Prove simple closed curves $f$'s exist, so $Gamma = C-sum_{i=1}^{k}{f_i}$ satisfies $...
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Let $C$ be the circle $C(0,2)$ traversed one time counter-clockwise. Prove that there exist $kin mathbb {Z}_+$ and $C^1$ simple closed cuves $f_1, dots ,f_k$ such that the cycle $Gamma = C-sum_{i=1}^{k}{f_i}$ satisfies
$$ int_{Gamma}{frac{z^3e^{1/z}}{(z^2 + z + 1)(z^2 + 1)}dz}=0$$
Note: $C^1$ means the curve has continuous derivatives for each $t$ within the curve's interval $[a,b]$.
What would be the simplest way to prove this?
complex-analysis complex-integration cauchy-integral-formula simple-functions
asked Nov 18 at 18:49
wtnmath
15212
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1
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Let $C$ be the circle $C(0,2)$ traversed one time counter-clockwise. Prove that there exist $kin mathbb {Z}_+$ and $C^1$ simple closed cuves $f_1, dots ,f_k$ such that the cycle $Gamma = C-sum_{i=1}^{k}{f_i}$ satisfies
$$ int_{Gamma}{frac{z^3e^{1/z}}{(z^2 + z + 1)(z^2 + 1)}dz}=0$$
Note: $C^1$ means the curve has continuous derivatives for each $t$ within the curve's interval $[a,b]$.
What would be the simplest way to prove this?
complex-analysis complex-integration cauchy-integral-formula simple-functions
asked Nov 18 at 18:49
wtnmath
15212
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $C$ be the circle $C(0,2)$ traversed one time counter-clockwise. Prove that there exist $kin mathbb {Z}_+$ and $C^1$ simple closed cuves $f_1, dots ,f_k$ such that the cycle $Gamma = C-sum_{i=1}^{k}{f_i}$ satisfies
$$ int_{Gamma}{frac{z^3e^{1/z}}{(z^2 + z + 1)(z^2 + 1)}dz}=0$$
Note: $C^1$ means the curve has continuous derivatives for each $t$ within the curve's interval $[a,b]$.
What would be the simplest way to prove this?
complex-analysis complex-integration cauchy-integral-formula simple-functions
asked Nov 18 at 18:49
wtnmath
15212
Let $C$ be the circle $C(0,2)$ traversed one time counter-clockwise. Prove that there exist $kin mathbb {Z}_+$ and $C^1$ simple closed cuves $f_1, dots ,f_k$ such that the cycle $Gamma = C-sum_{i=1}^{k}{f_i}$ satisfies
$$ int_{Gamma}{frac{z^3e^{1/z}}{(z^2 + z + 1)(z^2 + 1)}dz}=0$$
Note: $C^1$ means the curve has continuous derivatives for each $t$ within the curve's interval $[a,b]$.
What would be the simplest way to prove this?
complex-analysis complex-integration cauchy-integral-formula simple-functions
complex-analysis complex-integration cauchy-integral-formula simple-functions
asked Nov 18 at 18:49
wtnmath
15212
asked Nov 18 at 18:49
wtnmath
15212
asked Nov 18 at 18:49
wtnmath
15212
asked Nov 18 at 18:49
wtnmath
15212
asked Nov 18 at 18:49
wtnmath
15212
15212
add a comment |
add a comment |
1 Answer
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The simplest way to prove it is by providing an example.
Note that the singularities of the integrand are at $a_{1,2}=pm i,a_{3,4}=(-1pm isqrt3)/2,a_5=0$.
Thus, let $f_n=C(a_n,epsilon)$ where $1le nle 5$ and $epsilon$ being sufficiently small.
Then, clearly, $C-sum f_n$ encloses no singularities. Therefore, by Cauchy’s integral theorem,
$$oint_{Gamma} frac{z^3e^{1/z}}{(z^2 + z + 1)(z^2 + 1)}dz =0$$
answered Nov 19 at 23:54
Szeto
6,2292726
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
The simplest way to prove it is by providing an example.
Note that the singularities of the integrand are at $a_{1,2}=pm i,a_{3,4}=(-1pm isqrt3)/2,a_5=0$.
Thus, let $f_n=C(a_n,epsilon)$ where $1le nle 5$ and $epsilon$ being sufficiently small.
Then, clearly, $C-sum f_n$ encloses no singularities. Therefore, by Cauchy’s integral theorem,
$$oint_{Gamma} frac{z^3e^{1/z}}{(z^2 + z + 1)(z^2 + 1)}dz =0$$
answered Nov 19 at 23:54
Szeto
6,2292726
add a comment |
up vote
0
down vote
The simplest way to prove it is by providing an example.
Note that the singularities of the integrand are at $a_{1,2}=pm i,a_{3,4}=(-1pm isqrt3)/2,a_5=0$.
Thus, let $f_n=C(a_n,epsilon)$ where $1le nle 5$ and $epsilon$ being sufficiently small.
Then, clearly, $C-sum f_n$ encloses no singularities. Therefore, by Cauchy’s integral theorem,
$$oint_{Gamma} frac{z^3e^{1/z}}{(z^2 + z + 1)(z^2 + 1)}dz =0$$
answered Nov 19 at 23:54
Szeto
6,2292726
add a comment |
up vote
0
down vote
up vote
0
down vote
The simplest way to prove it is by providing an example.
Note that the singularities of the integrand are at $a_{1,2}=pm i,a_{3,4}=(-1pm isqrt3)/2,a_5=0$.
Thus, let $f_n=C(a_n,epsilon)$ where $1le nle 5$ and $epsilon$ being sufficiently small.
Then, clearly, $C-sum f_n$ encloses no singularities. Therefore, by Cauchy’s integral theorem,
$$oint_{Gamma} frac{z^3e^{1/z}}{(z^2 + z + 1)(z^2 + 1)}dz =0$$
answered Nov 19 at 23:54
Szeto
6,2292726
The simplest way to prove it is by providing an example.
Note that the singularities of the integrand are at $a_{1,2}=pm i,a_{3,4}=(-1pm isqrt3)/2,a_5=0$.
Thus, let $f_n=C(a_n,epsilon)$ where $1le nle 5$ and $epsilon$ being sufficiently small.
Then, clearly, $C-sum f_n$ encloses no singularities. Therefore, by Cauchy’s integral theorem,
$$oint_{Gamma} frac{z^3e^{1/z}}{(z^2 + z + 1)(z^2 + 1)}dz =0$$
answered Nov 19 at 23:54
Szeto
6,2292726
answered Nov 19 at 23:54
Szeto
6,2292726
answered Nov 19 at 23:54
Szeto
6,2292726
answered Nov 19 at 23:54
Szeto
6,2292726
6,2292726
add a comment |
add a comment |
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