# Linear fractional transformation

(Redirected from Fractional linear transformation)

In mathematics, the phrase linear fractional transformation usually refers to a Möbius transformation, which is a homography on the complex projective line P(C) where C is the field of complex numbers.

More generally in mathematics, C may be replaced by another ring (A, +, ×).[1] For example, the Cayley transform is a linear fractional transformation originally defined on the 3 x 3 real matrix ring.

In general, a linear fractional transformation refers to a homography over P(A), the projective line over a ring. When A is a commutative ring, then the linear fractional transformation has the familiar form

${\displaystyle z\mapsto {\frac {az+b}{cz+d}},\quad z,a,b,c,d\in A.}$

Otherwise homographies are expressed (az + b, cz + d) with homogeneous coordinates. The equivalence of such coordinates is expressed

${\displaystyle U(az+b,cz+d)\sim U((cz+d)^{-1}(az+b),1).}$

## Conformal property

The commutative rings of split-complex numbers and dual numbers join the ordinary complex numbers as rings that express angle. In each case the exponential map applied to the imaginary axis produces an isomorphism between one-parameter groups in (A, + ) and in the group of units (U, × ):

${\displaystyle \exp(yj)=\cosh y+j\sinh y,\quad j^{2}=+1,}$
${\displaystyle \exp(y\epsilon )=1+y\epsilon ,\quad \epsilon ^{2}=0,}$
${\displaystyle \exp(yi)=\cos y+i\sin y,\quad i^{2}=-1.}$

The "angle" y is hyperbolic angle, slope, or circular angle according to the host ring.

A linear fractional transformation can be generated by multiplicative inversion z → 1/z and affine transformations za z + b. Conformality can be confirmed by showing the generators are all conformal. The translation zz + b is a change of origin and makes no difference to angle. To see that zaz is conformal, consider the polar decomposition of a and z. In each case the angle of a is added to that of z giving a conformal map. Finally, inversion is conformal since z → 1/z sends ${\displaystyle \exp(yb)\mapsto \exp(-yb),\quad b^{2}=1,0,-1.}$