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Question Number 102598 by dw last updated on 10/Jul/20

Answered by mr W last updated on 10/Jul/20

z=2(cos θ+i sin θ) ∣z−i∣=(√(4cos^2 θ+(2 sin θ−1)^2 ))=(√(5−4 sin θ)) ∣z+i∣=(√(4cos^2 θ+(2 sin θ+1)^2 ))=(√(5+4 sin θ)) A=((∣z−i∣)/(∣z+i∣))=(√((5−4 sin θ)/(5+4 sin θ)))=(√(((10)/(5+4 sin θ))−1)) maximum is if sin θ=−1, A_(max) =(√(((10)/(5−4))−1))=3 minimum is if sin θ=1, A_(min) =(√(((10)/(5+4))−1))=(1/3)

$${z}=\mathrm{2}\left(\mathrm{cos}\:\theta+{i}\:\mathrm{sin}\:\theta\right) \\ $$$$\mid{z}−{i}\mid=\sqrt{\mathrm{4cos}^{\mathrm{2}} \:\theta+\left(\mathrm{2}\:\mathrm{sin}\:\theta−\mathrm{1}\right)^{\mathrm{2}} }=\sqrt{\mathrm{5}−\mathrm{4}\:\mathrm{sin}\:\theta} \\ $$$$\mid{z}+{i}\mid=\sqrt{\mathrm{4cos}^{\mathrm{2}} \:\theta+\left(\mathrm{2}\:\mathrm{sin}\:\theta+\mathrm{1}\right)^{\mathrm{2}} }=\sqrt{\mathrm{5}+\mathrm{4}\:\mathrm{sin}\:\theta} \\ $$$${A}=\frac{\mid{z}−{i}\mid}{\mid{z}+{i}\mid}=\sqrt{\frac{\mathrm{5}−\mathrm{4}\:\mathrm{sin}\:\theta}{\mathrm{5}+\mathrm{4}\:\mathrm{sin}\:\theta}}=\sqrt{\frac{\mathrm{10}}{\mathrm{5}+\mathrm{4}\:\mathrm{sin}\:\theta}−\mathrm{1}} \\ $$$${maximum}\:{is}\:{if}\:\mathrm{sin}\:\theta=−\mathrm{1}, \\ $$$${A}_{{max}} =\sqrt{\frac{\mathrm{10}}{\mathrm{5}−\mathrm{4}}−\mathrm{1}}=\mathrm{3} \\ $$$${minimum}\:{is}\:{if}\:\mathrm{sin}\:\theta=\mathrm{1}, \\ $$$${A}_{{min}} =\sqrt{\frac{\mathrm{10}}{\mathrm{5}+\mathrm{4}}−\mathrm{1}}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$

Answered by Dwaipayan Shikari last updated on 10/Jul/20

((∣z−i∣)/(∣z+i∣))=∣((z−i)/(z+i))∣=a {for maximum value ∣((z+i)/(z−i))∣=(1/a) ∣(z/i)∣=((1+a)/(1−a))⇒∣z∣=((1+a)/(1−a))=2 so a=(1/3)

$$\frac{\mid{z}−{i}\mid}{\mid{z}+{i}\mid}=\mid\frac{{z}−{i}}{{z}+{i}}\mid={a}\:\:\:\:\left\{{for}\:{maximum}\:{value}\right. \\ $$$$\mid\frac{{z}+{i}}{{z}−{i}}\mid=\frac{\mathrm{1}}{{a}} \\ $$$$\mid\frac{{z}}{{i}}\mid=\frac{\mathrm{1}+{a}}{\mathrm{1}−{a}}\Rightarrow\mid{z}\mid=\frac{\mathrm{1}+{a}}{\mathrm{1}−{a}}=\mathrm{2} \\ $$$${so}\:\:\:\:{a}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$

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