$int_gamma frac{e^z}{(z^2+1)^2} dz$ where $gamma$ is $C(0,2)$ traversed twice counter-clockwise.
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Evaluate $int_gamma frac{e^z}{(z^2+1)^2} dz$ where $gamma$ is $C(0,2)$ traversed twice counter-clockwise. This $gamma$ should be represented as a cycle I believe.
Hi all, I am wondering about how to evaluate this integral using the Cauchy Integral Formula.
I have tried to represent the integral as a linear combination of simpler integrals but I'm unsure how. Thank you.
complex-analysis complex-integration
asked Nov 17 at 1:53
Darkdub
8116
|
show 2 more comments
up vote
1
down vote
favorite
Evaluate $int_gamma frac{e^z}{(z^2+1)^2} dz$ where $gamma$ is $C(0,2)$ traversed twice counter-clockwise. This $gamma$ should be represented as a cycle I believe.
Hi all, I am wondering about how to evaluate this integral using the Cauchy Integral Formula.
I have tried to represent the integral as a linear combination of simpler integrals but I'm unsure how. Thank you.
complex-analysis complex-integration
asked Nov 17 at 1:53
Darkdub
8116
Do you know the Cauchy Integral Formula for derivatives?
– Apocalypse
Nov 17 at 1:58
@Apocalypse Yes.
– Darkdub
Nov 17 at 2:00
$frac{e^z}{(z^2+1)^2}=frac{e^z}{(z+i)^2(z-i)^2}$. Apply the formula.
– Apocalypse
Nov 17 at 2:02
What if z = i or z = -i? These are within the circle, no?
– Darkdub
Nov 17 at 2:10
1
That would be hard to handle since the Cauchy Integral Formula is only valid for functions with first order poles inside the contour, i.e., in the form $f(z)=frac{g(z)}{z-w}$ where $g(z)$ is holomorphic. If there are poles of higher orders you have to use derivatives.
– Apocalypse
Nov 17 at 2:30
|
show 2 more comments
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Evaluate $int_gamma frac{e^z}{(z^2+1)^2} dz$ where $gamma$ is $C(0,2)$ traversed twice counter-clockwise. This $gamma$ should be represented as a cycle I believe.
Hi all, I am wondering about how to evaluate this integral using the Cauchy Integral Formula.
I have tried to represent the integral as a linear combination of simpler integrals but I'm unsure how. Thank you.
complex-analysis complex-integration
asked Nov 17 at 1:53
Darkdub
8116
Evaluate $int_gamma frac{e^z}{(z^2+1)^2} dz$ where $gamma$ is $C(0,2)$ traversed twice counter-clockwise. This $gamma$ should be represented as a cycle I believe.
Hi all, I am wondering about how to evaluate this integral using the Cauchy Integral Formula.
I have tried to represent the integral as a linear combination of simpler integrals but I'm unsure how. Thank you.
complex-analysis complex-integration
complex-analysis complex-integration
asked Nov 17 at 1:53
Darkdub
8116
asked Nov 17 at 1:53
Darkdub
8116
asked Nov 17 at 1:53
Darkdub
8116
asked Nov 17 at 1:53
Darkdub
8116
asked Nov 17 at 1:53
Darkdub
8116
8116
Do you know the Cauchy Integral Formula for derivatives?
– Apocalypse
Nov 17 at 1:58
@Apocalypse Yes.
– Darkdub
Nov 17 at 2:00
$frac{e^z}{(z^2+1)^2}=frac{e^z}{(z+i)^2(z-i)^2}$. Apply the formula.
– Apocalypse
Nov 17 at 2:02
What if z = i or z = -i? These are within the circle, no?
– Darkdub
Nov 17 at 2:10
1
That would be hard to handle since the Cauchy Integral Formula is only valid for functions with first order poles inside the contour, i.e., in the form $f(z)=frac{g(z)}{z-w}$ where $g(z)$ is holomorphic. If there are poles of higher orders you have to use derivatives.
– Apocalypse
Nov 17 at 2:30
|
show 2 more comments
Do you know the Cauchy Integral Formula for derivatives?
– Apocalypse
Nov 17 at 1:58
@Apocalypse Yes.
– Darkdub
Nov 17 at 2:00
$frac{e^z}{(z^2+1)^2}=frac{e^z}{(z+i)^2(z-i)^2}$. Apply the formula.
– Apocalypse
Nov 17 at 2:02
What if z = i or z = -i? These are within the circle, no?
– Darkdub
Nov 17 at 2:10
1
That would be hard to handle since the Cauchy Integral Formula is only valid for functions with first order poles inside the contour, i.e., in the form $f(z)=frac{g(z)}{z-w}$ where $g(z)$ is holomorphic. If there are poles of higher orders you have to use derivatives.
– Apocalypse
Nov 17 at 2:30
Do you know the Cauchy Integral Formula for derivatives?
– Apocalypse
Nov 17 at 1:58
Do you know the Cauchy Integral Formula for derivatives?
– Apocalypse
Nov 17 at 1:58
@Apocalypse Yes.
– Darkdub
Nov 17 at 2:00
@Apocalypse Yes.
– Darkdub
Nov 17 at 2:00
$frac{e^z}{(z^2+1)^2}=frac{e^z}{(z+i)^2(z-i)^2}$. Apply the formula.
– Apocalypse
Nov 17 at 2:02
$frac{e^z}{(z^2+1)^2}=frac{e^z}{(z+i)^2(z-i)^2}$. Apply the formula.
– Apocalypse
Nov 17 at 2:02
What if z = i or z = -i? These are within the circle, no?
– Darkdub
Nov 17 at 2:10
What if z = i or z = -i? These are within the circle, no?
– Darkdub
Nov 17 at 2:10
1
1
That would be hard to handle since the Cauchy Integral Formula is only valid for functions with first order poles inside the contour, i.e., in the form $f(z)=frac{g(z)}{z-w}$ where $g(z)$ is holomorphic. If there are poles of higher orders you have to use derivatives.
– Apocalypse
Nov 17 at 2:30
That would be hard to handle since the Cauchy Integral Formula is only valid for functions with first order poles inside the contour, i.e., in the form $f(z)=frac{g(z)}{z-w}$ where $g(z)$ is holomorphic. If there are poles of higher orders you have to use derivatives.
– Apocalypse
Nov 17 at 2:30
|
show 2 more comments
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
Hint. Given partial fraction expansion
$$frac{1}{(z^2+1)^2}=-frac{1}{4(z-i)^2}-frac{i}{4(z-i)}-frac{1}{4(i+z)^2} + frac{i}{4(i + z)}$$
Cauchy's integral formula can be applied for each part individually, considering the winding number of course.
$$intlimits_{gamma}frac{e^z}{(z^2+1)^2} dz=-intlimits_{gamma}frac{e^z}{4(z-i)^2}dz-intlimits_{gamma}frac{ie^z}{4(z-i)}dz-intlimits_{gamma}frac{e^z}{4(i+z)^2}dz + intlimits_{gamma}frac{ie^z}{4(i + z)}dz$$
More details here (section 6) and here (pages 9 and 12).
answered Nov 17 at 10:43
rtybase
10k21433
1
Thank you very much!
– Darkdub
Nov 17 at 13:42
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Hint. Given partial fraction expansion
$$frac{1}{(z^2+1)^2}=-frac{1}{4(z-i)^2}-frac{i}{4(z-i)}-frac{1}{4(i+z)^2} + frac{i}{4(i + z)}$$
Cauchy's integral formula can be applied for each part individually, considering the winding number of course.
$$intlimits_{gamma}frac{e^z}{(z^2+1)^2} dz=-intlimits_{gamma}frac{e^z}{4(z-i)^2}dz-intlimits_{gamma}frac{ie^z}{4(z-i)}dz-intlimits_{gamma}frac{e^z}{4(i+z)^2}dz + intlimits_{gamma}frac{ie^z}{4(i + z)}dz$$
More details here (section 6) and here (pages 9 and 12).
answered Nov 17 at 10:43
rtybase
10k21433
1
Thank you very much!
– Darkdub
Nov 17 at 13:42
add a comment |
up vote
1
down vote
accepted
Hint. Given partial fraction expansion
$$frac{1}{(z^2+1)^2}=-frac{1}{4(z-i)^2}-frac{i}{4(z-i)}-frac{1}{4(i+z)^2} + frac{i}{4(i + z)}$$
Cauchy's integral formula can be applied for each part individually, considering the winding number of course.
$$intlimits_{gamma}frac{e^z}{(z^2+1)^2} dz=-intlimits_{gamma}frac{e^z}{4(z-i)^2}dz-intlimits_{gamma}frac{ie^z}{4(z-i)}dz-intlimits_{gamma}frac{e^z}{4(i+z)^2}dz + intlimits_{gamma}frac{ie^z}{4(i + z)}dz$$
More details here (section 6) and here (pages 9 and 12).
answered Nov 17 at 10:43
rtybase
10k21433
1
Thank you very much!
– Darkdub
Nov 17 at 13:42
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Hint. Given partial fraction expansion
$$frac{1}{(z^2+1)^2}=-frac{1}{4(z-i)^2}-frac{i}{4(z-i)}-frac{1}{4(i+z)^2} + frac{i}{4(i + z)}$$
Cauchy's integral formula can be applied for each part individually, considering the winding number of course.
$$intlimits_{gamma}frac{e^z}{(z^2+1)^2} dz=-intlimits_{gamma}frac{e^z}{4(z-i)^2}dz-intlimits_{gamma}frac{ie^z}{4(z-i)}dz-intlimits_{gamma}frac{e^z}{4(i+z)^2}dz + intlimits_{gamma}frac{ie^z}{4(i + z)}dz$$
More details here (section 6) and here (pages 9 and 12).
answered Nov 17 at 10:43
rtybase
10k21433
Hint. Given partial fraction expansion
$$frac{1}{(z^2+1)^2}=-frac{1}{4(z-i)^2}-frac{i}{4(z-i)}-frac{1}{4(i+z)^2} + frac{i}{4(i + z)}$$
Cauchy's integral formula can be applied for each part individually, considering the winding number of course.
$$intlimits_{gamma}frac{e^z}{(z^2+1)^2} dz=-intlimits_{gamma}frac{e^z}{4(z-i)^2}dz-intlimits_{gamma}frac{ie^z}{4(z-i)}dz-intlimits_{gamma}frac{e^z}{4(i+z)^2}dz + intlimits_{gamma}frac{ie^z}{4(i + z)}dz$$
More details here (section 6) and here (pages 9 and 12).
answered Nov 17 at 10:43
rtybase
10k21433
edited Nov 17 at 11:04
answered Nov 17 at 10:43
rtybase
10k21433
answered Nov 17 at 10:43
rtybase
10k21433
answered Nov 17 at 10:43
rtybase
10k21433
10k21433
1
Thank you very much!
– Darkdub
Nov 17 at 13:42
add a comment |
1
Thank you very much!
– Darkdub
Nov 17 at 13:42
1
1
Thank you very much!
– Darkdub
Nov 17 at 13:42
Thank you very much!
– Darkdub
Nov 17 at 13:42
add a comment |
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Do you know the Cauchy Integral Formula for derivatives?
– Apocalypse
Nov 17 at 1:58
@Apocalypse Yes.
– Darkdub
Nov 17 at 2:00
$frac{e^z}{(z^2+1)^2}=frac{e^z}{(z+i)^2(z-i)^2}$. Apply the formula.
– Apocalypse
Nov 17 at 2:02
What if z = i or z = -i? These are within the circle, no?
– Darkdub
Nov 17 at 2:10
1
That would be hard to handle since the Cauchy Integral Formula is only valid for functions with first order poles inside the contour, i.e., in the form $f(z)=frac{g(z)}{z-w}$ where $g(z)$ is holomorphic. If there are poles of higher orders you have to use derivatives.
– Apocalypse
Nov 17 at 2:30