Let $K$ be a compact subset of $\mathbb{C}$.
Let $c,d$ be integers such that $(c,d)=1$. Set $S_(c,d)=\max\{|cz+d|\mid z\in K\}$ (lower case (c,d) because it depends on the choice of $c$ and $d$, and existence because K is compact).
I want to show that #$\{(c,d):N\leq| S_(c,d)|\leq N+1\}=O(N)$ where # means the number of elements. Here $c$ and $d$ should be coprime.
This I think is true, but I don't know how to justify it. The difficulty for me is in dealing with $K$. Is this statement true? Can someone give me insight? Thanks!
$$S(u,v)=\sup_{z\in K} |uz+v|$$ is a norm on $\Bbb{R}^2$, $$S(tu,tv) = t S(u,v)$$ thus $$B_r=\iint_{S(u,v)< r} dudv= \iint_{S(u,v)< 1} d(ru)d(rv)=r^2B_1$$ Where $B_1$, the area of the unit ball of your norm, is finite non-zero.
For $S(c,d)\in [N,N+1]$ draw a unit square $[c,c+1]\times [d,d+1]$. There is $q=q_K$ such that every point of this square is in $N-q<S(u,v)<N+q$.
Whence the number of lattice points can be bounded by the area $$\sum_{c,d,N\le S(c,d)\le N+1} 1\le \int_{N-q<S(u,v)<N+q} dudv= (N+q)^2 B_1-(N-q)^2B_1=O(N)$$