 Solution

Choose one representative from $A$ above each point of $\pi (A)$ and collect them in $X$. Part (c) gives $|X|\le K^{O(1)}$. If $a\in A$ has the same image as $x\in X$, then $x^{-1}a\in A^2\cap N$, soThis is the required covering of $A$ by at most $K^{O(1)}$ left cosets of the abelian translation subgroup $N$.

The set $B=A^2\cap N$ is a $K^{O(1)}$-approximate group by the intersection of an approximate group power with a subgroup. Apply the Freiman-Green-Ruzsa theorem inside $N\cong (\mathbb{C},+)$. Because the additive group of the complex numbers is a torsion-free group, the finite subgroup part is trivial, so there is an abelian progression $P$ withSince $|B|\le |A^2|\le K|A|$, enlarging the implicit constant gives