Question
The maximum distance from the origin of coordinates to the point z satisfying the equation ∣∣z+1z∣∣=a is
Answer: Option C
:
C
let z=r (cosθ+isinθ)
Then ∣∣z+1z∣∣=a ⇒ ∣∣z+1z∣∣2=a2
⇒ r2+1r2+2cosθ = a2 ⋯ ⋯(i)
Differentiating w.r.t θ we get
2rdrdθ-2r3drdθ-4sin2θ
Putting drdθ=0, we get θ=0,π2
r is maximum for θ = 0, π2, therefore from (i)
r2+1r2−2=a2⇒r−1r=a⇒r=a+2√a2+42
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:
C
let z=r (cosθ+isinθ)
Then ∣∣z+1z∣∣=a ⇒ ∣∣z+1z∣∣2=a2
⇒ r2+1r2+2cosθ = a2 ⋯ ⋯(i)
Differentiating w.r.t θ we get
2rdrdθ-2r3drdθ-4sin2θ
Putting drdθ=0, we get θ=0,π2
r is maximum for θ = 0, π2, therefore from (i)
r2+1r2−2=a2⇒r−1r=a⇒r=a+2√a2+42
Was this answer helpful ?
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