Prove that $A_{r,s}$ $=[z\in \mathbb C : r<|z-z_0|

Question

Prove that $A_{r,s}$ $=[z\in \mathbb C : r<|z-z_0|<s ]$ is path connected.

I proved that $D_r(z_0)$ is pathh connected. But not this set of complex numbers.

Answer 1

If you have $r_1e^{i\theta_1}$ and $r_2e^{i\theta_2}$ in $A_{r,s}$, a path connecting these points can be the concatenation of the paths $\gamma(t)=r_1e^{i(\theta_1+t(\theta_2-\theta_1))}$ and $\lambda(t)=(r_1+t(r_2-r_1))e^{i\theta_2}$.

Answer 2

The comments seem to be helping you along quite swimmingly, so I have opted to draw an extremely crude picture for you!

The black circle is the disc with radius $r$ and the red circle is the disc with radius $s$. I have connected the two blue dots but first rotating the blue point so that it if we were to draw a straight line from the origin to the farther point, then the rotation would set the blue point at this line, then I have translated the point along this line until I reached the desired point.

Can you take it from here and turns this into actual mathematics?

Answer 3

There are two ways to see this. Take any two points $A$ and $B$ in the annulus $A_{r,s}$.

General approach: Take circles between $A$ and $B$ so that in each circle we have a path, since a circle is convex. Then take union of all the paths and we have a path between $A$ and $B$.

Specific approach: because of the special shape, there is another easy way to find the path between $A$ and $B$. We can take part of the circle with radius $|B-z_0|$ and then draw a straight line that connects the contour and $A$.