Let A = {z in mathbb{C} : 1 leq |z - (1 + i)| | vedclass

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(D) The set $A$ represents an annulus (a ring-shaped region) in the complex plane centered at $z_0 = 1 + i$ with inner radius $r_1 = 1$ and outer radi

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an infinite set

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(D) The set $A$ represents an annulus (a ring-shaped region) in the complex plane centered at $z_0 = 1 + i$ with inner radius $r_1 = 1$ and outer radius $r_2 = 2$.The set $B$ consists of points $z$ that lie within this annulus $A$ and also satisfy the equation $|z - (1 - i)| = 1$. This equation represents a circle centered at $z_1 = 1 - i$ with radius $r = 1$.Let us check the distance between the centers $z_0 = 1 + i$ and $z_1 = 1 - i$:$|z_0 - z_1| = |(1 + i) - (1 - i)| = |2i| = 2$.The circle defining $B$ has radius $1$. The points on this circle are at a distance of $1$ from the center $(1, -1)$.Consider the point $z = 1$. For $z = 1$,$|z - (1 + i)| = |1 - 1 - i| = |-i| = 1$,so $z = 1$ is on the inner boundary of $A$.Also,$|z - (1 - i)| = |1 - 1 + i| = |i| = 1$,so $z = 1$ is on the circle defining $B$.Thus,$z = 1$ is in $B$.Consider the point $z = 1 + 2i$. For $z = 1 + 2i$,$|z - (1 + i)| = |1 + 2i - 1 - i| = |i| = 1$,so $z = 1 + 2i$ is on the inner boundary of $A$.Also,$|z - (1 - i)| = |1 + 2i - 1 + i| = |3i| = 3 
eq 1$.Since the circle $|z - (1 - i)| = 1$ passes through the point $(1, 0)$ and $(1, -2)$ and $(0, -1)$ and $(2, -1)$,we can see that the circle intersects the region $A$ at more than one point. Specifically,the arc of the circle $|z - (1 - i)| = 1$ lies within the region $A$. Therefore,$B$ contains infinitely many points.

Let $A = \{z \in \mathbb{C} : 1 \leq |z - (1 + i)| \leq 2\}$ and $B = \{z \in A : |z - (1 - i)| = 1\}$. Then,$B$ is:

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