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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
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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
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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
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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
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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
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{
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
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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 .
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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
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{
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-
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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
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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.
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Additional Information about Complex Analysis (Part - 2) Solved Example for Mathematics Preparation
Complex Analysis (Part - 2) Solved Example Free PDF Download
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Complex Analysis (Part - 2) Solved Example Notes
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