The maximum distance from the origin of coordinates to the point satisfying the equation ∣∣z+1z∣∣=a is
A
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B
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C
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D
None of these
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Solution
The correct option is C letz=r(cosθ+isinθ)Then∣∣z+1z∣∣=a⇒∣∣z+1z∣∣2=a2⇒r2+1r2+2cos2θ=a2.........(i) Differntiatingw.r.t.θweget2rdrdθ−2r3drdθ−4sin2θ=0Puttingdrdθ=0,wegetθ=0,π2rismaximumforθ=π2,thereforefrom(i) r2+1r2−2=a2⇒r−1r=a⇒r=a+√a2+42