Q33 / 100 · BITSAT 2022 · Mathematics

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

  • A
    $\operatorname{Im}(z)=0$
  • B
    $|\operatorname{Im}(z)| \leq 1$
  • C
    $|\operatorname{Re}(z)| \leq 2$
  • Correct answer
    D
    $|z| \leq 3$

Solution

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.

Hence, the correct answer is option: 'D'.

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