Question Number 148951 by EDWIN88 last updated on 01/Aug/21
$${Let}\:{complex}\:{number}\:{z}=\left({a}+\mathrm{cos}\:\theta\right)+\left(\mathrm{2}{a}−\mathrm{sin}\:\theta\right){i}\:. \\ $$$${If}\:\mid{z}\mid\:\leqslant\mathrm{2}\:{for}\:{any}\:\theta\in{R}\:{then}\:{the} \\ $$$${range}\:{of}\:{real}\:{number}\:{a}\:{is}\:\_\_\_ \\ $$
Answered by iloveisrael last updated on 01/Aug/21
Answered by mr W last updated on 01/Aug/21
$$\left({a}+\mathrm{cos}\:\theta\right)^{\mathrm{2}} +\left(\mathrm{2}{a}−\mathrm{sin}\:\theta\right)^{\mathrm{2}} \leqslant\mathrm{2}^{\mathrm{2}} =\mathrm{4} \\ $$$$\mathrm{5}{a}^{\mathrm{2}} +\mathrm{2}{a}\left(\mathrm{cos}\:\theta−\mathrm{2sin}\:\theta\right)−\mathrm{3}\leqslant\mathrm{0} \\ $$$$\mathrm{5}{a}^{\mathrm{2}} +\mathrm{2}{ka}−\mathrm{3}\leqslant\mathrm{0} \\ $$$${k}=\mathrm{cos}\:\theta−\mathrm{2}\:\mathrm{sin}\:\theta=\sqrt{\mathrm{5}}\mathrm{cos}\:\left(\theta+\mathrm{tan}^{−\mathrm{1}} \mathrm{2}\right) \\ $$$$−\sqrt{\mathrm{5}}\leqslant{k}\leqslant\sqrt{\mathrm{5}} \\ $$$$\mathrm{5}{a}^{\mathrm{2}} +\mathrm{2}\sqrt{\mathrm{5}}{a}−\mathrm{3}\leqslant\mathrm{0} \\ $$$$\Rightarrow{a}\leqslant\frac{\sqrt{\mathrm{5}}}{\mathrm{5}} \\ $$$$\mathrm{5}{a}^{\mathrm{2}} −\mathrm{2}\sqrt{\mathrm{5}}{a}−\mathrm{3}\leqslant\mathrm{0} \\ $$$$\Rightarrow{a}\geqslant−\frac{\sqrt{\mathrm{5}}}{\mathrm{5}} \\ $$$$\Rightarrow−\frac{\sqrt{\mathrm{5}}}{\mathrm{5}}\leqslant{a}\leqslant\frac{\sqrt{\mathrm{5}}}{\mathrm{5}} \\ $$
Commented by puissant last updated on 01/Aug/21
$$\mathrm{nice}\:\mathrm{prof}.. \\ $$