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
asked Jan 5 at 10:37
calmcalm
1387
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add a comment |
$begingroup$
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
asked Jan 5 at 10:37
calmcalm
1387
$endgroup$
2
$begingroup$
Can you post the question perhaps?
$endgroup$
– Mariah
Jan 5 at 10:54
add a comment |
$begingroup$
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
asked Jan 5 at 10:37
calmcalm
1387
$endgroup$
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
asked Jan 5 at 10:37
calmcalm
1387
asked Jan 5 at 10:37
calmcalm
1387
edited Jan 5 at 11:38
calm
asked Jan 5 at 10:37
calmcalm
1387
asked Jan 5 at 10:37
calmcalm
1387
asked Jan 5 at 10:37
calmcalm
1387
1387
2
$begingroup$
Can you post the question perhaps?
$endgroup$
– Mariah
Jan 5 at 10:54
add a comment |
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
add a comment |
2 Answers
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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.
answered Jan 5 at 10:47
mathcounterexamples.netmathcounterexamples.net
27k22157
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add a comment |
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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.
answered Jan 5 at 13:40
LutzLLutzL
59.3k42057
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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.
answered Jan 5 at 10:47
mathcounterexamples.netmathcounterexamples.net
27k22157
$endgroup$
add a comment |
$begingroup$
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.
answered Jan 5 at 10:47
mathcounterexamples.netmathcounterexamples.net
27k22157
$endgroup$
add a comment |
$begingroup$
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.
answered Jan 5 at 10:47
mathcounterexamples.netmathcounterexamples.net
27k22157
$endgroup$
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.
answered Jan 5 at 10:47
mathcounterexamples.netmathcounterexamples.net
27k22157
edited Jan 5 at 17:32
answered Jan 5 at 10:47
mathcounterexamples.netmathcounterexamples.net
27k22157
answered Jan 5 at 10:47
mathcounterexamples.netmathcounterexamples.net
27k22157
answered Jan 5 at 10:47
mathcounterexamples.netmathcounterexamples.net
27k22157
27k22157
add a comment |
add a comment |
$begingroup$
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.
answered Jan 5 at 13:40
LutzLLutzL
59.3k42057
$endgroup$
add a comment |
$begingroup$
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.
answered Jan 5 at 13:40
LutzLLutzL
59.3k42057
$endgroup$
add a comment |
$begingroup$
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.
answered Jan 5 at 13:40
LutzLLutzL
59.3k42057
$endgroup$
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.
answered Jan 5 at 13:40
LutzLLutzL
59.3k42057
answered Jan 5 at 13:40
LutzLLutzL
59.3k42057
answered Jan 5 at 13:40
LutzLLutzL
59.3k42057
answered Jan 5 at 13:40
LutzLLutzL
59.3k42057
59.3k42057
add a comment |
add a comment |
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Can you post the question perhaps?
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– Mariah
Jan 5 at 10:54