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Prove that $int_gamma z^n,dz=0$ for any $textit{closed}$ path $gamma$ and integer $nneq -1$ [duplicate]
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This question already has an answer here:
Show that the complex integral is zero for any closed, piece wise smooth path
2 answers
We know that if $gamma$ is a curve in the complex plane parameterized by a function $z(t)$ which is a continuously differentiable function from the interval $[a,b]$ into $mathbb{C}$ and $f$ is analytic on an open set containing $tr(gamma)$, then $w(t)=f(z(t))$ is differentiable on $[a,b]$ and $w'(t)=f'(z(t))z'(t)$. Use this result to prove that $int_gamma z^n,dz=0$ for any textit{closed} path $gamma$ and integer $nneq -1$, assuming that $tr gamma$ does not contain the origin if $n<0$.
1st Try: Suppose that $ngeq 0$, then $z^n$ is analytic everywhere. By Goursat's Lemma, the integral is $0$. When $n<-1$, then $z^n$ is analytic everywhere except at the origin. But I am allowed to assume that $0notin trgamma$. So again the integral is $0$.
The problem is that I haven't use the assumption.
Note I am not allowed to use Cauchy's Integral Formula.
complex-analysis
edited Jan 30 at 21:28
MPW
31k12157
asked Jan 30 at 21:19
Username UnknownUsername Unknown
1,18242260
$endgroup$
marked as duplicate by Martin R, Shailesh, Lord Shark the Unknown, Lee David Chung Lin, metamorphy Jan 31 at 5:37
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 |
$begingroup$
This question already has an answer here:
Show that the complex integral is zero for any closed, piece wise smooth path
2 answers
We know that if $gamma$ is a curve in the complex plane parameterized by a function $z(t)$ which is a continuously differentiable function from the interval $[a,b]$ into $mathbb{C}$ and $f$ is analytic on an open set containing $tr(gamma)$, then $w(t)=f(z(t))$ is differentiable on $[a,b]$ and $w'(t)=f'(z(t))z'(t)$. Use this result to prove that $int_gamma z^n,dz=0$ for any textit{closed} path $gamma$ and integer $nneq -1$, assuming that $tr gamma$ does not contain the origin if $n<0$.
1st Try: Suppose that $ngeq 0$, then $z^n$ is analytic everywhere. By Goursat's Lemma, the integral is $0$. When $n<-1$, then $z^n$ is analytic everywhere except at the origin. But I am allowed to assume that $0notin trgamma$. So again the integral is $0$.
The problem is that I haven't use the assumption.
Note I am not allowed to use Cauchy's Integral Formula.
complex-analysis
edited Jan 30 at 21:28
MPW
31k12157
asked Jan 30 at 21:19
Username UnknownUsername Unknown
1,18242260
$endgroup$
marked as duplicate by Martin R, Shailesh, Lord Shark the Unknown, Lee David Chung Lin, metamorphy Jan 31 at 5:37
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.
1
$begingroup$
When you say that "[the function] $z^n$ is analytic everywhere except at the origin" and that, since $0$ is not on the path $gamma$, "the integral is $0$", the first part is clear but how you draw the consequence that the integral is $0$ is unclear. Certainly there is no theorem stating that if $gamma$ is a closed path and if $f$ is analytic on a neighborhood of $gamma$ then $oint_gamma f(z)dz=0$.
$endgroup$
– Did
Jan 30 at 21:32
add a comment |
$begingroup$
This question already has an answer here:
Show that the complex integral is zero for any closed, piece wise smooth path
2 answers
We know that if $gamma$ is a curve in the complex plane parameterized by a function $z(t)$ which is a continuously differentiable function from the interval $[a,b]$ into $mathbb{C}$ and $f$ is analytic on an open set containing $tr(gamma)$, then $w(t)=f(z(t))$ is differentiable on $[a,b]$ and $w'(t)=f'(z(t))z'(t)$. Use this result to prove that $int_gamma z^n,dz=0$ for any textit{closed} path $gamma$ and integer $nneq -1$, assuming that $tr gamma$ does not contain the origin if $n<0$.
1st Try: Suppose that $ngeq 0$, then $z^n$ is analytic everywhere. By Goursat's Lemma, the integral is $0$. When $n<-1$, then $z^n$ is analytic everywhere except at the origin. But I am allowed to assume that $0notin trgamma$. So again the integral is $0$.
The problem is that I haven't use the assumption.
Note I am not allowed to use Cauchy's Integral Formula.
complex-analysis
edited Jan 30 at 21:28
MPW
31k12157
asked Jan 30 at 21:19
Username UnknownUsername Unknown
1,18242260
$endgroup$
This question already has an answer here:
Show that the complex integral is zero for any closed, piece wise smooth path
2 answers
We know that if $gamma$ is a curve in the complex plane parameterized by a function $z(t)$ which is a continuously differentiable function from the interval $[a,b]$ into $mathbb{C}$ and $f$ is analytic on an open set containing $tr(gamma)$, then $w(t)=f(z(t))$ is differentiable on $[a,b]$ and $w'(t)=f'(z(t))z'(t)$. Use this result to prove that $int_gamma z^n,dz=0$ for any textit{closed} path $gamma$ and integer $nneq -1$, assuming that $tr gamma$ does not contain the origin if $n<0$.
1st Try: Suppose that $ngeq 0$, then $z^n$ is analytic everywhere. By Goursat's Lemma, the integral is $0$. When $n<-1$, then $z^n$ is analytic everywhere except at the origin. But I am allowed to assume that $0notin trgamma$. So again the integral is $0$.
The problem is that I haven't use the assumption.
Note I am not allowed to use Cauchy's Integral Formula.
This question already has an answer here:
Show that the complex integral is zero for any closed, piece wise smooth path
2 answers
complex-analysis
complex-analysis
edited Jan 30 at 21:28
MPW
31k12157
asked Jan 30 at 21:19
Username UnknownUsername Unknown
1,18242260
edited Jan 30 at 21:28
MPW
31k12157
asked Jan 30 at 21:19
Username UnknownUsername Unknown
1,18242260
edited Jan 30 at 21:28
MPW
31k12157
edited Jan 30 at 21:28
MPW
31k12157
edited Jan 30 at 21:28
MPW
31k12157
31k12157
asked Jan 30 at 21:19
Username UnknownUsername Unknown
1,18242260
asked Jan 30 at 21:19
Username UnknownUsername Unknown
1,18242260
asked Jan 30 at 21:19
Username UnknownUsername Unknown
1,18242260
1,18242260
marked as duplicate by Martin R, Shailesh, Lord Shark the Unknown, Lee David Chung Lin, metamorphy Jan 31 at 5:37
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, Shailesh, Lord Shark the Unknown, Lee David Chung Lin, metamorphy Jan 31 at 5:37
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.
1
$begingroup$
When you say that "[the function] $z^n$ is analytic everywhere except at the origin" and that, since $0$ is not on the path $gamma$, "the integral is $0$", the first part is clear but how you draw the consequence that the integral is $0$ is unclear. Certainly there is no theorem stating that if $gamma$ is a closed path and if $f$ is analytic on a neighborhood of $gamma$ then $oint_gamma f(z)dz=0$.
$endgroup$
– Did
Jan 30 at 21:32
add a comment |
1
$begingroup$
When you say that "[the function] $z^n$ is analytic everywhere except at the origin" and that, since $0$ is not on the path $gamma$, "the integral is $0$", the first part is clear but how you draw the consequence that the integral is $0$ is unclear. Certainly there is no theorem stating that if $gamma$ is a closed path and if $f$ is analytic on a neighborhood of $gamma$ then $oint_gamma f(z)dz=0$.
$endgroup$
– Did
Jan 30 at 21:32
1
1
$begingroup$
When you say that "[the function] $z^n$ is analytic everywhere except at the origin" and that, since $0$ is not on the path $gamma$, "the integral is $0$", the first part is clear but how you draw the consequence that the integral is $0$ is unclear. Certainly there is no theorem stating that if $gamma$ is a closed path and if $f$ is analytic on a neighborhood of $gamma$ then $oint_gamma f(z)dz=0$.
$endgroup$
– Did
Jan 30 at 21:32
$begingroup$
When you say that "[the function] $z^n$ is analytic everywhere except at the origin" and that, since $0$ is not on the path $gamma$, "the integral is $0$", the first part is clear but how you draw the consequence that the integral is $0$ is unclear. Certainly there is no theorem stating that if $gamma$ is a closed path and if $f$ is analytic on a neighborhood of $gamma$ then $oint_gamma f(z)dz=0$.
$endgroup$
– Did
Jan 30 at 21:32
add a comment |
1 Answer
1
active
oldest
votes
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If $nneq-1$, then $z^n$ has a primitive: $F(z)=dfrac{z^{n+1}}{n+1}$. So,$$int_gamma z^n,mathrm dz=Fbigl(gamma(b)bigr)-Fbigl(gamma(a)bigr)=0,$$since $gamma$ is a loop.
answered Jan 30 at 21:30
José Carlos SantosJosé Carlos Santos
172k23133241
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$begingroup$
This is the uninteresting part.
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– Did
Jan 30 at 21:45
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@JoseCarlosSantos That's is what I was doing in the beginning but it seemed to easy. Thank you!
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– Username Unknown
Jan 30 at 21:48
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If $nneq-1$, then $z^n$ has a primitive: $F(z)=dfrac{z^{n+1}}{n+1}$. So,$$int_gamma z^n,mathrm dz=Fbigl(gamma(b)bigr)-Fbigl(gamma(a)bigr)=0,$$since $gamma$ is a loop.
answered Jan 30 at 21:30
José Carlos SantosJosé Carlos Santos
172k23133241
$endgroup$
$begingroup$
This is the uninteresting part.
$endgroup$
– Did
Jan 30 at 21:45
$begingroup$
@JoseCarlosSantos That's is what I was doing in the beginning but it seemed to easy. Thank you!
$endgroup$
– Username Unknown
Jan 30 at 21:48
add a comment |
$begingroup$
If $nneq-1$, then $z^n$ has a primitive: $F(z)=dfrac{z^{n+1}}{n+1}$. So,$$int_gamma z^n,mathrm dz=Fbigl(gamma(b)bigr)-Fbigl(gamma(a)bigr)=0,$$since $gamma$ is a loop.
answered Jan 30 at 21:30
José Carlos SantosJosé Carlos Santos
172k23133241
$endgroup$
$begingroup$
This is the uninteresting part.
$endgroup$
– Did
Jan 30 at 21:45
$begingroup$
@JoseCarlosSantos That's is what I was doing in the beginning but it seemed to easy. Thank you!
$endgroup$
– Username Unknown
Jan 30 at 21:48
add a comment |
$begingroup$
If $nneq-1$, then $z^n$ has a primitive: $F(z)=dfrac{z^{n+1}}{n+1}$. So,$$int_gamma z^n,mathrm dz=Fbigl(gamma(b)bigr)-Fbigl(gamma(a)bigr)=0,$$since $gamma$ is a loop.
answered Jan 30 at 21:30
José Carlos SantosJosé Carlos Santos
172k23133241
$endgroup$
If $nneq-1$, then $z^n$ has a primitive: $F(z)=dfrac{z^{n+1}}{n+1}$. So,$$int_gamma z^n,mathrm dz=Fbigl(gamma(b)bigr)-Fbigl(gamma(a)bigr)=0,$$since $gamma$ is a loop.
answered Jan 30 at 21:30
José Carlos SantosJosé Carlos Santos
172k23133241
answered Jan 30 at 21:30
José Carlos SantosJosé Carlos Santos
172k23133241
answered Jan 30 at 21:30
José Carlos SantosJosé Carlos Santos
172k23133241
answered Jan 30 at 21:30
José Carlos SantosJosé Carlos Santos
172k23133241
172k23133241
$begingroup$
This is the uninteresting part.
$endgroup$
– Did
Jan 30 at 21:45
$begingroup$
@JoseCarlosSantos That's is what I was doing in the beginning but it seemed to easy. Thank you!
$endgroup$
– Username Unknown
Jan 30 at 21:48
add a comment |
$begingroup$
This is the uninteresting part.
$endgroup$
– Did
Jan 30 at 21:45
$begingroup$
@JoseCarlosSantos That's is what I was doing in the beginning but it seemed to easy. Thank you!
$endgroup$
– Username Unknown
Jan 30 at 21:48
$begingroup$
This is the uninteresting part.
$endgroup$
– Did
Jan 30 at 21:45
$begingroup$
This is the uninteresting part.
$endgroup$
– Did
Jan 30 at 21:45
$begingroup$
@JoseCarlosSantos That's is what I was doing in the beginning but it seemed to easy. Thank you!
$endgroup$
– Username Unknown
Jan 30 at 21:48
$begingroup$
@JoseCarlosSantos That's is what I was doing in the beginning but it seemed to easy. Thank you!
$endgroup$
– Username Unknown
Jan 30 at 21:48
add a comment |
1
$begingroup$
When you say that "[the function] $z^n$ is analytic everywhere except at the origin" and that, since $0$ is not on the path $gamma$, "the integral is $0$", the first part is clear but how you draw the consequence that the integral is $0$ is unclear. Certainly there is no theorem stating that if $gamma$ is a closed path and if $f$ is analytic on a neighborhood of $gamma$ then $oint_gamma f(z)dz=0$.
$endgroup$
– Did
Jan 30 at 21:32