Why $f(z)=\frac{az+b}{cz+d}$, $a,b,c,d \in \mathbb C$, is a linear transformation?

$z \to \dfrac{az+b}{cz+d}$ 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

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.

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 $GL(V)$).