Given a meromorphic function, $D_{r+delta}(w)backslash D_r(w)$ has no poles or zeros.
$begingroup$
Given a meromorphic function $f$ on $Omega$ that is not identically zero, and $f$ has no poles and never vanishes on $partial overline{D_r(w)}subseteq Omega$. I'm trying to prove that there exists $delta>0$ such that $D_{r+delta}(w)backslash D_r(w)$ has no poles or zeros.
Given the pieces in the problem, I think I need to use identity principle to prove by contradiction. But I'm not sure how to go about this. Any input is appreciated.
complex-analysis meromorphic-functions
edited Dec 1 '18 at 19:47
José Carlos Santos
154k22123226
asked Dec 1 '18 at 19:13
Ya GYa G
523210
$endgroup$
add a comment |
$begingroup$
Given a meromorphic function $f$ on $Omega$ that is not identically zero, and $f$ has no poles and never vanishes on $partial overline{D_r(w)}subseteq Omega$. I'm trying to prove that there exists $delta>0$ such that $D_{r+delta}(w)backslash D_r(w)$ has no poles or zeros.
Given the pieces in the problem, I think I need to use identity principle to prove by contradiction. But I'm not sure how to go about this. Any input is appreciated.
complex-analysis meromorphic-functions
edited Dec 1 '18 at 19:47
José Carlos Santos
154k22123226
asked Dec 1 '18 at 19:13
Ya GYa G
523210
$endgroup$
add a comment |
$begingroup$
Given a meromorphic function $f$ on $Omega$ that is not identically zero, and $f$ has no poles and never vanishes on $partial overline{D_r(w)}subseteq Omega$. I'm trying to prove that there exists $delta>0$ such that $D_{r+delta}(w)backslash D_r(w)$ has no poles or zeros.
Given the pieces in the problem, I think I need to use identity principle to prove by contradiction. But I'm not sure how to go about this. Any input is appreciated.
complex-analysis meromorphic-functions
edited Dec 1 '18 at 19:47
José Carlos Santos
154k22123226
asked Dec 1 '18 at 19:13
Ya GYa G
523210
$endgroup$
Given a meromorphic function $f$ on $Omega$ that is not identically zero, and $f$ has no poles and never vanishes on $partial overline{D_r(w)}subseteq Omega$. I'm trying to prove that there exists $delta>0$ such that $D_{r+delta}(w)backslash D_r(w)$ has no poles or zeros.
Given the pieces in the problem, I think I need to use identity principle to prove by contradiction. But I'm not sure how to go about this. Any input is appreciated.
complex-analysis meromorphic-functions
complex-analysis meromorphic-functions
edited Dec 1 '18 at 19:47
José Carlos Santos
154k22123226
asked Dec 1 '18 at 19:13
Ya GYa G
523210
edited Dec 1 '18 at 19:47
José Carlos Santos
154k22123226
asked Dec 1 '18 at 19:13
Ya GYa G
523210
edited Dec 1 '18 at 19:47
José Carlos Santos
154k22123226
edited Dec 1 '18 at 19:47
José Carlos Santos
154k22123226
edited Dec 1 '18 at 19:47
José Carlos Santos
154k22123226
154k22123226
asked Dec 1 '18 at 19:13
Ya GYa G
523210
asked Dec 1 '18 at 19:13
Ya GYa G
523210
asked Dec 1 '18 at 19:13
Ya GYa G
523210
523210
add a comment |
add a comment |
1 Answer
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$begingroup$
Suppose that no such $delta$ exists. Then, for each $ninmathbb N$, there is a $z_nin D_{r+frac1n}(w)setminus D_r(w)$ which is either a zero or pole of $f$. Passing to a subsequence, if needed, we can assume without loss of generality that each $z_n$ is a zero or that each $z_n$ is a pole. Since the sequence $(z_n)_{ninmathbb N}$ is bounded, it has a convergente subsequence; let $z_0$ be its limit. Then $z_0inpartialoverline{D_r(w)}$. We can assume, again without loss of generality, that $lim_{ntoinfty}z_n=z_0$. Now:
- if each $z_n$ is a zero of $f$, so is $z_0$, which is impossible, since $f$ has no zeros in $overline{D_r(w)}$;
- if each $z_n$ is a pole of $f$, then, since $f$ is analytic and it has no pole in $overline{D_r(w)}$, there is a neighborhhod of $z_0$ such that $f$ has no poles there, which contradicts the fact that $lim_{ntoinfty}z_n=z_0$ and that each $z_n$ is a pole.
answered Dec 1 '18 at 19:24
José Carlos SantosJosé Carlos Santos
154k22123226
$endgroup$
add a comment |
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1 Answer
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$begingroup$
Suppose that no such $delta$ exists. Then, for each $ninmathbb N$, there is a $z_nin D_{r+frac1n}(w)setminus D_r(w)$ which is either a zero or pole of $f$. Passing to a subsequence, if needed, we can assume without loss of generality that each $z_n$ is a zero or that each $z_n$ is a pole. Since the sequence $(z_n)_{ninmathbb N}$ is bounded, it has a convergente subsequence; let $z_0$ be its limit. Then $z_0inpartialoverline{D_r(w)}$. We can assume, again without loss of generality, that $lim_{ntoinfty}z_n=z_0$. Now:
- if each $z_n$ is a zero of $f$, so is $z_0$, which is impossible, since $f$ has no zeros in $overline{D_r(w)}$;
- if each $z_n$ is a pole of $f$, then, since $f$ is analytic and it has no pole in $overline{D_r(w)}$, there is a neighborhhod of $z_0$ such that $f$ has no poles there, which contradicts the fact that $lim_{ntoinfty}z_n=z_0$ and that each $z_n$ is a pole.
answered Dec 1 '18 at 19:24
José Carlos SantosJosé Carlos Santos
154k22123226
$endgroup$
add a comment |
$begingroup$
Suppose that no such $delta$ exists. Then, for each $ninmathbb N$, there is a $z_nin D_{r+frac1n}(w)setminus D_r(w)$ which is either a zero or pole of $f$. Passing to a subsequence, if needed, we can assume without loss of generality that each $z_n$ is a zero or that each $z_n$ is a pole. Since the sequence $(z_n)_{ninmathbb N}$ is bounded, it has a convergente subsequence; let $z_0$ be its limit. Then $z_0inpartialoverline{D_r(w)}$. We can assume, again without loss of generality, that $lim_{ntoinfty}z_n=z_0$. Now:
- if each $z_n$ is a zero of $f$, so is $z_0$, which is impossible, since $f$ has no zeros in $overline{D_r(w)}$;
- if each $z_n$ is a pole of $f$, then, since $f$ is analytic and it has no pole in $overline{D_r(w)}$, there is a neighborhhod of $z_0$ such that $f$ has no poles there, which contradicts the fact that $lim_{ntoinfty}z_n=z_0$ and that each $z_n$ is a pole.
answered Dec 1 '18 at 19:24
José Carlos SantosJosé Carlos Santos
154k22123226
$endgroup$
add a comment |
$begingroup$
Suppose that no such $delta$ exists. Then, for each $ninmathbb N$, there is a $z_nin D_{r+frac1n}(w)setminus D_r(w)$ which is either a zero or pole of $f$. Passing to a subsequence, if needed, we can assume without loss of generality that each $z_n$ is a zero or that each $z_n$ is a pole. Since the sequence $(z_n)_{ninmathbb N}$ is bounded, it has a convergente subsequence; let $z_0$ be its limit. Then $z_0inpartialoverline{D_r(w)}$. We can assume, again without loss of generality, that $lim_{ntoinfty}z_n=z_0$. Now:
- if each $z_n$ is a zero of $f$, so is $z_0$, which is impossible, since $f$ has no zeros in $overline{D_r(w)}$;
- if each $z_n$ is a pole of $f$, then, since $f$ is analytic and it has no pole in $overline{D_r(w)}$, there is a neighborhhod of $z_0$ such that $f$ has no poles there, which contradicts the fact that $lim_{ntoinfty}z_n=z_0$ and that each $z_n$ is a pole.
answered Dec 1 '18 at 19:24
José Carlos SantosJosé Carlos Santos
154k22123226
$endgroup$
Suppose that no such $delta$ exists. Then, for each $ninmathbb N$, there is a $z_nin D_{r+frac1n}(w)setminus D_r(w)$ which is either a zero or pole of $f$. Passing to a subsequence, if needed, we can assume without loss of generality that each $z_n$ is a zero or that each $z_n$ is a pole. Since the sequence $(z_n)_{ninmathbb N}$ is bounded, it has a convergente subsequence; let $z_0$ be its limit. Then $z_0inpartialoverline{D_r(w)}$. We can assume, again without loss of generality, that $lim_{ntoinfty}z_n=z_0$. Now:
- if each $z_n$ is a zero of $f$, so is $z_0$, which is impossible, since $f$ has no zeros in $overline{D_r(w)}$;
- if each $z_n$ is a pole of $f$, then, since $f$ is analytic and it has no pole in $overline{D_r(w)}$, there is a neighborhhod of $z_0$ such that $f$ has no poles there, which contradicts the fact that $lim_{ntoinfty}z_n=z_0$ and that each $z_n$ is a pole.
answered Dec 1 '18 at 19:24
José Carlos SantosJosé Carlos Santos
154k22123226
answered Dec 1 '18 at 19:24
José Carlos SantosJosé Carlos Santos
154k22123226
answered Dec 1 '18 at 19:24
José Carlos SantosJosé Carlos Santos
154k22123226
answered Dec 1 '18 at 19:24
José Carlos SantosJosé Carlos Santos
154k22123226
154k22123226
add a comment |
add a comment |
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