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Math Help - Opening mapping theorem in complex analysis

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    Opening mapping theorem in complex analysis

    There is a theorem in my book that says:

    If a function f is analytic on a connected open set U, then f is either a constant function or an opening mapping.

    I read the proof in wikipedia, but I really don't understand it. The book provided a proof that look easier, but I got really lost.

    What is the idea of the proof? Why is this true?

    Thank you.
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    Quote Originally Posted by tttcomrader View Post
    There is a theorem in my book that says:

    If a function f is analytic on a connected open set U, then f is either a constant function or an opening mapping.

    I read the proof in wikipedia, but I really don't understand it. The book provided a proof that look easier, but I got really lost.

    What is the idea of the proof? Why is this true?

    Thank you.
    Requirements] Here is a proof that I know and like. You need to know the minimum-modulos theorem. The maximum modulos theorem says (let us just say for disks for simplicity) that if f is a continous complex function on a closed disk that is analytic on the interior of the disk then f assumes it maximum on the boundary of the disk. The minimum-modulos theorem is similar, it says f (under the same assumptions) assumes it minimum value on the boundary provided that f is non-zero at any point inside the disk. You also need to know the identity-theorem, if f (under the same assumptions) has a sequence of distinct points \{z_n\} inside the disk that are zeros of f (i.e. f(z_n) = 0) and \{z_n\} converges to some point inside the disk then f must be the zero function on the disk.

    Basic Idea] Let f be an analytic function on an open set U. Remember we want to show that f(U) is open, this means for any point in f(U) we can find an open disk containing that point that lies wholly in f(U) (that is just the definition of being 'open'). Let \beta \in f(U), be some point. Therefore, \beta = f(\alpha) for some \alpha \in U. We need to show there is some disk containing \beta=f(\alpha) that lies in f(U). By using translations we can safely assume that f(\alpha) = 0 (why?). What remains now is that we have to show there is an open disk containing 0 that lies wholly in the image of f (i.e. f(U)).

    Special Circle] By the identity-theorem there must be a circle around \alpha (which is contained in U) so that f does not vanish on this circle. To see why this must be true, assume to the contrary that for any circle you draw around \alpha (that lies in U) there is always some point for which f vanishes. We know for some \tfrac{1}{r}>0 we can draw a circle around \alpha that is contained in U. On this circle (as we are assuming) there is a point z_0 so that f(z_0). Now draw a smaller circle around that point of radius \tfrac{1}{r+1}>0 and again by assumption there is a point on this circle z_1 so that f(z_1)=0. Now draw an even smaller circle around that point of radius \tfrac{1}{r+2}>0 and again there is a z_2 on the circle so that f(z_2) = 0. In general we can create z_n by drawing a circle of radius \tfrac{1}{r+n}>0 around \alpha. Thus, we have a sequence of point \{z_n\}. The limit of these points is \alpha (because they shrink down to \alpha in this limiting process). Now look at the identity theorem. We found a convergent sequence of distinct zeros that converge to a point ( \alpha) which is inside U and so it forces f to be the zero function on U (since U is an open connected set, I realized I stated the identity theorem for disks above but it works for open connected sets as well). But f is not constant. Therefore, f must have a circle, call it C, on which it does not vanish.

    Minimum] We know that |f| attains a mimimum on C (it is a compact set and continous functions on compact sets attain minimums and maximums). Let 2m = \min \{ |f(z)| : z \in C \}>0. We will show that the disk D around 0 of radius m lies in the image of f on the disk bounded by C around \alpha (and so definitely an open disk around 0 is contained in f(U)). Let \xi \in D and if z\in C then |f(z) - \xi| \geq ||f(z)|-|\xi|| \geq |f(z)| - |\xi| \geq 2m - m = m. However, |f(\alpha)  - \xi| = |-\xi| < m, so |f(\alpha) - \xi| < |f(z) - \xi|. Therefore, on the closed disk around \alpha with boundary C we see that |f(z) - \xi| does not attain a minimum on the boundary, and so by minimum-modulos theorem it means |f(z^*) - \xi| = 0 for some z^* inside this disk, and so \xi = f(z^*). Thus, we have proven that any point in D is the image of some point in that disk bounded by C and centered at \alpha. Thus, \xi \in f(U). Thus, f(U) is open.
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