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Question Number 19690 by Tinkutara last updated on 14/Aug/17

If ∣z − (4/z)∣ = 2, then find the maximum  value of ∣z∣.

$$\mathrm{If}\:\mid{z}\:−\:\frac{\mathrm{4}}{{z}}\mid\:=\:\mathrm{2},\:\mathrm{then}\:\mathrm{find}\:\mathrm{the}\:\mathrm{maximum} \\ $$$$\mathrm{value}\:\mathrm{of}\:\mid{z}\mid. \\ $$

Answered by ajfour last updated on 15/Aug/17

  ∣(∣z∣−(4/(∣z∣)))∣≤2  let ∣z∣=t  ⇒ ∣t−(4/t)∣≤2  ⇒  −2≤t−(4/t)≤2  ⇒     −2t≤t^2 −4≤2t        (t≥0)  case I:       t^2 +2t≥4      (t+1)^2 ≥5    ⇒   t≥(√5)−1  case II:      t^2 −2t≤4  ⇒    (t−1)^2 ≤5  ⇒     t≤(√5)+1  so  ∣z∣_(max) =(√5)+1 .

$$\:\:\mid\left(\mid\mathrm{z}\mid−\frac{\mathrm{4}}{\mid\mathrm{z}\mid}\right)\mid\leqslant\mathrm{2} \\ $$$$\mathrm{let}\:\mid\mathrm{z}\mid=\mathrm{t} \\ $$$$\Rightarrow\:\mid\mathrm{t}−\frac{\mathrm{4}}{\mathrm{t}}\mid\leqslant\mathrm{2} \\ $$$$\Rightarrow\:\:−\mathrm{2}\leqslant\mathrm{t}−\frac{\mathrm{4}}{\mathrm{t}}\leqslant\mathrm{2} \\ $$$$\Rightarrow\:\:\:\:\:−\mathrm{2t}\leqslant\mathrm{t}^{\mathrm{2}} −\mathrm{4}\leqslant\mathrm{2t}\:\:\:\:\:\:\:\:\left(\mathrm{t}\geqslant\mathrm{0}\right) \\ $$$$\mathrm{case}\:\mathrm{I}: \\ $$$$\:\:\:\:\:\mathrm{t}^{\mathrm{2}} +\mathrm{2t}\geqslant\mathrm{4} \\ $$$$\:\:\:\:\left(\mathrm{t}+\mathrm{1}\right)^{\mathrm{2}} \geqslant\mathrm{5}\:\:\:\:\Rightarrow\:\:\:\mathrm{t}\geqslant\sqrt{\mathrm{5}}−\mathrm{1} \\ $$$$\mathrm{case}\:\mathrm{II}: \\ $$$$\:\:\:\:\mathrm{t}^{\mathrm{2}} −\mathrm{2t}\leqslant\mathrm{4} \\ $$$$\Rightarrow\:\:\:\:\left(\mathrm{t}−\mathrm{1}\right)^{\mathrm{2}} \leqslant\mathrm{5} \\ $$$$\Rightarrow\:\:\:\:\:\mathrm{t}\leqslant\sqrt{\mathrm{5}}+\mathrm{1} \\ $$$$\mathrm{so}\:\:\mid\mathrm{z}\mid_{\mathrm{max}} =\sqrt{\mathrm{5}}+\mathrm{1}\:. \\ $$

Commented by Tinkutara last updated on 15/Aug/17

Thank you very much Sir!

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$

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