Page 75 - DMTH402_COMPLEX_ANALYSIS_AND_DIFFERENTIAL_GEOMETRY

Complex Analysis and Differential Geometry

Notes a
i
 = e i  a = ce , where  is real.
c
Thus, we get

i  z   
w = e   (3)
 z   
Since z =  gives w = 0,  must be a point of the upper half plane i.e. Im > 0. With this condition,

(3) gives the required transformation. In (3), if z is real, obviously | w | = 1 and if Im z > 0, then
z is nearer to  than to  and so | w | < 1. Hence, the general linear transformation of the half
plane Im z  0 on the circle | w |  1 is

 z   
i
w = e   z    , Im  > 0.


Example: Find all bilinear transformations of the unit | z |  1 into the unit circle | w |
 1.

OR
Find the general homographic transformations which leaves the unit circle invariant.
Solution. Let the required transformation be

az b a (z b/a)


w =  (1)
cz d c (z d/c)


Here, w = 0 and w = , correspond to inverse points
b d
z =  , z =  , so we may write
a c
b d 1
 = ,  = such that | a | < 1.
a c 

So, w = a   z      a   z     (2)
z 1 
 
c z 1/  c   

The point z = 1 on the boundary of the unit circle in z-plane must correspond to a point on the
boundary of the unit circle in w-plane so that

1 = | w | = a 1    a
c   1 c
or a  = c e , where  is real.
i
Hence (2) becomes,

il
w = e   z     , | a | < 1 (3)
z 1 
  

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