Mixing
Quick way of solving the contour integral $oint frac{1}{1+z^5} dz$
$begingroup$
Consider the contour integral in the complex plane:
$$oint frac{1}{1+z^5} dz$$
Here the contour is a circle with radius $3$ with centre in the origin. If we look at the poles, they need to satisfy $z^5 = -1$. So the solutions of the poles are given by:
begin{align*}
z_0 &= cos(frac{pi}{5}) + i sin(frac{pi}{5})\
z_1 &= cos(frac{3pi}{5}) + i sin(frac{3pi}{5})\
z_2 &= cos(pi) + i sin(pi) = -1\
z_3 &= cos(frac{7pi}{5}) + i sin(frac{7pi}{5})\
z_4 &= cos(frac{9pi}{5}) + i sin(frac{9pi}{5})
end{align*}
So one can use Cauchy's formula or the residue theorem to calculate for every solution the integral and then adding them up to get the full integral. But I have the feeling that there needs to be a more simple way of calculating the full contour integral. Can one just calculate the integral for one solution $z_i$ (like the simple solution $-1$) and then multiply by it $5$, suggesting that the others have the same value. This would make the calculation much efficienter.
EDIT:
I now see that $4$ solutions are symmetric (the solutions except $z=-1$) in the complex plane. If one approximates the solutions of the poles in decimals, one finds:
begin{align*}
z_0 &= 0.81 + 0.58i\
z_1 &= -0.31 + 0.95i\
z_2 &= -1\
z_3 &= -0.31 -0.95i\
z_4 &= 0.81 -0.58i
end{align*}
So there are four symmetric solutions. For instance $z_0$ is symmetric with $z_4$, they are mirrored around the x-axis. Could this mean they cancell each other out so we only need to calculate the integral for $z_2 = -1$?
integration complex-analysis contour-integration complex-integration
asked Jan 30 at 18:55
Belgium_PhysicsBelgium_Physics
322110
$endgroup$
add a comment |
$begingroup$
Consider the contour integral in the complex plane:
$$oint frac{1}{1+z^5} dz$$
Here the contour is a circle with radius $3$ with centre in the origin. If we look at the poles, they need to satisfy $z^5 = -1$. So the solutions of the poles are given by:
begin{align*}
z_0 &= cos(frac{pi}{5}) + i sin(frac{pi}{5})\
z_1 &= cos(frac{3pi}{5}) + i sin(frac{3pi}{5})\
z_2 &= cos(pi) + i sin(pi) = -1\
z_3 &= cos(frac{7pi}{5}) + i sin(frac{7pi}{5})\
z_4 &= cos(frac{9pi}{5}) + i sin(frac{9pi}{5})
end{align*}
So one can use Cauchy's formula or the residue theorem to calculate for every solution the integral and then adding them up to get the full integral. But I have the feeling that there needs to be a more simple way of calculating the full contour integral. Can one just calculate the integral for one solution $z_i$ (like the simple solution $-1$) and then multiply by it $5$, suggesting that the others have the same value. This would make the calculation much efficienter.
EDIT:
I now see that $4$ solutions are symmetric (the solutions except $z=-1$) in the complex plane. If one approximates the solutions of the poles in decimals, one finds:
begin{align*}
z_0 &= 0.81 + 0.58i\
z_1 &= -0.31 + 0.95i\
z_2 &= -1\
z_3 &= -0.31 -0.95i\
z_4 &= 0.81 -0.58i
end{align*}
So there are four symmetric solutions. For instance $z_0$ is symmetric with $z_4$, they are mirrored around the x-axis. Could this mean they cancell each other out so we only need to calculate the integral for $z_2 = -1$?
integration complex-analysis contour-integration complex-integration
asked Jan 30 at 18:55
Belgium_PhysicsBelgium_Physics
322110
$endgroup$
$begingroup$
No. You don't have any net residues at all. See below ... .
$endgroup$
– Oscar Lanzi
Jan 30 at 20:59
add a comment |
$begingroup$
Consider the contour integral in the complex plane:
$$oint frac{1}{1+z^5} dz$$
Here the contour is a circle with radius $3$ with centre in the origin. If we look at the poles, they need to satisfy $z^5 = -1$. So the solutions of the poles are given by:
begin{align*}
z_0 &= cos(frac{pi}{5}) + i sin(frac{pi}{5})\
z_1 &= cos(frac{3pi}{5}) + i sin(frac{3pi}{5})\
z_2 &= cos(pi) + i sin(pi) = -1\
z_3 &= cos(frac{7pi}{5}) + i sin(frac{7pi}{5})\
z_4 &= cos(frac{9pi}{5}) + i sin(frac{9pi}{5})
end{align*}
So one can use Cauchy's formula or the residue theorem to calculate for every solution the integral and then adding them up to get the full integral. But I have the feeling that there needs to be a more simple way of calculating the full contour integral. Can one just calculate the integral for one solution $z_i$ (like the simple solution $-1$) and then multiply by it $5$, suggesting that the others have the same value. This would make the calculation much efficienter.
EDIT:
I now see that $4$ solutions are symmetric (the solutions except $z=-1$) in the complex plane. If one approximates the solutions of the poles in decimals, one finds:
begin{align*}
z_0 &= 0.81 + 0.58i\
z_1 &= -0.31 + 0.95i\
z_2 &= -1\
z_3 &= -0.31 -0.95i\
z_4 &= 0.81 -0.58i
end{align*}
So there are four symmetric solutions. For instance $z_0$ is symmetric with $z_4$, they are mirrored around the x-axis. Could this mean they cancell each other out so we only need to calculate the integral for $z_2 = -1$?
integration complex-analysis contour-integration complex-integration
asked Jan 30 at 18:55
Belgium_PhysicsBelgium_Physics
322110
$endgroup$
Consider the contour integral in the complex plane:
$$oint frac{1}{1+z^5} dz$$
Here the contour is a circle with radius $3$ with centre in the origin. If we look at the poles, they need to satisfy $z^5 = -1$. So the solutions of the poles are given by:
begin{align*}
z_0 &= cos(frac{pi}{5}) + i sin(frac{pi}{5})\
z_1 &= cos(frac{3pi}{5}) + i sin(frac{3pi}{5})\
z_2 &= cos(pi) + i sin(pi) = -1\
z_3 &= cos(frac{7pi}{5}) + i sin(frac{7pi}{5})\
z_4 &= cos(frac{9pi}{5}) + i sin(frac{9pi}{5})
end{align*}
So one can use Cauchy's formula or the residue theorem to calculate for every solution the integral and then adding them up to get the full integral. But I have the feeling that there needs to be a more simple way of calculating the full contour integral. Can one just calculate the integral for one solution $z_i$ (like the simple solution $-1$) and then multiply by it $5$, suggesting that the others have the same value. This would make the calculation much efficienter.
EDIT:
I now see that $4$ solutions are symmetric (the solutions except $z=-1$) in the complex plane. If one approximates the solutions of the poles in decimals, one finds:
begin{align*}
z_0 &= 0.81 + 0.58i\
z_1 &= -0.31 + 0.95i\
z_2 &= -1\
z_3 &= -0.31 -0.95i\
z_4 &= 0.81 -0.58i
end{align*}
So there are four symmetric solutions. For instance $z_0$ is symmetric with $z_4$, they are mirrored around the x-axis. Could this mean they cancell each other out so we only need to calculate the integral for $z_2 = -1$?
integration complex-analysis contour-integration complex-integration
integration complex-analysis contour-integration complex-integration
asked Jan 30 at 18:55
Belgium_PhysicsBelgium_Physics
322110
asked Jan 30 at 18:55
Belgium_PhysicsBelgium_Physics
322110
edited Jan 30 at 19:46
Belgium_Physics
asked Jan 30 at 18:55
Belgium_PhysicsBelgium_Physics
322110
asked Jan 30 at 18:55
Belgium_PhysicsBelgium_Physics
322110
asked Jan 30 at 18:55
Belgium_PhysicsBelgium_Physics
322110
322110
$begingroup$
No. You don't have any net residues at all. See below ... .
$endgroup$
– Oscar Lanzi
Jan 30 at 20:59
add a comment |
$begingroup$
No. You don't have any net residues at all. See below ... .
$endgroup$
– Oscar Lanzi
Jan 30 at 20:59
$begingroup$
No. You don't have any net residues at all. See below ... .
$endgroup$
– Oscar Lanzi
Jan 30 at 20:59
$begingroup$
No. You don't have any net residues at all. See below ... .
$endgroup$
– Oscar Lanzi
Jan 30 at 20:59
add a comment |
5 Answers
5
active
oldest
votes
$begingroup$
HINT:
All of the poles lie on the unit circle. Hence, for any values of $R_1>1$ and $R_2>1$, we have
$$oint_{|z|=R_1}frac1{1+z^5},dz=oint_{|z|=R_2}frac1{1+z^5},dz$$
Now, what happens for $R_1=3$ and $R_2toinfty$? The answer is zero. This approach is tantamount to invoking the residue at infinity.
If one insists on summing the residues, then one may proceed in general as follows. First note that if $z^n+1=0$, then $z=z_k=e^{i(2k-1)pi/n}$ for $k=1,2,dots,n$.
Next, we calculate the residues of $frac{1}{z^n+1}$ at $z_k$ using L'Hospital's Rule as follows:
$$begin{align}
text{Res}left(frac1{z^n+1},z=z_kright)&=lim_{zto z_k}left(frac{z-z_k}{z^n+1}right)\\
&=frac{1}{nz_k^{n-1}}\\
&=frac1n e^{-i(2k-1)pi(n-1)/n}\\
&=-frac1n e^{-ipi/n}left(e^{i2pi/n}right)^k
end{align}$$
Summing all of the residues reveals for $n>1$
$$begin{align}
sum_{k=1}^n text{Res}left(frac1{1+z^n},z=z_kright)&=-frac1ne^{-ipi/n}sum_{k=1}^nleft(e^{i2pi/n}right)^k\\
&=-frac1ne^{=ipi/n} left(frac{e^{i2pi/n}-left(e^{i2pi/n}right)^{n+1}}{1-e^{i2pi/n}}right)\\
&=0
end{align}$$
Hence, for $n>1$ the residues of $frac1{1+z^n}$ add to zero. And we are done!
answered Jan 30 at 22:03
Mark ViolaMark Viola
134k1278177
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add a comment |
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Do you know what the residue at the infinity is?
There is a relation between the residues at complex numbers and the residue at infinity when you have finitely many poles.
The relation is $$sum_{k=0}^{n} operatorname{Res}_{a_k}(f) = -operatorname{Res}_{infty}(f),$$ where ${ a_{k} : n(Gamma,a_k)ne 0}_{k=0}^{n}$ are the poles (inside the path $Gamma$) of the function you want to compute the integral and $f$ is the function you are integrating. Here $n(gamma,z)$ is the index of the complex number $z$ with respect to the path $gamma$. I'm supposing that the curve only winds one time around the origin.
answered Jan 30 at 19:13
AlgebraicallyClosedAlgebraicallyClosed
709114
$endgroup$
$begingroup$
This can help and I will try it, but in my edit I suggest making use of the symmetrie of the solutions, could that be working?
$endgroup$
– Belgium_Physics
Jan 30 at 19:26
2
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The symmetries betweem the complex poles let you to compute fastly that integral, but introducing the infinity you are studying one more symmetrie, a symmetrie in the Riemann Sphere. What I want to say you is that the symmetries between the poles/zeros can appear by more than one way. But yes, you will be correct if you use the symmetries avaliable at the complex plane.
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– AlgebraicallyClosed
Jan 30 at 19:36
add a comment |
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Here is a subtle approach that completely gets around having to determine the residues.
Cut the plane along the positive real axis. Do your integration around a contour that goes like this:
Part 1: Counterclockwise from $z=3$ on the upper side of the cut to $x=3$ on the lower side of the cut.
Part 2: Follow the lower side of the cut out to $z=R$, where we will take the limit as $Rtoinfty$.
Part 3: Clockwise around the circle $|z|=R$ ending back on the upper side of the cut.
Part 4: Back to $z=3$ along the upper side of the cut.
This contour encloses no singularities, therefore the integral around it is zero. But also:
Parts 2 and 4: The real integral converges between $x=3$ and $x=infty$ by comparison with $1/x^5$ whose antiderivative $(-1/(4x^4))+C$ also converges. Therefore Part 2 and the additive inverse of Part 4 must have a common value, forcing these parts to cancel each other.
Part 3: The integrand is $O(1/R^5)$ and the length of the contour is $2pi R$, so by the triangle inequality the Part 3 integral is $O(1/R^4)to 0$.
So the Part 1 integral which is your original integral must also be zero.
In general: When you integrate a rational function around a contour that covers all of its poles and the function decreases faster than $1/z$ as $ztoinfty$, you must get zero. The residues inside the contour are forced to add up to zero, to conform.
answered Jan 30 at 20:50
Oscar LanziOscar Lanzi
13.5k12136
$endgroup$
$begingroup$
Is there a theorem or proof for this in general property? It looks something interesting that I have not encountered yet.
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– Belgium_Physics
Jan 30 at 22:17
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If you use the type of contour described here, for tge functions I include, you get this result. Parts 2 and 4 always cancel by convergence, and Part 3 always is $O(1/R)$ or less.
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– Oscar Lanzi
Jan 30 at 22:30
add a comment |
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Another way to go about this (but this is actually closely related to the solutions above invoking the residue at infinity or arguing that extending the radius of the contour doesn't affect the integral and taking the limit of infinite radius), is to substitute $z = frac{1}{w}$. The integral then becomes:
$$oint_C frac{w^3 dw}{1+w^5}$$
where $C$ is now a counterclockwise contour with radius $dfrac{1}{3}$ (note that the transform makes the original contour a clockwise contour and we can then make that a counterclockwise one by using the minus sign in $dz = -frac{dw}{w^2} $). Since there are now no poles inside the contour, the integral is zero.
You can use this transform to derive the result that a counterclockwise contour integral with winding number 1 is also given by minus the sum of all the residues outside the contour where you also need to include a suitably defined "residue at infinity".
answered Jan 30 at 22:01
Count IblisCount Iblis
8,52221534
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Partial fractions can work. I think this approach is pretty much equivalent to Mark Viola's residue one, but it's nice to see the agreement.
Let $zeta$ be a primitive fifth root of unity. If you want to be explicit, let $zeta = e^{2pi/5}$. But the important thing is that
- the roots of $z^5+1$ are $-1$, $-zeta$, $-zeta^2$, $-zeta^3$, and $-zeta^4$.
- Since $(z^5 - 1) = (z-1)(z^4+z^3+z^2+z+1)$, we know that $1 + zeta + zeta^2 + zeta^3 + zeta^4 = 0$.
Now for something about partial fractions that I didn't know until just now.
If $f(z) = (z-alpha_1) cdots (z-alpha_n)$, and all the roots are distinct, then
$$
frac{1}{f(z)} = frac{1}{f'(alpha_1)(z-alpha_1)} + frac{1}{f'(alpha_2)(z-alpha_2)} + dots + frac{1}{f'(alpha_n)(z-alpha_n)}
$$
To show this, start from
$$
frac{1}{(z-alpha_1) cdots (z-alpha_n)} = sum_{i=1}^n frac{A_i}{z-alpha_i}
$$
for some constants $A_1, dots, A_n$. Clear out the denominators, and you get
$$
1 = sum_{i=1}^n frac{A_i f(z)}{z-alpha_i}
= sum_{i=1}^n A_i frac{f(z)-f(alpha_i)}{z-alpha_i}
tag{$*$}
$$
Now for any $i$ and $j$,
$$
lim_{zto alpha_j} frac{f(z)-f(alpha_i)}{z-alpha_i} =
begin{cases} 0 & i neq j \ f'(alpha_i) & i = j end{cases}
$$
So if we take the limit of both sides of ($*$) as $zto z_j$ we get $1 = A_j f'(alpha_j)$ for each $j$.
This means that
begin{align*}
frac{1}{z^5+1}
&= frac{1}{5(z+1)} + frac{1}{5(-zeta)^4(z+zeta)} + frac{1}{5(-zeta^2)^4(z+zeta^2)} + frac{1}{5(-zeta^3)^4(z+zeta^3)} + frac{1}{5(-zeta^4)^4(z+zeta^4)}
\&= frac{1}{5}left(frac{1}{z+1} + frac{1}{zeta^4(z + zeta)} + frac{1}{zeta^3(z + zeta^2)} + frac{1}{zeta^2(z + zeta^3)} + frac{1}{zeta(z + zeta^4)}right)
\&= frac{1}{5}left(frac{1}{z+1} + frac{zeta}{z + zeta} + frac{zeta^2}{z + zeta^2} + frac{zeta^3}{z + zeta^3} + frac{zeta^4}{z + zeta^4}right)
end{align*}
You wanted an argument involving symmetry; what could be more symmetric than that equation?
Now $C$ encloses all those roots, so by the Cauchy Integral Formula,
$$
oint_C frac{dz}{z^5+1} = frac{2pi i}{5}left(1 + zeta + zeta^2 + zeta^3 + zeta^4right)
$$
and as we showed above, the latter factor is zero.
answered Jan 30 at 19:04
Matthew LeingangMatthew Leingang
16.7k12244
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$begingroup$
One can check this, but this would be hard. I've made an edit to my question where I make it simpler by using symmetries.
$endgroup$
– Belgium_Physics
Jan 30 at 19:25
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Unfortunately, if $P(X)=(X-a_1)....(X-a_n)$ has distinct roots, then $$sum_{k=1}^n frac{1}{X-a_k}=frac{P'(X)}{P(X)}$$ This follows immediately from the product rule:$$P'(X)=((X-a_1)....(X-a_n))'=(X-a_2)(X-a_3)....(X-a_n)+(X-a_1)(X-a_3)....(X-a_n)+...+(X-a_1)(X-a_2)....(X-a_{n_1})$$ Now divide both sides by $(X -a_1)....(X-a_n)$.
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– N. S.
Jan 31 at 3:44
$begingroup$
Therefore $$ frac{1}{z-z_0} + frac{1}{z-z_1} + frac{1}{z-z_2} + frac{1}{z-z_3} + frac{1}{z-z_4}=frac{5z^4}{z^5+1} neq frac{1}{z^5+1}$$
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– N. S.
Jan 31 at 3:45
$begingroup$
@NS you’re right, of course. Like I said, it was a clue to try partial fractions. And although the numerators aren’t one, there is a decomposition of that form with the “1”s replaces by constants.
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– Matthew Leingang
Jan 31 at 11:35
$begingroup$
@N.S. I've replaced my hint with a full answer.
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– Matthew Leingang
Feb 1 at 22:06
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
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$begingroup$
HINT:
All of the poles lie on the unit circle. Hence, for any values of $R_1>1$ and $R_2>1$, we have
$$oint_{|z|=R_1}frac1{1+z^5},dz=oint_{|z|=R_2}frac1{1+z^5},dz$$
Now, what happens for $R_1=3$ and $R_2toinfty$? The answer is zero. This approach is tantamount to invoking the residue at infinity.
If one insists on summing the residues, then one may proceed in general as follows. First note that if $z^n+1=0$, then $z=z_k=e^{i(2k-1)pi/n}$ for $k=1,2,dots,n$.
Next, we calculate the residues of $frac{1}{z^n+1}$ at $z_k$ using L'Hospital's Rule as follows:
$$begin{align}
text{Res}left(frac1{z^n+1},z=z_kright)&=lim_{zto z_k}left(frac{z-z_k}{z^n+1}right)\\
&=frac{1}{nz_k^{n-1}}\\
&=frac1n e^{-i(2k-1)pi(n-1)/n}\\
&=-frac1n e^{-ipi/n}left(e^{i2pi/n}right)^k
end{align}$$
Summing all of the residues reveals for $n>1$
$$begin{align}
sum_{k=1}^n text{Res}left(frac1{1+z^n},z=z_kright)&=-frac1ne^{-ipi/n}sum_{k=1}^nleft(e^{i2pi/n}right)^k\\
&=-frac1ne^{=ipi/n} left(frac{e^{i2pi/n}-left(e^{i2pi/n}right)^{n+1}}{1-e^{i2pi/n}}right)\\
&=0
end{align}$$
Hence, for $n>1$ the residues of $frac1{1+z^n}$ add to zero. And we are done!
answered Jan 30 at 22:03
Mark ViolaMark Viola
134k1278177
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add a comment |
$begingroup$
HINT:
All of the poles lie on the unit circle. Hence, for any values of $R_1>1$ and $R_2>1$, we have
$$oint_{|z|=R_1}frac1{1+z^5},dz=oint_{|z|=R_2}frac1{1+z^5},dz$$
Now, what happens for $R_1=3$ and $R_2toinfty$? The answer is zero. This approach is tantamount to invoking the residue at infinity.
If one insists on summing the residues, then one may proceed in general as follows. First note that if $z^n+1=0$, then $z=z_k=e^{i(2k-1)pi/n}$ for $k=1,2,dots,n$.
Next, we calculate the residues of $frac{1}{z^n+1}$ at $z_k$ using L'Hospital's Rule as follows:
$$begin{align}
text{Res}left(frac1{z^n+1},z=z_kright)&=lim_{zto z_k}left(frac{z-z_k}{z^n+1}right)\\
&=frac{1}{nz_k^{n-1}}\\
&=frac1n e^{-i(2k-1)pi(n-1)/n}\\
&=-frac1n e^{-ipi/n}left(e^{i2pi/n}right)^k
end{align}$$
Summing all of the residues reveals for $n>1$
$$begin{align}
sum_{k=1}^n text{Res}left(frac1{1+z^n},z=z_kright)&=-frac1ne^{-ipi/n}sum_{k=1}^nleft(e^{i2pi/n}right)^k\\
&=-frac1ne^{=ipi/n} left(frac{e^{i2pi/n}-left(e^{i2pi/n}right)^{n+1}}{1-e^{i2pi/n}}right)\\
&=0
end{align}$$
Hence, for $n>1$ the residues of $frac1{1+z^n}$ add to zero. And we are done!
answered Jan 30 at 22:03
Mark ViolaMark Viola
134k1278177
$endgroup$
add a comment |
$begingroup$
HINT:
All of the poles lie on the unit circle. Hence, for any values of $R_1>1$ and $R_2>1$, we have
$$oint_{|z|=R_1}frac1{1+z^5},dz=oint_{|z|=R_2}frac1{1+z^5},dz$$
Now, what happens for $R_1=3$ and $R_2toinfty$? The answer is zero. This approach is tantamount to invoking the residue at infinity.
If one insists on summing the residues, then one may proceed in general as follows. First note that if $z^n+1=0$, then $z=z_k=e^{i(2k-1)pi/n}$ for $k=1,2,dots,n$.
Next, we calculate the residues of $frac{1}{z^n+1}$ at $z_k$ using L'Hospital's Rule as follows:
$$begin{align}
text{Res}left(frac1{z^n+1},z=z_kright)&=lim_{zto z_k}left(frac{z-z_k}{z^n+1}right)\\
&=frac{1}{nz_k^{n-1}}\\
&=frac1n e^{-i(2k-1)pi(n-1)/n}\\
&=-frac1n e^{-ipi/n}left(e^{i2pi/n}right)^k
end{align}$$
Summing all of the residues reveals for $n>1$
$$begin{align}
sum_{k=1}^n text{Res}left(frac1{1+z^n},z=z_kright)&=-frac1ne^{-ipi/n}sum_{k=1}^nleft(e^{i2pi/n}right)^k\\
&=-frac1ne^{=ipi/n} left(frac{e^{i2pi/n}-left(e^{i2pi/n}right)^{n+1}}{1-e^{i2pi/n}}right)\\
&=0
end{align}$$
Hence, for $n>1$ the residues of $frac1{1+z^n}$ add to zero. And we are done!
answered Jan 30 at 22:03
Mark ViolaMark Viola
134k1278177
$endgroup$
HINT:
All of the poles lie on the unit circle. Hence, for any values of $R_1>1$ and $R_2>1$, we have
$$oint_{|z|=R_1}frac1{1+z^5},dz=oint_{|z|=R_2}frac1{1+z^5},dz$$
Now, what happens for $R_1=3$ and $R_2toinfty$? The answer is zero. This approach is tantamount to invoking the residue at infinity.
If one insists on summing the residues, then one may proceed in general as follows. First note that if $z^n+1=0$, then $z=z_k=e^{i(2k-1)pi/n}$ for $k=1,2,dots,n$.
Next, we calculate the residues of $frac{1}{z^n+1}$ at $z_k$ using L'Hospital's Rule as follows:
$$begin{align}
text{Res}left(frac1{z^n+1},z=z_kright)&=lim_{zto z_k}left(frac{z-z_k}{z^n+1}right)\\
&=frac{1}{nz_k^{n-1}}\\
&=frac1n e^{-i(2k-1)pi(n-1)/n}\\
&=-frac1n e^{-ipi/n}left(e^{i2pi/n}right)^k
end{align}$$
Summing all of the residues reveals for $n>1$
$$begin{align}
sum_{k=1}^n text{Res}left(frac1{1+z^n},z=z_kright)&=-frac1ne^{-ipi/n}sum_{k=1}^nleft(e^{i2pi/n}right)^k\\
&=-frac1ne^{=ipi/n} left(frac{e^{i2pi/n}-left(e^{i2pi/n}right)^{n+1}}{1-e^{i2pi/n}}right)\\
&=0
end{align}$$
Hence, for $n>1$ the residues of $frac1{1+z^n}$ add to zero. And we are done!
answered Jan 30 at 22:03
Mark ViolaMark Viola
134k1278177
edited Jan 31 at 14:21
answered Jan 30 at 22:03
Mark ViolaMark Viola
134k1278177
answered Jan 30 at 22:03
Mark ViolaMark Viola
134k1278177
answered Jan 30 at 22:03
Mark ViolaMark Viola
134k1278177
134k1278177
add a comment |
add a comment |
$begingroup$
Do you know what the residue at the infinity is?
There is a relation between the residues at complex numbers and the residue at infinity when you have finitely many poles.
The relation is $$sum_{k=0}^{n} operatorname{Res}_{a_k}(f) = -operatorname{Res}_{infty}(f),$$ where ${ a_{k} : n(Gamma,a_k)ne 0}_{k=0}^{n}$ are the poles (inside the path $Gamma$) of the function you want to compute the integral and $f$ is the function you are integrating. Here $n(gamma,z)$ is the index of the complex number $z$ with respect to the path $gamma$. I'm supposing that the curve only winds one time around the origin.
answered Jan 30 at 19:13
AlgebraicallyClosedAlgebraicallyClosed
709114
$endgroup$
$begingroup$
This can help and I will try it, but in my edit I suggest making use of the symmetrie of the solutions, could that be working?
$endgroup$
– Belgium_Physics
Jan 30 at 19:26
2
$begingroup$
The symmetries betweem the complex poles let you to compute fastly that integral, but introducing the infinity you are studying one more symmetrie, a symmetrie in the Riemann Sphere. What I want to say you is that the symmetries between the poles/zeros can appear by more than one way. But yes, you will be correct if you use the symmetries avaliable at the complex plane.
$endgroup$
– AlgebraicallyClosed
Jan 30 at 19:36
add a comment |
$begingroup$
Do you know what the residue at the infinity is?
There is a relation between the residues at complex numbers and the residue at infinity when you have finitely many poles.
The relation is $$sum_{k=0}^{n} operatorname{Res}_{a_k}(f) = -operatorname{Res}_{infty}(f),$$ where ${ a_{k} : n(Gamma,a_k)ne 0}_{k=0}^{n}$ are the poles (inside the path $Gamma$) of the function you want to compute the integral and $f$ is the function you are integrating. Here $n(gamma,z)$ is the index of the complex number $z$ with respect to the path $gamma$. I'm supposing that the curve only winds one time around the origin.
answered Jan 30 at 19:13
AlgebraicallyClosedAlgebraicallyClosed
709114
$endgroup$
$begingroup$
This can help and I will try it, but in my edit I suggest making use of the symmetrie of the solutions, could that be working?
$endgroup$
– Belgium_Physics
Jan 30 at 19:26
2
$begingroup$
The symmetries betweem the complex poles let you to compute fastly that integral, but introducing the infinity you are studying one more symmetrie, a symmetrie in the Riemann Sphere. What I want to say you is that the symmetries between the poles/zeros can appear by more than one way. But yes, you will be correct if you use the symmetries avaliable at the complex plane.
$endgroup$
– AlgebraicallyClosed
Jan 30 at 19:36
add a comment |
$begingroup$
Do you know what the residue at the infinity is?
There is a relation between the residues at complex numbers and the residue at infinity when you have finitely many poles.
The relation is $$sum_{k=0}^{n} operatorname{Res}_{a_k}(f) = -operatorname{Res}_{infty}(f),$$ where ${ a_{k} : n(Gamma,a_k)ne 0}_{k=0}^{n}$ are the poles (inside the path $Gamma$) of the function you want to compute the integral and $f$ is the function you are integrating. Here $n(gamma,z)$ is the index of the complex number $z$ with respect to the path $gamma$. I'm supposing that the curve only winds one time around the origin.
answered Jan 30 at 19:13
AlgebraicallyClosedAlgebraicallyClosed
709114
$endgroup$
Do you know what the residue at the infinity is?
There is a relation between the residues at complex numbers and the residue at infinity when you have finitely many poles.
The relation is $$sum_{k=0}^{n} operatorname{Res}_{a_k}(f) = -operatorname{Res}_{infty}(f),$$ where ${ a_{k} : n(Gamma,a_k)ne 0}_{k=0}^{n}$ are the poles (inside the path $Gamma$) of the function you want to compute the integral and $f$ is the function you are integrating. Here $n(gamma,z)$ is the index of the complex number $z$ with respect to the path $gamma$. I'm supposing that the curve only winds one time around the origin.
answered Jan 30 at 19:13
AlgebraicallyClosedAlgebraicallyClosed
709114
edited Jan 30 at 19:30
answered Jan 30 at 19:13
AlgebraicallyClosedAlgebraicallyClosed
709114
answered Jan 30 at 19:13
AlgebraicallyClosedAlgebraicallyClosed
709114
answered Jan 30 at 19:13
AlgebraicallyClosedAlgebraicallyClosed
709114
709114
$begingroup$
This can help and I will try it, but in my edit I suggest making use of the symmetrie of the solutions, could that be working?
$endgroup$
– Belgium_Physics
Jan 30 at 19:26
2
$begingroup$
The symmetries betweem the complex poles let you to compute fastly that integral, but introducing the infinity you are studying one more symmetrie, a symmetrie in the Riemann Sphere. What I want to say you is that the symmetries between the poles/zeros can appear by more than one way. But yes, you will be correct if you use the symmetries avaliable at the complex plane.
$endgroup$
– AlgebraicallyClosed
Jan 30 at 19:36
add a comment |
$begingroup$
This can help and I will try it, but in my edit I suggest making use of the symmetrie of the solutions, could that be working?
$endgroup$
– Belgium_Physics
Jan 30 at 19:26
2
$begingroup$
The symmetries betweem the complex poles let you to compute fastly that integral, but introducing the infinity you are studying one more symmetrie, a symmetrie in the Riemann Sphere. What I want to say you is that the symmetries between the poles/zeros can appear by more than one way. But yes, you will be correct if you use the symmetries avaliable at the complex plane.
$endgroup$
– AlgebraicallyClosed
Jan 30 at 19:36
$begingroup$
This can help and I will try it, but in my edit I suggest making use of the symmetrie of the solutions, could that be working?
$endgroup$
– Belgium_Physics
Jan 30 at 19:26
$begingroup$
This can help and I will try it, but in my edit I suggest making use of the symmetrie of the solutions, could that be working?
$endgroup$
– Belgium_Physics
Jan 30 at 19:26
2
2
$begingroup$
The symmetries betweem the complex poles let you to compute fastly that integral, but introducing the infinity you are studying one more symmetrie, a symmetrie in the Riemann Sphere. What I want to say you is that the symmetries between the poles/zeros can appear by more than one way. But yes, you will be correct if you use the symmetries avaliable at the complex plane.
$endgroup$
– AlgebraicallyClosed
Jan 30 at 19:36
$begingroup$
The symmetries betweem the complex poles let you to compute fastly that integral, but introducing the infinity you are studying one more symmetrie, a symmetrie in the Riemann Sphere. What I want to say you is that the symmetries between the poles/zeros can appear by more than one way. But yes, you will be correct if you use the symmetries avaliable at the complex plane.
$endgroup$
– AlgebraicallyClosed
Jan 30 at 19:36
add a comment |
$begingroup$
Here is a subtle approach that completely gets around having to determine the residues.
Cut the plane along the positive real axis. Do your integration around a contour that goes like this:
Part 1: Counterclockwise from $z=3$ on the upper side of the cut to $x=3$ on the lower side of the cut.
Part 2: Follow the lower side of the cut out to $z=R$, where we will take the limit as $Rtoinfty$.
Part 3: Clockwise around the circle $|z|=R$ ending back on the upper side of the cut.
Part 4: Back to $z=3$ along the upper side of the cut.
This contour encloses no singularities, therefore the integral around it is zero. But also:
Parts 2 and 4: The real integral converges between $x=3$ and $x=infty$ by comparison with $1/x^5$ whose antiderivative $(-1/(4x^4))+C$ also converges. Therefore Part 2 and the additive inverse of Part 4 must have a common value, forcing these parts to cancel each other.
Part 3: The integrand is $O(1/R^5)$ and the length of the contour is $2pi R$, so by the triangle inequality the Part 3 integral is $O(1/R^4)to 0$.
So the Part 1 integral which is your original integral must also be zero.
In general: When you integrate a rational function around a contour that covers all of its poles and the function decreases faster than $1/z$ as $ztoinfty$, you must get zero. The residues inside the contour are forced to add up to zero, to conform.
answered Jan 30 at 20:50
Oscar LanziOscar Lanzi
13.5k12136
$endgroup$
$begingroup$
Is there a theorem or proof for this in general property? It looks something interesting that I have not encountered yet.
$endgroup$
– Belgium_Physics
Jan 30 at 22:17
$begingroup$
If you use the type of contour described here, for tge functions I include, you get this result. Parts 2 and 4 always cancel by convergence, and Part 3 always is $O(1/R)$ or less.
$endgroup$
– Oscar Lanzi
Jan 30 at 22:30
add a comment |
$begingroup$
Here is a subtle approach that completely gets around having to determine the residues.
Cut the plane along the positive real axis. Do your integration around a contour that goes like this:
Part 1: Counterclockwise from $z=3$ on the upper side of the cut to $x=3$ on the lower side of the cut.
Part 2: Follow the lower side of the cut out to $z=R$, where we will take the limit as $Rtoinfty$.
Part 3: Clockwise around the circle $|z|=R$ ending back on the upper side of the cut.
Part 4: Back to $z=3$ along the upper side of the cut.
This contour encloses no singularities, therefore the integral around it is zero. But also:
Parts 2 and 4: The real integral converges between $x=3$ and $x=infty$ by comparison with $1/x^5$ whose antiderivative $(-1/(4x^4))+C$ also converges. Therefore Part 2 and the additive inverse of Part 4 must have a common value, forcing these parts to cancel each other.
Part 3: The integrand is $O(1/R^5)$ and the length of the contour is $2pi R$, so by the triangle inequality the Part 3 integral is $O(1/R^4)to 0$.
So the Part 1 integral which is your original integral must also be zero.
In general: When you integrate a rational function around a contour that covers all of its poles and the function decreases faster than $1/z$ as $ztoinfty$, you must get zero. The residues inside the contour are forced to add up to zero, to conform.
answered Jan 30 at 20:50
Oscar LanziOscar Lanzi
13.5k12136
$endgroup$
$begingroup$
Is there a theorem or proof for this in general property? It looks something interesting that I have not encountered yet.
$endgroup$
– Belgium_Physics
Jan 30 at 22:17
$begingroup$
If you use the type of contour described here, for tge functions I include, you get this result. Parts 2 and 4 always cancel by convergence, and Part 3 always is $O(1/R)$ or less.
$endgroup$
– Oscar Lanzi
Jan 30 at 22:30
add a comment |
$begingroup$
Here is a subtle approach that completely gets around having to determine the residues.
Cut the plane along the positive real axis. Do your integration around a contour that goes like this:
Part 1: Counterclockwise from $z=3$ on the upper side of the cut to $x=3$ on the lower side of the cut.
Part 2: Follow the lower side of the cut out to $z=R$, where we will take the limit as $Rtoinfty$.
Part 3: Clockwise around the circle $|z|=R$ ending back on the upper side of the cut.
Part 4: Back to $z=3$ along the upper side of the cut.
This contour encloses no singularities, therefore the integral around it is zero. But also:
Parts 2 and 4: The real integral converges between $x=3$ and $x=infty$ by comparison with $1/x^5$ whose antiderivative $(-1/(4x^4))+C$ also converges. Therefore Part 2 and the additive inverse of Part 4 must have a common value, forcing these parts to cancel each other.
Part 3: The integrand is $O(1/R^5)$ and the length of the contour is $2pi R$, so by the triangle inequality the Part 3 integral is $O(1/R^4)to 0$.
So the Part 1 integral which is your original integral must also be zero.
In general: When you integrate a rational function around a contour that covers all of its poles and the function decreases faster than $1/z$ as $ztoinfty$, you must get zero. The residues inside the contour are forced to add up to zero, to conform.
answered Jan 30 at 20:50
Oscar LanziOscar Lanzi
13.5k12136
$endgroup$
Here is a subtle approach that completely gets around having to determine the residues.
Cut the plane along the positive real axis. Do your integration around a contour that goes like this:
Part 1: Counterclockwise from $z=3$ on the upper side of the cut to $x=3$ on the lower side of the cut.
Part 2: Follow the lower side of the cut out to $z=R$, where we will take the limit as $Rtoinfty$.
Part 3: Clockwise around the circle $|z|=R$ ending back on the upper side of the cut.
Part 4: Back to $z=3$ along the upper side of the cut.
This contour encloses no singularities, therefore the integral around it is zero. But also:
Parts 2 and 4: The real integral converges between $x=3$ and $x=infty$ by comparison with $1/x^5$ whose antiderivative $(-1/(4x^4))+C$ also converges. Therefore Part 2 and the additive inverse of Part 4 must have a common value, forcing these parts to cancel each other.
Part 3: The integrand is $O(1/R^5)$ and the length of the contour is $2pi R$, so by the triangle inequality the Part 3 integral is $O(1/R^4)to 0$.
So the Part 1 integral which is your original integral must also be zero.
In general: When you integrate a rational function around a contour that covers all of its poles and the function decreases faster than $1/z$ as $ztoinfty$, you must get zero. The residues inside the contour are forced to add up to zero, to conform.
answered Jan 30 at 20:50
Oscar LanziOscar Lanzi
13.5k12136
edited Jan 30 at 21:57
answered Jan 30 at 20:50
Oscar LanziOscar Lanzi
13.5k12136
answered Jan 30 at 20:50
Oscar LanziOscar Lanzi
13.5k12136
answered Jan 30 at 20:50
Oscar LanziOscar Lanzi
13.5k12136
13.5k12136
$begingroup$
Is there a theorem or proof for this in general property? It looks something interesting that I have not encountered yet.
$endgroup$
– Belgium_Physics
Jan 30 at 22:17
$begingroup$
If you use the type of contour described here, for tge functions I include, you get this result. Parts 2 and 4 always cancel by convergence, and Part 3 always is $O(1/R)$ or less.
$endgroup$
– Oscar Lanzi
Jan 30 at 22:30
add a comment |
$begingroup$
Is there a theorem or proof for this in general property? It looks something interesting that I have not encountered yet.
$endgroup$
– Belgium_Physics
Jan 30 at 22:17
$begingroup$
If you use the type of contour described here, for tge functions I include, you get this result. Parts 2 and 4 always cancel by convergence, and Part 3 always is $O(1/R)$ or less.
$endgroup$
– Oscar Lanzi
Jan 30 at 22:30
$begingroup$
Is there a theorem or proof for this in general property? It looks something interesting that I have not encountered yet.
$endgroup$
– Belgium_Physics
Jan 30 at 22:17
$begingroup$
Is there a theorem or proof for this in general property? It looks something interesting that I have not encountered yet.
$endgroup$
– Belgium_Physics
Jan 30 at 22:17
$begingroup$
If you use the type of contour described here, for tge functions I include, you get this result. Parts 2 and 4 always cancel by convergence, and Part 3 always is $O(1/R)$ or less.
$endgroup$
– Oscar Lanzi
Jan 30 at 22:30
$begingroup$
If you use the type of contour described here, for tge functions I include, you get this result. Parts 2 and 4 always cancel by convergence, and Part 3 always is $O(1/R)$ or less.
$endgroup$
– Oscar Lanzi
Jan 30 at 22:30
add a comment |
$begingroup$
Another way to go about this (but this is actually closely related to the solutions above invoking the residue at infinity or arguing that extending the radius of the contour doesn't affect the integral and taking the limit of infinite radius), is to substitute $z = frac{1}{w}$. The integral then becomes:
$$oint_C frac{w^3 dw}{1+w^5}$$
where $C$ is now a counterclockwise contour with radius $dfrac{1}{3}$ (note that the transform makes the original contour a clockwise contour and we can then make that a counterclockwise one by using the minus sign in $dz = -frac{dw}{w^2} $). Since there are now no poles inside the contour, the integral is zero.
You can use this transform to derive the result that a counterclockwise contour integral with winding number 1 is also given by minus the sum of all the residues outside the contour where you also need to include a suitably defined "residue at infinity".
answered Jan 30 at 22:01
Count IblisCount Iblis
8,52221534
$endgroup$
add a comment |
$begingroup$
Another way to go about this (but this is actually closely related to the solutions above invoking the residue at infinity or arguing that extending the radius of the contour doesn't affect the integral and taking the limit of infinite radius), is to substitute $z = frac{1}{w}$. The integral then becomes:
$$oint_C frac{w^3 dw}{1+w^5}$$
where $C$ is now a counterclockwise contour with radius $dfrac{1}{3}$ (note that the transform makes the original contour a clockwise contour and we can then make that a counterclockwise one by using the minus sign in $dz = -frac{dw}{w^2} $). Since there are now no poles inside the contour, the integral is zero.
You can use this transform to derive the result that a counterclockwise contour integral with winding number 1 is also given by minus the sum of all the residues outside the contour where you also need to include a suitably defined "residue at infinity".
answered Jan 30 at 22:01
Count IblisCount Iblis
8,52221534
$endgroup$
add a comment |
$begingroup$
Another way to go about this (but this is actually closely related to the solutions above invoking the residue at infinity or arguing that extending the radius of the contour doesn't affect the integral and taking the limit of infinite radius), is to substitute $z = frac{1}{w}$. The integral then becomes:
$$oint_C frac{w^3 dw}{1+w^5}$$
where $C$ is now a counterclockwise contour with radius $dfrac{1}{3}$ (note that the transform makes the original contour a clockwise contour and we can then make that a counterclockwise one by using the minus sign in $dz = -frac{dw}{w^2} $). Since there are now no poles inside the contour, the integral is zero.
You can use this transform to derive the result that a counterclockwise contour integral with winding number 1 is also given by minus the sum of all the residues outside the contour where you also need to include a suitably defined "residue at infinity".
answered Jan 30 at 22:01
Count IblisCount Iblis
8,52221534
$endgroup$
Another way to go about this (but this is actually closely related to the solutions above invoking the residue at infinity or arguing that extending the radius of the contour doesn't affect the integral and taking the limit of infinite radius), is to substitute $z = frac{1}{w}$. The integral then becomes:
$$oint_C frac{w^3 dw}{1+w^5}$$
where $C$ is now a counterclockwise contour with radius $dfrac{1}{3}$ (note that the transform makes the original contour a clockwise contour and we can then make that a counterclockwise one by using the minus sign in $dz = -frac{dw}{w^2} $). Since there are now no poles inside the contour, the integral is zero.
You can use this transform to derive the result that a counterclockwise contour integral with winding number 1 is also given by minus the sum of all the residues outside the contour where you also need to include a suitably defined "residue at infinity".
answered Jan 30 at 22:01
Count IblisCount Iblis
8,52221534
answered Jan 30 at 22:01
Count IblisCount Iblis
8,52221534
answered Jan 30 at 22:01
Count IblisCount Iblis
8,52221534
answered Jan 30 at 22:01
Count IblisCount Iblis
8,52221534
8,52221534
add a comment |
add a comment |
$begingroup$
Partial fractions can work. I think this approach is pretty much equivalent to Mark Viola's residue one, but it's nice to see the agreement.
Let $zeta$ be a primitive fifth root of unity. If you want to be explicit, let $zeta = e^{2pi/5}$. But the important thing is that
- the roots of $z^5+1$ are $-1$, $-zeta$, $-zeta^2$, $-zeta^3$, and $-zeta^4$.
- Since $(z^5 - 1) = (z-1)(z^4+z^3+z^2+z+1)$, we know that $1 + zeta + zeta^2 + zeta^3 + zeta^4 = 0$.
Now for something about partial fractions that I didn't know until just now.
If $f(z) = (z-alpha_1) cdots (z-alpha_n)$, and all the roots are distinct, then
$$
frac{1}{f(z)} = frac{1}{f'(alpha_1)(z-alpha_1)} + frac{1}{f'(alpha_2)(z-alpha_2)} + dots + frac{1}{f'(alpha_n)(z-alpha_n)}
$$
To show this, start from
$$
frac{1}{(z-alpha_1) cdots (z-alpha_n)} = sum_{i=1}^n frac{A_i}{z-alpha_i}
$$
for some constants $A_1, dots, A_n$. Clear out the denominators, and you get
$$
1 = sum_{i=1}^n frac{A_i f(z)}{z-alpha_i}
= sum_{i=1}^n A_i frac{f(z)-f(alpha_i)}{z-alpha_i}
tag{$*$}
$$
Now for any $i$ and $j$,
$$
lim_{zto alpha_j} frac{f(z)-f(alpha_i)}{z-alpha_i} =
begin{cases} 0 & i neq j \ f'(alpha_i) & i = j end{cases}
$$
So if we take the limit of both sides of ($*$) as $zto z_j$ we get $1 = A_j f'(alpha_j)$ for each $j$.
This means that
begin{align*}
frac{1}{z^5+1}
&= frac{1}{5(z+1)} + frac{1}{5(-zeta)^4(z+zeta)} + frac{1}{5(-zeta^2)^4(z+zeta^2)} + frac{1}{5(-zeta^3)^4(z+zeta^3)} + frac{1}{5(-zeta^4)^4(z+zeta^4)}
\&= frac{1}{5}left(frac{1}{z+1} + frac{1}{zeta^4(z + zeta)} + frac{1}{zeta^3(z + zeta^2)} + frac{1}{zeta^2(z + zeta^3)} + frac{1}{zeta(z + zeta^4)}right)
\&= frac{1}{5}left(frac{1}{z+1} + frac{zeta}{z + zeta} + frac{zeta^2}{z + zeta^2} + frac{zeta^3}{z + zeta^3} + frac{zeta^4}{z + zeta^4}right)
end{align*}
You wanted an argument involving symmetry; what could be more symmetric than that equation?
Now $C$ encloses all those roots, so by the Cauchy Integral Formula,
$$
oint_C frac{dz}{z^5+1} = frac{2pi i}{5}left(1 + zeta + zeta^2 + zeta^3 + zeta^4right)
$$
and as we showed above, the latter factor is zero.
answered Jan 30 at 19:04
Matthew LeingangMatthew Leingang
16.7k12244
$endgroup$
$begingroup$
One can check this, but this would be hard. I've made an edit to my question where I make it simpler by using symmetries.
$endgroup$
– Belgium_Physics
Jan 30 at 19:25
$begingroup$
Unfortunately, if $P(X)=(X-a_1)....(X-a_n)$ has distinct roots, then $$sum_{k=1}^n frac{1}{X-a_k}=frac{P'(X)}{P(X)}$$ This follows immediately from the product rule:$$P'(X)=((X-a_1)....(X-a_n))'=(X-a_2)(X-a_3)....(X-a_n)+(X-a_1)(X-a_3)....(X-a_n)+...+(X-a_1)(X-a_2)....(X-a_{n_1})$$ Now divide both sides by $(X -a_1)....(X-a_n)$.
$endgroup$
– N. S.
Jan 31 at 3:44
$begingroup$
Therefore $$ frac{1}{z-z_0} + frac{1}{z-z_1} + frac{1}{z-z_2} + frac{1}{z-z_3} + frac{1}{z-z_4}=frac{5z^4}{z^5+1} neq frac{1}{z^5+1}$$
$endgroup$
– N. S.
Jan 31 at 3:45
$begingroup$
@NS you’re right, of course. Like I said, it was a clue to try partial fractions. And although the numerators aren’t one, there is a decomposition of that form with the “1”s replaces by constants.
$endgroup$
– Matthew Leingang
Jan 31 at 11:35
$begingroup$
@N.S. I've replaced my hint with a full answer.
$endgroup$
– Matthew Leingang
Feb 1 at 22:06
add a comment |
$begingroup$
Partial fractions can work. I think this approach is pretty much equivalent to Mark Viola's residue one, but it's nice to see the agreement.
Let $zeta$ be a primitive fifth root of unity. If you want to be explicit, let $zeta = e^{2pi/5}$. But the important thing is that
- the roots of $z^5+1$ are $-1$, $-zeta$, $-zeta^2$, $-zeta^3$, and $-zeta^4$.
- Since $(z^5 - 1) = (z-1)(z^4+z^3+z^2+z+1)$, we know that $1 + zeta + zeta^2 + zeta^3 + zeta^4 = 0$.
Now for something about partial fractions that I didn't know until just now.
If $f(z) = (z-alpha_1) cdots (z-alpha_n)$, and all the roots are distinct, then
$$
frac{1}{f(z)} = frac{1}{f'(alpha_1)(z-alpha_1)} + frac{1}{f'(alpha_2)(z-alpha_2)} + dots + frac{1}{f'(alpha_n)(z-alpha_n)}
$$
To show this, start from
$$
frac{1}{(z-alpha_1) cdots (z-alpha_n)} = sum_{i=1}^n frac{A_i}{z-alpha_i}
$$
for some constants $A_1, dots, A_n$. Clear out the denominators, and you get
$$
1 = sum_{i=1}^n frac{A_i f(z)}{z-alpha_i}
= sum_{i=1}^n A_i frac{f(z)-f(alpha_i)}{z-alpha_i}
tag{$*$}
$$
Now for any $i$ and $j$,
$$
lim_{zto alpha_j} frac{f(z)-f(alpha_i)}{z-alpha_i} =
begin{cases} 0 & i neq j \ f'(alpha_i) & i = j end{cases}
$$
So if we take the limit of both sides of ($*$) as $zto z_j$ we get $1 = A_j f'(alpha_j)$ for each $j$.
This means that
begin{align*}
frac{1}{z^5+1}
&= frac{1}{5(z+1)} + frac{1}{5(-zeta)^4(z+zeta)} + frac{1}{5(-zeta^2)^4(z+zeta^2)} + frac{1}{5(-zeta^3)^4(z+zeta^3)} + frac{1}{5(-zeta^4)^4(z+zeta^4)}
\&= frac{1}{5}left(frac{1}{z+1} + frac{1}{zeta^4(z + zeta)} + frac{1}{zeta^3(z + zeta^2)} + frac{1}{zeta^2(z + zeta^3)} + frac{1}{zeta(z + zeta^4)}right)
\&= frac{1}{5}left(frac{1}{z+1} + frac{zeta}{z + zeta} + frac{zeta^2}{z + zeta^2} + frac{zeta^3}{z + zeta^3} + frac{zeta^4}{z + zeta^4}right)
end{align*}
You wanted an argument involving symmetry; what could be more symmetric than that equation?
Now $C$ encloses all those roots, so by the Cauchy Integral Formula,
$$
oint_C frac{dz}{z^5+1} = frac{2pi i}{5}left(1 + zeta + zeta^2 + zeta^3 + zeta^4right)
$$
and as we showed above, the latter factor is zero.
answered Jan 30 at 19:04
Matthew LeingangMatthew Leingang
16.7k12244
$endgroup$
$begingroup$
One can check this, but this would be hard. I've made an edit to my question where I make it simpler by using symmetries.
$endgroup$
– Belgium_Physics
Jan 30 at 19:25
$begingroup$
Unfortunately, if $P(X)=(X-a_1)....(X-a_n)$ has distinct roots, then $$sum_{k=1}^n frac{1}{X-a_k}=frac{P'(X)}{P(X)}$$ This follows immediately from the product rule:$$P'(X)=((X-a_1)....(X-a_n))'=(X-a_2)(X-a_3)....(X-a_n)+(X-a_1)(X-a_3)....(X-a_n)+...+(X-a_1)(X-a_2)....(X-a_{n_1})$$ Now divide both sides by $(X -a_1)....(X-a_n)$.
$endgroup$
– N. S.
Jan 31 at 3:44
$begingroup$
Therefore $$ frac{1}{z-z_0} + frac{1}{z-z_1} + frac{1}{z-z_2} + frac{1}{z-z_3} + frac{1}{z-z_4}=frac{5z^4}{z^5+1} neq frac{1}{z^5+1}$$
$endgroup$
– N. S.
Jan 31 at 3:45
$begingroup$
@NS you’re right, of course. Like I said, it was a clue to try partial fractions. And although the numerators aren’t one, there is a decomposition of that form with the “1”s replaces by constants.
$endgroup$
– Matthew Leingang
Jan 31 at 11:35
$begingroup$
@N.S. I've replaced my hint with a full answer.
$endgroup$
– Matthew Leingang
Feb 1 at 22:06
add a comment |
$begingroup$
Partial fractions can work. I think this approach is pretty much equivalent to Mark Viola's residue one, but it's nice to see the agreement.
Let $zeta$ be a primitive fifth root of unity. If you want to be explicit, let $zeta = e^{2pi/5}$. But the important thing is that
- the roots of $z^5+1$ are $-1$, $-zeta$, $-zeta^2$, $-zeta^3$, and $-zeta^4$.
- Since $(z^5 - 1) = (z-1)(z^4+z^3+z^2+z+1)$, we know that $1 + zeta + zeta^2 + zeta^3 + zeta^4 = 0$.
Now for something about partial fractions that I didn't know until just now.
If $f(z) = (z-alpha_1) cdots (z-alpha_n)$, and all the roots are distinct, then
$$
frac{1}{f(z)} = frac{1}{f'(alpha_1)(z-alpha_1)} + frac{1}{f'(alpha_2)(z-alpha_2)} + dots + frac{1}{f'(alpha_n)(z-alpha_n)}
$$
To show this, start from
$$
frac{1}{(z-alpha_1) cdots (z-alpha_n)} = sum_{i=1}^n frac{A_i}{z-alpha_i}
$$
for some constants $A_1, dots, A_n$. Clear out the denominators, and you get
$$
1 = sum_{i=1}^n frac{A_i f(z)}{z-alpha_i}
= sum_{i=1}^n A_i frac{f(z)-f(alpha_i)}{z-alpha_i}
tag{$*$}
$$
Now for any $i$ and $j$,
$$
lim_{zto alpha_j} frac{f(z)-f(alpha_i)}{z-alpha_i} =
begin{cases} 0 & i neq j \ f'(alpha_i) & i = j end{cases}
$$
So if we take the limit of both sides of ($*$) as $zto z_j$ we get $1 = A_j f'(alpha_j)$ for each $j$.
This means that
begin{align*}
frac{1}{z^5+1}
&= frac{1}{5(z+1)} + frac{1}{5(-zeta)^4(z+zeta)} + frac{1}{5(-zeta^2)^4(z+zeta^2)} + frac{1}{5(-zeta^3)^4(z+zeta^3)} + frac{1}{5(-zeta^4)^4(z+zeta^4)}
\&= frac{1}{5}left(frac{1}{z+1} + frac{1}{zeta^4(z + zeta)} + frac{1}{zeta^3(z + zeta^2)} + frac{1}{zeta^2(z + zeta^3)} + frac{1}{zeta(z + zeta^4)}right)
\&= frac{1}{5}left(frac{1}{z+1} + frac{zeta}{z + zeta} + frac{zeta^2}{z + zeta^2} + frac{zeta^3}{z + zeta^3} + frac{zeta^4}{z + zeta^4}right)
end{align*}
You wanted an argument involving symmetry; what could be more symmetric than that equation?
Now $C$ encloses all those roots, so by the Cauchy Integral Formula,
$$
oint_C frac{dz}{z^5+1} = frac{2pi i}{5}left(1 + zeta + zeta^2 + zeta^3 + zeta^4right)
$$
and as we showed above, the latter factor is zero.
answered Jan 30 at 19:04
Matthew LeingangMatthew Leingang
16.7k12244
$endgroup$
Partial fractions can work. I think this approach is pretty much equivalent to Mark Viola's residue one, but it's nice to see the agreement.
Let $zeta$ be a primitive fifth root of unity. If you want to be explicit, let $zeta = e^{2pi/5}$. But the important thing is that
- the roots of $z^5+1$ are $-1$, $-zeta$, $-zeta^2$, $-zeta^3$, and $-zeta^4$.
- Since $(z^5 - 1) = (z-1)(z^4+z^3+z^2+z+1)$, we know that $1 + zeta + zeta^2 + zeta^3 + zeta^4 = 0$.
Now for something about partial fractions that I didn't know until just now.
If $f(z) = (z-alpha_1) cdots (z-alpha_n)$, and all the roots are distinct, then
$$
frac{1}{f(z)} = frac{1}{f'(alpha_1)(z-alpha_1)} + frac{1}{f'(alpha_2)(z-alpha_2)} + dots + frac{1}{f'(alpha_n)(z-alpha_n)}
$$
To show this, start from
$$
frac{1}{(z-alpha_1) cdots (z-alpha_n)} = sum_{i=1}^n frac{A_i}{z-alpha_i}
$$
for some constants $A_1, dots, A_n$. Clear out the denominators, and you get
$$
1 = sum_{i=1}^n frac{A_i f(z)}{z-alpha_i}
= sum_{i=1}^n A_i frac{f(z)-f(alpha_i)}{z-alpha_i}
tag{$*$}
$$
Now for any $i$ and $j$,
$$
lim_{zto alpha_j} frac{f(z)-f(alpha_i)}{z-alpha_i} =
begin{cases} 0 & i neq j \ f'(alpha_i) & i = j end{cases}
$$
So if we take the limit of both sides of ($*$) as $zto z_j$ we get $1 = A_j f'(alpha_j)$ for each $j$.
This means that
begin{align*}
frac{1}{z^5+1}
&= frac{1}{5(z+1)} + frac{1}{5(-zeta)^4(z+zeta)} + frac{1}{5(-zeta^2)^4(z+zeta^2)} + frac{1}{5(-zeta^3)^4(z+zeta^3)} + frac{1}{5(-zeta^4)^4(z+zeta^4)}
\&= frac{1}{5}left(frac{1}{z+1} + frac{1}{zeta^4(z + zeta)} + frac{1}{zeta^3(z + zeta^2)} + frac{1}{zeta^2(z + zeta^3)} + frac{1}{zeta(z + zeta^4)}right)
\&= frac{1}{5}left(frac{1}{z+1} + frac{zeta}{z + zeta} + frac{zeta^2}{z + zeta^2} + frac{zeta^3}{z + zeta^3} + frac{zeta^4}{z + zeta^4}right)
end{align*}
You wanted an argument involving symmetry; what could be more symmetric than that equation?
Now $C$ encloses all those roots, so by the Cauchy Integral Formula,
$$
oint_C frac{dz}{z^5+1} = frac{2pi i}{5}left(1 + zeta + zeta^2 + zeta^3 + zeta^4right)
$$
and as we showed above, the latter factor is zero.
answered Jan 30 at 19:04
Matthew LeingangMatthew Leingang
16.7k12244
edited Feb 1 at 22:05
answered Jan 30 at 19:04
Matthew LeingangMatthew Leingang
16.7k12244
answered Jan 30 at 19:04
Matthew LeingangMatthew Leingang
16.7k12244
answered Jan 30 at 19:04
Matthew LeingangMatthew Leingang
16.7k12244
16.7k12244
$begingroup$
One can check this, but this would be hard. I've made an edit to my question where I make it simpler by using symmetries.
$endgroup$
– Belgium_Physics
Jan 30 at 19:25
$begingroup$
Unfortunately, if $P(X)=(X-a_1)....(X-a_n)$ has distinct roots, then $$sum_{k=1}^n frac{1}{X-a_k}=frac{P'(X)}{P(X)}$$ This follows immediately from the product rule:$$P'(X)=((X-a_1)....(X-a_n))'=(X-a_2)(X-a_3)....(X-a_n)+(X-a_1)(X-a_3)....(X-a_n)+...+(X-a_1)(X-a_2)....(X-a_{n_1})$$ Now divide both sides by $(X -a_1)....(X-a_n)$.
$endgroup$
– N. S.
Jan 31 at 3:44
$begingroup$
Therefore $$ frac{1}{z-z_0} + frac{1}{z-z_1} + frac{1}{z-z_2} + frac{1}{z-z_3} + frac{1}{z-z_4}=frac{5z^4}{z^5+1} neq frac{1}{z^5+1}$$
$endgroup$
– N. S.
Jan 31 at 3:45
$begingroup$
@NS you’re right, of course. Like I said, it was a clue to try partial fractions. And although the numerators aren’t one, there is a decomposition of that form with the “1”s replaces by constants.
$endgroup$
– Matthew Leingang
Jan 31 at 11:35
$begingroup$
@N.S. I've replaced my hint with a full answer.
$endgroup$
– Matthew Leingang
Feb 1 at 22:06
add a comment |
$begingroup$
One can check this, but this would be hard. I've made an edit to my question where I make it simpler by using symmetries.
$endgroup$
– Belgium_Physics
Jan 30 at 19:25
$begingroup$
Unfortunately, if $P(X)=(X-a_1)....(X-a_n)$ has distinct roots, then $$sum_{k=1}^n frac{1}{X-a_k}=frac{P'(X)}{P(X)}$$ This follows immediately from the product rule:$$P'(X)=((X-a_1)....(X-a_n))'=(X-a_2)(X-a_3)....(X-a_n)+(X-a_1)(X-a_3)....(X-a_n)+...+(X-a_1)(X-a_2)....(X-a_{n_1})$$ Now divide both sides by $(X -a_1)....(X-a_n)$.
$endgroup$
– N. S.
Jan 31 at 3:44
$begingroup$
Therefore $$ frac{1}{z-z_0} + frac{1}{z-z_1} + frac{1}{z-z_2} + frac{1}{z-z_3} + frac{1}{z-z_4}=frac{5z^4}{z^5+1} neq frac{1}{z^5+1}$$
$endgroup$
– N. S.
Jan 31 at 3:45
$begingroup$
@NS you’re right, of course. Like I said, it was a clue to try partial fractions. And although the numerators aren’t one, there is a decomposition of that form with the “1”s replaces by constants.
$endgroup$
– Matthew Leingang
Jan 31 at 11:35
$begingroup$
@N.S. I've replaced my hint with a full answer.
$endgroup$
– Matthew Leingang
Feb 1 at 22:06
$begingroup$
One can check this, but this would be hard. I've made an edit to my question where I make it simpler by using symmetries.
$endgroup$
– Belgium_Physics
Jan 30 at 19:25
$begingroup$
One can check this, but this would be hard. I've made an edit to my question where I make it simpler by using symmetries.
$endgroup$
– Belgium_Physics
Jan 30 at 19:25
$begingroup$
Unfortunately, if $P(X)=(X-a_1)....(X-a_n)$ has distinct roots, then $$sum_{k=1}^n frac{1}{X-a_k}=frac{P'(X)}{P(X)}$$ This follows immediately from the product rule:$$P'(X)=((X-a_1)....(X-a_n))'=(X-a_2)(X-a_3)....(X-a_n)+(X-a_1)(X-a_3)....(X-a_n)+...+(X-a_1)(X-a_2)....(X-a_{n_1})$$ Now divide both sides by $(X -a_1)....(X-a_n)$.
$endgroup$
– N. S.
Jan 31 at 3:44
$begingroup$
Unfortunately, if $P(X)=(X-a_1)....(X-a_n)$ has distinct roots, then $$sum_{k=1}^n frac{1}{X-a_k}=frac{P'(X)}{P(X)}$$ This follows immediately from the product rule:$$P'(X)=((X-a_1)....(X-a_n))'=(X-a_2)(X-a_3)....(X-a_n)+(X-a_1)(X-a_3)....(X-a_n)+...+(X-a_1)(X-a_2)....(X-a_{n_1})$$ Now divide both sides by $(X -a_1)....(X-a_n)$.
$endgroup$
– N. S.
Jan 31 at 3:44
$begingroup$
Therefore $$ frac{1}{z-z_0} + frac{1}{z-z_1} + frac{1}{z-z_2} + frac{1}{z-z_3} + frac{1}{z-z_4}=frac{5z^4}{z^5+1} neq frac{1}{z^5+1}$$
$endgroup$
– N. S.
Jan 31 at 3:45
$begingroup$
Therefore $$ frac{1}{z-z_0} + frac{1}{z-z_1} + frac{1}{z-z_2} + frac{1}{z-z_3} + frac{1}{z-z_4}=frac{5z^4}{z^5+1} neq frac{1}{z^5+1}$$
$endgroup$
– N. S.
Jan 31 at 3:45
$begingroup$
@NS you’re right, of course. Like I said, it was a clue to try partial fractions. And although the numerators aren’t one, there is a decomposition of that form with the “1”s replaces by constants.
$endgroup$
– Matthew Leingang
Jan 31 at 11:35
$begingroup$
@NS you’re right, of course. Like I said, it was a clue to try partial fractions. And although the numerators aren’t one, there is a decomposition of that form with the “1”s replaces by constants.
$endgroup$
– Matthew Leingang
Jan 31 at 11:35
$begingroup$
@N.S. I've replaced my hint with a full answer.
$endgroup$
– Matthew Leingang
Feb 1 at 22:06
$begingroup$
@N.S. I've replaced my hint with a full answer.
$endgroup$
– Matthew Leingang
Feb 1 at 22:06
add a comment |
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No. You don't have any net residues at all. See below ... .
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– Oscar Lanzi
Jan 30 at 20:59