Proof Check of a Complex Analysis Result
$begingroup$
I have attempted to prove this fact:
Let $f:mathbb{C}-mathbb{Z}$ an injective, holomorphic function. Prove that $f$ is a Möbius transformation.
This is my attempt of proof:
I observe that the limits
$$lim_{ztoinfty}{f(z)},qquad lim_{zto m}{f(z)}quad (minmathbb{Z})$$
exist, otherwise, by Casorati-Weierstrass theorem, $f$ would not be injective. Hence $infty$ and the integers must be poles or removable singularities. If they were all removable singularities, then, by Liouville theorem, $f$ would be constant, and not injective. Hence there is at least a pole. But, since $f$ is continuous injective, there is at most one pole, thus there is exactly one pole. We have to cases:
If the pole is at infinity, $f$ is a polynomial, and since it is also injective it must be of the form $f(z)=az+b.$
If the pole is in one $minmathbb{Z},$ we can assume that the pole is at $m=0$ (if not we can translate and come back to this case). We can observe, always because $f$ is injective, that the pole is of order $1.$ We have, in the annulus $0<z<infty,$ the following Laurent expansion of $f(z)$ at $0:$
$$f(z)=sum_{k=-1}^{0}{c_kz^k}=frac{c_{-1}}{z}+c_0=frac{c_0z+c_{-1}}{z}.$$
$$———$$
Is this proof correct, or are there some wrong steps? In this case, why are they wrong, and how can this result be proved?
complex-analysis proof-verification
asked Dec 17 '18 at 0:05
MatPMatP
1367
$endgroup$
add a comment |
$begingroup$
I have attempted to prove this fact:
Let $f:mathbb{C}-mathbb{Z}$ an injective, holomorphic function. Prove that $f$ is a Möbius transformation.
This is my attempt of proof:
I observe that the limits
$$lim_{ztoinfty}{f(z)},qquad lim_{zto m}{f(z)}quad (minmathbb{Z})$$
exist, otherwise, by Casorati-Weierstrass theorem, $f$ would not be injective. Hence $infty$ and the integers must be poles or removable singularities. If they were all removable singularities, then, by Liouville theorem, $f$ would be constant, and not injective. Hence there is at least a pole. But, since $f$ is continuous injective, there is at most one pole, thus there is exactly one pole. We have to cases:
If the pole is at infinity, $f$ is a polynomial, and since it is also injective it must be of the form $f(z)=az+b.$
If the pole is in one $minmathbb{Z},$ we can assume that the pole is at $m=0$ (if not we can translate and come back to this case). We can observe, always because $f$ is injective, that the pole is of order $1.$ We have, in the annulus $0<z<infty,$ the following Laurent expansion of $f(z)$ at $0:$
$$f(z)=sum_{k=-1}^{0}{c_kz^k}=frac{c_{-1}}{z}+c_0=frac{c_0z+c_{-1}}{z}.$$
$$———$$
Is this proof correct, or are there some wrong steps? In this case, why are they wrong, and how can this result be proved?
complex-analysis proof-verification
asked Dec 17 '18 at 0:05
MatPMatP
1367
$endgroup$
$begingroup$
Apriori, infinity is not an isolated singularity as the integers accumulate there; however, applying your argument for the integers (which are isolated for sure), you remove all but at most one and then the proof is correct.
$endgroup$
– Conrad
Dec 17 '18 at 2:22
add a comment |
$begingroup$
I have attempted to prove this fact:
Let $f:mathbb{C}-mathbb{Z}$ an injective, holomorphic function. Prove that $f$ is a Möbius transformation.
This is my attempt of proof:
I observe that the limits
$$lim_{ztoinfty}{f(z)},qquad lim_{zto m}{f(z)}quad (minmathbb{Z})$$
exist, otherwise, by Casorati-Weierstrass theorem, $f$ would not be injective. Hence $infty$ and the integers must be poles or removable singularities. If they were all removable singularities, then, by Liouville theorem, $f$ would be constant, and not injective. Hence there is at least a pole. But, since $f$ is continuous injective, there is at most one pole, thus there is exactly one pole. We have to cases:
If the pole is at infinity, $f$ is a polynomial, and since it is also injective it must be of the form $f(z)=az+b.$
If the pole is in one $minmathbb{Z},$ we can assume that the pole is at $m=0$ (if not we can translate and come back to this case). We can observe, always because $f$ is injective, that the pole is of order $1.$ We have, in the annulus $0<z<infty,$ the following Laurent expansion of $f(z)$ at $0:$
$$f(z)=sum_{k=-1}^{0}{c_kz^k}=frac{c_{-1}}{z}+c_0=frac{c_0z+c_{-1}}{z}.$$
$$———$$
Is this proof correct, or are there some wrong steps? In this case, why are they wrong, and how can this result be proved?
complex-analysis proof-verification
asked Dec 17 '18 at 0:05
MatPMatP
1367
$endgroup$
I have attempted to prove this fact:
Let $f:mathbb{C}-mathbb{Z}$ an injective, holomorphic function. Prove that $f$ is a Möbius transformation.
This is my attempt of proof:
I observe that the limits
$$lim_{ztoinfty}{f(z)},qquad lim_{zto m}{f(z)}quad (minmathbb{Z})$$
exist, otherwise, by Casorati-Weierstrass theorem, $f$ would not be injective. Hence $infty$ and the integers must be poles or removable singularities. If they were all removable singularities, then, by Liouville theorem, $f$ would be constant, and not injective. Hence there is at least a pole. But, since $f$ is continuous injective, there is at most one pole, thus there is exactly one pole. We have to cases:
If the pole is at infinity, $f$ is a polynomial, and since it is also injective it must be of the form $f(z)=az+b.$
If the pole is in one $minmathbb{Z},$ we can assume that the pole is at $m=0$ (if not we can translate and come back to this case). We can observe, always because $f$ is injective, that the pole is of order $1.$ We have, in the annulus $0<z<infty,$ the following Laurent expansion of $f(z)$ at $0:$
$$f(z)=sum_{k=-1}^{0}{c_kz^k}=frac{c_{-1}}{z}+c_0=frac{c_0z+c_{-1}}{z}.$$
$$———$$
Is this proof correct, or are there some wrong steps? In this case, why are they wrong, and how can this result be proved?
complex-analysis proof-verification
complex-analysis proof-verification
asked Dec 17 '18 at 0:05
MatPMatP
1367
asked Dec 17 '18 at 0:05
MatPMatP
1367
asked Dec 17 '18 at 0:05
MatPMatP
1367
asked Dec 17 '18 at 0:05
MatPMatP
1367
asked Dec 17 '18 at 0:05
MatPMatP
1367
1367
$begingroup$
Apriori, infinity is not an isolated singularity as the integers accumulate there; however, applying your argument for the integers (which are isolated for sure), you remove all but at most one and then the proof is correct.
$endgroup$
– Conrad
Dec 17 '18 at 2:22
add a comment |
$begingroup$
Apriori, infinity is not an isolated singularity as the integers accumulate there; however, applying your argument for the integers (which are isolated for sure), you remove all but at most one and then the proof is correct.
$endgroup$
– Conrad
Dec 17 '18 at 2:22
$begingroup$
Apriori, infinity is not an isolated singularity as the integers accumulate there; however, applying your argument for the integers (which are isolated for sure), you remove all but at most one and then the proof is correct.
$endgroup$
– Conrad
Dec 17 '18 at 2:22
$begingroup$
Apriori, infinity is not an isolated singularity as the integers accumulate there; however, applying your argument for the integers (which are isolated for sure), you remove all but at most one and then the proof is correct.
$endgroup$
– Conrad
Dec 17 '18 at 2:22
add a comment |
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$begingroup$
Apriori, infinity is not an isolated singularity as the integers accumulate there; however, applying your argument for the integers (which are isolated for sure), you remove all but at most one and then the proof is correct.
$endgroup$
– Conrad
Dec 17 '18 at 2:22