The maximum distance from the origin of coordinates to the point z satisf

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

Options:

A .

\frac{1}{2}

(

\sqrt[2]{(a^{2} + 1)}

+a)

B .

\frac{1}{2}

(

\sqrt[2]{(a^{2} + 2)}

+a)

C .

\frac{1}{2}

(

\sqrt[2]{(a^{2} + 4)}

+a)

D . None of these

Answer: Option C
:
C

let z=r (cos $θ$ +isin $θ$ )

Then $∣ ∣ z + \frac{1}{z} ∣ ∣$ =a $\Rightarrow$ ${∣ ∣ z + \frac{1}{z} ∣ ∣}^{2}$ = $a^{2}$

$\Rightarrow$ $r^{2}$ + $\frac{1}{r^{2}}$ +2cos $θ$ = $a^{2}$ $\dots$ $\dots$ (i)

Differentiating w.r.t $θ$ we get

2r $\frac{d r}{d θ}$ - $\frac{2}{r^{3}}$ $\frac{d r}{d θ}$ -4sin2 $θ$

Putting $\frac{d r}{d θ}$ =0, we get $θ$ =0, $\frac{π}{2}$

r is maximum for $θ$ = 0, $\frac{π}{2}$ , therefore from (i)

$r^{2} + \frac{1}{r^{2}} - 2 = a^{2} \Rightarrow r - \frac{1}{r}$ = $a \Rightarrow r$ = $\frac{a + \sqrt[2]{a^{2} + 4}}{2}$

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