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I am working on a previous year's complex analysis qual, and I am having a hard time with the area-related problems. Specifically, the following:

Let $f(z)$ be an injective holomorphic function on the punctured disk $D_0=\{z: 0<|z|<1\}$ such that the area of its mapping image $f(D_0)$ is finite. Prove that the length of the image by $f$ of the interval $(0,1/2]$ on the $x$-axis is finite.

To me it seems like if the area of the image is finite, then the area of the image of a subset of a region is obviously finite, but I guess as long as some other subset of the region has area $-\infty$, it could work?

Please help me build intuition or approach a solution on this problem!

qualsqualsquals
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    the area thing is a bit of a red herring in the sense that injectivity is crucial since that implies the function has at most a pole of order $1$ at $0$ (why?) but then if $f(z)=a/z+g(z)$ with $g$ holomorphic in the unit disc, it is obvious that the area of a small punctured disc is infinite under $f$ unless $a=0$ (as $1/z$ sends it to a disc exterior, while $g$ is continuous on the small closed disc so its image is compact) so $a=0$ hence $f$ is analytic in the unit disc, hence continuously differentiable on $[0,1/2]$ and the claim is obvious – Conrad Aug 17 '24 at 04:50
  • @Conrad Concerning "why?": $f$ can have at most one zero, and if we assume the origin is not a removable singularity we gather that on some small open neighbourhood of the origin $g:=1/f$ is analytic and injective. If $f$ has a pole at the origin, $g$ will have a zero of finite order and by the argument from here that order must be $1$. If $f$ has an essential singularity at the origin, then $g$ would have a zero of infinite order and would be constantly zero by the identity theorem, contradicting injectivity. Is that what you had in mind? – FShrike Aug 17 '24 at 09:29
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    @FShrike essential singularity is excluded by Picard - but otherwise is correct - injective on a disc allows only for one simple inside or two simple poles or a double pole on the boundary – Conrad Aug 17 '24 at 13:56
  • @Conrad does the specified interval matter at all here? It doesn't seem like it would have been any different if they asked about, say, [1/2, 1). It seems like they're giving lots of hypotheses that aren't necessarily being used. – qualsqualsquals Aug 20 '24 at 14:52
  • The interval needs to be included in a disc of radius strictly smaller than $1$ so $(1/2,3/4)$ works but $[1/2,1)$ not necessarily; for example think of a map that sends the unit disc to the domain under $1/x^2$ and over the positive axis and to the right of $1$; RMT gives the existence of a conformal isomorphism, the area of the image is finite but one easily can construct such so the segment $(0,1)$ goes to an infinite length curve - of course the infinite length coming from the $(r,1)$ part – Conrad Aug 20 '24 at 15:31

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