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Q.

Let $S = {z = x + iy : | z - 1 + i | ⩾ | z |, | z | < 2, | z + i | = | z - 1 |}$ . Then the set of all values of x, for which $w = 2 x + iy \in S$ for some $y \in ℝ$ , is

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a

$(- \sqrt{2}, \frac{1}{2 \sqrt{2}}]$

b

$(- \frac{1}{\sqrt{2}}, \frac{1}{4}]$

c

$(\sqrt{2}, \frac{1}{2}]$

d

$(- \frac{1}{\sqrt{2}}, \frac{1}{2 \sqrt{2}}]$

answer is B.

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Detailed Solution

$| z - 1 + i | ⩾ | z |; | z | < 2; | z + i | = | z - 1 |$
Question Image

Hence
$\begin{array}{l} w = 2 x + iy \in S \\ 2 x ⩽ \frac{1}{2} ∴ x ⩽ \frac{1}{4} \end{array}$
Now
$\begin{array}{l} (2 x)^{2} + (2 x)^{2} < 4 \\ x^{2} < \frac{1}{2} \Rightarrow x \in (\frac{- 1}{\sqrt{2}}, \frac{1}{\sqrt{2}}) \\ ∴ x \in (\frac{- 1}{\sqrt{2}}, \frac{1}{4}] \end{array}$

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Let S={z=x+iy:|z−1+i|⩾|z|,|z|<2,|z+i|=|z−1|}. Then the set of all values of x, for which w=2x+iy∈S for some y∈ℝ, is