )
=u(x, y)+iv(x, y) say
? u x = 3 x
3 y
v y = 3 x
y
u y xy v x = 6xy
i.e p
u
i v
p
3 x
3 y
i ( xy )
p
u
i v
p
xy i ( 3 x
3 y
)
i p
i ( 3 x
3 y
) i
( xy )
xy i ( 3 x
3 y
) p
i.e. p(x+iy) is analytic (poly.)
here , is not need to check analycity b/c it is given
now , p ( x iy ) ( x
3x y
x ) i ( 3 x
y
y )
x
3 c
x
( iy ) 3 c
x ( iy )
( iy )
( x iy )
( x iy )
( x iy )
p ( z ) z
z w her e z x iy
? p
( z ) 3 z
Read More
|
27 docs|150 tests
|
FAQs on Complex Analysis (Part - 2) Solved Example - Topic-wise Tests & Solved Examples for Mathematics
| 1. What is complex analysis? |  |
Ans. Complex analysis is a branch of mathematics that deals with the study of complex numbers and their functions. It involves the analysis of functions that are defined on complex numbers and the properties of these functions, such as differentiation, integration, and series representations.
| 2. How is complex analysis used in real life? | |
Ans. Complex analysis has various applications in real-life problems, particularly in physics and engineering. It is used in the study of fluid dynamics, electromagnetism, quantum mechanics, and signal processing. It also finds applications in finance, where complex analysis techniques are used to model and analyze stock market behavior.
| 3. What are the key concepts in complex analysis? | |
Ans. Some key concepts in complex analysis include analytic functions, Cauchy-Riemann equations, complex integration, residues, and the theory of conformal mappings. These concepts are fundamental in understanding the behavior of complex functions and their geometric properties.
| 4. What are the Cauchy-Riemann equations? | |
Ans. The Cauchy-Riemann equations are a set of partial differential equations that determine whether a complex-valued function is analytic. They express the conditions for a function to be differentiable in terms of its real and imaginary parts. If a function satisfies the Cauchy-Riemann equations, it is said to be holomorphic or analytic.
| 5. How is complex integration different from real integration? | |
Ans. Complex integration involves integrating complex-valued functions along a curve in the complex plane. Unlike real integration, where the path of integration does not affect the value of the integral, complex integration can be path-dependent. This means that the value of the integral can vary depending on the chosen path. Complex integration also introduces the concept of residues, which play a crucial role in evaluating certain types of integrals.
Related Exams
About this Document
|
4.90/5 Rating |
|
Dec 22, 2024 Last updated |
Document Description: Complex Analysis (Part - 2) Solved Example for Mathematics 2024 is part of Topic-wise Tests & Solved Examples for Mathematics preparation.
The notes and questions for Complex Analysis (Part - 2) Solved Example have been prepared according to the Mathematics exam syllabus. Information about Complex Analysis (Part - 2) Solved Example covers topics
like and Complex Analysis (Part - 2) Solved Example Example, for Mathematics 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for Complex Analysis (Part - 2) Solved Example.
Introduction of Complex Analysis (Part - 2) Solved Example in English is available as part of our Topic-wise Tests & Solved Examples for Mathematics
for Mathematics & Complex Analysis (Part - 2) Solved Example 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: Complex Analysis (Part - 2) Solved Example | Topic-wise Tests & Solved Examples for Mathematics
Description
Full syllabus notes, lecture & questions for Complex Analysis (Part - 2) Solved Example | 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 Complex Analysis (Part - 2) Solved Example
In this doc you can find the meaning of Complex Analysis (Part - 2) Solved Example defined & explained in the simplest way possible. Besides explaining types of
Complex Analysis (Part - 2) Solved Example theory, EduRev gives you an ample number of questions to practice Complex Analysis (Part - 2) Solved Example 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
,study material
,shortcuts and tricks
,MCQs
,Viva Questions
,Exam
,Complex Analysis (Part - 2) Solved Example | Topic-wise Tests & Solved Examples for Mathematics
,Important questions
,Free
,Complex Analysis (Part - 2) Solved Example | Topic-wise Tests & Solved Examples for Mathematics
,video lectures
,Extra Questions
,Summary
,Semester Notes
,mock tests for examination
,past year papers
,Previous Year Questions with Solutions
,Complex Analysis (Part - 2) Solved Example | Topic-wise Tests & Solved Examples for Mathematics
,Sample Paper
,ppt
,practice quizzes
;
Additional Information about Complex Analysis (Part - 2) Solved Example for Mathematics Preparation
Complex Analysis (Part - 2) Solved Example Free PDF Download
The Complex Analysis (Part - 2) Solved Example 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 Complex Analysis (Part - 2) Solved Example now and kickstart your journey towards success in the Mathematics exam.
Importance of Complex Analysis (Part - 2) Solved Example
The importance of Complex Analysis (Part - 2) Solved Example 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.
Complex Analysis (Part - 2) Solved Example Notes
Complex Analysis (Part - 2) Solved Example Notes offer in-depth insights into the specific topic to help you master it with ease.
This comprehensive document covers all aspects related to Complex Analysis (Part - 2) Solved Example.
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, Complex Analysis (Part - 2) Solved Example Notes on EduRev are your ultimate resource for success.
Complex Analysis (Part - 2) Solved Example Mathematics Questions
The "Complex Analysis (Part - 2) Solved Example 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 Complex Analysis (Part - 2) Solved Example on the App
Students of Mathematics can study Complex Analysis (Part - 2) Solved Example alongwith tests & analysis from the EduRev app,
which will help them while preparing for their exam. Apart from the Complex Analysis (Part - 2) Solved Example,
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 Complex Analysis (Part - 2) Solved Example 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