Conditions for Rouché's theorem












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For the statement of Rouché's theorem, I've always seen that both $f$ and $g$ have to be holomorphic on and inside a simple closed curve $ C $. However, I am solving a problem which seems to suggest that I should use Rouché's theorem even though I only know that $ f $ is holomorphic in the unit disk $ D $ and continuous in $ bar{D} $. I also check Wikipedia's page on Rouché's theorem which says that $ f $ and $ g $ only need to be holomorphic inside the region, not on the boundary. Is this sufficient?






complex-analysis







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    $begingroup$
    Can you post the question perhaps?
    $endgroup$
    – Mariah
    Jan 5 at 10:54
















4












$begingroup$


For the statement of Rouché's theorem, I've always seen that both $f$ and $g$ have to be holomorphic on and inside a simple closed curve $ C $. However, I am solving a problem which seems to suggest that I should use Rouché's theorem even though I only know that $ f $ is holomorphic in the unit disk $ D $ and continuous in $ bar{D} $. I also check Wikipedia's page on Rouché's theorem which says that $ f $ and $ g $ only need to be holomorphic inside the region, not on the boundary. Is this sufficient?






complex-analysis







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Can you post the question perhaps?
    $endgroup$
    – Mariah
    Jan 5 at 10:54














4












4








4





$begingroup$


For the statement of Rouché's theorem, I've always seen that both $f$ and $g$ have to be holomorphic on and inside a simple closed curve $ C $. However, I am solving a problem which seems to suggest that I should use Rouché's theorem even though I only know that $ f $ is holomorphic in the unit disk $ D $ and continuous in $ bar{D} $. I also check Wikipedia's page on Rouché's theorem which says that $ f $ and $ g $ only need to be holomorphic inside the region, not on the boundary. Is this sufficient?






complex-analysis







share|cite|improve this question











$endgroup$




For the statement of Rouché's theorem, I've always seen that both $f$ and $g$ have to be holomorphic on and inside a simple closed curve $ C $. However, I am solving a problem which seems to suggest that I should use Rouché's theorem even though I only know that $ f $ is holomorphic in the unit disk $ D $ and continuous in $ bar{D} $. I also check Wikipedia's page on Rouché's theorem which says that $ f $ and $ g $ only need to be holomorphic inside the region, not on the boundary. Is this sufficient?






complex-analysis




complex-analysis






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edited Jan 5 at 11:38







calm

















asked Jan 5 at 10:37









calmcalm

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  • 2




    $begingroup$
    Can you post the question perhaps?
    $endgroup$
    – Mariah
    Jan 5 at 10:54














  • 2




    $begingroup$
    Can you post the question perhaps?
    $endgroup$
    – Mariah
    Jan 5 at 10:54








2




2




$begingroup$
Can you post the question perhaps?
$endgroup$
– Mariah
Jan 5 at 10:54




$begingroup$
Can you post the question perhaps?
$endgroup$
– Mariah
Jan 5 at 10:54










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Yes that also looks strange to me. Rouché’s theorem hypothesis is to have a simply connected open subset $Usubseteq mathbb C$ and a compact $K subset U$ whose boundary is a closed simple curve positively oriented.






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    $begingroup$

    Yes, it is sufficient since by continuity the inequality assumption of Rouché's theorem extends to some neighborhood inside the boundary curve, and thus inside the holomorphic domain. In other words, shift the curve along some inside normal vector field a little bit, which is possible because of the compactness of the curve, to get a situation that is conform with the version of the theorem as you know it.






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      2 Answers
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      2 Answers
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      $begingroup$

      Yes that also looks strange to me. Rouché’s theorem hypothesis is to have a simply connected open subset $Usubseteq mathbb C$ and a compact $K subset U$ whose boundary is a closed simple curve positively oriented.






      share|cite|improve this answer











      $endgroup$


















        2












        $begingroup$

        Yes that also looks strange to me. Rouché’s theorem hypothesis is to have a simply connected open subset $Usubseteq mathbb C$ and a compact $K subset U$ whose boundary is a closed simple curve positively oriented.






        share|cite|improve this answer











        $endgroup$
















          2












          2








          2





          $begingroup$

          Yes that also looks strange to me. Rouché’s theorem hypothesis is to have a simply connected open subset $Usubseteq mathbb C$ and a compact $K subset U$ whose boundary is a closed simple curve positively oriented.






          share|cite|improve this answer











          $endgroup$



          Yes that also looks strange to me. Rouché’s theorem hypothesis is to have a simply connected open subset $Usubseteq mathbb C$ and a compact $K subset U$ whose boundary is a closed simple curve positively oriented.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 5 at 17:32

























          answered Jan 5 at 10:47









          mathcounterexamples.netmathcounterexamples.net

          27k22157




          27k22157























              1












              $begingroup$

              Yes, it is sufficient since by continuity the inequality assumption of Rouché's theorem extends to some neighborhood inside the boundary curve, and thus inside the holomorphic domain. In other words, shift the curve along some inside normal vector field a little bit, which is possible because of the compactness of the curve, to get a situation that is conform with the version of the theorem as you know it.






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                Yes, it is sufficient since by continuity the inequality assumption of Rouché's theorem extends to some neighborhood inside the boundary curve, and thus inside the holomorphic domain. In other words, shift the curve along some inside normal vector field a little bit, which is possible because of the compactness of the curve, to get a situation that is conform with the version of the theorem as you know it.






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  Yes, it is sufficient since by continuity the inequality assumption of Rouché's theorem extends to some neighborhood inside the boundary curve, and thus inside the holomorphic domain. In other words, shift the curve along some inside normal vector field a little bit, which is possible because of the compactness of the curve, to get a situation that is conform with the version of the theorem as you know it.






                  share|cite|improve this answer









                  $endgroup$



                  Yes, it is sufficient since by continuity the inequality assumption of Rouché's theorem extends to some neighborhood inside the boundary curve, and thus inside the holomorphic domain. In other words, shift the curve along some inside normal vector field a little bit, which is possible because of the compactness of the curve, to get a situation that is conform with the version of the theorem as you know it.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 5 at 13:40









                  LutzLLutzL

                  59.3k42057




                  59.3k42057






























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