Prove that number of poles in $D_r(w)$ is finite.
$begingroup$
Let $f$ be meromorphic on $Omega$ open and assume that $fnotequiv0$ in any neighborhood of any point in $Omega$. Suppose $overline{D_r(w)}subseteqOmega$, and suppose that $f$ has no poles and never vanishes on the circle $C=partialoverline{D_r(w)}$. Prove that the number of poles inside $D_r(w)$ is finite. Also, prove that there exists $delta>0$ such that $D_{r+delta}(w)subseteqOmega$ and there are no poles or zeros of $f$ inside $D_{r+delta}(w)backslash D_r(w)$.
I know the easiest way to tackle this is to assume there exists some radius $r>0$ such that the number of poles inside $D_r(w)$ is infinite, and use identity principle to show the contradiction, but I'm quite lost. Any help is appreciated.
complex-analysis
asked Dec 1 '18 at 0:57
Ya GYa G
514210
$endgroup$
add a comment |
$begingroup$
Let $f$ be meromorphic on $Omega$ open and assume that $fnotequiv0$ in any neighborhood of any point in $Omega$. Suppose $overline{D_r(w)}subseteqOmega$, and suppose that $f$ has no poles and never vanishes on the circle $C=partialoverline{D_r(w)}$. Prove that the number of poles inside $D_r(w)$ is finite. Also, prove that there exists $delta>0$ such that $D_{r+delta}(w)subseteqOmega$ and there are no poles or zeros of $f$ inside $D_{r+delta}(w)backslash D_r(w)$.
I know the easiest way to tackle this is to assume there exists some radius $r>0$ such that the number of poles inside $D_r(w)$ is infinite, and use identity principle to show the contradiction, but I'm quite lost. Any help is appreciated.
complex-analysis
asked Dec 1 '18 at 0:57
Ya GYa G
514210
$endgroup$
add a comment |
$begingroup$
Let $f$ be meromorphic on $Omega$ open and assume that $fnotequiv0$ in any neighborhood of any point in $Omega$. Suppose $overline{D_r(w)}subseteqOmega$, and suppose that $f$ has no poles and never vanishes on the circle $C=partialoverline{D_r(w)}$. Prove that the number of poles inside $D_r(w)$ is finite. Also, prove that there exists $delta>0$ such that $D_{r+delta}(w)subseteqOmega$ and there are no poles or zeros of $f$ inside $D_{r+delta}(w)backslash D_r(w)$.
I know the easiest way to tackle this is to assume there exists some radius $r>0$ such that the number of poles inside $D_r(w)$ is infinite, and use identity principle to show the contradiction, but I'm quite lost. Any help is appreciated.
complex-analysis
asked Dec 1 '18 at 0:57
Ya GYa G
514210
$endgroup$
Let $f$ be meromorphic on $Omega$ open and assume that $fnotequiv0$ in any neighborhood of any point in $Omega$. Suppose $overline{D_r(w)}subseteqOmega$, and suppose that $f$ has no poles and never vanishes on the circle $C=partialoverline{D_r(w)}$. Prove that the number of poles inside $D_r(w)$ is finite. Also, prove that there exists $delta>0$ such that $D_{r+delta}(w)subseteqOmega$ and there are no poles or zeros of $f$ inside $D_{r+delta}(w)backslash D_r(w)$.
I know the easiest way to tackle this is to assume there exists some radius $r>0$ such that the number of poles inside $D_r(w)$ is infinite, and use identity principle to show the contradiction, but I'm quite lost. Any help is appreciated.
complex-analysis
complex-analysis
asked Dec 1 '18 at 0:57
Ya GYa G
514210
asked Dec 1 '18 at 0:57
Ya GYa G
514210
asked Dec 1 '18 at 0:57
Ya GYa G
514210
asked Dec 1 '18 at 0:57
Ya GYa G
514210
asked Dec 1 '18 at 0:57
Ya GYa G
514210
514210
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
f is meromorphic in D if it is holomorphic in $D-${$z_0,z_1.....$}
where
1){$z_0,z_1.....$} is the sequence without limit point
2)Poles occure at {$z_0,z_1.....$}
I.e this says
Meromorphic function are the function with given domain which is holomorphic everywhere except at isolated point in which poles occurs
So if you have bounded domain then by Weierstrass theorem any infinite set has limit point then contradiction to the meromorphic function definition
answered Dec 1 '18 at 13:37
SRJSRJ
1,6121520
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add a comment |
$begingroup$
Hint: Define $g(z)=frac{1}{f(z)}, $. Suppose that $f$ has infinitely many poles inside your domain. What this will mean for $g$?
answered Dec 1 '18 at 20:04
Beslikas ThanosBeslikas Thanos
758314
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2 Answers
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active
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2 Answers
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active
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$begingroup$
f is meromorphic in D if it is holomorphic in $D-${$z_0,z_1.....$}
where
1){$z_0,z_1.....$} is the sequence without limit point
2)Poles occure at {$z_0,z_1.....$}
I.e this says
Meromorphic function are the function with given domain which is holomorphic everywhere except at isolated point in which poles occurs
So if you have bounded domain then by Weierstrass theorem any infinite set has limit point then contradiction to the meromorphic function definition
answered Dec 1 '18 at 13:37
SRJSRJ
1,6121520
$endgroup$
add a comment |
$begingroup$
f is meromorphic in D if it is holomorphic in $D-${$z_0,z_1.....$}
where
1){$z_0,z_1.....$} is the sequence without limit point
2)Poles occure at {$z_0,z_1.....$}
I.e this says
Meromorphic function are the function with given domain which is holomorphic everywhere except at isolated point in which poles occurs
So if you have bounded domain then by Weierstrass theorem any infinite set has limit point then contradiction to the meromorphic function definition
answered Dec 1 '18 at 13:37
SRJSRJ
1,6121520
$endgroup$
add a comment |
$begingroup$
f is meromorphic in D if it is holomorphic in $D-${$z_0,z_1.....$}
where
1){$z_0,z_1.....$} is the sequence without limit point
2)Poles occure at {$z_0,z_1.....$}
I.e this says
Meromorphic function are the function with given domain which is holomorphic everywhere except at isolated point in which poles occurs
So if you have bounded domain then by Weierstrass theorem any infinite set has limit point then contradiction to the meromorphic function definition
answered Dec 1 '18 at 13:37
SRJSRJ
1,6121520
$endgroup$
f is meromorphic in D if it is holomorphic in $D-${$z_0,z_1.....$}
where
1){$z_0,z_1.....$} is the sequence without limit point
2)Poles occure at {$z_0,z_1.....$}
I.e this says
Meromorphic function are the function with given domain which is holomorphic everywhere except at isolated point in which poles occurs
So if you have bounded domain then by Weierstrass theorem any infinite set has limit point then contradiction to the meromorphic function definition
answered Dec 1 '18 at 13:37
SRJSRJ
1,6121520
answered Dec 1 '18 at 13:37
SRJSRJ
1,6121520
answered Dec 1 '18 at 13:37
SRJSRJ
1,6121520
answered Dec 1 '18 at 13:37
SRJSRJ
1,6121520
1,6121520
add a comment |
add a comment |
$begingroup$
Hint: Define $g(z)=frac{1}{f(z)}, $. Suppose that $f$ has infinitely many poles inside your domain. What this will mean for $g$?
answered Dec 1 '18 at 20:04
Beslikas ThanosBeslikas Thanos
758314
$endgroup$
add a comment |
$begingroup$
Hint: Define $g(z)=frac{1}{f(z)}, $. Suppose that $f$ has infinitely many poles inside your domain. What this will mean for $g$?
answered Dec 1 '18 at 20:04
Beslikas ThanosBeslikas Thanos
758314
$endgroup$
add a comment |
$begingroup$
Hint: Define $g(z)=frac{1}{f(z)}, $. Suppose that $f$ has infinitely many poles inside your domain. What this will mean for $g$?
answered Dec 1 '18 at 20:04
Beslikas ThanosBeslikas Thanos
758314
$endgroup$
Hint: Define $g(z)=frac{1}{f(z)}, $. Suppose that $f$ has infinitely many poles inside your domain. What this will mean for $g$?
answered Dec 1 '18 at 20:04
Beslikas ThanosBeslikas Thanos
758314
answered Dec 1 '18 at 20:04
Beslikas ThanosBeslikas Thanos
758314
answered Dec 1 '18 at 20:04
Beslikas ThanosBeslikas Thanos
758314
answered Dec 1 '18 at 20:04
Beslikas ThanosBeslikas Thanos
758314
758314
add a comment |
add a comment |
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