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Suppose $f(z)$ is meromorphic on $\mathbb C$, and there are two circles $K$, $K'$, so that $f(K)⊂K'$, prove that $f(z)$ is a rational function.

I think $f(z)$ has to extend to $\mathbb{C}∪\{∞\}$ and then uses Liouvile's theorem. But how to prove that $f(z)$ can extend to $\mathbb{C}∪\{∞\}$, by polynomial function? Thank you very much.

HeroZhang001
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Kane
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  • Sketch of how to go about it - use Mobius transforms to send both circles to the unit circle and compose appropriately so we can assume wlog $|f(z)|=1$ when $|z|=1$; show that this implies that $f$ must be analytic either inside or outside the unit disc (this is essentially topological stuff plus open mapping theorem for nonconstant meromorphic functions) and then if $f$ analytic outside use $f(1/z)$ so assume $f$ analytic inside and then extracting the finite zeroes inside the unit disc show $f$ is a finite Blaschke product hence rational – Conrad May 25 '23 at 03:11
  • This question https://math.stackexchange.com/q/3408/42969 is about entire functions, but this answer https://math.stackexchange.com/a/3436/42969 applies to meromorphic functions as well: “If a function $f$ is holomorphic in a neighborhood of the closed disk and has modulus $1$ on the circle, then $f$ is a finite Blaschke product.” – Therefore I would suggest to close this as a duplicate. – Martin R May 31 '23 at 06:59

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