Mixing
Residue of $f(z)=frac{z}{sin{left(frac{pi}{z+1}right)}}$ in all isolated singularities
$begingroup$
I have this complex function:
$$f(z)=frac{z}{sinleft(frac{pi}{z+1}right)}$$
I'd like to compute residues in all isolate singularities.
If I'm not mistaken $f$ has poles in $z=frac{1}{k}-1$ and a non isolated singularity in $z=-1$, because it is an accumulation point of poles.
I tried to do something similar to this answer, but I don't seem to get a clean expression in terms of $xi$, where $xi$ is $z-frac{1}{k}+1$.
The best I can obtain is this:
$$frac{z}{sinleft(frac{kxi+1-k}{kxi+1}right)}$$
Can you help me?
complex-analysis residue-calculus laurent-series
edited Jan 15 at 22:14
rtybase
10.8k21533
asked Jan 7 at 13:30
EugenioDiPaolaEugenioDiPaola
513
$endgroup$
add a comment |
$begingroup$
I have this complex function:
$$f(z)=frac{z}{sinleft(frac{pi}{z+1}right)}$$
I'd like to compute residues in all isolate singularities.
If I'm not mistaken $f$ has poles in $z=frac{1}{k}-1$ and a non isolated singularity in $z=-1$, because it is an accumulation point of poles.
I tried to do something similar to this answer, but I don't seem to get a clean expression in terms of $xi$, where $xi$ is $z-frac{1}{k}+1$.
The best I can obtain is this:
$$frac{z}{sinleft(frac{kxi+1-k}{kxi+1}right)}$$
Can you help me?
complex-analysis residue-calculus laurent-series
edited Jan 15 at 22:14
rtybase
10.8k21533
asked Jan 7 at 13:30
EugenioDiPaolaEugenioDiPaola
513
$endgroup$
add a comment |
$begingroup$
I have this complex function:
$$f(z)=frac{z}{sinleft(frac{pi}{z+1}right)}$$
I'd like to compute residues in all isolate singularities.
If I'm not mistaken $f$ has poles in $z=frac{1}{k}-1$ and a non isolated singularity in $z=-1$, because it is an accumulation point of poles.
I tried to do something similar to this answer, but I don't seem to get a clean expression in terms of $xi$, where $xi$ is $z-frac{1}{k}+1$.
The best I can obtain is this:
$$frac{z}{sinleft(frac{kxi+1-k}{kxi+1}right)}$$
Can you help me?
complex-analysis residue-calculus laurent-series
edited Jan 15 at 22:14
rtybase
10.8k21533
asked Jan 7 at 13:30
EugenioDiPaolaEugenioDiPaola
513
$endgroup$
I have this complex function:
$$f(z)=frac{z}{sinleft(frac{pi}{z+1}right)}$$
I'd like to compute residues in all isolate singularities.
If I'm not mistaken $f$ has poles in $z=frac{1}{k}-1$ and a non isolated singularity in $z=-1$, because it is an accumulation point of poles.
I tried to do something similar to this answer, but I don't seem to get a clean expression in terms of $xi$, where $xi$ is $z-frac{1}{k}+1$.
The best I can obtain is this:
$$frac{z}{sinleft(frac{kxi+1-k}{kxi+1}right)}$$
Can you help me?
complex-analysis residue-calculus laurent-series
complex-analysis residue-calculus laurent-series
edited Jan 15 at 22:14
rtybase
10.8k21533
asked Jan 7 at 13:30
EugenioDiPaolaEugenioDiPaola
513
edited Jan 15 at 22:14
rtybase
10.8k21533
asked Jan 7 at 13:30
EugenioDiPaolaEugenioDiPaola
513
edited Jan 15 at 22:14
rtybase
10.8k21533
edited Jan 15 at 22:14
rtybase
10.8k21533
edited Jan 15 at 22:14
rtybase
10.8k21533
10.8k21533
asked Jan 7 at 13:30
EugenioDiPaolaEugenioDiPaola
513
asked Jan 7 at 13:30
EugenioDiPaolaEugenioDiPaola
513
asked Jan 7 at 13:30
EugenioDiPaolaEugenioDiPaola
513
513
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let $f(z)=sinleft(fracpi{z+1}right)$. Then$$f'(z)=-frac{picosleft(fracpi{z+1}right)}{(z+1)^2}$$and therefore$$f'left(frac1k-1right)=(-1)^{k-1}k^2pi.$$So,$$operatorname{res}_{z=frac1k-1}left(frac z{sinleft(fracpi{z+1}right)}right)=frac{frac1k-1}{(-1)^{k-1}k^2pi}=(-1)^{k-1}frac{1-k}{k^3pi}.$$
answered Jan 7 at 13:37
José Carlos SantosJosé Carlos Santos
157k22126227
$endgroup$
$begingroup$
thank you, but i don't understand the last equality; why is the residue equal to $frac{frac{1}{k}-1}{f^{'}(frac{1}{k}-1)}$?
$endgroup$
– EugenioDiPaola
Jan 7 at 14:53
1
$begingroup$
If $f$ and $g$ are analytic functions, $f(z_0)neq0$, and $z_0$ is a simple pole of $g$, then$$operatorname{res}_{z=z_0}left(frac{f(z)}{g(z)}right)=frac{f(z_0)}{g'(z_0)}.$$
$endgroup$
– José Carlos Santos
Jan 7 at 14:56
1
$begingroup$
@EugenioDiPaola For completeness, you should include the residue at $z = infty$, which is $pi/6$. The sum of all residues will be zero (if we choose smaller and smaller circles around $z = -1$ between the poles, the sequence of integrals tends to zero).
$endgroup$
– Maxim
Jan 16 at 2:29
$begingroup$
right, i forgot about that :) That's interesting, my course material states: "if a function has ONLY isolated singularities the sum of all residues, taken in account also the eventual residue at infinity, is zero". You're saying that the integral that contains only -1 inside and all the isolated singularities outside is zero. So, is there a stronger statement such as "the sum of all residues of isolated singularities is zero", even if the function has not isolated singularities? Or we're just lucky for this particular function?
$endgroup$
– EugenioDiPaola
Jan 16 at 22:37
1
$begingroup$
My guess is that you were just lucky with this particular function.
$endgroup$
– José Carlos Santos
Jan 16 at 23:26
|
show 2 more comments
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1 Answer
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active
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1 Answer
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active
oldest
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$begingroup$
Let $f(z)=sinleft(fracpi{z+1}right)$. Then$$f'(z)=-frac{picosleft(fracpi{z+1}right)}{(z+1)^2}$$and therefore$$f'left(frac1k-1right)=(-1)^{k-1}k^2pi.$$So,$$operatorname{res}_{z=frac1k-1}left(frac z{sinleft(fracpi{z+1}right)}right)=frac{frac1k-1}{(-1)^{k-1}k^2pi}=(-1)^{k-1}frac{1-k}{k^3pi}.$$
answered Jan 7 at 13:37
José Carlos SantosJosé Carlos Santos
157k22126227
$endgroup$
$begingroup$
thank you, but i don't understand the last equality; why is the residue equal to $frac{frac{1}{k}-1}{f^{'}(frac{1}{k}-1)}$?
$endgroup$
– EugenioDiPaola
Jan 7 at 14:53
1
$begingroup$
If $f$ and $g$ are analytic functions, $f(z_0)neq0$, and $z_0$ is a simple pole of $g$, then$$operatorname{res}_{z=z_0}left(frac{f(z)}{g(z)}right)=frac{f(z_0)}{g'(z_0)}.$$
$endgroup$
– José Carlos Santos
Jan 7 at 14:56
1
$begingroup$
@EugenioDiPaola For completeness, you should include the residue at $z = infty$, which is $pi/6$. The sum of all residues will be zero (if we choose smaller and smaller circles around $z = -1$ between the poles, the sequence of integrals tends to zero).
$endgroup$
– Maxim
Jan 16 at 2:29
$begingroup$
right, i forgot about that :) That's interesting, my course material states: "if a function has ONLY isolated singularities the sum of all residues, taken in account also the eventual residue at infinity, is zero". You're saying that the integral that contains only -1 inside and all the isolated singularities outside is zero. So, is there a stronger statement such as "the sum of all residues of isolated singularities is zero", even if the function has not isolated singularities? Or we're just lucky for this particular function?
$endgroup$
– EugenioDiPaola
Jan 16 at 22:37
1
$begingroup$
My guess is that you were just lucky with this particular function.
$endgroup$
– José Carlos Santos
Jan 16 at 23:26
|
show 2 more comments
$begingroup$
Let $f(z)=sinleft(fracpi{z+1}right)$. Then$$f'(z)=-frac{picosleft(fracpi{z+1}right)}{(z+1)^2}$$and therefore$$f'left(frac1k-1right)=(-1)^{k-1}k^2pi.$$So,$$operatorname{res}_{z=frac1k-1}left(frac z{sinleft(fracpi{z+1}right)}right)=frac{frac1k-1}{(-1)^{k-1}k^2pi}=(-1)^{k-1}frac{1-k}{k^3pi}.$$
answered Jan 7 at 13:37
José Carlos SantosJosé Carlos Santos
157k22126227
$endgroup$
$begingroup$
thank you, but i don't understand the last equality; why is the residue equal to $frac{frac{1}{k}-1}{f^{'}(frac{1}{k}-1)}$?
$endgroup$
– EugenioDiPaola
Jan 7 at 14:53
1
$begingroup$
If $f$ and $g$ are analytic functions, $f(z_0)neq0$, and $z_0$ is a simple pole of $g$, then$$operatorname{res}_{z=z_0}left(frac{f(z)}{g(z)}right)=frac{f(z_0)}{g'(z_0)}.$$
$endgroup$
– José Carlos Santos
Jan 7 at 14:56
1
$begingroup$
@EugenioDiPaola For completeness, you should include the residue at $z = infty$, which is $pi/6$. The sum of all residues will be zero (if we choose smaller and smaller circles around $z = -1$ between the poles, the sequence of integrals tends to zero).
$endgroup$
– Maxim
Jan 16 at 2:29
$begingroup$
right, i forgot about that :) That's interesting, my course material states: "if a function has ONLY isolated singularities the sum of all residues, taken in account also the eventual residue at infinity, is zero". You're saying that the integral that contains only -1 inside and all the isolated singularities outside is zero. So, is there a stronger statement such as "the sum of all residues of isolated singularities is zero", even if the function has not isolated singularities? Or we're just lucky for this particular function?
$endgroup$
– EugenioDiPaola
Jan 16 at 22:37
1
$begingroup$
My guess is that you were just lucky with this particular function.
$endgroup$
– José Carlos Santos
Jan 16 at 23:26
|
show 2 more comments
$begingroup$
Let $f(z)=sinleft(fracpi{z+1}right)$. Then$$f'(z)=-frac{picosleft(fracpi{z+1}right)}{(z+1)^2}$$and therefore$$f'left(frac1k-1right)=(-1)^{k-1}k^2pi.$$So,$$operatorname{res}_{z=frac1k-1}left(frac z{sinleft(fracpi{z+1}right)}right)=frac{frac1k-1}{(-1)^{k-1}k^2pi}=(-1)^{k-1}frac{1-k}{k^3pi}.$$
answered Jan 7 at 13:37
José Carlos SantosJosé Carlos Santos
157k22126227
$endgroup$
Let $f(z)=sinleft(fracpi{z+1}right)$. Then$$f'(z)=-frac{picosleft(fracpi{z+1}right)}{(z+1)^2}$$and therefore$$f'left(frac1k-1right)=(-1)^{k-1}k^2pi.$$So,$$operatorname{res}_{z=frac1k-1}left(frac z{sinleft(fracpi{z+1}right)}right)=frac{frac1k-1}{(-1)^{k-1}k^2pi}=(-1)^{k-1}frac{1-k}{k^3pi}.$$
answered Jan 7 at 13:37
José Carlos SantosJosé Carlos Santos
157k22126227
answered Jan 7 at 13:37
José Carlos SantosJosé Carlos Santos
157k22126227
answered Jan 7 at 13:37
José Carlos SantosJosé Carlos Santos
157k22126227
answered Jan 7 at 13:37
José Carlos SantosJosé Carlos Santos
157k22126227
157k22126227
$begingroup$
thank you, but i don't understand the last equality; why is the residue equal to $frac{frac{1}{k}-1}{f^{'}(frac{1}{k}-1)}$?
$endgroup$
– EugenioDiPaola
Jan 7 at 14:53
1
$begingroup$
If $f$ and $g$ are analytic functions, $f(z_0)neq0$, and $z_0$ is a simple pole of $g$, then$$operatorname{res}_{z=z_0}left(frac{f(z)}{g(z)}right)=frac{f(z_0)}{g'(z_0)}.$$
$endgroup$
– José Carlos Santos
Jan 7 at 14:56
1
$begingroup$
@EugenioDiPaola For completeness, you should include the residue at $z = infty$, which is $pi/6$. The sum of all residues will be zero (if we choose smaller and smaller circles around $z = -1$ between the poles, the sequence of integrals tends to zero).
$endgroup$
– Maxim
Jan 16 at 2:29
$begingroup$
right, i forgot about that :) That's interesting, my course material states: "if a function has ONLY isolated singularities the sum of all residues, taken in account also the eventual residue at infinity, is zero". You're saying that the integral that contains only -1 inside and all the isolated singularities outside is zero. So, is there a stronger statement such as "the sum of all residues of isolated singularities is zero", even if the function has not isolated singularities? Or we're just lucky for this particular function?
$endgroup$
– EugenioDiPaola
Jan 16 at 22:37
1
$begingroup$
My guess is that you were just lucky with this particular function.
$endgroup$
– José Carlos Santos
Jan 16 at 23:26
|
show 2 more comments
$begingroup$
thank you, but i don't understand the last equality; why is the residue equal to $frac{frac{1}{k}-1}{f^{'}(frac{1}{k}-1)}$?
$endgroup$
– EugenioDiPaola
Jan 7 at 14:53
1
$begingroup$
If $f$ and $g$ are analytic functions, $f(z_0)neq0$, and $z_0$ is a simple pole of $g$, then$$operatorname{res}_{z=z_0}left(frac{f(z)}{g(z)}right)=frac{f(z_0)}{g'(z_0)}.$$
$endgroup$
– José Carlos Santos
Jan 7 at 14:56
1
$begingroup$
@EugenioDiPaola For completeness, you should include the residue at $z = infty$, which is $pi/6$. The sum of all residues will be zero (if we choose smaller and smaller circles around $z = -1$ between the poles, the sequence of integrals tends to zero).
$endgroup$
– Maxim
Jan 16 at 2:29
$begingroup$
right, i forgot about that :) That's interesting, my course material states: "if a function has ONLY isolated singularities the sum of all residues, taken in account also the eventual residue at infinity, is zero". You're saying that the integral that contains only -1 inside and all the isolated singularities outside is zero. So, is there a stronger statement such as "the sum of all residues of isolated singularities is zero", even if the function has not isolated singularities? Or we're just lucky for this particular function?
$endgroup$
– EugenioDiPaola
Jan 16 at 22:37
1
$begingroup$
My guess is that you were just lucky with this particular function.
$endgroup$
– José Carlos Santos
Jan 16 at 23:26
$begingroup$
thank you, but i don't understand the last equality; why is the residue equal to $frac{frac{1}{k}-1}{f^{'}(frac{1}{k}-1)}$?
$endgroup$
– EugenioDiPaola
Jan 7 at 14:53
$begingroup$
thank you, but i don't understand the last equality; why is the residue equal to $frac{frac{1}{k}-1}{f^{'}(frac{1}{k}-1)}$?
$endgroup$
– EugenioDiPaola
Jan 7 at 14:53
1
1
$begingroup$
If $f$ and $g$ are analytic functions, $f(z_0)neq0$, and $z_0$ is a simple pole of $g$, then$$operatorname{res}_{z=z_0}left(frac{f(z)}{g(z)}right)=frac{f(z_0)}{g'(z_0)}.$$
$endgroup$
– José Carlos Santos
Jan 7 at 14:56
$begingroup$
If $f$ and $g$ are analytic functions, $f(z_0)neq0$, and $z_0$ is a simple pole of $g$, then$$operatorname{res}_{z=z_0}left(frac{f(z)}{g(z)}right)=frac{f(z_0)}{g'(z_0)}.$$
$endgroup$
– José Carlos Santos
Jan 7 at 14:56
1
1
$begingroup$
@EugenioDiPaola For completeness, you should include the residue at $z = infty$, which is $pi/6$. The sum of all residues will be zero (if we choose smaller and smaller circles around $z = -1$ between the poles, the sequence of integrals tends to zero).
$endgroup$
– Maxim
Jan 16 at 2:29
$begingroup$
@EugenioDiPaola For completeness, you should include the residue at $z = infty$, which is $pi/6$. The sum of all residues will be zero (if we choose smaller and smaller circles around $z = -1$ between the poles, the sequence of integrals tends to zero).
$endgroup$
– Maxim
Jan 16 at 2:29
$begingroup$
right, i forgot about that :) That's interesting, my course material states: "if a function has ONLY isolated singularities the sum of all residues, taken in account also the eventual residue at infinity, is zero". You're saying that the integral that contains only -1 inside and all the isolated singularities outside is zero. So, is there a stronger statement such as "the sum of all residues of isolated singularities is zero", even if the function has not isolated singularities? Or we're just lucky for this particular function?
$endgroup$
– EugenioDiPaola
Jan 16 at 22:37
$begingroup$
right, i forgot about that :) That's interesting, my course material states: "if a function has ONLY isolated singularities the sum of all residues, taken in account also the eventual residue at infinity, is zero". You're saying that the integral that contains only -1 inside and all the isolated singularities outside is zero. So, is there a stronger statement such as "the sum of all residues of isolated singularities is zero", even if the function has not isolated singularities? Or we're just lucky for this particular function?
$endgroup$
– EugenioDiPaola
Jan 16 at 22:37
1
1
$begingroup$
My guess is that you were just lucky with this particular function.
$endgroup$
– José Carlos Santos
Jan 16 at 23:26
$begingroup$
My guess is that you were just lucky with this particular function.
$endgroup$
– José Carlos Santos
Jan 16 at 23:26
|
show 2 more comments
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