Plotting odd complex functions w/o computer help
$begingroup$
I am evaluating a complex integral that utilises the Cauchy Integral Formula and its properties.
In the book I'm reading, they give examples of evaluating integrals using CIT by graphing them, which really does help to see if points of a function is analytic in a certain domain.
For example, in evaluating this integral - how would I be able to plot the function shown in this integral, where C is a circle $|z|=1$ traversed once counter clockwise. $$int_{C}frac{z+i}{z^3+2z^2}$$
Thanks.
complex-analysis complex-numbers
asked Jan 7 at 1:25
Dr.DoofusDr.Doofus
12612
$endgroup$
add a comment |
$begingroup$
I am evaluating a complex integral that utilises the Cauchy Integral Formula and its properties.
In the book I'm reading, they give examples of evaluating integrals using CIT by graphing them, which really does help to see if points of a function is analytic in a certain domain.
For example, in evaluating this integral - how would I be able to plot the function shown in this integral, where C is a circle $|z|=1$ traversed once counter clockwise. $$int_{C}frac{z+i}{z^3+2z^2}$$
Thanks.
complex-analysis complex-numbers
asked Jan 7 at 1:25
Dr.DoofusDr.Doofus
12612
$endgroup$
$begingroup$
You may be able to find ways of visualizing complex mappings but in general there is no way to plot the graph of a function from $mathbb{C}to mathbb{C}$ analogously to graphs of functions from $mathbb{R} to mathbb{R}$ (or up to 3-space). The point $(z, f(z))$ is in some sense a point of $mathbb{R}^4$ which our brains find difficult to visualize...
$endgroup$
– LoveTooNap29
Jan 7 at 1:34
$begingroup$
I understand. This is something that I will come to terms with. I'm used to plotting basic real functions. Thanks for that. So in a case like this, how would one go about solving it? I understand that the integrand fails to be analytic point z = 0, but where to from there?
$endgroup$
– Dr.Doofus
Jan 7 at 1:43
$begingroup$
Indeed, it would be relatively straightforward to apply residue theorem (although probably tedious) but it appears your only given tool is Cauchy’s integral formula, correct? Just want to make sure before I offer a sketch of a solution.
$endgroup$
– LoveTooNap29
Jan 7 at 2:02
$begingroup$
Yes, sir. It's how they want us to go about it.
$endgroup$
– Dr.Doofus
Jan 7 at 2:14
add a comment |
$begingroup$
I am evaluating a complex integral that utilises the Cauchy Integral Formula and its properties.
In the book I'm reading, they give examples of evaluating integrals using CIT by graphing them, which really does help to see if points of a function is analytic in a certain domain.
For example, in evaluating this integral - how would I be able to plot the function shown in this integral, where C is a circle $|z|=1$ traversed once counter clockwise. $$int_{C}frac{z+i}{z^3+2z^2}$$
Thanks.
complex-analysis complex-numbers
asked Jan 7 at 1:25
Dr.DoofusDr.Doofus
12612
$endgroup$
I am evaluating a complex integral that utilises the Cauchy Integral Formula and its properties.
In the book I'm reading, they give examples of evaluating integrals using CIT by graphing them, which really does help to see if points of a function is analytic in a certain domain.
For example, in evaluating this integral - how would I be able to plot the function shown in this integral, where C is a circle $|z|=1$ traversed once counter clockwise. $$int_{C}frac{z+i}{z^3+2z^2}$$
Thanks.
complex-analysis complex-numbers
complex-analysis complex-numbers
asked Jan 7 at 1:25
Dr.DoofusDr.Doofus
12612
asked Jan 7 at 1:25
Dr.DoofusDr.Doofus
12612
asked Jan 7 at 1:25
Dr.DoofusDr.Doofus
12612
asked Jan 7 at 1:25
Dr.DoofusDr.Doofus
12612
asked Jan 7 at 1:25
Dr.DoofusDr.Doofus
12612
12612
$begingroup$
You may be able to find ways of visualizing complex mappings but in general there is no way to plot the graph of a function from $mathbb{C}to mathbb{C}$ analogously to graphs of functions from $mathbb{R} to mathbb{R}$ (or up to 3-space). The point $(z, f(z))$ is in some sense a point of $mathbb{R}^4$ which our brains find difficult to visualize...
$endgroup$
– LoveTooNap29
Jan 7 at 1:34
$begingroup$
I understand. This is something that I will come to terms with. I'm used to plotting basic real functions. Thanks for that. So in a case like this, how would one go about solving it? I understand that the integrand fails to be analytic point z = 0, but where to from there?
$endgroup$
– Dr.Doofus
Jan 7 at 1:43
$begingroup$
Indeed, it would be relatively straightforward to apply residue theorem (although probably tedious) but it appears your only given tool is Cauchy’s integral formula, correct? Just want to make sure before I offer a sketch of a solution.
$endgroup$
– LoveTooNap29
Jan 7 at 2:02
$begingroup$
Yes, sir. It's how they want us to go about it.
$endgroup$
– Dr.Doofus
Jan 7 at 2:14
add a comment |
$begingroup$
You may be able to find ways of visualizing complex mappings but in general there is no way to plot the graph of a function from $mathbb{C}to mathbb{C}$ analogously to graphs of functions from $mathbb{R} to mathbb{R}$ (or up to 3-space). The point $(z, f(z))$ is in some sense a point of $mathbb{R}^4$ which our brains find difficult to visualize...
$endgroup$
– LoveTooNap29
Jan 7 at 1:34
$begingroup$
I understand. This is something that I will come to terms with. I'm used to plotting basic real functions. Thanks for that. So in a case like this, how would one go about solving it? I understand that the integrand fails to be analytic point z = 0, but where to from there?
$endgroup$
– Dr.Doofus
Jan 7 at 1:43
$begingroup$
Indeed, it would be relatively straightforward to apply residue theorem (although probably tedious) but it appears your only given tool is Cauchy’s integral formula, correct? Just want to make sure before I offer a sketch of a solution.
$endgroup$
– LoveTooNap29
Jan 7 at 2:02
$begingroup$
Yes, sir. It's how they want us to go about it.
$endgroup$
– Dr.Doofus
Jan 7 at 2:14
$begingroup$
You may be able to find ways of visualizing complex mappings but in general there is no way to plot the graph of a function from $mathbb{C}to mathbb{C}$ analogously to graphs of functions from $mathbb{R} to mathbb{R}$ (or up to 3-space). The point $(z, f(z))$ is in some sense a point of $mathbb{R}^4$ which our brains find difficult to visualize...
$endgroup$
– LoveTooNap29
Jan 7 at 1:34
$begingroup$
You may be able to find ways of visualizing complex mappings but in general there is no way to plot the graph of a function from $mathbb{C}to mathbb{C}$ analogously to graphs of functions from $mathbb{R} to mathbb{R}$ (or up to 3-space). The point $(z, f(z))$ is in some sense a point of $mathbb{R}^4$ which our brains find difficult to visualize...
$endgroup$
– LoveTooNap29
Jan 7 at 1:34
$begingroup$
I understand. This is something that I will come to terms with. I'm used to plotting basic real functions. Thanks for that. So in a case like this, how would one go about solving it? I understand that the integrand fails to be analytic point z = 0, but where to from there?
$endgroup$
– Dr.Doofus
Jan 7 at 1:43
$begingroup$
I understand. This is something that I will come to terms with. I'm used to plotting basic real functions. Thanks for that. So in a case like this, how would one go about solving it? I understand that the integrand fails to be analytic point z = 0, but where to from there?
$endgroup$
– Dr.Doofus
Jan 7 at 1:43
$begingroup$
Indeed, it would be relatively straightforward to apply residue theorem (although probably tedious) but it appears your only given tool is Cauchy’s integral formula, correct? Just want to make sure before I offer a sketch of a solution.
$endgroup$
– LoveTooNap29
Jan 7 at 2:02
$begingroup$
Indeed, it would be relatively straightforward to apply residue theorem (although probably tedious) but it appears your only given tool is Cauchy’s integral formula, correct? Just want to make sure before I offer a sketch of a solution.
$endgroup$
– LoveTooNap29
Jan 7 at 2:02
$begingroup$
Yes, sir. It's how they want us to go about it.
$endgroup$
– Dr.Doofus
Jan 7 at 2:14
$begingroup$
Yes, sir. It's how they want us to go about it.
$endgroup$
– Dr.Doofus
Jan 7 at 2:14
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The idea is that you only need to know where the function is singular
In your case it is at $z= 0$ and $z = -2$, but the second point is outside the region of integration, so we do not care about it
$$
oint_C frac{z + i}{z^2(z + 2)} {rm d}z= oint_C frac{f(z)}{z^2}{rm d}z
$$
where the function
$$
f(z) = frac{z + i}{z + 2}
$$
is well behaved in the region $|z| leq 1$. The problem is at $z = 0$, where you have second order singularity, so you can use Cauchy's integral form
$$
left.frac{{rm d}^n f}{{rm d}z^n}right|_{z = 0} = frac{n!}{2pi i}oint_C frac{f(z)}{z^{n + 1}}{rm d}z
$$
with $n = 1$ which yields
$$
oint_Cfrac{z + i}{z^2(z + 2)} = 2 pi i frac{{rm d}}{{rm d}z}left(frac{z + i}{z + 2}right)_{z = 0} = 2pi i left(frac{1}{2} - frac{i}{4} right)
$$
answered Jan 7 at 2:24
caveraccaverac
14.8k31130
$endgroup$
$begingroup$
This is great. Thank you. Better than the provided solutions!
$endgroup$
– Dr.Doofus
Jan 7 at 2:27
$begingroup$
Ah it looks like you beat me to it! Excellent visual too. Good to at least confirm by someone else's work that I can still recall this material correctly.
$endgroup$
– LoveTooNap29
Jan 7 at 2:30
add a comment |
$begingroup$
Sketch:
Note that the pole at $z=0$ has order $2$. Use the slightly generalized Cauchy formula for derivatives,
$$f^{(n)}(a)=frac{n!}{2pi i}oint frac{f(z)}{(z-a)^{n+1}} mathrm{d}z,$$
Write the original integrand as $g(z)$ and identify $f(z)$ from $g(z)=f(z)/z^2$ in your case for $n=1$ and $a=0$. Then compute the first derivative of $f$ at $a=0$ to compute the original integral (remember to multiply this by $2pi i$). This yields $pi (frac12 +i)$.
Be sure to check all hypotheses of the Cauchy integral formula though to justify this rigorously, though (that $f$ is holomorphic inside the curve is required, not $g$, etc). Comment if you want some steps expanded, but preferably with where you got stuck also, or if you spot any typos, etc. It has been some time since I've gotten my hands dirty with contour integrals with only Cauchy's integral formula, however, the example on wikipedia is rather illustrative, though the poles there are only of order 1, unlike here.
answered Jan 7 at 2:28
LoveTooNap29LoveTooNap29
1,1421614
$endgroup$
$begingroup$
Thank you. I genuinely appreciate it.
$endgroup$
– Dr.Doofus
Jan 7 at 2:32
$begingroup$
Sorry, didn't know you were working on an answer as well (+1)
$endgroup$
– caverac
Jan 7 at 2:33
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The idea is that you only need to know where the function is singular
In your case it is at $z= 0$ and $z = -2$, but the second point is outside the region of integration, so we do not care about it
$$
oint_C frac{z + i}{z^2(z + 2)} {rm d}z= oint_C frac{f(z)}{z^2}{rm d}z
$$
where the function
$$
f(z) = frac{z + i}{z + 2}
$$
is well behaved in the region $|z| leq 1$. The problem is at $z = 0$, where you have second order singularity, so you can use Cauchy's integral form
$$
left.frac{{rm d}^n f}{{rm d}z^n}right|_{z = 0} = frac{n!}{2pi i}oint_C frac{f(z)}{z^{n + 1}}{rm d}z
$$
with $n = 1$ which yields
$$
oint_Cfrac{z + i}{z^2(z + 2)} = 2 pi i frac{{rm d}}{{rm d}z}left(frac{z + i}{z + 2}right)_{z = 0} = 2pi i left(frac{1}{2} - frac{i}{4} right)
$$
answered Jan 7 at 2:24
caveraccaverac
14.8k31130
$endgroup$
$begingroup$
This is great. Thank you. Better than the provided solutions!
$endgroup$
– Dr.Doofus
Jan 7 at 2:27
$begingroup$
Ah it looks like you beat me to it! Excellent visual too. Good to at least confirm by someone else's work that I can still recall this material correctly.
$endgroup$
– LoveTooNap29
Jan 7 at 2:30
add a comment |
$begingroup$
The idea is that you only need to know where the function is singular
In your case it is at $z= 0$ and $z = -2$, but the second point is outside the region of integration, so we do not care about it
$$
oint_C frac{z + i}{z^2(z + 2)} {rm d}z= oint_C frac{f(z)}{z^2}{rm d}z
$$
where the function
$$
f(z) = frac{z + i}{z + 2}
$$
is well behaved in the region $|z| leq 1$. The problem is at $z = 0$, where you have second order singularity, so you can use Cauchy's integral form
$$
left.frac{{rm d}^n f}{{rm d}z^n}right|_{z = 0} = frac{n!}{2pi i}oint_C frac{f(z)}{z^{n + 1}}{rm d}z
$$
with $n = 1$ which yields
$$
oint_Cfrac{z + i}{z^2(z + 2)} = 2 pi i frac{{rm d}}{{rm d}z}left(frac{z + i}{z + 2}right)_{z = 0} = 2pi i left(frac{1}{2} - frac{i}{4} right)
$$
answered Jan 7 at 2:24
caveraccaverac
14.8k31130
$endgroup$
$begingroup$
This is great. Thank you. Better than the provided solutions!
$endgroup$
– Dr.Doofus
Jan 7 at 2:27
$begingroup$
Ah it looks like you beat me to it! Excellent visual too. Good to at least confirm by someone else's work that I can still recall this material correctly.
$endgroup$
– LoveTooNap29
Jan 7 at 2:30
add a comment |
$begingroup$
The idea is that you only need to know where the function is singular
In your case it is at $z= 0$ and $z = -2$, but the second point is outside the region of integration, so we do not care about it
$$
oint_C frac{z + i}{z^2(z + 2)} {rm d}z= oint_C frac{f(z)}{z^2}{rm d}z
$$
where the function
$$
f(z) = frac{z + i}{z + 2}
$$
is well behaved in the region $|z| leq 1$. The problem is at $z = 0$, where you have second order singularity, so you can use Cauchy's integral form
$$
left.frac{{rm d}^n f}{{rm d}z^n}right|_{z = 0} = frac{n!}{2pi i}oint_C frac{f(z)}{z^{n + 1}}{rm d}z
$$
with $n = 1$ which yields
$$
oint_Cfrac{z + i}{z^2(z + 2)} = 2 pi i frac{{rm d}}{{rm d}z}left(frac{z + i}{z + 2}right)_{z = 0} = 2pi i left(frac{1}{2} - frac{i}{4} right)
$$
answered Jan 7 at 2:24
caveraccaverac
14.8k31130
$endgroup$
The idea is that you only need to know where the function is singular
In your case it is at $z= 0$ and $z = -2$, but the second point is outside the region of integration, so we do not care about it
$$
oint_C frac{z + i}{z^2(z + 2)} {rm d}z= oint_C frac{f(z)}{z^2}{rm d}z
$$
where the function
$$
f(z) = frac{z + i}{z + 2}
$$
is well behaved in the region $|z| leq 1$. The problem is at $z = 0$, where you have second order singularity, so you can use Cauchy's integral form
$$
left.frac{{rm d}^n f}{{rm d}z^n}right|_{z = 0} = frac{n!}{2pi i}oint_C frac{f(z)}{z^{n + 1}}{rm d}z
$$
with $n = 1$ which yields
$$
oint_Cfrac{z + i}{z^2(z + 2)} = 2 pi i frac{{rm d}}{{rm d}z}left(frac{z + i}{z + 2}right)_{z = 0} = 2pi i left(frac{1}{2} - frac{i}{4} right)
$$
answered Jan 7 at 2:24
caveraccaverac
14.8k31130
answered Jan 7 at 2:24
caveraccaverac
14.8k31130
answered Jan 7 at 2:24
caveraccaverac
14.8k31130
answered Jan 7 at 2:24
caveraccaverac
14.8k31130
14.8k31130
$begingroup$
This is great. Thank you. Better than the provided solutions!
$endgroup$
– Dr.Doofus
Jan 7 at 2:27
$begingroup$
Ah it looks like you beat me to it! Excellent visual too. Good to at least confirm by someone else's work that I can still recall this material correctly.
$endgroup$
– LoveTooNap29
Jan 7 at 2:30
add a comment |
$begingroup$
This is great. Thank you. Better than the provided solutions!
$endgroup$
– Dr.Doofus
Jan 7 at 2:27
$begingroup$
Ah it looks like you beat me to it! Excellent visual too. Good to at least confirm by someone else's work that I can still recall this material correctly.
$endgroup$
– LoveTooNap29
Jan 7 at 2:30
$begingroup$
This is great. Thank you. Better than the provided solutions!
$endgroup$
– Dr.Doofus
Jan 7 at 2:27
$begingroup$
This is great. Thank you. Better than the provided solutions!
$endgroup$
– Dr.Doofus
Jan 7 at 2:27
$begingroup$
Ah it looks like you beat me to it! Excellent visual too. Good to at least confirm by someone else's work that I can still recall this material correctly.
$endgroup$
– LoveTooNap29
Jan 7 at 2:30
$begingroup$
Ah it looks like you beat me to it! Excellent visual too. Good to at least confirm by someone else's work that I can still recall this material correctly.
$endgroup$
– LoveTooNap29
Jan 7 at 2:30
add a comment |
$begingroup$
Sketch:
Note that the pole at $z=0$ has order $2$. Use the slightly generalized Cauchy formula for derivatives,
$$f^{(n)}(a)=frac{n!}{2pi i}oint frac{f(z)}{(z-a)^{n+1}} mathrm{d}z,$$
Write the original integrand as $g(z)$ and identify $f(z)$ from $g(z)=f(z)/z^2$ in your case for $n=1$ and $a=0$. Then compute the first derivative of $f$ at $a=0$ to compute the original integral (remember to multiply this by $2pi i$). This yields $pi (frac12 +i)$.
Be sure to check all hypotheses of the Cauchy integral formula though to justify this rigorously, though (that $f$ is holomorphic inside the curve is required, not $g$, etc). Comment if you want some steps expanded, but preferably with where you got stuck also, or if you spot any typos, etc. It has been some time since I've gotten my hands dirty with contour integrals with only Cauchy's integral formula, however, the example on wikipedia is rather illustrative, though the poles there are only of order 1, unlike here.
answered Jan 7 at 2:28
LoveTooNap29LoveTooNap29
1,1421614
$endgroup$
$begingroup$
Thank you. I genuinely appreciate it.
$endgroup$
– Dr.Doofus
Jan 7 at 2:32
$begingroup$
Sorry, didn't know you were working on an answer as well (+1)
$endgroup$
– caverac
Jan 7 at 2:33
add a comment |
$begingroup$
Sketch:
Note that the pole at $z=0$ has order $2$. Use the slightly generalized Cauchy formula for derivatives,
$$f^{(n)}(a)=frac{n!}{2pi i}oint frac{f(z)}{(z-a)^{n+1}} mathrm{d}z,$$
Write the original integrand as $g(z)$ and identify $f(z)$ from $g(z)=f(z)/z^2$ in your case for $n=1$ and $a=0$. Then compute the first derivative of $f$ at $a=0$ to compute the original integral (remember to multiply this by $2pi i$). This yields $pi (frac12 +i)$.
Be sure to check all hypotheses of the Cauchy integral formula though to justify this rigorously, though (that $f$ is holomorphic inside the curve is required, not $g$, etc). Comment if you want some steps expanded, but preferably with where you got stuck also, or if you spot any typos, etc. It has been some time since I've gotten my hands dirty with contour integrals with only Cauchy's integral formula, however, the example on wikipedia is rather illustrative, though the poles there are only of order 1, unlike here.
answered Jan 7 at 2:28
LoveTooNap29LoveTooNap29
1,1421614
$endgroup$
$begingroup$
Thank you. I genuinely appreciate it.
$endgroup$
– Dr.Doofus
Jan 7 at 2:32
$begingroup$
Sorry, didn't know you were working on an answer as well (+1)
$endgroup$
– caverac
Jan 7 at 2:33
add a comment |
$begingroup$
Sketch:
Note that the pole at $z=0$ has order $2$. Use the slightly generalized Cauchy formula for derivatives,
$$f^{(n)}(a)=frac{n!}{2pi i}oint frac{f(z)}{(z-a)^{n+1}} mathrm{d}z,$$
Write the original integrand as $g(z)$ and identify $f(z)$ from $g(z)=f(z)/z^2$ in your case for $n=1$ and $a=0$. Then compute the first derivative of $f$ at $a=0$ to compute the original integral (remember to multiply this by $2pi i$). This yields $pi (frac12 +i)$.
Be sure to check all hypotheses of the Cauchy integral formula though to justify this rigorously, though (that $f$ is holomorphic inside the curve is required, not $g$, etc). Comment if you want some steps expanded, but preferably with where you got stuck also, or if you spot any typos, etc. It has been some time since I've gotten my hands dirty with contour integrals with only Cauchy's integral formula, however, the example on wikipedia is rather illustrative, though the poles there are only of order 1, unlike here.
answered Jan 7 at 2:28
LoveTooNap29LoveTooNap29
1,1421614
$endgroup$
Sketch:
Note that the pole at $z=0$ has order $2$. Use the slightly generalized Cauchy formula for derivatives,
$$f^{(n)}(a)=frac{n!}{2pi i}oint frac{f(z)}{(z-a)^{n+1}} mathrm{d}z,$$
Write the original integrand as $g(z)$ and identify $f(z)$ from $g(z)=f(z)/z^2$ in your case for $n=1$ and $a=0$. Then compute the first derivative of $f$ at $a=0$ to compute the original integral (remember to multiply this by $2pi i$). This yields $pi (frac12 +i)$.
Be sure to check all hypotheses of the Cauchy integral formula though to justify this rigorously, though (that $f$ is holomorphic inside the curve is required, not $g$, etc). Comment if you want some steps expanded, but preferably with where you got stuck also, or if you spot any typos, etc. It has been some time since I've gotten my hands dirty with contour integrals with only Cauchy's integral formula, however, the example on wikipedia is rather illustrative, though the poles there are only of order 1, unlike here.
answered Jan 7 at 2:28
LoveTooNap29LoveTooNap29
1,1421614
answered Jan 7 at 2:28
LoveTooNap29LoveTooNap29
1,1421614
answered Jan 7 at 2:28
LoveTooNap29LoveTooNap29
1,1421614
answered Jan 7 at 2:28
LoveTooNap29LoveTooNap29
1,1421614
1,1421614
$begingroup$
Thank you. I genuinely appreciate it.
$endgroup$
– Dr.Doofus
Jan 7 at 2:32
$begingroup$
Sorry, didn't know you were working on an answer as well (+1)
$endgroup$
– caverac
Jan 7 at 2:33
add a comment |
$begingroup$
Thank you. I genuinely appreciate it.
$endgroup$
– Dr.Doofus
Jan 7 at 2:32
$begingroup$
Sorry, didn't know you were working on an answer as well (+1)
$endgroup$
– caverac
Jan 7 at 2:33
$begingroup$
Thank you. I genuinely appreciate it.
$endgroup$
– Dr.Doofus
Jan 7 at 2:32
$begingroup$
Thank you. I genuinely appreciate it.
$endgroup$
– Dr.Doofus
Jan 7 at 2:32
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Sorry, didn't know you were working on an answer as well (+1)
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– caverac
Jan 7 at 2:33
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Sorry, didn't know you were working on an answer as well (+1)
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– caverac
Jan 7 at 2:33
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You may be able to find ways of visualizing complex mappings but in general there is no way to plot the graph of a function from $mathbb{C}to mathbb{C}$ analogously to graphs of functions from $mathbb{R} to mathbb{R}$ (or up to 3-space). The point $(z, f(z))$ is in some sense a point of $mathbb{R}^4$ which our brains find difficult to visualize...
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– LoveTooNap29
Jan 7 at 1:34
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I understand. This is something that I will come to terms with. I'm used to plotting basic real functions. Thanks for that. So in a case like this, how would one go about solving it? I understand that the integrand fails to be analytic point z = 0, but where to from there?
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– Dr.Doofus
Jan 7 at 1:43
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Indeed, it would be relatively straightforward to apply residue theorem (although probably tedious) but it appears your only given tool is Cauchy’s integral formula, correct? Just want to make sure before I offer a sketch of a solution.
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– LoveTooNap29
Jan 7 at 2:02
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Yes, sir. It's how they want us to go about it.
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– Dr.Doofus
Jan 7 at 2:14