Why $T(z)=az+b$ is called a Linear Transformation? As we always read in Complex Analysis, Linear Transformation (L.T.) is a combination of Translation, Rotation and Magnification i.e. $T(z)=az+b$ is a L.T. in complex. However, it doesn't satisfy the linearity properties
(1) $T(z_1+z_2)=T(z_1)+T(z_2)$
(2) $T(az)=aT(z)$. 
So, my question is that if $T(z)=az+b$ doesn't satisfy the properties of L.T. then why it is called a L.T.?
Please make it clear to me. It's very urgent. 
 A: There are two reasonable definitions of "linear", if you imagine that you're the first mathematician to define this term. 


*

*A transformation that sends lines to lines. A good example is "translation". Lines form natural subsets of euclidean spaces, esp. if you've studied geometry a la Euclid, so preserving this structure leads to interesting characteristics for a transformation; it needs a name, so you pick "linear." One might call this the "geometric" definition.

*A transformation that sends subspaces of a vector space to subspaces. In a vector space, which is a sort of natural generalization of the real numbers in which you have scalar multiplication instead of multiplication, not all lines are the same: the ones through the origin have special properties (closed under addition and scalar multiplication). So they tend to be the ones that you study a lot. And transformations that preserve these important lines get called "linear." One might call this the "algebraic" definition. 
The first usage is the one informally used in complex vars; the second is the one used in linear algebra.(*) 
There's a unification, however. If you imagine the $z = 1$ plane of $xyz$-space, you can look at the special class of linear (in the second sense) transformations of $xyz$-space that send the $z = 1$ plane to itself. These turn out to be exactly the linear (in the first sense) transformations of the $z = 1$ plane. 
You can even generalize a bit: look at any transformation $T$ of $xyz$-space, and follow it by the map 
$$
H: \mathbb R^3 - \{ (x, y, 0) | x, y \in \mathbb R \} \to \mathbb R^3: (x, y, z) \mapsto (x/z, y/z, 1),
$$
which folks in computer graphics call "homogenization" (which is exactly the wrong word). The transformation $H \circ T$ will send the $z = 1$ plane to itself, with just a tiny problem: If $T(a, b, 1) = (p, q, 0)$, then $H \circ T$ won't be defined for the point $(a, b, 1)$. Or you might say that because you divided by zero, the result is a "point at infinity." That sounds like nonsense, but is in fact the starting point of projective geometry; for a lovely introduction, once you know a little abstract algebra, see Hartshorne's book "Foundations of Projective Geometry."  
(*) It's not enough to say "sends line-subspaces to line-subspaces". Consider the map on the plane that defines $T(x, y) = (x, y)$ if $x/y$ is rational, and $(0,0)$ otherwise. That satisfies $T(\alpha \mathbf u) = \alpha T(\mathbf u)$, but not the addition rule (although it satisfies the addition rule on each line-subspace of the plane!). But it's certainly not linear! 
A: The proper name should be linear fractional transformation, not just "linear transformation", but some textbooks are a little sloppy.
As you point out, they are not linear in the usual sense of the words. However, they are projective transformations and in fact form the so called projective linear group of $\mathbb{C}$, which also happens to be the automorphism group of the Riemann sphere.
A: Thanks for your answer. Actually the linear fractional transformations are different from linear transformations and linear fractional transformations are defined as $$T(z)=\dfrac{az+b}{cz+d}$$ which is a generalization of linear transformation.
So the conclusion is that we should not say that $T(z)=az+b$ is a linear transformation. 
