Page 74 - DMTH402_COMPLEX_ANALYSIS_AND_DIFFERENTIAL_GEOMETRY

Unit 7: Transformations and Conformal Mappings

Notes
ap b

w(d cp) (ap b) w  cp d cq d





 w(d cq) (aq b) = K  aq b = K cp d (1)




w  
cq d

The form of equation (1) shows that it represents a circle in the w-plane whose inverse points are
ap b and aq b . Thus, a circle in the z-plane transforms into a circle in the w-plane and the


cp d cq d


inverse points transform into the inverse points.
cq d

Also if K = 1, then equation (1) represents a straight line bisecting at right angle the join
cp d

ap b aq b


of the points and so that these points are symmetric about this line. Thus, in a
cp d cq d


particular case, a circle in the z-plane transforms into a straight line in the w-plane and the
inverse points transform into points symmetrical about the line.
Example: Find all bilinear transformations of the half plane Im z  0 into the unit circle
| w |  1.
Solution. We know that two points z, z , symmetrical about the real z-axis(Im z = 0) correspond
to points w, 1 , inverse w.r.t. the unit wcircle. (|w| 1 = 1). In particular, the origin and the
w |w|
point at infinity in the w-plane correspond to conjugate values of z.
Let
az b a( z b/a)


w = = (1)
cz d c (z d/c)


be the required transformation.
Clearly c  0, otherwise points at  in the two planes would correspond.
Also, w = 0 and w =  are the inverse points w.r.t. | w | = 1. Since in (1), w = 0, w =  correspond
b d
respectively to z =  , z =  , therefore, these two values of zplane must be conjugate to each
a c
other. Hence, we may write
b d
 = ,  =  so that
a c

w = a z   (2)
c z  
The point z = 0 on the boundary of the half plane Im z  0 must correspond to a point on the
boundary of the circle | w | = 1, so that

1 = | w | = a 0    a
c 0   c

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