1. Interpret the sets geometrically

Let $A=\{z\in \mathbb{C}: 1\le |z-(1+i)|\le 2\}.$ This is the closed annulus centered at the point $(1,1)$ with inner radius $1$ and outer radius $2$.

Now $B=\{z\in A: |z-(1-i)|=1\}.$ So $B$ consists of those points of the annulus $A$ which also lie on the circle centered at $(1,-1)$ with radius $1$.

Thus, we need to find how this circle intersects the annulus.

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2. Parametrize the circle defining $B$

The circle $|z-(1-i)|=1$ has center $C_2=(1,-1)$ and radius $1$.

Take any point $P$ on this circle. We want to know whether it satisfies $1\le |P-(1+i)|\le 2.$

The point $(1+i)$ is $C_1=(1,1)$.

Distance between the centers: $C_1C_2=|(1+i)-(1-i)|=|2i|=2.$

So we have two circles:

• circle 1: center $C_1=(1,1)$
• circle 2: center $C_2=(1,-1)$, radius $1$
• distance between centers = $2$

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3. Find possible values of $|P-C_1|$ for points $P$ on the circle centered at $C_2$

Since $P$ lies on the circle centered at $C_2$ of radius $1$, by triangle inequality, $|C_1C_2|-|PC_2| \le |PC_1| \le |C_1C_2|+|PC_2|.$ That is, $2-1 \le |P-C_1| \le 2+1,$ so $1\le |P-(1+i)| \le 3.$

But for membership in $A$, we need $1\le |P-(1+i)| \le 2.$

So among points on the circle $|z-(1-i)|=1$, we must keep only those for which distance from $(1,1)$ is at most $2$.

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4. Check whether there are finitely many or infinitely many such points

The circle centered at $(1,-1)$ with radius $1$ is not entirely inside the disk $|z-(1+i)|\le 2,$ because its farthest point from $(1,1)$ has distance $3$.

It is also not entirely outside, because its nearest point from $(1,1)$ has distance $1$.

Hence the circle passes through the disk $|z-(1+i)|\le 2$ in an arc, not just isolated points.

More precisely:

• the condition $|z-(1-i)|=1$ gives a whole circle,
• the additional condition $1\le |z-(1+i)|\le 2$ selects a nonzero arc of that circle.

A nonzero arc contains infinitely many points.

Therefore, $B$ is an infinite set.

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5. Optional algebraic verification

Let $z=x+iy.$ From $|z-(1-i)|=1$ we get $(x-1)^2+(y+1)^2=1.$ Now check points on this circle that also satisfy $1\le \sqrt{(x-1)^2+(y-1)^2}\le 2.$ Squaring: $1\le (x-1)^2+(y-1)^2\le 4.$

Using $(x-1)^2+(y+1)^2=1,$ we get

\implies (x-1)^2+y^2+2y=0.$$ Then $$(x-1)^2+(y-1)^2=(x-1)^2+y^2-2y+1.

Since $(x-1)^2+y^2=-2y$, this becomes $-2y-2y+1=1-4y.$ So the annulus condition becomes

$$$$

Thus

\implies -\frac34 \le y \le 0.$$ On the circle $$(x-1)^2+(y+1)^2=1,$$ there are infinitely many points with $y\in\left[-\frac34,0\right]$ (indeed an arc of the circle). Hence $B$ is infinite. --- 6. **Conclusion** Therefore, the correct option is $$\boxed{\text{D: } B \text{ is an infinite set}}.$$