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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.

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    $\begingroup$ Welcome to MSE bond =) just a tip for asking questions in the future : Try not asking questions with an imperative attitude, you are not a math manual and we are not a solution manual. Being more friendly makes it more human. Also, try explaining why you're having trouble with this question : if you write what you've tried we're more willing to help. $\endgroup$ – Patrick Da Silva Nov 22 '11 at 11:31
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    $\begingroup$ Grammar (such as use of the imperative mood) is irrelevant. The relevant criticism is that the question consists of the text of an exercise without any thoughts of the asker's own added. That's not good, but it is independent of what grammatical mood is used. Rewriting the question to use different grammar without adding any independent work or thoughts wouldn't solve anything. $\endgroup$ – Henning Makholm Nov 22 '11 at 15:03
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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.

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  • $\begingroup$ Now, the question is really more about topology than complex analysis, but since it's tagged as such, I am guessing it's from a textbook on CA. Hence a CA-themed answer. $\endgroup$ – Arthur Nov 22 '11 at 11:54
  • $\begingroup$ I haven't done enough complex analysis yet to think about that kind of proof... That is one nice way to do it, find a "pole" that is not in the domain. =) Good job $\endgroup$ – Patrick Da Silva Nov 22 '11 at 20:06
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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,

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