Let $D$ be the open unit disk. Show that $D - [-1/2, 1/2]$ and $D - \{0\}$ are not conformally equivalent.
Thoughts so far: I'm not sure where to begin, but a hint would most helpful to get me started.
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$\begingroup$ Conformality is related with mapping in which angles between curves are conserved. But what is meant by "conformally equivalent" ? $\endgroup$– SalihcyilmazCommented Aug 27, 2015 at 18:30
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1$\begingroup$ Do you have a stock of examples of things that are not conformally equivalent, so you can try to morph your situation into one of those by biholomorphic transformations? $\endgroup$– Eric AuldCommented Aug 27, 2015 at 18:34
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2$\begingroup$ @Salihcyilmaz 'conformally equivalent' means there is a conformal mapping between the two spaces. $\endgroup$– user19817Commented Aug 27, 2015 at 18:37
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$\begingroup$ @Eric Auld I don't have a stock of examples, but I should look for this, thanks! $\endgroup$– user19817Commented Aug 27, 2015 at 18:38
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1$\begingroup$ An observation to get you started: A holomorphic map $f\colon D-\{0\}\to D - [-1/2, 1/2]$ is bounded near $0$. $\endgroup$– froggieCommented Aug 27, 2015 at 18:41
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1 Answer
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Any bounded one-to-one analytic function $\phi$ on $ D \backslash \{0\}$ has a removable singularity at $0$. Thus there is $\widetilde{\phi}$ analytic on $D$ whose restriction to $D \backslash \{0\}$ is $\phi$. Now $\widetilde{\phi}$ must also be one-to-one, because if $\widetilde{\phi}(0) = w_0 \in \phi(D)$ there would be at least two members of $\phi^{-1}(w)$ for all $w$ in a deleted neighbourhood of $w_0$. Moreover, $\widetilde{\phi}(D) = \phi(D) \cup \{\widetilde{\phi}(0)\}$ is open. So any bounded domain $\tilde{D}$ conformally equivalent to $D \backslash \{0\}$ has the property that $\tilde{D} \cup \{w\}$ is open for some $w \notin \tilde{D}$. This is not the case for $D \backslash [-1/2,1/2]$.
answered Aug 27, 2015 at 18:41
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$\begingroup$ Thank you for the post! A couple questions: why does the one-to-one function have to be bounded? And why would there be at least two member of $\phi^{-1}(w)$ for all $w$ in a deleted neighbourhood of $w_0$? $\endgroup$ Commented Aug 27, 2015 at 23:43
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1$\begingroup$ 1) $D \backslash [-1/2, 1/2]$ is bounded. 2) Argument principle. $\endgroup$ Commented Aug 28, 2015 at 0:07
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$\begingroup$ I'm not sure why the argument principle gives this as a consequence, could you give me a hint? $\endgroup$ Commented Aug 28, 2015 at 0:46
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$\begingroup$ The number of zeros of $f(z) - w$ (counted by multiplicity) inside a simple positively oriented closed contour can be expressed as $(2\pi i)^{-1}$ times the integral of $f'/(f-w)$ over the contour, and is constant as long as none of the zeros hits the contour. $\endgroup$ Commented Aug 28, 2015 at 0:59
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$\begingroup$ Do you mean that the integral is constant as $w$ varies? So the integral will be the same weather the denominator is $f-w$ or $f-w_0$? $\endgroup$ Commented Aug 28, 2015 at 17:28