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A-If-w-2-then-the-set-of-points-z-w-1-w-is-contained-in-or-equal-to-B-If-w-1-then-the-set-of-points-z-w-1-w-is-contained-in-or-equal-to-Options-for-both-A-and-B-p-An-ellip

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Question Number 21219 by Tinkutara last updated on 16/Sep/17
(A) If ∣w∣ = 2, then the set of points z = w − (1/w) is contained in or equal to (B) If ∣w∣ = 1, then the set of points z = w + (1/w) is contained in or equal to Options for both A and B: (p) An ellipse with eccentricity (4/5) (q) The set of points z satisfying Im z = 0 (r) The set of points z satisfying ∣Im z∣ ≤ 1 (s) The set of points z satisfying ∣Re z∣ ≤ 2 (t) The set of points z satisfying ∣z∣ ≤ 3
$$\left(\mathrm{A}\right)\:\mathrm{If}\:\mid{w}\mid\:=\:\mathrm{2},\:\mathrm{then}\:\mathrm{the}\:\mathrm{set}\:\mathrm{of}\:\mathrm{points} \\ $$$${z}\:=\:{w}\:−\:\frac{\mathrm{1}}{{w}}\:\mathrm{is}\:\mathrm{contained}\:\mathrm{in}\:\mathrm{or}\:\mathrm{equal}\:\mathrm{to} \\ $$$$\left(\mathrm{B}\right)\:\mathrm{If}\:\mid{w}\mid\:=\:\mathrm{1},\:\mathrm{then}\:\mathrm{the}\:\mathrm{set}\:\mathrm{of}\:\mathrm{points} \\ $$$${z}\:=\:{w}\:+\:\frac{\mathrm{1}}{{w}}\:\mathrm{is}\:\mathrm{contained}\:\mathrm{in}\:\mathrm{or}\:\mathrm{equal}\:\mathrm{to} \\ $$$$\mathrm{Options}\:\mathrm{for}\:\mathrm{both}\:\mathrm{A}\:\mathrm{and}\:\mathrm{B}: \\ $$$$\left(\mathrm{p}\right)\:\mathrm{An}\:\mathrm{ellipse}\:\mathrm{with}\:\mathrm{eccentricity}\:\frac{\mathrm{4}}{\mathrm{5}} \\ $$$$\left(\mathrm{q}\right)\:\mathrm{The}\:\mathrm{set}\:\mathrm{of}\:\mathrm{points}\:{z}\:\mathrm{satisfying}\:\mathrm{Im}\:{z} \\ $$$$=\:\mathrm{0} \\ $$$$\left(\mathrm{r}\right)\:\mathrm{The}\:\mathrm{set}\:\mathrm{of}\:\mathrm{points}\:{z}\:\mathrm{satisfying}\:\mid\mathrm{Im}\:{z}\mid \\ $$$$\leqslant\:\mathrm{1} \\ $$$$\left(\mathrm{s}\right)\:\mathrm{The}\:\mathrm{set}\:\mathrm{of}\:\mathrm{points}\:{z}\:\mathrm{satisfying}\:\mid\mathrm{Re}\:{z}\mid \\ $$$$\leqslant\:\mathrm{2} \\ $$$$\left(\mathrm{t}\right)\:\mathrm{The}\:\mathrm{set}\:\mathrm{of}\:\mathrm{points}\:{z}\:\mathrm{satisfying}\:\mid{z}\mid\:\leqslant\:\mathrm{3} \\ $$
Answered by Tinkutara last updated on 17/Sep/17
(A) Let w=2(cos θ+isin θ) Then (1/w)=(1/2)(cos θ−isin θ) w−(1/w)=(3/2)cos θ+(5/2)isin θ=x+iy(=z) ⇒cos θ=((2x)/3); sin θ=((2y)/5) ((4x^2 )/9)+((4y^2 )/(25))=1⇒Ellipse with e = (4/5) (B) w=cos θ+isin θ; (1/w)=cos θ−isin θ z=2cos θ ∴ ∣z∣≤2 and Im(z)=0.
$$\left(\boldsymbol{\mathrm{A}}\right)\:{Let}\:{w}=\mathrm{2}\left(\mathrm{cos}\:\theta+{i}\mathrm{sin}\:\theta\right) \\ $$$${Then}\:\frac{\mathrm{1}}{{w}}=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{cos}\:\theta−{i}\mathrm{sin}\:\theta\right) \\ $$$${w}−\frac{\mathrm{1}}{{w}}=\frac{\mathrm{3}}{\mathrm{2}}\mathrm{cos}\:\theta+\frac{\mathrm{5}}{\mathrm{2}}{i}\mathrm{sin}\:\theta={x}+{iy}\left(={z}\right) \\ $$$$\Rightarrow\mathrm{cos}\:\theta=\frac{\mathrm{2}{x}}{\mathrm{3}};\:\mathrm{sin}\:\theta=\frac{\mathrm{2}{y}}{\mathrm{5}} \\ $$$$\frac{\mathrm{4}{x}^{\mathrm{2}} }{\mathrm{9}}+\frac{\mathrm{4}{y}^{\mathrm{2}} }{\mathrm{25}}=\mathrm{1}\Rightarrow{Ellipse}\:{with}\:{e}\:=\:\frac{\mathrm{4}}{\mathrm{5}} \\ $$$$\left(\boldsymbol{\mathrm{B}}\right)\:{w}=\mathrm{cos}\:\theta+{i}\mathrm{sin}\:\theta;\:\frac{\mathrm{1}}{{w}}=\mathrm{cos}\:\theta−{i}\mathrm{sin}\:\theta \\ $$$${z}=\mathrm{2cos}\:\theta \\ $$$$\therefore\:\mid{z}\mid\leqslant\mathrm{2}\:{and}\:{Im}\left({z}\right)=\mathrm{0}. \\ $$

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