20 
[892 
893] 
892. 
NOTE ON THE ORTHOMORPHIC TRANSFORMATION OF A 
CIRCLE INTO ITSELF. 
[From the Proceedings of the Edinburgh Mathematical Society, vol. vm. (1890), 
pp. 91, 92.] 
The following is, of course, substantially well known, but it strikes me as rather 
pretty:—to find the orthomorphic transformation of the circle 
+ y 2 — 1 = 0 
into itself. Assume this to be 
A (x + iy) + B 
0B 1 + lVi= -t -n, • 
1 J1 1 + G(x + iy) 
Then, writing A', B', C' for the conjugates of A, B, G, we have 
A' (x — iy) + B' 
x '~ ly '= l + G'(x-iy) ; 
and then 
A A' (,x 2 + y 2 ) + AB' (x + iy) + A'B (x — iy) + BB' 
Xl +yi ~ T+C(x + iy) + G' (x - iy) + CC {a? + f) 
which should be an identity for « 2 + y 2 = l, x^ + y^— 1. 
Evidently G=AB', whence G' = A'B\ and the equation then is 
1 +AA'BB' = AA' + BB > , 
that is, 
(1 - A A') (1 - BB') = 0. 
But BB' = 1 gives the illusory result 
+ iyi = B, 
therefore 
and the required solution thus is 
1 -AA'= 0; 
, . A (x + iy) + B 
Xl + njl ~l+AB'(x + iyy 
where A is a unit-vector (say A = cos A, + i sin \) and B, B! are conjugate vectors. Or, 
writing B = b + i/3, B' = b — i/3, the constants are \, b, /3; three constants as it should be. 
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