Prove region is not simply connected Let $A=\{z=x+iy\in C | 1\lt |z-z_0 | \lt 2 \}$ . ($z_0$ is any point on the plane).
Prove that $A$ is not a simply connected region in $C$.
How can I prove this?  The only idea that comes to my mind is to compute an integral over a closed curve contained in $A$ ,supposing $A$ was simply connected, and then use Cauchy's theorem to get to a contradiction. 
Any help will be appreciatted.

A region $A \subseteq \mathbb{C}$ is simply connected (with respect to the standard topology induced by the Euclidean distance) if and only if $A^c = \overline{\mathbb{C}} \backslash A$ is connected, see for example Ahlfors' book "Complex Analysis", definition 1 on page 139. In your case, you have $A^c = \{|z - z_0| \leq 1\} \cup \{|z - z_0| \geq 2\}$, and this set is obviously not connected, since it consists of two separate point sets. Hence, $A$ is not simply connected. In fact, $A$ is doubly connected, since its complement $A^c$ consists of two components.
