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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=2(cosθ+isinθ)∣z−i∣=4cos2θ+(2sinθ−1)2=5−4sinθ∣z+i∣=4cos2θ+(2sinθ+1)2=5+4sinθA=∣z−i∣∣z+i∣=5−4sinθ5+4sinθ=105+4sinθ−1maximumisifsinθ=−1,Amax=105−4−1=3minimumisifsinθ=1,Amin=105+4−1=13

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)

∣z−i∣∣z+i∣=∣z−iz+i∣=a{formaximumvalue∣z+iz−i∣=1a∣zi∣=1+a1−a⇒∣z∣=1+a1−a=2soa=13

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