Integral of the reciprocal of a complex polynomial [duplicate]
This question already has an answer here:
Proof that $frac {1} {2pi i} oint frac {{rm d}z} {P(z)} $ over a closed curve is zero.
2 answers
For a polynomial P(z) of degree $ngeq2$, show that there exists some $R_{1}>0$ such that for $R>R_{1}$ it holds that:
$$int_{C_{R}}frac{1}{P(z)}dz=0$$ where $C_{R}$ is a circle of radius R with the centre $0$.
I am already aware of the fact that $$lim_{Rto infty}int_{C_{R}}frac{1}{P(z)}dz=0$$ and how to go about proving it but I'm not sure how to evaluate the first integral written above without using any limit. Would I maybe use a similar approach as was used to show the latter integral?
complex-analysis complex-integration
asked Dec 8 at 17:10
basic_ceremony
53
marked as duplicate by Martin R, Lord Shark the Unknown, user10354138, Cesareo, Rebellos Dec 9 at 10:24
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
This question already has an answer here:
Proof that $frac {1} {2pi i} oint frac {{rm d}z} {P(z)} $ over a closed curve is zero.
2 answers
For a polynomial P(z) of degree $ngeq2$, show that there exists some $R_{1}>0$ such that for $R>R_{1}$ it holds that:
$$int_{C_{R}}frac{1}{P(z)}dz=0$$ where $C_{R}$ is a circle of radius R with the centre $0$.
I am already aware of the fact that $$lim_{Rto infty}int_{C_{R}}frac{1}{P(z)}dz=0$$ and how to go about proving it but I'm not sure how to evaluate the first integral written above without using any limit. Would I maybe use a similar approach as was used to show the latter integral?
complex-analysis complex-integration
asked Dec 8 at 17:10
basic_ceremony
53
marked as duplicate by Martin R, Lord Shark the Unknown, user10354138, Cesareo, Rebellos Dec 9 at 10:24
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
Yes, I apologise if it is. I made a lot of research before I posted this question but I found nothing that was smiliar + simple enough for me to understand. Sorry!
– basic_ceremony
Dec 8 at 17:19
add a comment |
This question already has an answer here:
Proof that $frac {1} {2pi i} oint frac {{rm d}z} {P(z)} $ over a closed curve is zero.
2 answers
For a polynomial P(z) of degree $ngeq2$, show that there exists some $R_{1}>0$ such that for $R>R_{1}$ it holds that:
$$int_{C_{R}}frac{1}{P(z)}dz=0$$ where $C_{R}$ is a circle of radius R with the centre $0$.
I am already aware of the fact that $$lim_{Rto infty}int_{C_{R}}frac{1}{P(z)}dz=0$$ and how to go about proving it but I'm not sure how to evaluate the first integral written above without using any limit. Would I maybe use a similar approach as was used to show the latter integral?
complex-analysis complex-integration
asked Dec 8 at 17:10
basic_ceremony
53
This question already has an answer here:
Proof that $frac {1} {2pi i} oint frac {{rm d}z} {P(z)} $ over a closed curve is zero.
2 answers
For a polynomial P(z) of degree $ngeq2$, show that there exists some $R_{1}>0$ such that for $R>R_{1}$ it holds that:
$$int_{C_{R}}frac{1}{P(z)}dz=0$$ where $C_{R}$ is a circle of radius R with the centre $0$.
I am already aware of the fact that $$lim_{Rto infty}int_{C_{R}}frac{1}{P(z)}dz=0$$ and how to go about proving it but I'm not sure how to evaluate the first integral written above without using any limit. Would I maybe use a similar approach as was used to show the latter integral?
This question already has an answer here:
Proof that $frac {1} {2pi i} oint frac {{rm d}z} {P(z)} $ over a closed curve is zero.
2 answers
complex-analysis complex-integration
complex-analysis complex-integration
asked Dec 8 at 17:10
basic_ceremony
53
asked Dec 8 at 17:10
basic_ceremony
53
asked Dec 8 at 17:10
basic_ceremony
53
asked Dec 8 at 17:10
basic_ceremony
53
asked Dec 8 at 17:10
basic_ceremony
53
53
marked as duplicate by Martin R, Lord Shark the Unknown, user10354138, Cesareo, Rebellos Dec 9 at 10:24
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Martin R, Lord Shark the Unknown, user10354138, Cesareo, Rebellos Dec 9 at 10:24
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
Yes, I apologise if it is. I made a lot of research before I posted this question but I found nothing that was smiliar + simple enough for me to understand. Sorry!
– basic_ceremony
Dec 8 at 17:19
add a comment |
Yes, I apologise if it is. I made a lot of research before I posted this question but I found nothing that was smiliar + simple enough for me to understand. Sorry!
– basic_ceremony
Dec 8 at 17:19
Yes, I apologise if it is. I made a lot of research before I posted this question but I found nothing that was smiliar + simple enough for me to understand. Sorry!
– basic_ceremony
Dec 8 at 17:19
Yes, I apologise if it is. I made a lot of research before I posted this question but I found nothing that was smiliar + simple enough for me to understand. Sorry!
– basic_ceremony
Dec 8 at 17:19
add a comment |
1 Answer
1
active
oldest
votes
If $R$ and $S$ are greater than the absolute value of every root of $P$, then, by Cauchy's theorem$$oint_{C_R}frac1{P(z)},mathrm dz=oint_{C_S}frac1{P(z)},mathrm dz.$$This, together with the fact that that limit that you mentioned is $0$; is enogh to prove that $oint_{C_R}frac1{P(z)},mathrm dz$ is $0$ if $R$ is large enough.
answered Dec 8 at 17:14
José Carlos Santos
149k22117219
Would 'If R and S are greater than the absolute value of every root of P' be another way of saying that all roots of P are contained in R and S? I'm confident the answer is yes but I'd like to be certain
– basic_ceremony
Dec 8 at 17:17
Yes, you are right.
– José Carlos Santos
Dec 8 at 17:30
Thanks for the great responses + respone times, as usual :)
– basic_ceremony
Dec 8 at 17:33
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
If $R$ and $S$ are greater than the absolute value of every root of $P$, then, by Cauchy's theorem$$oint_{C_R}frac1{P(z)},mathrm dz=oint_{C_S}frac1{P(z)},mathrm dz.$$This, together with the fact that that limit that you mentioned is $0$; is enogh to prove that $oint_{C_R}frac1{P(z)},mathrm dz$ is $0$ if $R$ is large enough.
answered Dec 8 at 17:14
José Carlos Santos
149k22117219
Would 'If R and S are greater than the absolute value of every root of P' be another way of saying that all roots of P are contained in R and S? I'm confident the answer is yes but I'd like to be certain
– basic_ceremony
Dec 8 at 17:17
Yes, you are right.
– José Carlos Santos
Dec 8 at 17:30
Thanks for the great responses + respone times, as usual :)
– basic_ceremony
Dec 8 at 17:33
add a comment |
If $R$ and $S$ are greater than the absolute value of every root of $P$, then, by Cauchy's theorem$$oint_{C_R}frac1{P(z)},mathrm dz=oint_{C_S}frac1{P(z)},mathrm dz.$$This, together with the fact that that limit that you mentioned is $0$; is enogh to prove that $oint_{C_R}frac1{P(z)},mathrm dz$ is $0$ if $R$ is large enough.
answered Dec 8 at 17:14
José Carlos Santos
149k22117219
Would 'If R and S are greater than the absolute value of every root of P' be another way of saying that all roots of P are contained in R and S? I'm confident the answer is yes but I'd like to be certain
– basic_ceremony
Dec 8 at 17:17
Yes, you are right.
– José Carlos Santos
Dec 8 at 17:30
Thanks for the great responses + respone times, as usual :)
– basic_ceremony
Dec 8 at 17:33
add a comment |
If $R$ and $S$ are greater than the absolute value of every root of $P$, then, by Cauchy's theorem$$oint_{C_R}frac1{P(z)},mathrm dz=oint_{C_S}frac1{P(z)},mathrm dz.$$This, together with the fact that that limit that you mentioned is $0$; is enogh to prove that $oint_{C_R}frac1{P(z)},mathrm dz$ is $0$ if $R$ is large enough.
answered Dec 8 at 17:14
José Carlos Santos
149k22117219
If $R$ and $S$ are greater than the absolute value of every root of $P$, then, by Cauchy's theorem$$oint_{C_R}frac1{P(z)},mathrm dz=oint_{C_S}frac1{P(z)},mathrm dz.$$This, together with the fact that that limit that you mentioned is $0$; is enogh to prove that $oint_{C_R}frac1{P(z)},mathrm dz$ is $0$ if $R$ is large enough.
answered Dec 8 at 17:14
José Carlos Santos
149k22117219
answered Dec 8 at 17:14
José Carlos Santos
149k22117219
answered Dec 8 at 17:14
José Carlos Santos
149k22117219
answered Dec 8 at 17:14
José Carlos Santos
149k22117219
149k22117219
Would 'If R and S are greater than the absolute value of every root of P' be another way of saying that all roots of P are contained in R and S? I'm confident the answer is yes but I'd like to be certain
– basic_ceremony
Dec 8 at 17:17
Yes, you are right.
– José Carlos Santos
Dec 8 at 17:30
Thanks for the great responses + respone times, as usual :)
– basic_ceremony
Dec 8 at 17:33
add a comment |
Would 'If R and S are greater than the absolute value of every root of P' be another way of saying that all roots of P are contained in R and S? I'm confident the answer is yes but I'd like to be certain
– basic_ceremony
Dec 8 at 17:17
Yes, you are right.
– José Carlos Santos
Dec 8 at 17:30
Thanks for the great responses + respone times, as usual :)
– basic_ceremony
Dec 8 at 17:33
Would 'If R and S are greater than the absolute value of every root of P' be another way of saying that all roots of P are contained in R and S? I'm confident the answer is yes but I'd like to be certain
– basic_ceremony
Dec 8 at 17:17
Would 'If R and S are greater than the absolute value of every root of P' be another way of saying that all roots of P are contained in R and S? I'm confident the answer is yes but I'd like to be certain
– basic_ceremony
Dec 8 at 17:17
Yes, you are right.
– José Carlos Santos
Dec 8 at 17:30
Yes, you are right.
– José Carlos Santos
Dec 8 at 17:30
Thanks for the great responses + respone times, as usual :)
– basic_ceremony
Dec 8 at 17:33
Thanks for the great responses + respone times, as usual :)
– basic_ceremony
Dec 8 at 17:33
add a comment |
Yes, I apologise if it is. I made a lot of research before I posted this question but I found nothing that was smiliar + simple enough for me to understand. Sorry!
– basic_ceremony
Dec 8 at 17:19