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Consider elements $x,z \in \mathbb{Z}[i]$, the Gaussian integers. Let $S(T_1, T_2)$ be the subset of $\mathbb{Z}[i] \times \mathbb{Z}[i]$ consisting of those elements $(x,z)$ such that $T_1 < |x| \leq 2T_1, T_2 < |z| \leq 2T_2$. I want to estimate the cardinality of the set

$$\displaystyle S_k(T_1, T_2) =\{(x,z) \in S(T_1, T_2) : \left \lvert \Re(xz^k) \right \rvert \leq M \}, k \geq 2.$$

Note that if $M$ is close in size to $T_1 T_2^k$, then the restriction that $\lvert \Re(xz^k) \rvert \leq M$ is hardly a restriction at all and the obvious bound of $|S(T_1, T_2)|$ will give the correct order of magnitude. However when $M$ is very small compared to $T_1 T_2^k$ then $|S_k(T_1, T_2)|$ should be substantially smaller than $|S(T_1, T_2)|$.

If we write $x = r_1 e^{i \theta_1}$ and $z = r_2 e^{i \theta_2}$, then the condition $|\Re(xz^k)| \leq M$ can be written as $|r_1 r_2^k \cos(\theta_1 + k \theta_2)| \leq M$. If $M$ is small compared to $T_1 T_2^k$, then this is essentially saying that the angle $\theta_1 + k\theta_2 \pmod{\pi}$ is close to $\pi/2$.

Is there a good way to estimate the size of $S_k(T_1, T_2)$?

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If we consider $\mathbb{C}$ instead of $\mathbb{Z}[i]$, we can answer this well. In that case we are interested in the integral

$$\iiiint {\large\chi}\!\left[|r_1 r_2^k \cos(\theta_1 + k \theta_2)| \leq M\right](r_1 dr_1 d\theta_1)(r_2 dr_2 d\theta_2)$$

and the challenge is to do the integrations and the approximation in the right order.

We integrate first over $0 \le \theta_1 \le 2\pi$. Whatever the value of $\theta_2$, there are four ranges of $\theta$ in which the inequality holds, each of width $\frac{\pi}{2}-\arccos(M/r_1 r_2^k)$, or $\arcsin(M/r_1 r_2^k)$. So the integral reduces to

$$\iiint 4\arcsin(M/r_1 r_2^k) (r_1 dr_1)(r_2 dr_2 d\theta_2)$$

We then integrate over $0 \le \theta_2 \le 2\pi$ to get

$$\iint 8\pi\arcsin(M/r_1 r_2^k) (r_1 dr_1)(r_2 dr_2)$$

We integrate this over $T_1 \le r_1 \le 2 T_1$ to get

$$\int 4\pi r_2 T_1^2\ f(r_2^k T_1/M)\ dr_2,$$ $$f(u) = (4\arcsin{\frac1{2u}}+\sqrt{\frac{4}{u^2}-\frac{1}{u^4}})-(\arcsin{\frac1u}+\sqrt{\frac{1}{u^2}-\frac{1}{u^4}})$$ Since $f(u)\sim 5/(2u)$, we can approximate the last integral over $T_2 < r_2 < 2T_2$ as

$$\int 4\pi r_2 T_1^2 \frac{5M}{2r_2^k T_1} dr_2$$ $$=\frac{2^{1-k}(2^k-4)}{k-2}5\pi M T_1 T_2^{2-k}$$

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    $\begingroup$ This is a nice calculation but the integral case is substantially harder, since one would need to deal with the case when lower dimensional volumes are larger than the expected count. For instance, the co-dimension one projections are of size $\sim T_1 T_2 (T_1 + T_2)$, so a naive application of the ideas of geometry of numbers fail to capture the truth whenever this error term is larger than the volume indicated above... or roughly when $M \ll T_2^{k-1} (T_1 + T_2)$. $\endgroup$ Commented Jan 6, 2020 at 0:30
  • $\begingroup$ I was approximating $f(u)$ for large $u$, but I see now it would be more useful to approximate it for small $u$ instead, where $f(u) \sim (\sqrt{3}-\pi/3) + 2\sqrt{2}\sqrt{u-1}-\sqrt{3}(u-1)$. That might give better results for the region of interest to you. $\endgroup$
    – user44143
    Commented Jan 6, 2020 at 15:54

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