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If $|w|=2$, then the set of points $z=w-\frac{1}{w}$ is contained in or equal to the set of points $z$ satisfying
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Verified Answer
The correct answer is:
$|z| \leq 3$
$|z| \leq 3$
We have, $|w|=2 \Rightarrow w=2(\cos \theta+i \sin \theta)$
$\therefore \quad z=w-\frac{1}{w}$
$=2(\cos \theta+i \sin \theta)-\frac{1}{2}(\cos \theta-i \sin \theta)$
$\Rightarrow \quad z=\frac{3}{2} \cos \theta+\frac{5}{2} i \sin \theta$
$\Rightarrow \quad|z|=\sqrt{\frac{9}{4}+\frac{25}{4}}=\frac{\sqrt{17}}{2} < 3$
Hence, option (d) is correct.
$\therefore \quad z=w-\frac{1}{w}$
$=2(\cos \theta+i \sin \theta)-\frac{1}{2}(\cos \theta-i \sin \theta)$
$\Rightarrow \quad z=\frac{3}{2} \cos \theta+\frac{5}{2} i \sin \theta$
$\Rightarrow \quad|z|=\sqrt{\frac{9}{4}+\frac{25}{4}}=\frac{\sqrt{17}}{2} < 3$
Hence, option (d) is correct.
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