Use The Fundamental Theorem of Contour Integration or otherwise to evaluate the following integrals. (If it...
a) $int|z|dz$, where $gamma(t) = 3e^{it} (0 le t le pi)$;
b) $int cos z - zsin z dz$, where $gamma(t) = (−1 + 2t) + it (0 le t le 1).$
integration complex-analysis analysis contour-integration
edited Dec 13 '18 at 11:15
Christoph
11.9k1542
asked Dec 13 '18 at 10:31
Aoife CoyleAoife Coyle
123
closed as off-topic by Henning Makholm, Christoph, rtybase, user10354138, Jyrki Lahtonen Dec 13 '18 at 14:28
This question appears to be off-topic. The users who voted to close gave this specific reason:
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a) $int|z|dz$, where $gamma(t) = 3e^{it} (0 le t le pi)$;
b) $int cos z - zsin z dz$, where $gamma(t) = (−1 + 2t) + it (0 le t le 1).$
integration complex-analysis analysis contour-integration
edited Dec 13 '18 at 11:15
Christoph
11.9k1542
asked Dec 13 '18 at 10:31
Aoife CoyleAoife Coyle
123
closed as off-topic by Henning Makholm, Christoph, rtybase, user10354138, Jyrki Lahtonen Dec 13 '18 at 14:28
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Henning Makholm, Christoph, rtybase, user10354138, Jyrki Lahtonen
If this question can be reworded to fit the rules in the help center, please edit the question.
What did you try? Can you see, for example, that in (1) we have $;|z|=3;$ when $;zingamma(z);$ ?
– DonAntonio
Dec 13 '18 at 10:47
I know that to use the Theorem that an anti-derivative must exist but I am having difficulty trying to find this. I am also unsure where the value |Z|=3 has come from
– Aoife Coyle
Dec 13 '18 at 10:54
If you're going to do line integral on the canonical circle of radius $;3;$ , namely: $;{zinBbb C;|;z=3e^{it};,;;tin[0,2pi]};$ , then on this line $;|z|=3;$ ...right?
– DonAntonio
Dec 13 '18 at 11:05
Yes I can see this. Do I need to show ∫ f(γ(z).γ'(z) dz ?
– Aoife Coyle
Dec 13 '18 at 11:11
Note that there is no $gamma(z)$. (I edited the question accordingly). $gamma$ is a curve assigning to a real number $t$ the complex number $gamma(t)$. — When @DonAntonio wrote "$zingamma(z)$" he ment $z=gamma(t)$.
– Christoph
Dec 13 '18 at 11:16
add a comment |
a) $int|z|dz$, where $gamma(t) = 3e^{it} (0 le t le pi)$;
b) $int cos z - zsin z dz$, where $gamma(t) = (−1 + 2t) + it (0 le t le 1).$
integration complex-analysis analysis contour-integration
edited Dec 13 '18 at 11:15
Christoph
11.9k1542
asked Dec 13 '18 at 10:31
Aoife CoyleAoife Coyle
123
a) $int|z|dz$, where $gamma(t) = 3e^{it} (0 le t le pi)$;
b) $int cos z - zsin z dz$, where $gamma(t) = (−1 + 2t) + it (0 le t le 1).$
integration complex-analysis analysis contour-integration
integration complex-analysis analysis contour-integration
edited Dec 13 '18 at 11:15
Christoph
11.9k1542
asked Dec 13 '18 at 10:31
Aoife CoyleAoife Coyle
123
edited Dec 13 '18 at 11:15
Christoph
11.9k1542
asked Dec 13 '18 at 10:31
Aoife CoyleAoife Coyle
123
edited Dec 13 '18 at 11:15
Christoph
11.9k1542
edited Dec 13 '18 at 11:15
Christoph
11.9k1542
edited Dec 13 '18 at 11:15
Christoph
11.9k1542
11.9k1542
asked Dec 13 '18 at 10:31
Aoife CoyleAoife Coyle
123
asked Dec 13 '18 at 10:31
Aoife CoyleAoife Coyle
123
asked Dec 13 '18 at 10:31
Aoife CoyleAoife Coyle
123
123
closed as off-topic by Henning Makholm, Christoph, rtybase, user10354138, Jyrki Lahtonen Dec 13 '18 at 14:28
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Henning Makholm, Christoph, rtybase, user10354138, Jyrki Lahtonen
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Henning Makholm, Christoph, rtybase, user10354138, Jyrki Lahtonen Dec 13 '18 at 14:28
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Henning Makholm, Christoph, rtybase, user10354138, Jyrki Lahtonen
If this question can be reworded to fit the rules in the help center, please edit the question.
What did you try? Can you see, for example, that in (1) we have $;|z|=3;$ when $;zingamma(z);$ ?
– DonAntonio
Dec 13 '18 at 10:47
I know that to use the Theorem that an anti-derivative must exist but I am having difficulty trying to find this. I am also unsure where the value |Z|=3 has come from
– Aoife Coyle
Dec 13 '18 at 10:54
If you're going to do line integral on the canonical circle of radius $;3;$ , namely: $;{zinBbb C;|;z=3e^{it};,;;tin[0,2pi]};$ , then on this line $;|z|=3;$ ...right?
– DonAntonio
Dec 13 '18 at 11:05
Yes I can see this. Do I need to show ∫ f(γ(z).γ'(z) dz ?
– Aoife Coyle
Dec 13 '18 at 11:11
Note that there is no $gamma(z)$. (I edited the question accordingly). $gamma$ is a curve assigning to a real number $t$ the complex number $gamma(t)$. — When @DonAntonio wrote "$zingamma(z)$" he ment $z=gamma(t)$.
– Christoph
Dec 13 '18 at 11:16
add a comment |
What did you try? Can you see, for example, that in (1) we have $;|z|=3;$ when $;zingamma(z);$ ?
– DonAntonio
Dec 13 '18 at 10:47
I know that to use the Theorem that an anti-derivative must exist but I am having difficulty trying to find this. I am also unsure where the value |Z|=3 has come from
– Aoife Coyle
Dec 13 '18 at 10:54
If you're going to do line integral on the canonical circle of radius $;3;$ , namely: $;{zinBbb C;|;z=3e^{it};,;;tin[0,2pi]};$ , then on this line $;|z|=3;$ ...right?
– DonAntonio
Dec 13 '18 at 11:05
Yes I can see this. Do I need to show ∫ f(γ(z).γ'(z) dz ?
– Aoife Coyle
Dec 13 '18 at 11:11
Note that there is no $gamma(z)$. (I edited the question accordingly). $gamma$ is a curve assigning to a real number $t$ the complex number $gamma(t)$. — When @DonAntonio wrote "$zingamma(z)$" he ment $z=gamma(t)$.
– Christoph
Dec 13 '18 at 11:16
What did you try? Can you see, for example, that in (1) we have $;|z|=3;$ when $;zingamma(z);$ ?
– DonAntonio
Dec 13 '18 at 10:47
What did you try? Can you see, for example, that in (1) we have $;|z|=3;$ when $;zingamma(z);$ ?
– DonAntonio
Dec 13 '18 at 10:47
I know that to use the Theorem that an anti-derivative must exist but I am having difficulty trying to find this. I am also unsure where the value |Z|=3 has come from
– Aoife Coyle
Dec 13 '18 at 10:54
I know that to use the Theorem that an anti-derivative must exist but I am having difficulty trying to find this. I am also unsure where the value |Z|=3 has come from
– Aoife Coyle
Dec 13 '18 at 10:54
If you're going to do line integral on the canonical circle of radius $;3;$ , namely: $;{zinBbb C;|;z=3e^{it};,;;tin[0,2pi]};$ , then on this line $;|z|=3;$ ...right?
– DonAntonio
Dec 13 '18 at 11:05
If you're going to do line integral on the canonical circle of radius $;3;$ , namely: $;{zinBbb C;|;z=3e^{it};,;;tin[0,2pi]};$ , then on this line $;|z|=3;$ ...right?
– DonAntonio
Dec 13 '18 at 11:05
Yes I can see this. Do I need to show ∫ f(γ(z).γ'(z) dz ?
– Aoife Coyle
Dec 13 '18 at 11:11
Yes I can see this. Do I need to show ∫ f(γ(z).γ'(z) dz ?
– Aoife Coyle
Dec 13 '18 at 11:11
Note that there is no $gamma(z)$. (I edited the question accordingly). $gamma$ is a curve assigning to a real number $t$ the complex number $gamma(t)$. — When @DonAntonio wrote "$zingamma(z)$" he ment $z=gamma(t)$.
– Christoph
Dec 13 '18 at 11:16
Note that there is no $gamma(z)$. (I edited the question accordingly). $gamma$ is a curve assigning to a real number $t$ the complex number $gamma(t)$. — When @DonAntonio wrote "$zingamma(z)$" he ment $z=gamma(t)$.
– Christoph
Dec 13 '18 at 11:16
add a comment |
1 Answer
1
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Some highlights:
$$z=3e^{it}implies |z|=3,,,,dz=3ie^{it},dtimplies int_gamma |z|dz=int_0^pi 3cdot3ie^{it},dt=ldots$$
The other one is a segment of straight line joining the point $;(-i,0),,,,(1,1);$ in the complex plane, or if you prefer the complex notation: the point $;-1,,,,1+i;$ , so directly:
$$(zcos z)'=cos z-zsin zimpliesleft.int_gamma (cos z-zsin z)dz=zcos zright|_{-1}^{1+i}=ldots$$
answered Dec 13 '18 at 14:44
DonAntonioDonAntonio
177k1492225
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Some highlights:
$$z=3e^{it}implies |z|=3,,,,dz=3ie^{it},dtimplies int_gamma |z|dz=int_0^pi 3cdot3ie^{it},dt=ldots$$
The other one is a segment of straight line joining the point $;(-i,0),,,,(1,1);$ in the complex plane, or if you prefer the complex notation: the point $;-1,,,,1+i;$ , so directly:
$$(zcos z)'=cos z-zsin zimpliesleft.int_gamma (cos z-zsin z)dz=zcos zright|_{-1}^{1+i}=ldots$$
answered Dec 13 '18 at 14:44
DonAntonioDonAntonio
177k1492225
add a comment |
Some highlights:
$$z=3e^{it}implies |z|=3,,,,dz=3ie^{it},dtimplies int_gamma |z|dz=int_0^pi 3cdot3ie^{it},dt=ldots$$
The other one is a segment of straight line joining the point $;(-i,0),,,,(1,1);$ in the complex plane, or if you prefer the complex notation: the point $;-1,,,,1+i;$ , so directly:
$$(zcos z)'=cos z-zsin zimpliesleft.int_gamma (cos z-zsin z)dz=zcos zright|_{-1}^{1+i}=ldots$$
answered Dec 13 '18 at 14:44
DonAntonioDonAntonio
177k1492225
add a comment |
Some highlights:
$$z=3e^{it}implies |z|=3,,,,dz=3ie^{it},dtimplies int_gamma |z|dz=int_0^pi 3cdot3ie^{it},dt=ldots$$
The other one is a segment of straight line joining the point $;(-i,0),,,,(1,1);$ in the complex plane, or if you prefer the complex notation: the point $;-1,,,,1+i;$ , so directly:
$$(zcos z)'=cos z-zsin zimpliesleft.int_gamma (cos z-zsin z)dz=zcos zright|_{-1}^{1+i}=ldots$$
answered Dec 13 '18 at 14:44
DonAntonioDonAntonio
177k1492225
Some highlights:
$$z=3e^{it}implies |z|=3,,,,dz=3ie^{it},dtimplies int_gamma |z|dz=int_0^pi 3cdot3ie^{it},dt=ldots$$
The other one is a segment of straight line joining the point $;(-i,0),,,,(1,1);$ in the complex plane, or if you prefer the complex notation: the point $;-1,,,,1+i;$ , so directly:
$$(zcos z)'=cos z-zsin zimpliesleft.int_gamma (cos z-zsin z)dz=zcos zright|_{-1}^{1+i}=ldots$$
answered Dec 13 '18 at 14:44
DonAntonioDonAntonio
177k1492225
edited Dec 13 '18 at 14:58
answered Dec 13 '18 at 14:44
DonAntonioDonAntonio
177k1492225
answered Dec 13 '18 at 14:44
DonAntonioDonAntonio
177k1492225
answered Dec 13 '18 at 14:44
DonAntonioDonAntonio
177k1492225
177k1492225
add a comment |
add a comment |
What did you try? Can you see, for example, that in (1) we have $;|z|=3;$ when $;zingamma(z);$ ?
– DonAntonio
Dec 13 '18 at 10:47
I know that to use the Theorem that an anti-derivative must exist but I am having difficulty trying to find this. I am also unsure where the value |Z|=3 has come from
– Aoife Coyle
Dec 13 '18 at 10:54
If you're going to do line integral on the canonical circle of radius $;3;$ , namely: $;{zinBbb C;|;z=3e^{it};,;;tin[0,2pi]};$ , then on this line $;|z|=3;$ ...right?
– DonAntonio
Dec 13 '18 at 11:05
Yes I can see this. Do I need to show ∫ f(γ(z).γ'(z) dz ?
– Aoife Coyle
Dec 13 '18 at 11:11
Note that there is no $gamma(z)$. (I edited the question accordingly). $gamma$ is a curve assigning to a real number $t$ the complex number $gamma(t)$. — When @DonAntonio wrote "$zingamma(z)$" he ment $z=gamma(t)$.
– Christoph
Dec 13 '18 at 11:16