# Why the name linear fractional map?

Fractional linear transformation is a map from extended complex plane to itself, defined by: $$$$z\to \frac{az+b}{cz+d}$$$$ with $$ad-bc\ne0$$.

Wikipedia says that "a linear fractional transformation is a transformation that is represented by a fraction whose numerator and denominator are linear". Either in the numerator or denominator there are translations and translations aren't linear transformation (they can't map origin in the origin). So, why "linear" in the name?

I know that for an inversion:

Is for this reason the name?

• Equations of the form $ax+b$ used to be called linear in the past but now they're called affine. So it's an archaic term. I prefer to call them Mobius transformations instead for this reason but you'll see both. Feb 29 at 15:28
• there is an isomorphism from these fractional linear transformations with a group of 2x2 matrices, maybe this is the reason for the "linear" in the name. Moreover, the fractioanal linear transformations maps circles to circles (in the Riemann sphere), so if you think of a circle as a "line" in this sphere then these maps are linear in this sense Feb 29 at 15:52
• take a look here Feb 29 at 15:58

Functions of the form $$f(x) = mx+b$$ are often called "linear" because their graphs are lines. This usage differs from the more restrictive notion of a linear function $$F$$ being one satisfying $$F(\alpha x+\beta y)=\alpha F(x)+\beta F(y)$$.
I am not a historian, but: The Riemann sphere may be identified with the complex projective line, the set of complex lines through the origin in the "complex plane" $$\mathbf{C}^{2}$$. Fractional linear transformations on the sphere are precisely those induced by invertible complex-linear transformations.
Geometrically, the invertible linear transformation with standard matrix $$\left[\begin{array}{@{}cc@{}}a & b \\ c & d \\ \end{array}\right]$$ on the (complex) Cartesian plane acts on the set of lines through the origin. Viewed as a mapping on slopes, the line of slope $$z$$ maps to the line of slope $$\dfrac{az + b}{cz + d}$$. Algebraically, let $$(Z, W)$$ denote Cartesian coordinates, and write $$z = Z/W$$. Up to scaling, i.e., in projective coordinates, the point $$\left[\begin{array}{@{}c@{}} z \\ 1 \\ \end{array}\right] = \left[\begin{array}{@{}c@{}} Z/W \\ 1 \\ \end{array}\right] \simeq \left[\begin{array}{@{}c@{}} Z \\ W \\ \end{array}\right]$$ maps to $$\left[\begin{array}{@{}cc@{}}a & b \\ c & d \\ \end{array}\right] \left[\begin{array}{@{}c@{}} Z \\ W \\ \end{array}\right] = \left[\begin{array}{@{}c@{}} aZ + bW \\ cZ + dW \\ \end{array}\right] \simeq \left[\begin{array}{@{}c@{}} \dfrac{aZ + bW}{cZ + dW} \\ 1 \\ \end{array}\right] = \left[\begin{array}{@{}c@{}} \dfrac{az + b}{cz + d} \\ 1 \\ \end{array}\right].$$