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