proving that $\{z\in \mathbb{C}: 1<|z-1|<2\}$ is polygonally connected

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I have the following set : $$ A=\{z \in \mathbb{C} \mid 1 < |z-1| < 2\}.$$ It's obvious that any two points in A can be connected by at least two line segments .

Note :(I am also Considering the case that if 2 points can be connected by one segment I will divide that segment also to two )

So, $$\forall z_{1},z_{2} \in \mathbb{C}$$ It is enough to find z such that : $$ |z-z_{1}|=|z-z_{2}| \ , [z_{1},z]\cup[z,z_{2}]\subseteq A $$ Is there any other idea to prove that A is polygonally connected algebraically ?and if f is continuous and B connected then f(B) is connected .

How Can I find such f and B ?such that f(B)=A

I found out that z must be $$ z=x+i\frac{2xa-a^2-b^2+c^2+d^2}{2d-2c}$$ where $$z_{1}=(a,b) , z_{2}=(c,d) , x \in \mathbb{(-2,-1)\cup(2,3)}$$

Thank you .

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Let me describe a geometric process using words. Perhaps you should read this with pencil and paper drawing things as I describe.

Let A,B be the two points in the annulus. Draw a circle concentric to the boundary of the annuli and passing though B.

Now draw any line segment from A to this newly drawn circle intersecting it at B1. Now B and B1 are in the same circle. Now a sequence of chords of sufficiently short length (so that it won't intersect the inner boundary of the annulus) can take one from B1 to B; so the initial line segment from A to B1 followed by this sequence of chords is a polygonal path lying completely inside the annulus.