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Let-complex-number-z-a-cos-2a-sin-i-If-z-2-for-any-R-then-the-range-of-real-number-a-is-




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

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