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Find the residues at the singularities of $frac{z^2 - z}{1-sin{z}}$.
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I've been set a question where I'm asked to classify all of the singularities of
$$f(z) = frac{z^2 - z}{1-sin{z}}$$
and then calculate the residue of each of its singularities. I've found the singularities, all of which are of the form $ z_k =(2k + frac{1}{2})pi$, where $k$ is an integer, and shown each of these to be double poles.
I have the formula for the residue of $f$ at $z_k$ as
$$limlimits_{z to z_k} (frac{d}{dz}((z-z_k)^2f(z))$$
(since $z_k$ is a pole of order 2) but it ends up being really complicated and I can't seem to make it work. Is there a better way to do it?
complex-analysis limits analysis derivatives residue-calculus
asked 9 hours ago
pixuj
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up vote
0
down vote
favorite
I've been set a question where I'm asked to classify all of the singularities of
$$f(z) = frac{z^2 - z}{1-sin{z}}$$
and then calculate the residue of each of its singularities. I've found the singularities, all of which are of the form $ z_k =(2k + frac{1}{2})pi$, where $k$ is an integer, and shown each of these to be double poles.
I have the formula for the residue of $f$ at $z_k$ as
$$limlimits_{z to z_k} (frac{d}{dz}((z-z_k)^2f(z))$$
(since $z_k$ is a pole of order 2) but it ends up being really complicated and I can't seem to make it work. Is there a better way to do it?
complex-analysis limits analysis derivatives residue-calculus
asked 9 hours ago
pixuj
11
New contributor
pixuj is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
The formula you have only works for poles of order 2. It might be a better idea to find the coefficient of $z^{-1}$ on the Laurent series of $f$ around $z_k$. You can do it by finding the Taylor series of $z^2 -z$ and $1-sin(z)$ around $z_k$ and then chasing the wanted coefficient by performing “long division”.
– Alonso Delfín
7 hours ago
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I've been set a question where I'm asked to classify all of the singularities of
$$f(z) = frac{z^2 - z}{1-sin{z}}$$
and then calculate the residue of each of its singularities. I've found the singularities, all of which are of the form $ z_k =(2k + frac{1}{2})pi$, where $k$ is an integer, and shown each of these to be double poles.
I have the formula for the residue of $f$ at $z_k$ as
$$limlimits_{z to z_k} (frac{d}{dz}((z-z_k)^2f(z))$$
(since $z_k$ is a pole of order 2) but it ends up being really complicated and I can't seem to make it work. Is there a better way to do it?
complex-analysis limits analysis derivatives residue-calculus
asked 9 hours ago
pixuj
11
New contributor
pixuj is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I've been set a question where I'm asked to classify all of the singularities of
$$f(z) = frac{z^2 - z}{1-sin{z}}$$
and then calculate the residue of each of its singularities. I've found the singularities, all of which are of the form $ z_k =(2k + frac{1}{2})pi$, where $k$ is an integer, and shown each of these to be double poles.
I have the formula for the residue of $f$ at $z_k$ as
$$limlimits_{z to z_k} (frac{d}{dz}((z-z_k)^2f(z))$$
(since $z_k$ is a pole of order 2) but it ends up being really complicated and I can't seem to make it work. Is there a better way to do it?
complex-analysis limits analysis derivatives residue-calculus
complex-analysis limits analysis derivatives residue-calculus
asked 9 hours ago
pixuj
11
New contributor
pixuj is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 9 hours ago
pixuj
11
New contributor
pixuj is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 12 mins ago
asked 9 hours ago
pixuj
11
New contributor
pixuj is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 9 hours ago
pixuj
11
asked 9 hours ago
pixuj
11
11
New contributor
pixuj is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
pixuj is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
pixuj is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
The formula you have only works for poles of order 2. It might be a better idea to find the coefficient of $z^{-1}$ on the Laurent series of $f$ around $z_k$. You can do it by finding the Taylor series of $z^2 -z$ and $1-sin(z)$ around $z_k$ and then chasing the wanted coefficient by performing “long division”.
– Alonso Delfín
7 hours ago
add a comment |
The formula you have only works for poles of order 2. It might be a better idea to find the coefficient of $z^{-1}$ on the Laurent series of $f$ around $z_k$. You can do it by finding the Taylor series of $z^2 -z$ and $1-sin(z)$ around $z_k$ and then chasing the wanted coefficient by performing “long division”.
– Alonso Delfín
7 hours ago
The formula you have only works for poles of order 2. It might be a better idea to find the coefficient of $z^{-1}$ on the Laurent series of $f$ around $z_k$. You can do it by finding the Taylor series of $z^2 -z$ and $1-sin(z)$ around $z_k$ and then chasing the wanted coefficient by performing “long division”.
– Alonso Delfín
7 hours ago
The formula you have only works for poles of order 2. It might be a better idea to find the coefficient of $z^{-1}$ on the Laurent series of $f$ around $z_k$. You can do it by finding the Taylor series of $z^2 -z$ and $1-sin(z)$ around $z_k$ and then chasing the wanted coefficient by performing “long division”.
– Alonso Delfín
7 hours ago
add a comment |
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pixuj is a new contributor. Be nice, and check out our Code of Conduct.
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The formula you have only works for poles of order 2. It might be a better idea to find the coefficient of $z^{-1}$ on the Laurent series of $f$ around $z_k$. You can do it by finding the Taylor series of $z^2 -z$ and $1-sin(z)$ around $z_k$ and then chasing the wanted coefficient by performing “long division”.
– Alonso Delfín
7 hours ago