Relating a limit to to the concrete case (complex integration)
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I have shown that when $Rrightarrowinfty$, it holds that $int_{C_R}frac{1}{P(z)}dz =0$ for any (complex) polynomial with degree $ngeq2$, where $C_R$ is the circle centered at $0$ with radius R. However, I would like to prove that there exists an $r>0$ such that, when $R>r$, it follows that $int_{C_R}frac{1}{P(z)}dz =0$.
I cannot see any justification for this claim, based on the limit that I have shown to be true. Where can I draw the conclusion from?
complex-analysis limits
asked Dec 10 '18 at 15:26
user569579user569579
755
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add a comment |
$begingroup$
I have shown that when $Rrightarrowinfty$, it holds that $int_{C_R}frac{1}{P(z)}dz =0$ for any (complex) polynomial with degree $ngeq2$, where $C_R$ is the circle centered at $0$ with radius R. However, I would like to prove that there exists an $r>0$ such that, when $R>r$, it follows that $int_{C_R}frac{1}{P(z)}dz =0$.
I cannot see any justification for this claim, based on the limit that I have shown to be true. Where can I draw the conclusion from?
complex-analysis limits
asked Dec 10 '18 at 15:26
user569579user569579
755
$endgroup$
add a comment |
$begingroup$
I have shown that when $Rrightarrowinfty$, it holds that $int_{C_R}frac{1}{P(z)}dz =0$ for any (complex) polynomial with degree $ngeq2$, where $C_R$ is the circle centered at $0$ with radius R. However, I would like to prove that there exists an $r>0$ such that, when $R>r$, it follows that $int_{C_R}frac{1}{P(z)}dz =0$.
I cannot see any justification for this claim, based on the limit that I have shown to be true. Where can I draw the conclusion from?
complex-analysis limits
asked Dec 10 '18 at 15:26
user569579user569579
755
$endgroup$
I have shown that when $Rrightarrowinfty$, it holds that $int_{C_R}frac{1}{P(z)}dz =0$ for any (complex) polynomial with degree $ngeq2$, where $C_R$ is the circle centered at $0$ with radius R. However, I would like to prove that there exists an $r>0$ such that, when $R>r$, it follows that $int_{C_R}frac{1}{P(z)}dz =0$.
I cannot see any justification for this claim, based on the limit that I have shown to be true. Where can I draw the conclusion from?
complex-analysis limits
complex-analysis limits
asked Dec 10 '18 at 15:26
user569579user569579
755
asked Dec 10 '18 at 15:26
user569579user569579
755
asked Dec 10 '18 at 15:26
user569579user569579
755
asked Dec 10 '18 at 15:26
user569579user569579
755
asked Dec 10 '18 at 15:26
user569579user569579
755
755
add a comment |
add a comment |
1 Answer
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Use the residue theorem. Your integral must be the same for all $R$ large enough to contain each zero of $P(z).$ As such, let $r$ be the distance to the furthest zero. Since $lim_{R to infty} int_{C_R}frac{1}{P(z)} = 0,$ and $int_{C_{R_1}}frac{1}{P(z)} = int_{C_{R_2}}frac{1}{P(z)}$ for all $R_1, R_2 > r,$ we conclude $int_{C_{R}}frac{1}{P(z)} = 0$ for all $R > r.$
answered Dec 11 '18 at 11:16
Brevan EllefsenBrevan Ellefsen
11.7k31649
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$begingroup$
Use the residue theorem. Your integral must be the same for all $R$ large enough to contain each zero of $P(z).$ As such, let $r$ be the distance to the furthest zero. Since $lim_{R to infty} int_{C_R}frac{1}{P(z)} = 0,$ and $int_{C_{R_1}}frac{1}{P(z)} = int_{C_{R_2}}frac{1}{P(z)}$ for all $R_1, R_2 > r,$ we conclude $int_{C_{R}}frac{1}{P(z)} = 0$ for all $R > r.$
answered Dec 11 '18 at 11:16
Brevan EllefsenBrevan Ellefsen
11.7k31649
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add a comment |
$begingroup$
Use the residue theorem. Your integral must be the same for all $R$ large enough to contain each zero of $P(z).$ As such, let $r$ be the distance to the furthest zero. Since $lim_{R to infty} int_{C_R}frac{1}{P(z)} = 0,$ and $int_{C_{R_1}}frac{1}{P(z)} = int_{C_{R_2}}frac{1}{P(z)}$ for all $R_1, R_2 > r,$ we conclude $int_{C_{R}}frac{1}{P(z)} = 0$ for all $R > r.$
answered Dec 11 '18 at 11:16
Brevan EllefsenBrevan Ellefsen
11.7k31649
$endgroup$
add a comment |
$begingroup$
Use the residue theorem. Your integral must be the same for all $R$ large enough to contain each zero of $P(z).$ As such, let $r$ be the distance to the furthest zero. Since $lim_{R to infty} int_{C_R}frac{1}{P(z)} = 0,$ and $int_{C_{R_1}}frac{1}{P(z)} = int_{C_{R_2}}frac{1}{P(z)}$ for all $R_1, R_2 > r,$ we conclude $int_{C_{R}}frac{1}{P(z)} = 0$ for all $R > r.$
answered Dec 11 '18 at 11:16
Brevan EllefsenBrevan Ellefsen
11.7k31649
$endgroup$
Use the residue theorem. Your integral must be the same for all $R$ large enough to contain each zero of $P(z).$ As such, let $r$ be the distance to the furthest zero. Since $lim_{R to infty} int_{C_R}frac{1}{P(z)} = 0,$ and $int_{C_{R_1}}frac{1}{P(z)} = int_{C_{R_2}}frac{1}{P(z)}$ for all $R_1, R_2 > r,$ we conclude $int_{C_{R}}frac{1}{P(z)} = 0$ for all $R > r.$
answered Dec 11 '18 at 11:16
Brevan EllefsenBrevan Ellefsen
11.7k31649
answered Dec 11 '18 at 11:16
Brevan EllefsenBrevan Ellefsen
11.7k31649
answered Dec 11 '18 at 11:16
Brevan EllefsenBrevan Ellefsen
11.7k31649
answered Dec 11 '18 at 11:16
Brevan EllefsenBrevan Ellefsen
11.7k31649
11.7k31649
add a comment |
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