Question Number 226635 by Kassista last updated on 08/Dec/25
Answered by mingski last updated on 08/Dec/25
$${C}:{z}^{\mathrm{2}} +\mathrm{6}{z}+\mathrm{10}=\mathrm{0} \\ $$$$\left({z}+\mathrm{3}\right)^{\mathrm{2}} =−\mathrm{1} \\ $$$${z}+\mathrm{3}=\pm\mathrm{i},{z}=−\mathrm{3}\pm\mathrm{i}. \\ $$$${for}\:{z}_{\mathrm{1}} =−\mathrm{3}+{i}.\mid−\mathrm{3}+\mathrm{2}{i}\mid>\frac{\mathrm{7}}{\mathrm{2}}{soz}_{\mathrm{1}} \notin{B}. \\ $$$${and}\:\mid{z}_{\mathrm{1}} +\mathrm{2}−\mathrm{3i}\mid=\mid−\mathrm{1}−\mathrm{2i}\mid<\sqrt{\mathrm{19}}{so}\:{z}_{\mathrm{1}} \in{A}. \\ $$$${z}_{\mathrm{2}} =−\mathrm{3}−{i},{z}_{\mathrm{2}} \in{B}\: \\ $$$${so}\:{the}\:{answer}\:{is}\:−\mathrm{3}+\mathrm{i}. \\ $$
Commented by Kassista last updated on 08/Dec/25
thank you