Page 77 - DMTH402_COMPLEX_ANALYSIS_AND_DIFFERENTIAL_GEOMETRY

Complex Analysis and Differential Geometry

Notes a 1
= , | z  a | = | z   |
c 

i
  = a  a =  e ,  being real.
c c
Thus, the required transformation becomes

 z   
2  ,
i
w =  e    z    | a | < .

Example: Find the bilinear transformation which maps the point 2, i, 2 onto the points
1, i, 1.
Solution. Under the concept of crossratio, the required transformation is given by


(w w )(w  w )  (z z )(z  z )

1
2
1
2
3
3
(w  w )(w  w) (z  z )(z  z)
3
1
2
1
2
3
Using the values of z and w , we get
i
i
(w 1)(i 1) (z 2)(i 2)




(1 i)( 1 w) = (2 i)( 2 z)










or w 1 =   z 2   2 i   1 i  
 
 
w 1   2 i   
1 i 
z 2   

w 1 4 3i z 2



or =
w 1 5 z 2


w 1 w 1 (4 3i)(z 2) 5(z 2)







or =
w 1 (w 1) (4 3i)(z 2) 5(z 2)







3z(3 i) 2i(3 i) 3z 2i




or w = 
 iz(z i) 6(3 i)  (iz 6)





or w = 3z 2i
iz 6

which is the required transformation.
7.2 Conformal Mappings
Let S be a domain in a plane in which x and y are taken as rectangular Cartesian co-ordinates. Let
us suppose that the functions u(x, y) and v(x, y) are continuous and possess continuous partial
derivatives of the first order at each point of the domain S. The equations
u = u(x, y), v = v(x, y)
set up a correspondence between the points of S and the points of a set T in the (u, v) plane. The
set T is evidently a domain and is called a map of S. Moreover, since the first order partial
derivatives of u and v are continuous, a curve in S which has a continuously turning tangent is
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