Prove that this domain isn't simply connected Let $Q_r = \{ z=x+iy : x,y \in (-r,r) \} \subseteq \mathbb C$
Show that $Q_4 \backslash Q_1$ isn't simply connected domain.
 A: The circle $\gamma$ with radius $2$ and center at $0$ is completely within the domain, and the function $f(z) = \frac{1}{z}$ is holomorphic there. $\int_\gamma f(z) dz\neq 0$, so $\gamma$ cannot be continuously deformed to a point. Thus the domain isn't simply connected.
A: I don't know to what extent you are rigorous about this, but if you are in a calculus course usually the notion of simply connected goes with intuition, i.e. is defined like this : any simple closed curve within the domain can be shrunk continuously to a point by remaining inside the domain during the shrinking. Now if you consider the curve that stays in $Q_4$ but goes around $Q_1$, for instance the curve 
$$
C = \{ (2 \cos \theta, 2 \sin \theta) \, | \, \theta \in [0, 2\pi] \}
$$
(You can readily see that this curve is simple and closed (it's a circle), it is in $Q_4$ since $|2\cos \theta| < 4$ and $|2 \sin \theta| < 4$ but is not in $Q_1$ since we cannot have $|\cos \theta| \le \frac 12$ and $|\sin \theta| \le \frac 12$ (I'm just stating things that are obvious geometrical facts. I basically just took an explicit example but if you make the drawing of $Q_4 \backslash Q_1$ you're convinced ; it's a square of side length 8 with a square of side length 2 missing in the middle.)
If you are not happy with this kind-of drawing proof, then notice this : every simple continuous closed curve that goes around $Q_1$ has arc length greater than $4 \cdots 2$, because that is the "shortest" curve you can get inside $Q_4 \backslash Q_1$. Since to shrink a curve like $C$ to a point you would need to keep it around $Q_1$ and make its length go to $0$, it's impossible to do such a thing. 
I must admit I am not quite happy with this argument though... too "sketchy". If anyone knows how to make it right I'd be pleased.
Hope that helps,
