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(D) The given inequality is $|z - (1 - i)| \leq 2$. This represents a disk in the Argand plane with center $(1, -1)$ and radius $r = 2$.We check each

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(D) The given inequality is $|z - (1 - i)| \leq 2$. This represents a disk in the Argand plane with center $(1, -1)$ and radius $r = 2$.We check each point $z$ to see if $|z - (1 - i)| \leq 2$ holds:$(A)$ For $z = \frac{1 - i}{2}$,$|z - (1 - i)| = |\frac{1 - i}{2} - 1 + i| = |\frac{1 - i - 2 + 2i}{2}| = |\frac{-1 + i}{2}| = \frac{\sqrt{(-1)^2 + 1^2}}{2} = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}} \approx 0.707 \leq 2$. (Inside)$(B)$ For $z = 1$,$|z - (1 - i)| = |1 - 1 + i| = |i| = 1 \leq 2$. (Inside)$(C)$ For $z = \frac{1 - i}{4}$,$|z - (1 - i)| = |\frac{1 - i}{4} - 1 + i| = |\frac{1 - i - 4 + 4i}{4}| = |\frac{-3 + 3i}{4}| = \frac{3}{4} \sqrt{1^2 + (-1)^2} = \frac{3\sqrt{2}}{4} \approx 1.06 \leq 2$. (Inside)$(D)$ For $z = i$,$|z - (1 - i)| = |i - 1 + i| = |-1 + 2i| = \sqrt{(-1)^2 + 2^2} = \sqrt{1 + 4} = \sqrt{5} \approx 2.236 > 2$. (Outside)Thus,the point $i$ is not in the region.

If $z = x + iy$ represents a point in the Argand plane,then a point which is not in the region represented by $|z - 1 + i| \leq 2$ is

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