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The maximum distance from origin to the point z satisfying the equation $\big|z+\large\frac{1}{z}\big|=a$ is

$\begin{array}{1 1}(A)\;\large\frac{1}{2}\normalsize (\sqrt{a^2+1}+a)&(B)\;\large\frac{1}{2}\normalsize (\sqrt{a^2+2}+a)\\(C)\;\large\frac{1}{2}\normalsize (\sqrt{a^2+4}+a)&(D)\;\text{None of these}\end{array} $

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Let $z=r(\cos\theta+i\sin\theta)$
$\big|z+\large\frac{1}{z}\big|$$=a\Rightarrow \big|z+\large\frac{1}{z}\big|^2$$=a^2$
$\Rightarrow r^2+\large\frac{1}{r^2}$$+2\cos 2\theta=a^2$
Clearly r is maximum for $\theta=\large\frac{\pi}{2}$
$\Rightarrow r=\large\frac{a+\sqrt{a^2+4}}{2}$
Hence (C) is the correct answer.
answered Apr 10, 2014 by sreemathi.v
 

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