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Title: conformal mapping problem Post by immanuel78 on Feb 11th, 2009, 6:21am Show that there is an analytic function defined on G=ann(0;0;1) such that f'(prime) never vanishes and f(G)=B(0;1). This problem comes from the textbook by John. Conway-In particular, Riemann mapping theorem section. |
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Title: Re: conformal mapping problem Post by Eigenray on Feb 11th, 2009, 5:17pm That's a good question. If we didn't require f to be locally injective we could simply take an automorphism of the unit disk http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/cdelta.gif that moves 0, and then compose with squaring. By the Riemann mapping theorem, it suffices to show that there is a surjective conformal map from X \ {p} to Y, where X,Y are proper simply connected domains, and p http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/in.gif X. Then you can compose with the biholomorphic maps between X,Y, and http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/cdelta.gif, and an automorphism taking 0 to the appropriate point. In fact we can replace http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/cdelta.gif \ {p} with http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/cdelta.gif \ {p1,...,pn}, where n http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/ge.gif 1. That is, for any p1,...,pn http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/in.gif http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/cdelta.gif, we can find q1,...,qn-1 http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/in.gifhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/cdelta.gif and a surjective conformal map from http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/cdelta.gif \ {q1,...,qn-1} to http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/cdelta.gif\ {p1,...,pn}. To see this, apply an automorphism to take pn = 0, and pick J an interval of length > 2http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif which contains exactly one argument of each of p1,...,pn-1. Then exp takes U = { x + i y | x < 0, y http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/in.gif J } surjectively onto http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/cdelta.gif\ {0}, and we just remove the preimages of p1,...,pn-1. The simplest example I can think of is to take X a large disk around the origin, and f(z) = z2(z-1). We need to remove {0, 2/3} since these are the points where f'(z)=0. But f(1) = f(0) and f(-1/3) = f(2/3), so as long as X is big enough, f(X \ {0,2/3}) = f(X). It remains to show that for X large enough, f(X) is simply connected. Actually I'm not sure how to do this, but it should be true for any polynomial. |
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Title: Re: conformal mapping problem Post by Eigenray on Feb 12th, 2009, 3:20am Maybe this is easier: Let http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/varphi.gif(z) = z+1/z, and for r > 1, let Ar = { z : 1/r < |z| < r }. Then http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/varphi.gif(Ar) is the interior of the ellipse (x/(r+1/r))2 + (y/(r-1/r))2 = 1. In particular, it is simply connected. Now, let F(z) = z3 - 3z. Since F(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/varphi.gif(z)) = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/varphi.gif(z3), F(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/varphi.gif(Ar)) = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/varphi.gif(Ar^3) is also simply connected. F is conformal on http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/varphi.gif(Ar), except at z = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pm.gif 1. But we can remove these points without changing the image because F(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/mp.gif2) = F(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pm.gif1). To sum up: (disk minus a point) http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/cong.gif (half-strip minus a point) http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/twoheadrightarrow.gif (disk minus 2 points) http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/cong.gif (http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/varphi.gif(Ar) \ {http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pm.gif1}) http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/twoheadrightarrow.gif http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/varphi.gif(Ar^3) http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/cong.gif disk, where the http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/twoheadrightarrow.gif maps are exp and F, respectively. |
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Title: Re: conformal mapping problem Post by Michael Dagg on Feb 15th, 2009, 10:09am Clever example! How did you come with it? I have done a similar thing to produce a locally univalent but not univalent holomorphic map of the unit disk onto a convex image. You should be able to see how to do that from what you did here. |
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Title: Re: conformal mapping problem Post by Eigenray on Feb 15th, 2009, 10:49am I had the same idea: try to find a non-injective conformal map between two simply connected regions. The exponential won't work (since the image of a path between z and z + 2http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gifi k has winding number k), so I thought about taking a polynomial and removing the roots of the derivative, since we can add as many holes as we want. A quadratic won't work, so I tried a cubic. I wasn't sure how to show the image was simply connected though so I checked the section of Ahlfors on conformal mappings, where he uses the trick of pulling back via z = w+1/w to reduce a general cubic to w http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/mapsto.gifw3. |
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Title: Re: conformal mapping problem Post by Michael Dagg on Feb 18th, 2009, 8:31am Thanks. For some a, 0 < a < 1 , then z(z^2 - a^2)/(a^2 z^2 - 1) maps D onto D and branches at exactly two points z1,z2 . (D \ {z1,z2}) maps to the unit disk. We can get a conformal map from D onto D by considering a composed universal covering map of the domain. Now just restrict the domain to the annulus. |
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Title: Re: conformal mapping problem Post by Michael Dagg on Mar 9th, 2009, 6:46pm Here are some similar constructions to yours: 1. D.S.Greenstein, Monthly 65 ( 1958 ) , p.214. 2. Due to John V.Ryff (in a totally inaccessible 1969 journal (Math Reviews, vol. 42, review #1981. |
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