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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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For more 8130648819
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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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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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? 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
For more 8130648819
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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FAQs on Introduction To Conformal Mapping - Topic-wise Tests & Solved Examples for Mathematics
| 1. What is conformal mapping in mathematics? |  |
Ans. Conformal mapping is a mathematical technique used to map one geometric shape onto another in a way that preserves angles locally. It is a type of transformation that maintains the shape and angles of objects, but can change their size and position.
| 2. Are conformal maps only applicable to two-dimensional shapes? | |
Ans. Yes, conformal maps are primarily used for two-dimensional shapes. They are particularly useful in complex analysis and the study of complex functions, where the preservation of angles is important.
| 3. How are conformal mappings used in real-life applications? | |
Ans. Conformal mappings have various applications in physics, engineering, and computer science. For example, they are used in fluid dynamics to analyze flow fields, in cartography to accurately represent the Earth's surface on a flat map, and in computer graphics to create realistic deformations of objects.
| 4. Can conformal mappings distort distances between points? | |
Ans. Yes, conformal mappings can distort distances between points. While angles are preserved, the scale or size of objects can be altered in the mapping process. Therefore, distances between points on the original shape may not be preserved in the transformed shape.
| 5. What are some common examples of conformal mappings? | |
Ans. Some common examples of conformal mappings include the stereographic projection, which maps the surface of a sphere onto a plane, and the exponential map, which maps a strip in the complex plane onto a half-plane. These mappings have important applications in various fields of mathematics and physics.
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