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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
  
Free coaching of B.Sc (h) maths & JAM 
For more 8130648819 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
  
    
  
 
 
  
      
      
       
              
              
 
   
   
   
 
 
         
 
 
 
 
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For more 8130648819 
 
 ? 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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 ? 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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