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?
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$\begingroup$ For all $\,z\,$? There are no such $\alpha, \beta\,$. $\endgroup$– dxivJan 20, 2018 at 18:15
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$\begingroup$ Note: $|z|^2=x^2 + y^2$, so you are just looking for conditions for the real inequality $x^2 + y^2 > \alpha x + \beta y$ $\endgroup$– 57JimmyJan 20, 2018 at 18:16
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$\begingroup$ @dxiv.....You are right......this is not true for all $z$, Is is true in some domain? $\endgroup$– skorpionJan 20, 2018 at 20:37
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$\begingroup$ @57Jimmy....I want to prove it ....I know what you have written.....$x^2+y^2=\alpha x+\beta y$ iff $x=\alpha$ and $y=\beta$. What about inequality? $\endgroup$– skorpionJan 20, 2018 at 20:40
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$\begingroup$ @skorpion If $(x,y)\not=(0,0)$ then 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 $(2x,2y)$. $\endgroup$– Robert ZJan 20, 2018 at 22:22
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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 $\beta=y$).
answered Jan 20, 2018 at 18:15
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$\begingroup$ Thanks :)....How to prove it....that is what I am looking for..... $\endgroup$– skorpionJan 20, 2018 at 20:49