Page 73 - DMTH402_COMPLEX_ANALYSIS_AND_DIFFERENTIAL_GEOMETRY

Complex Analysis and Differential Geometry

Notes Further, we observe that in the form (1), p and q are inverse points w.r.t. the circle. For this, if the
circle is |z  z | = r and p and q are inverse points w.r.t. it, then
0
z  z = re , p  z = qe ,
i
i
0
0
q  z =  a 2 e i
0
Therefore,

z p =  e  ae i a e  ae i
i
i


z q i  2 i   ae   e i
i

 e  e
a



= K  (cos  isin ) a(cos  isin ) , K  a
a(cos  isin )    (cos  isin )  





= K ( cos  acos ) i( sin  asin )
(acos   cos ) i(asin   sin )



2
2
 ( cos  acos )  ( sin  asin )  1/2




= K  2 2 
 (acos   cos )   (asin   sin )  
= K, where K  1, since a  r
Thus, if p and q are inverse points w.r.t. a circle, then its equation can be written as
z p

z q = K, K  1, K being a real constant.

Theorem
In a bilinear transformation, a circle transforms into a circle and inverse points transform into
inverse points. In the particular case in which the circle becomes a straight line, inverse points
become points symmetric about the line.
z p

Proof : We know that = K represents a circle in the z-plane with p and q as inverse points,
z q

where K  1. Let the bilinear transformation be
az b dw b


w = cz d so that z =  cw a


Then under this bilinear transformation, the circle transforms into
dw b p




 cw a  dw b p(q cw)

dw b q = K  dw b q(a cw) = K





 cw a 
66 LOVELY PROFESSIONAL UNIVERSITY