Absolute Value of a Complex Number Inequality

Question

Let $z=x+ i y$ , $x,y>0$ be a complex number. What conditions on $\alpha, \beta \in \mathbb{R}$ (or on $x$ and $y$) must be imposed so that the following inequality holds. \begin{align*} \vert z \vert^2>\alpha \mathrm{Re}(z)+\beta \mathrm{Im}(z). \end{align*} Is there any result which can help in this regard?

Answer 1

Note that the given inequality is equivalent to $$x^2-\alpha x+y^2-\beta y>0$$ that is, after multiplying both sides by $4$, $$(2x-\alpha)^2+(2y-\beta)^2>(0-\alpha)^2+(0-\beta)^2,$$ which means that the point $(\alpha,\beta)$ is closer to the point $(2x,2y)$ than to the origin $(0,0)$.

What is, from the geometric point of view, the set of such points $(\alpha,\beta)$?

Give it a try by making a drawing!

P.S. There are infinite $(\alpha,\beta)$ such that $x^2+y^2=\alpha x+\beta y$. All the points along the line bisector of the segment with endpoints $(0,0)$ and $(2x,2y)$ (which includes $\alpha=x$ and $\beta=y$).