Plotting odd complex functions w/o computer help












1














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.










share|cite|improve this question






















  • 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...
    – LoveTooNap29
    4 hours ago










  • 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?
    – Dr.Doofus
    4 hours ago










  • 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.
    – LoveTooNap29
    4 hours ago










  • Yes, sir. It's how they want us to go about it.
    – Dr.Doofus
    4 hours ago
















1














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.










share|cite|improve this question






















  • 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...
    – LoveTooNap29
    4 hours ago










  • 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?
    – Dr.Doofus
    4 hours ago










  • 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.
    – LoveTooNap29
    4 hours ago










  • Yes, sir. It's how they want us to go about it.
    – Dr.Doofus
    4 hours ago














1












1








1







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.










share|cite|improve this question













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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked 5 hours ago









Dr.Doofus

9210




9210












  • 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...
    – LoveTooNap29
    4 hours ago










  • 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?
    – Dr.Doofus
    4 hours ago










  • 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.
    – LoveTooNap29
    4 hours ago










  • Yes, sir. It's how they want us to go about it.
    – Dr.Doofus
    4 hours ago


















  • 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...
    – LoveTooNap29
    4 hours ago










  • 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?
    – Dr.Doofus
    4 hours ago










  • 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.
    – LoveTooNap29
    4 hours ago










  • Yes, sir. It's how they want us to go about it.
    – Dr.Doofus
    4 hours ago
















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...
– LoveTooNap29
4 hours ago




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...
– LoveTooNap29
4 hours ago












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?
– Dr.Doofus
4 hours ago




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?
– Dr.Doofus
4 hours ago












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.
– LoveTooNap29
4 hours ago




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.
– LoveTooNap29
4 hours ago












Yes, sir. It's how they want us to go about it.
– Dr.Doofus
4 hours ago




Yes, sir. It's how they want us to go about it.
– Dr.Doofus
4 hours ago










2 Answers
2






active

oldest

votes


















2














The idea is that you only need to know where the function is singular



enter image description here



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



$$
oint_Cfrac{z + i}{z^2(z + 2)} = 2 pi i frac{{rm d}}{{rm d}z}left(frac{z + i}{z + 2}right)_{z = 0} = 2pi i left(frac{1}{2} - frac{i}{4} right)
$$






share|cite|improve this answer





















  • This is great. Thank you. Better than the provided solutions!
    – Dr.Doofus
    4 hours ago










  • 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.
    – LoveTooNap29
    3 hours ago



















2














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.






share|cite|improve this answer





















  • Thank you. I genuinely appreciate it.
    – Dr.Doofus
    3 hours ago










  • Sorry, didn't know you were working on an answer as well (+1)
    – caverac
    3 hours ago











Your Answer





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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









2














The idea is that you only need to know where the function is singular



enter image description here



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



$$
oint_Cfrac{z + i}{z^2(z + 2)} = 2 pi i frac{{rm d}}{{rm d}z}left(frac{z + i}{z + 2}right)_{z = 0} = 2pi i left(frac{1}{2} - frac{i}{4} right)
$$






share|cite|improve this answer





















  • This is great. Thank you. Better than the provided solutions!
    – Dr.Doofus
    4 hours ago










  • 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.
    – LoveTooNap29
    3 hours ago
















2














The idea is that you only need to know where the function is singular



enter image description here



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



$$
oint_Cfrac{z + i}{z^2(z + 2)} = 2 pi i frac{{rm d}}{{rm d}z}left(frac{z + i}{z + 2}right)_{z = 0} = 2pi i left(frac{1}{2} - frac{i}{4} right)
$$






share|cite|improve this answer





















  • This is great. Thank you. Better than the provided solutions!
    – Dr.Doofus
    4 hours ago










  • 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.
    – LoveTooNap29
    3 hours ago














2












2








2






The idea is that you only need to know where the function is singular



enter image description here



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



$$
oint_Cfrac{z + i}{z^2(z + 2)} = 2 pi i frac{{rm d}}{{rm d}z}left(frac{z + i}{z + 2}right)_{z = 0} = 2pi i left(frac{1}{2} - frac{i}{4} right)
$$






share|cite|improve this answer












The idea is that you only need to know where the function is singular



enter image description here



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



$$
oint_Cfrac{z + i}{z^2(z + 2)} = 2 pi i frac{{rm d}}{{rm d}z}left(frac{z + i}{z + 2}right)_{z = 0} = 2pi i left(frac{1}{2} - frac{i}{4} right)
$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 4 hours ago









caverac

13.9k21130




13.9k21130












  • This is great. Thank you. Better than the provided solutions!
    – Dr.Doofus
    4 hours ago










  • 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.
    – LoveTooNap29
    3 hours ago


















  • This is great. Thank you. Better than the provided solutions!
    – Dr.Doofus
    4 hours ago










  • 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.
    – LoveTooNap29
    3 hours ago
















This is great. Thank you. Better than the provided solutions!
– Dr.Doofus
4 hours ago




This is great. Thank you. Better than the provided solutions!
– Dr.Doofus
4 hours ago












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.
– LoveTooNap29
3 hours ago




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.
– LoveTooNap29
3 hours ago











2














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.






share|cite|improve this answer





















  • Thank you. I genuinely appreciate it.
    – Dr.Doofus
    3 hours ago










  • Sorry, didn't know you were working on an answer as well (+1)
    – caverac
    3 hours ago
















2














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.






share|cite|improve this answer





















  • Thank you. I genuinely appreciate it.
    – Dr.Doofus
    3 hours ago










  • Sorry, didn't know you were working on an answer as well (+1)
    – caverac
    3 hours ago














2












2








2






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.






share|cite|improve this answer












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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 4 hours ago









LoveTooNap29

1,0241613




1,0241613












  • Thank you. I genuinely appreciate it.
    – Dr.Doofus
    3 hours ago










  • Sorry, didn't know you were working on an answer as well (+1)
    – caverac
    3 hours ago


















  • Thank you. I genuinely appreciate it.
    – Dr.Doofus
    3 hours ago










  • Sorry, didn't know you were working on an answer as well (+1)
    – caverac
    3 hours ago
















Thank you. I genuinely appreciate it.
– Dr.Doofus
3 hours ago




Thank you. I genuinely appreciate it.
– Dr.Doofus
3 hours ago












Sorry, didn't know you were working on an answer as well (+1)
– caverac
3 hours ago




Sorry, didn't know you were working on an answer as well (+1)
– caverac
3 hours ago


















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