currently I am studying quasiconformal maps and the following question came up during my studies. In complex analysis one learns that $$ \{f \colon \mathbb C \to \mathbb C : f \text{ is conformal }\} = \{z \mapsto az+b \mid a,b \in \mathbb C\}, $$ so all the conformal maps (surjective, injective with non-vanishing derivative) maps are given by the linear map up to two constants. Now for quasiconformal maps can one built a similiar equivalence, i.e. $$ \{f \colon \mathbb C \to \mathbb C : f \text{ is quasiconformal} \} = \quad ? $$ Since conformal maps $f \colon \mathbb C \to \mathbb C$ are already included on the right hand side by defintion but so far my gues is that there should be much more maps than simply the conformal ones. For example taking some $s<-1$ then $z \mapsto z |z|^{-s}$ is $K$-quasiconformal with $K = \max(1+s,\frac{1}{1+s})$ and would already give an example for a $K$-quasiconformal map that is not conformal. Even worse in some sense, we could now compose this with any conformal map and still stay in the LHS, so maybe the question is not 100% percent correctly poised, still is there an more or less concrete answer to this?
Thanks in advance!