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? 1 1 ß 1
? 2 1 ß 2 w her e a r g f ( z 0)
S o, ? 2 ? 1 ß 2 – ß 1
Prob : f(z) = e
z
f ( z ) e
z
? z
? f is conformal everywhere (by thm 13.4)
Example /158 f(z) = z
2
where f is conformal, not conformal ?
Sol
n
f(z) = z
2
f ( z ) z ? z
f is confor mal thr ou g hou t z
But what about at z = 0 not conformal at z = 0
Sol
n
image at real axis under f is +ve real axis
Image at +ve imaginary axis under f is –ve real axis .
Def
n
13.3 ( a) f is s. t. b l ocally at z 0 if ? ?? > 0 s.t. for distinct z 1, z 2 D( z 0; ?? ),
f(z 1 ) f ( z 2)
( b ) f is s . t. b . loc ally thr ou g hou t a r egion D if f is locally at ever y z D.
( c) f is f un
c
in a region D, if for every distinct z 1, z 2 D, f ( z 1 ) f ( z 2)
Example/158 f(z) = z
2
is n ot locally at r e g ion D( , ?? )
Let f(z) = z
2
Consider any disk D(0, ?? ) center at origin.
? disti nct p oin ts z z in D( , ?? ) s.t.
f(z) = z
2
f( z ) z
2
fig sand 25
? f ( z ) f ( z ) b ut z z
Hence, f(z) = z
2
is n ot l ocally at or i g in .
Que. 2/ex 1/159 find the image under w = e
z
of the lines x = constant & y = constant.
Sol
n
let w = e
z
I t i s an aly ti c f ( z ) e
z
? z ( her e f( z ) w )
? w f ( z ) e
z
is confor mal locally ? z
f is con f or mal f or z .
T he image of or tho g ona l line s x consta nt y consta nt ly in g in z p lane are also or thog ona l in w p lane .
Consider, the vertical line x = constant in z p lane
? I t i mag e i n w p lane i s w = f(z)
= e
x+iy
= e
e
iy
( x )
|w| = e
|e
iy
|
= e
( | e
iy
| = 1)
Wh ich r ep r ese nt a ci r cle in w p lane w it h cent er at orig in .
C onsider the hor iz ont al line y ß in z p la ne
? I t i mag e und er w p lane is w e
z
? w e
x+iy
= e
x i ß
= e
x
e
iß
= e
x
( cos ß i sin ß )
Wh ich r ep r ese nt a ray thr oug h the o r igin with arg um en t ß .
?(C 1, C 2) ?( G 1, G 2)
? f is not conformal at z
= 0.
C
X
axis
C
Y
axis }
f ( C
) X
f ( C ) X
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? 1 1 ß 1
? 2 1 ß 2 w her e a r g f ( z 0)
S o, ? 2 ? 1 ß 2 – ß 1
Prob : f(z) = e
z
f ( z ) e
z
? z
? f is conformal everywhere (by thm 13.4)
Example /158 f(z) = z
2
where f is conformal, not conformal ?
Sol
n
f(z) = z
2
f ( z ) z ? z
f is confor mal thr ou g hou t z
But what about at z = 0 not conformal at z = 0
Sol
n
image at real axis under f is +ve real axis
Image at +ve imaginary axis under f is –ve real axis .
Def
n
13.3 ( a) f is s. t. b l ocally at z 0 if ? ?? > 0 s.t. for distinct z 1, z 2 D( z 0; ?? ),
f(z 1 ) f ( z 2)
( b ) f is s . t. b . loc ally thr ou g hou t a r egion D if f is locally at ever y z D.
( c) f is f un
c
in a region D, if for every distinct z 1, z 2 D, f ( z 1 ) f ( z 2)
Example/158 f(z) = z
2
is n ot locally at r e g ion D( , ?? )
Let f(z) = z
2
Consider any disk D(0, ?? ) center at origin.
? disti nct p oin ts z z in D( , ?? ) s.t.
f(z) = z
2
f( z ) z
2
fig sand 25
? f ( z ) f ( z ) b ut z z
Hence, f(z) = z
2
is n ot l ocally at or i g in .
Que. 2/ex 1/159 find the image under w = e
z
of the lines x = constant & y = constant.
Sol
n
let w = e
z
I t i s an aly ti c f ( z ) e
z
? z ( her e f( z ) w )
? w f ( z ) e
z
is confor mal locally ? z
f is con f or mal f or z .
T he image of or tho g ona l line s x consta nt y consta nt ly in g in z p lane are also or thog ona l in w p lane .
Consider, the vertical line x = constant in z p lane
? I t i mag e i n w p lane i s w = f(z)
= e
x+iy
= e
e
iy
( x )
|w| = e
|e
iy
|
= e
( | e
iy
| = 1)
Wh ich r ep r ese nt a ci r cle in w p lane w it h cent er at orig in .
C onsider the hor iz ont al line y ß in z p la ne
? I t i mag e und er w p lane is w e
z
? w e
x+iy
= e
x i ß
= e
x
e
iß
= e
x
( cos ß i sin ß )
Wh ich r ep r ese nt a ray thr oug h the o r igin with arg um en t ß .
?(C 1, C 2) ?( G 1, G 2)
? f is not conformal at z
= 0.
C
X
axis
C
Y
axis }
f ( C
) X
f ( C ) X
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? 1 1 ß 1
? 2 1 ß 2 w her e a r g f ( z 0)
S o, ? 2 ? 1 ß 2 – ß 1
Prob : f(z) = e
z
f ( z ) e
z
? z
? f is conformal everywhere (by thm 13.4)
Example /158 f(z) = z
2
where f is conformal, not conformal ?
Sol
n
f(z) = z
2
f ( z ) z ? z
f is confor mal thr ou g hou t z
But what about at z = 0 not conformal at z = 0
Sol
n
image at real axis under f is +ve real axis
Image at +ve imaginary axis under f is –ve real axis .
Def
n
13.3 ( a) f is s. t. b l ocally at z 0 if ? ?? > 0 s.t. for distinct z 1, z 2 D( z 0; ?? ),
f(z 1 ) f ( z 2)
( b ) f is s . t. b . loc ally thr ou g hou t a r egion D if f is locally at ever y z D.
( c) f is f un
c
in a region D, if for every distinct z 1, z 2 D, f ( z 1 ) f ( z 2)
Example/158 f(z) = z
2
is n ot locally at r e g ion D( , ?? )
Let f(z) = z
2
Consider any disk D(0, ?? ) center at origin.
? disti nct p oin ts z z in D( , ?? ) s.t.
f(z) = z
2
f( z ) z
2
fig sand 25
? f ( z ) f ( z ) b ut z z
Hence, f(z) = z
2
is n ot l ocally at or i g in .
Que. 2/ex 1/159 find the image under w = e
z
of the lines x = constant & y = constant.
Sol
n
let w = e
z
I t i s an aly ti c f ( z ) e
z
? z ( her e f( z ) w )
? w f ( z ) e
z
is confor mal locally ? z
f is con f or mal f or z .
T he image of or tho g ona l line s x consta nt y consta nt ly in g in z p lane are also or thog ona l in w p lane .
Consider, the vertical line x = constant in z p lane
? I t i mag e i n w p lane i s w = f(z)
= e
x+iy
= e
e
iy
( x )
|w| = e
|e
iy
|
= e
( | e
iy
| = 1)
Wh ich r ep r ese nt a ci r cle in w p lane w it h cent er at orig in .
C onsider the hor iz ont al line y ß in z p la ne
? I t i mag e und er w p lane is w e
z
? w e
x+iy
= e
x i ß
= e
x
e
iß
= e
x
( cos ß i sin ß )
Wh ich r ep r ese nt a ray thr oug h the o r igin with arg um en t ß .
?(C 1, C 2) ?( G 1, G 2)
? f is not conformal at z
= 0.
C
X
axis
C
Y
axis }
f ( C
) X
f ( C ) X
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Special mapping
Elementary Transformation T he line ar m ap p i n g w f ( z ) b y an aly ti c m ap p in g in the
entire plane onto itself.
Magnification A map of the form w=kz , k>0 is called a magnification if it send each point onto another point
along the same ray from the origin multiplying its magnitude by a factor of k .
(ii) Rotation : The mapping w = e
i z is counterclockwise r ota ti on t hr o ug h an a ngl e .
(iii) Translation The map. w= z +b translate each point by a complex no. b. Therefore it is called translation.
Therefore, A linear mapping is composition of w = w 3 °w 2 °w 1 i.e. magnification , rotation , translation where
w 1 = kz
w 2 = e
i
z
w 3 = z+b
Bi-linear transformation (Mobious transformation) A mapping of the form w = f(z) = =
, ad bc is
called Bi-linear transformation.
Note ad b c f ( z ) f is not a const a nt map .
Obs: the Bi-linear transfor mati on map s 2
3 onto
2
3
Since, the eq
n
= w has the explicit sol
n
z =
, ? w a/c
also, f ( ) & f . /
Properties of bi-linear transformation
(i) A bi- line ar m ap p in g c an ’t b e i d en ti cally const an t.
Let f(z) =
, ad b c
? f ( z ) ( ) ( ) ( )
=
( )
=
( )
f ( z ) ? z ( of ad b c )
f ( z ) is n ot a const an t .
(ii) A Bi- line ar m ap p in g is gl oba lly : let f(z 1)= f(z 2) [TS: z 1 =z 2
( az 1+b) (cz 2+d) = (az 2+b)(cz 1+d)
az 1cz 2+adz 1+bcz 2+bd = acz 1z 2+adz 2+bcz 1+bd
ad z 1+bcz 2 = adz 2+bcz 1
ad ( z 1 z 2)+bc(z 2 z 1) = 0
ad ( z 1 z 2 ) b c ( z 1 z 2) = 0
( ad b c ) ( z 1 z 2) = 0
z 1 z 2 ( of ad b c
z 1= z 2
? f ( z ) is gl oba lly
(iii) Inverse of Bi-linear mapping is also a Bi-linear mapping.
Let w = f(z) =
, ad b c
Be a bi-linear mapping
w ( cz d ) az b
w cz w d az b
z ( w c a ) b w d
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? 1 1 ß 1
? 2 1 ß 2 w her e a r g f ( z 0)
S o, ? 2 ? 1 ß 2 – ß 1
Prob : f(z) = e
z
f ( z ) e
z
? z
? f is conformal everywhere (by thm 13.4)
Example /158 f(z) = z
2
where f is conformal, not conformal ?
Sol
n
f(z) = z
2
f ( z ) z ? z
f is confor mal thr ou g hou t z
But what about at z = 0 not conformal at z = 0
Sol
n
image at real axis under f is +ve real axis
Image at +ve imaginary axis under f is –ve real axis .
Def
n
13.3 ( a) f is s. t. b l ocally at z 0 if ? ?? > 0 s.t. for distinct z 1, z 2 D( z 0; ?? ),
f(z 1 ) f ( z 2)
( b ) f is s . t. b . loc ally thr ou g hou t a r egion D if f is locally at ever y z D.
( c) f is f un
c
in a region D, if for every distinct z 1, z 2 D, f ( z 1 ) f ( z 2)
Example/158 f(z) = z
2
is n ot locally at r e g ion D( , ?? )
Let f(z) = z
2
Consider any disk D(0, ?? ) center at origin.
? disti nct p oin ts z z in D( , ?? ) s.t.
f(z) = z
2
f( z ) z
2
fig sand 25
? f ( z ) f ( z ) b ut z z
Hence, f(z) = z
2
is n ot l ocally at or i g in .
Que. 2/ex 1/159 find the image under w = e
z
of the lines x = constant & y = constant.
Sol
n
let w = e
z
I t i s an aly ti c f ( z ) e
z
? z ( her e f( z ) w )
? w f ( z ) e
z
is confor mal locally ? z
f is con f or mal f or z .
T he image of or tho g ona l line s x consta nt y consta nt ly in g in z p lane are also or thog ona l in w p lane .
Consider, the vertical line x = constant in z p lane
? I t i mag e i n w p lane i s w = f(z)
= e
x+iy
= e
e
iy
( x )
|w| = e
|e
iy
|
= e
( | e
iy
| = 1)
Wh ich r ep r ese nt a ci r cle in w p lane w it h cent er at orig in .
C onsider the hor iz ont al line y ß in z p la ne
? I t i mag e und er w p lane is w e
z
? w e
x+iy
= e
x i ß
= e
x
e
iß
= e
x
( cos ß i sin ß )
Wh ich r ep r ese nt a ray thr oug h the o r igin with arg um en t ß .
?(C 1, C 2) ?( G 1, G 2)
? f is not conformal at z
= 0.
C
X
axis
C
Y
axis }
f ( C
) X
f ( C ) X
Free coaching of B.Sc (h) maths & JAM
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Free coaching of B.Sc (h) maths & JAM
For more 8130648819
Special mapping
Elementary Transformation T he line ar m ap p i n g w f ( z ) b y an aly ti c m ap p in g in the
entire plane onto itself.
Magnification A map of the form w=kz , k>0 is called a magnification if it send each point onto another point
along the same ray from the origin multiplying its magnitude by a factor of k .
(ii) Rotation : The mapping w = e
i z is counterclockwise r ota ti on t hr o ug h an a ngl e .
(iii) Translation The map. w= z +b translate each point by a complex no. b. Therefore it is called translation.
Therefore, A linear mapping is composition of w = w 3 °w 2 °w 1 i.e. magnification , rotation , translation where
w 1 = kz
w 2 = e
i
z
w 3 = z+b
Bi-linear transformation (Mobious transformation) A mapping of the form w = f(z) = =
, ad bc is
called Bi-linear transformation.
Note ad b c f ( z ) f is not a const a nt map .
Obs: the Bi-linear transfor mati on map s 2
3 onto
2
3
Since, the eq
n
= w has the explicit sol
n
z =
, ? w a/c
also, f ( ) & f . /
Properties of bi-linear transformation
(i) A bi- line ar m ap p in g c an ’t b e i d en ti cally const an t.
Let f(z) =
, ad b c
? f ( z ) ( ) ( ) ( )
=
( )
=
( )
f ( z ) ? z ( of ad b c )
f ( z ) is n ot a const an t .
(ii) A Bi- line ar m ap p in g is gl oba lly : let f(z 1)= f(z 2) [TS: z 1 =z 2
( az 1+b) (cz 2+d) = (az 2+b)(cz 1+d)
az 1cz 2+adz 1+bcz 2+bd = acz 1z 2+adz 2+bcz 1+bd
ad z 1+bcz 2 = adz 2+bcz 1
ad ( z 1 z 2)+bc(z 2 z 1) = 0
ad ( z 1 z 2 ) b c ( z 1 z 2) = 0
( ad b c ) ( z 1 z 2) = 0
z 1 z 2 ( of ad b c
z 1= z 2
? f ( z ) is gl oba lly
(iii) Inverse of Bi-linear mapping is also a Bi-linear mapping.
Let w = f(z) =
, ad b c
Be a bi-linear mapping
w ( cz d ) az b
w cz w d az b
z ( w c a ) b w d
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z w her e, ad b c .
f
1
(w)=
, ad b c
( d a ( b ) ( c ) ad b c )
in v erse is a lso a i -linear mapping.
(iv) f map the wh ole com p lex pla ne minus the p oint ( d /c) to t he w hol e comple x plan e min us the p t ( a/c ) or
The Bi-linear map can be treated from C - to C
Q: 6/176 Prove that the Bi-linear mapping form a group under composition.
Sol
n
let R denote the set of all Bi-linear transformation (or Bi-linear mapping)
(i)Closure let f 1, f 2 R (TS: f 2 °f 1 R)
? f
az b
cz d
, ad bc f z Cz D
, D C
C onsider , f
° f
( z ) f
( f
( z ) ) f
4
az b
cz d
5 (
az b
cz d
* C (
az b
cz d
* D
( az b ) ( cz d )
C ( az b ) D ( cz d )
( a c ) z ( b d )
( aC Dc ) z ( Cb Dd )
( ) z ( )
( 3 ) z ( )
,TS ( ) ( ) ( ) ( 3 ) f or line ar map . -
Consider,
( a c) ( C b Dd ) ( b d ) ( aC Dc)
= C ab ? +ADad+BCbc+ D cd
? C ab ? Dbc C d a D cd
?
Dad C b c D b c Cd a
= (a d b c) ( D C )
? f 2 °f 1 R
? clo sur e e xists.
Closure property, shows that the composition of two Bi-linear maps is again Bi-linear.
Associativity : we can easily verify that, If f 1, f 2, f 3 R then
f 1 °(f 2 °f 3)(z)= (f 1 °f 2 ) °f 3(z)
existence of identity: I(z) = z =
. . w her e . .
? I ( z ) is i -linear mapping.
or eov er, I °f f ° I f ? f R
Inverse: by property (3)
? ( R, ° ) fr om a gr p .
Remark : Every linear map is Bi-linear map.
Proof : let f ( z ) az b w her e a b e a l inear map.
Then f(z) =
. p r ov id ed a. b . a
Hence, every linear map is Bi-linear map
L emma 3 . let S be a ci r cle or a line f ( z ) z
then f ( S ) is a cir cle or a line
i . e . p / v that f ( z ) z
map s cir cle line s in ont o other cir cl e and line s .
Proof : (a proof involving the Riemann sphere is outline in exercise 21 and 22 of chapter 1. The following
proof is more direct
let f(z) =
Case 1 let S = C(a, r) be a circle with center a & radius r .
Therefore , the eq
n
of circle S is
| z a| r
| z a|
2
= r
2
( z a) ( z a )
¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯
= r
2
( z a) ( z a ) r
2
z z z a az a a r
2
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