Why $f(z)=\frac{az+b}{cz+d}$, $a,b,c,d \in \mathbb C$, is a linear transformation? Now I'm confused with what "a linear transformation" means.
In linear algebra textbook, I learned that a linear transformation is $T:V \to W$, where V,W are vector spaces, which satisfies additivity and homogeneity, in other words, $T(u+v)=Tu+Tv, T(av)=aTv$ for all $u,v \in V$ and $a \in F$
But in my complex analysis textbook, $\displaystyle f(z)=\frac{az+b}{cz+d}$, $a,b,c,d \in \mathbb C$, is introduced as an example of a linear transformation. 
However, this function $f$ doesn't seem to follow the definition from linear algebra. Indeed, $f(0) \neq 0$.
Is it like there are two kinds of linear transformations in mathematics, or they are actually the same thing but I don't get it well?
 A: Let $T=f$, then we have $$f(u+v)=\frac{a(u+v)+b}{c(u+v)+d}.$$
Likewise, $f(u)+f(v)=\frac{au+b}{cu+d}+\frac{av+b}{cv+d}.$ It is clear that $f(u+v)\ne f(u)+f(v)$, hence this is not a linear transform. However, does it satisfy the second property? Let's see:
$$f(\alpha u)=\frac{a(\alpha u)+b}{c(\alpha u)+d},$$
and $$\alpha f(u)=\alpha \cdot \frac{au+b}{cu+d}=\frac{a\alpha u+\alpha b}{cu+d}.$$
Since $f(\alpha u)\ne \alpha f(u)$, $f$ does not satisfy the second property either.
As Robert Israel first pointed out, this is the very definition of a linear fractional transform.
A: $z \to \dfrac{az+b}{cz+d}$ is more properly called a fractional linear transformation (or linear fractional transformation, or  Möbius transformation).  It is not the same as a linear transformation, although abuse of the language sometimes does take place.
EDIT: For example, Ford's "Automorphic Functions", first published 1929, defines $z' = \dfrac{az+b}{cz+d}$ as a "linear transformation": in a footnote he says 'This is more properly called a "linear fractional transformation"; but we shall use the briefer designation.'
 http://books.google.ca/books?id=aqPvo173YIIC&pg=PA1
A: The linear transformation you described in terms of linear algebra is a distinct notion in this case. If you are familiar with category theory, a linear transformation is a morphism in the category of vector spaces (if not disregard this sentence).
In the context of complex analysis, the type of transformation you have described is a called a Möbius transformation,
$$f(z) = \frac{az +b}{cz + d}$$
with $a,b,c,d \in \mathbb{C}$, and traditionally, $ad - bc \neq 0$, since we don't want to consider the constant function. A Möbius transformation is a projective linear transformation (also called a homography) of the complex projective line. That is to say, it is a non-linear transformation in terms of Cartesian coordinates and generates a different group-theoretic structure (compare the projective linear group $PGL(V)$ to the general linear group $GL(V)$).
A: I think perhaps one of the reasons it is called this is that composition of Möbius transformations acts like matrix multiplication on the coefficients.  That is, if we have $$f(z) = \frac{a z + b}{c z + d},$$ we associate it with the matrix $$M_f=\begin{pmatrix} a & b \\ c & d \end{pmatrix}.$$  Then, if $f(z)$ and $g(z)$ are Möbius transformations, then $$M_{f\circ g}=M_fM_g.$$ That is, the coefficients of $f(g(z))$ are exactly the elements of the product of the coefficients of $f(z)$ and $g(z)$ as matrices.  
