Animated Solution for Mathematics - Complex Numbers: Let $A=\{z\in \mathbb{C}:1\le |z-(1+i)|\le 2\}$ and $B=\{z\in A:|z-(1-i)|=1\}$. Then, $B$ :

The set $A$ is defined by the condition $1\le |z-(1+i)|\le 2$. In the complex plane, the expression $|z-z_0|=r$ represents a circle centered at $z_0$ with radius $r$.

Here, the center is $C_1=1+i$, corresponding to the coordinate $(1,1)$. The inequality defines an annulus (a ring-shaped region) centered at $(1,1)$ with an inner radius of $1$ and an outer radius of $2$.

Set $B$ is defined as the intersection of set $A$ with the condition $|z-(1-i)|=1$. This condition represents a circle centered at $C_2=1-i$, which corresponds to the coordinate $(1,-1)$, with a radius of $1$.

To understand the interaction between these two sets, we calculate the distance $d$ between the centers $C_1(1,1)$ and $C_2(1,-1)$:

The distance between the centers is exactly $2$, which is equal to the outer radius of the annulus $A$. This implies that the center $C_2$ lies exactly on the outer boundary of the annulus.

For any point $z$ on the circle $|z-C_2|=1$, the distance from $C_1$ is constrained by the triangle inequality. The distance $|z-C_1|$ must fall within the range:

Set $A$ requires that the distance from $C_1$ must be in the interval $[1,2]$. Consequently, the intersection of the circle and the annulus consists of all points on the circle that satisfy:

Since the range of distances for points on the circle is $[1,3]$, the subset of the circle that falls within the annulus $[1,2]$ forms a continuous arc.

Because this arc contains an infinite number of points, we conclude that set $B$ is an infinite set.