The maximum distance from the origin of coordinates to the point satisfying the equation - z -1- z - a isA- None of these B- 1-2-a2-1-aC- 1-2-a2-2-aD- 1-2-a2-4-a

Let z=r (cos θ+isin θ) Then |z+1/z|=a⇒ |z+1/z|^2=a^2 ⇒ r^2+1/r^2+2 cos 2 θ = a^2 .........(i) Differntiating w.r.t. θ we get 2rdr/dθ 2/r^3dr/dθ 4sin 2θ=0 Pu ...

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Standard XII

Mathematics

Polar Representation of a Complex Number

The maximum d...

Question

The maximum distance from the origin of coordinates to the point satisfying the equation $∣ ∣ z + \frac{1}{z} ∣ ∣ = a$ is

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None of these

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Solution

The correct option is C
$l e t z = r (c o s θ + i s i n θ) T h e n ∣ ∣ z + \frac{1}{z} ∣ ∣ = a \Rightarrow {∣ ∣ z + \frac{1}{z} ∣ ∣}^{2} = a^{2} \Rightarrow r^{2} + \frac{1}{r^{2}} + 2 c o s 2 θ = a^{2} . . . . . . . . . (i)$
$D i f f e r n t i a t i n g w . r . t . θ w e g e t 2 r \frac{d r}{d θ} - \frac{2}{r^{3}} \frac{d r}{d θ} - 4 s i n 2 θ = 0 P u t t i n g \frac{d r}{d θ} = 0, w e g e t θ = 0, \frac{π}{2} r i s m a x i m u m f o r θ = \frac{π}{2}, t h e r e f o r e f r o m (i)$
$r^{2} + \frac{1}{r^{2}} - 2 = a^{2} \Rightarrow r - \frac{1}{r} = a \Rightarrow r = \frac{a + \sqrt{a^{2} + 4}}{2}$

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