# Asymptotic Behaviour of Rational Approximations of an Irrational Real Number

This question is inspired by this thread. I used Mathematica to verify that the answer to the question below seems to be yes for several irrational real numbers.

Let $r\in\mathbb{R}$ be an irrational real number. Write $$\gamma_b(r):=\max\,\left\{\frac{1}{b^2\,\left|r-\frac{a}{b}\right|}\,\Big|\,a\in\mathbb{Z}\right\}$$ for every positive integer $b$. Define $$\Gamma(r):=\liminf_{b\to\infty}\,b\cdot\gamma_b(r)\,.$$ Does it always hold that $$\Gamma(r)=\inf\,\big\{b\cdot\gamma_b(r)\,\big|\,b=1,2,3,\ldots\big\}=2\,?$$

What if we allow $r$ to be in $\mathbb{C}$ and $a$ to be in $\mathbb{Z}[\text{i}]$, where $\text{i}$ is the imaginary unit $\sqrt{-1}$? We say that $r$ is an irrational complex number if $r\notin\mathbb{Q}[\text{i}]$, or equivalently, $r\neq \frac{a}{b}$ for all $a\in\mathbb{Z}[\text{i}]$ and $b\in\mathbb{N}$. That is, what is the answer to the reformulation below?

Let $r\in\mathbb{C}$ be an irrational complex number. Write $$\gamma_b(r):=\max\,\left\{\frac{1}{b^2\,\left|r-\frac{a}{b}\right|}\,\Big|\,a\in\mathbb{Z}[i]\right\}$$ for every positive integer $b$. Define $$\Gamma(r):=\liminf_{b\to\infty}\,b\cdot\gamma_b(r)\,.$$ Does it always hold that $$\Gamma(r)=\inf\,\big\{b\cdot\gamma_b(r)\,\big|\,b=1,2,3,\ldots\big\}\,?$$ What is the value of $\Gamma(r)$?

Due to Mathematica, it seems to be the case that $\Gamma(r)$ can be lower than $2$ when $r$ is not real. For example, it appears that $\Gamma\left(\text{e}+\text{i}\pi\right)=\sqrt{2}$ (of course, this is a guess). I think the following is true: $$\Gamma(r)=\inf\,\big\{b\cdot\gamma_b(r)\,\big|\,b=1,2,3,\ldots\big\}=\begin{cases}2\,,&\text{if }r\notin\mathbb{Q}[\text{i}]\,,\text{ but }\text{Re}(r)\in\mathbb{Q}\text{ or }\text{Im}(r)\in\mathbb{Q}\,, \\ \sqrt{2}\,,&\text{if }\text{Re}(r)\notin\mathbb{Q}\text{ and }\text{Im}(r)\notin\mathbb{Q}\,. \end{cases}$$

The user quid has made a remark in the aforementioned thread. I think the same remark answers both of my questions. However, I will leave this thread open for someone to try.