# Finding a conformal map between annuli

Put $A_1=\{z\,:\, 0<|z|<1\}$ and $A_2=\{z\,:\, r<|z|<R\}$, with $R>r>0$. Find a conformal map from $A_1$ to $A_2$.

This is tricky to me, because $A_1$ has two boundary components. Still though, it seems like we out to do this in one smooth maneuver with a linear fractional transformation. Since the second annulus is symmetrically centered at the origin, I was trying to think of a map sending $\pm\, R\mapsto \pm\, 1$, and $\pm\, r\mapsto 0$. I can't figure it out, though. And maybe you can't do it with a single map, I was just hoping it'd be nice and clean

Thanks

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Hint: there isn't one. Solution here. –  user53153 Jan 15 '13 at 1:35
No such conformal mapping! –  Riemann Jan 15 '13 at 1:40
This question was taken from a qualifying exam. Perhaps there is some mix up? The second part of the problem asks one to show that there is no bijective conformal mapping from $A_1$ to $A_2$. –  Bey Jan 15 '13 at 1:43
Well, if Riemann himself says so... // Hm, it seems that "conformal map from $A_1$ to $A_2$" is not required to be surjective, contrary to my expectation. Well, if it does not have to be surjective, you can use the linear map $f(z)=az+b$ where $a$ is small enough so that $f(A_1)$ can fit inside the domain $A_2$. The shape of $A_1,A_2$ plays no role here (hence, the problem is not very interesting). –  user53153 Jan 15 '13 at 1:55