# Proving two domains are not conformally equivalent

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.

• Conformality is related with mapping in which angles between curves are conserved. But what is meant by "conformally equivalent" ? Commented Aug 27, 2015 at 18:30
• 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? Commented Aug 27, 2015 at 18:34
• @Salihcyilmaz 'conformally equivalent' means there is a conformal mapping between the two spaces. Commented Aug 27, 2015 at 18:37
• @Eric Auld I don't have a stock of examples, but I should look for this, thanks! Commented Aug 27, 2015 at 18:38
• An observation to get you started: A holomorphic map $f\colon D-\{0\}\to D - [-1/2, 1/2]$ is bounded near $0$. Commented Aug 27, 2015 at 18:41

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 $D \backslash [-1/2,1/2]$.
• 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 $w_0$? Commented Aug 27, 2015 at 23:43
• 1) $D \backslash [-1/2, 1/2]$ is bounded. 2) Argument principle. Commented Aug 28, 2015 at 0:07
• 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. Commented Aug 28, 2015 at 0:59
• 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 $f-w_0$? Commented Aug 28, 2015 at 17:28