Given a meromorphic function, $D_{r+delta}(w)backslash D_r(w)$ has no poles or zeros.












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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







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    $begingroup$


    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







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      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







      share|cite|improve this question











      $endgroup$




      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






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      share|cite|improve this question













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      edited Dec 1 '18 at 19:47









      José Carlos Santos

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      asked Dec 1 '18 at 19:13









      Ya GYa G

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          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.






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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.






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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.






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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.






                share|cite|improve this answer









                $endgroup$



                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.







                share|cite|improve this answer












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                share|cite|improve this answer










                answered Dec 1 '18 at 19:24









                José Carlos SantosJosé Carlos Santos

                154k22123226




                154k22123226






























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