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I am trying to construct a nonempty open subset $D$ of the annulus $\{z:1<|z|<2\}$ such that (i) $D$ is connected and so is its boundary, (ii) a holomorphic branch of $\log z$ can be defined on $D$, (iii) $|\log z|$ is unbounded on $D$.

I was thinking that since $\ln|z|$ will be bounded on $D$, I will need $\arg z$ to be unbounded on $D$ to get (iii). To get (ii), I think I just need to be able to define $\arg z$ continuously on $D$. What came to my mind was to have a boundary curve that somehow spirals endlessly inside the annulus but I don't know how to make it precise and also to consider condition (i).

Update: I have arrived at the spiral curve given by $r=1+\frac{1}{\theta-1}$ for $\theta\geq\pi$. This curve is contained in the annulus and spirals around the circle $|z|=1$, getting closer to the circle as it does so. I can take a suitable neighborhood of the curve as $D$. Now what I am missing is the justification that the boundary of $D$ will be connected.

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To guarantee that $\partial D$ is connected, you must require your spiral curve "connecting" the two boundary circles $|z|=1,2$ of the annulus. That is to say, the curve cannot be contained in $r<|z|<R$ for any $1<r<R<2$. So in your example, $\theta\ge\pi$ is not a good choice, and you should replace it with $\theta>2$. –  23rd Nov 11 '12 at 18:20
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I think your basic idea can be realized as follows. Denote $A=\{z\in\mathbb{C}:1<|z|<2\}$. Let $r:\mathbb{R}\to(1,2)$ be any continuous and strictly increasing function with $\lim_{\theta\to-\infty}r(\theta)=1$ and $\lim_{\theta\to+\infty}r(\theta)=2$. Let $C:=\{r(\theta)e^{i\theta}:\theta\in\mathbb{R}\}\subset A$ and let $D=A\setminus C$. Then $D$ is a connected and simply connected open subset of $A$ and $\partial D=C\cup\partial A$ is connected. Therefore, $\log z$ can be well defined on $D$ and every branch of it is unbounded.

Remark: Actually, $D$ is homeomorphic to $A\setminus\mathbb{R}_+$. To see this, note that the map $r$ is a homeomorphism, and denote its inverse by $\Theta:(1,2)\to\mathbb{R}$. Then it is easy to check that $$h:A\setminus\mathbb{R}_+\to D,\quad re^{i\theta}\mapsto re^{i(\theta+\Theta(r))}$$ gives a homeomorphism between $A\setminus\mathbb{R}_+$ and $D$.

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Would you mind explaining why $D$ will be connected and simply connected? And if instead of allowing $\theta\rightarrow -\infty$, I start from a fixed $\theta$ and only allow $\theta\rightarrow\infty$ (see the update to my question), will your argument still work? –  user44532 Nov 11 '12 at 17:12
    
@CYC: my argument works when the range of the $r$ function is unbounded, so in particular, it works with your example. See my update of answer later to see the topological properties of $D$. –  23rd Nov 11 '12 at 17:23
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