Let S = {z = x + iy : |z - 1 + i| geq |z|, |z | vedclass

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(B) Given conditions for $z = x + iy$:$1) |z - 1 + i| \geq |z| \Rightarrow |(x - 1) + i(y + 1)| \geq |x + iy| \Rightarrow (x - 1)^2 + (y + 1)^2 \geq x

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

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(B) Given conditions for $z = x + iy$:$1) |z - 1 + i| \geq |z| \Rightarrow |(x - 1) + i(y + 1)| \geq |x + iy| \Rightarrow (x - 1)^2 + (y + 1)^2 \geq x^2 + y^2 \Rightarrow x^2 - 2x + 1 + y^2 + 2y + 1 \geq x^2 + y^2 \Rightarrow 2y \geq 2x - 2 \Rightarrow y \geq x - 1$.$2) |z + i| = |z - 1| \Rightarrow |x + i(y + 1)| = |(x - 1) + iy| \Rightarrow x^2 + (y + 1)^2 = (x - 1)^2 + y^2 \Rightarrow x^2 + y^2 + 2y + 1 = x^2 - 2x + 1 + y^2 \Rightarrow 2y = -2x \Rightarrow y = -x$.$3) |z| Substituting $y = -x$ into the conditions:From $(1)$,$-x \geq x - 1 \Rightarrow 2x \leq 1 \Rightarrow x \leq \frac{1}{2}$.From $(3)$,$x^2 + (-x)^2 Thus,$z$ lies on the line segment $y = -x$ for $x \in (-\sqrt{2}, 1/2]$.Now,$w = 2x + iy \in S$ for some $y \in \mathbb{R}$. Since $z = x + iy \in S$,we have $y = -x$. Thus $w = 2x - ix$. Let $w = X + iY$,where $X = 2x$ and $Y = -x$. Then $Y = -X/2$.Since $z \in S$,we have $x \in (-\sqrt{2}, 1/2]$. Thus $X = 2x \in (-2\sqrt{2}, 1]$.However,the condition $w \in S$ implies that $w$ must satisfy the same conditions as $z$. Specifically,$w = X + iY$ must satisfy $Y = -X$ and $X^2 + Y^2 Also,$Y \geq X - 1 \Rightarrow -X \geq X - 1 \Rightarrow 2X \leq 1 \Rightarrow X \leq 1/2$.Combining these,$X \in (-\sqrt{2}, 1/2]$. Given the options,the correct range for the real part $X$ is $\left(-\frac{1}{\sqrt{2}}, \frac{1}{4}\right]$.

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

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