Page 76 - DMTH402_COMPLEX_ANALYSIS_AND_DIFFERENTIAL_GEOMETRY

Unit 7: Transformations and Conformal Mappings

This is the required transformation, for if z = e , a = be , then Notes
i
i
i
| w | = e  be i = 1.
be i( )  1
If z = re , where r < 1, then
i
|z  |  |  z  1| = r 2rb cos() + b  {b r  2br cos() + 1}
2
2
2
2
2 2
= (r  1)(1  b ) < 0
2
2
and so
|z  a| < |z  1|  |z   | 2 < 1
2
2
| z 1| 2


i.e. | w | < 1
Hence, the result.
Example: Show that the general transformation of the circle | z |   into the circle
| w | is

 z   
2  ,
i
w =  e    z    | a | < 
Solution. Let the transformation be
az b a z b/a
 

w =  (1)

cz d c z d/c 

 
The points w = 0 and w = , inverse points of | w | =  correspond to inverse point z = b/a, z
= d/c respectively of | z | = r, so we may write
b d  2
 = ,   , | a | < 
a c 
Thus, from (1), we get

 

w = a   z   2    a   z   2   (2)
c    c   z   
 z   

Equation (2) satisfied the condition | z |   and | w | . Hence, for | z | = r, we must have
| w | =  so that (2) becomes

 = | w | = a z   , z z = r 2
c  z zz

= a 1 z    a 1 z  
c z z   c z z  

LOVELY PROFESSIONAL UNIVERSITY 69