Use The Fundamental Theorem of Contour Integration or otherwise to evaluate the following integrals. (If it...
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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
12.5k1642
asked Dec 13 '18 at 10:31
Aoife CoyleAoife Coyle
125
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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.
add a comment |
$begingroup$
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
12.5k1642
asked Dec 13 '18 at 10:31
Aoife CoyleAoife Coyle
125
$endgroup$
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.
$begingroup$
What did you try? Can you see, for example, that in (1) we have $;|z|=3;$ when $;zingamma(z);$ ?
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– DonAntonio
Dec 13 '18 at 10:47
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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
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– Aoife Coyle
Dec 13 '18 at 10:54
$begingroup$
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?
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– DonAntonio
Dec 13 '18 at 11:05
$begingroup$
Yes I can see this. Do I need to show ∫ f(γ(z).γ'(z) dz ?
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– Aoife Coyle
Dec 13 '18 at 11:11
$begingroup$
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 $z=gamma(t)$.
$endgroup$
– Christoph
Dec 13 '18 at 11:16
add a comment |
$begingroup$
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
12.5k1642
asked Dec 13 '18 at 10:31
Aoife CoyleAoife Coyle
125
$endgroup$
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
12.5k1642
asked Dec 13 '18 at 10:31
Aoife CoyleAoife Coyle
125
edited Dec 13 '18 at 11:15
Christoph
12.5k1642
asked Dec 13 '18 at 10:31
Aoife CoyleAoife Coyle
125
edited Dec 13 '18 at 11:15
Christoph
12.5k1642
edited Dec 13 '18 at 11:15
Christoph
12.5k1642
edited Dec 13 '18 at 11:15
Christoph
12.5k1642
12.5k1642
asked Dec 13 '18 at 10:31
Aoife CoyleAoife Coyle
125
asked Dec 13 '18 at 10:31
Aoife CoyleAoife Coyle
125
asked Dec 13 '18 at 10:31
Aoife CoyleAoife Coyle
125
125
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.
$begingroup$
What did you try? Can you see, for example, that in (1) we have $;|z|=3;$ when $;zingamma(z);$ ?
$endgroup$
– DonAntonio
Dec 13 '18 at 10:47
$begingroup$
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
$endgroup$
– Aoife Coyle
Dec 13 '18 at 10:54
$begingroup$
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?
$endgroup$
– DonAntonio
Dec 13 '18 at 11:05
$begingroup$
Yes I can see this. Do I need to show ∫ f(γ(z).γ'(z) dz ?
$endgroup$
– Aoife Coyle
Dec 13 '18 at 11:11
$begingroup$
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 $z=gamma(t)$.
$endgroup$
– Christoph
Dec 13 '18 at 11:16
add a comment |
$begingroup$
What did you try? Can you see, for example, that in (1) we have $;|z|=3;$ when $;zingamma(z);$ ?
$endgroup$
– DonAntonio
Dec 13 '18 at 10:47
$begingroup$
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
$endgroup$
– Aoife Coyle
Dec 13 '18 at 10:54
$begingroup$
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?
$endgroup$
– DonAntonio
Dec 13 '18 at 11:05
$begingroup$
Yes I can see this. Do I need to show ∫ f(γ(z).γ'(z) dz ?
$endgroup$
– Aoife Coyle
Dec 13 '18 at 11:11
$begingroup$
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 $z=gamma(t)$.
$endgroup$
– Christoph
Dec 13 '18 at 11:16
$begingroup$
What did you try? Can you see, for example, that in (1) we have $;|z|=3;$ when $;zingamma(z);$ ?
$endgroup$
– DonAntonio
Dec 13 '18 at 10:47
$begingroup$
What did you try? Can you see, for example, that in (1) we have $;|z|=3;$ when $;zingamma(z);$ ?
$endgroup$
– DonAntonio
Dec 13 '18 at 10:47
$begingroup$
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
$endgroup$
– Aoife Coyle
Dec 13 '18 at 10:54
$begingroup$
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
$endgroup$
– Aoife Coyle
Dec 13 '18 at 10:54
$begingroup$
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?
$endgroup$
– DonAntonio
Dec 13 '18 at 11:05
$begingroup$
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?
$endgroup$
– DonAntonio
Dec 13 '18 at 11:05
$begingroup$
Yes I can see this. Do I need to show ∫ f(γ(z).γ'(z) dz ?
$endgroup$
– Aoife Coyle
Dec 13 '18 at 11:11
$begingroup$
Yes I can see this. Do I need to show ∫ f(γ(z).γ'(z) dz ?
$endgroup$
– Aoife Coyle
Dec 13 '18 at 11:11
$begingroup$
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 $z=gamma(t)$.
$endgroup$
– Christoph
Dec 13 '18 at 11:16
$begingroup$
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 $z=gamma(t)$.
$endgroup$
– Christoph
Dec 13 '18 at 11:16
add a comment |
1 Answer
1
active
oldest
votes
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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
180k1494233
$endgroup$
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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
180k1494233
$endgroup$
add a comment |
$begingroup$
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
180k1494233
$endgroup$
add a comment |
$begingroup$
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
180k1494233
$endgroup$
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
180k1494233
edited Dec 13 '18 at 14:58
answered Dec 13 '18 at 14:44
DonAntonioDonAntonio
180k1494233
answered Dec 13 '18 at 14:44
DonAntonioDonAntonio
180k1494233
answered Dec 13 '18 at 14:44
DonAntonioDonAntonio
180k1494233
180k1494233
add a comment |
add a comment |
$begingroup$
What did you try? Can you see, for example, that in (1) we have $;|z|=3;$ when $;zingamma(z);$ ?
$endgroup$
– DonAntonio
Dec 13 '18 at 10:47
$begingroup$
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
$endgroup$
– Aoife Coyle
Dec 13 '18 at 10:54
$begingroup$
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?
$endgroup$
– DonAntonio
Dec 13 '18 at 11:05
$begingroup$
Yes I can see this. Do I need to show ∫ f(γ(z).γ'(z) dz ?
$endgroup$
– Aoife Coyle
Dec 13 '18 at 11:11
$begingroup$
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 $z=gamma(t)$.
$endgroup$
– Christoph
Dec 13 '18 at 11:16