Finding the residues of $frac{cos z -1}{(e^z-1)^2}$.
I've found the poles of $frac{cos z -1}{(e^z-1)^2}$ to be double poles at each $z_k = 2kpi i$, where $kinmathbb{Z}$ and $kneq 0$. (At $k=0$ this is a removable singularity instead.)
I have no idea how to find out the residues at each $z_k$ - I tried using the formula $limlimits_{zrightarrow z_{0}}dfrac{d}{dz}(z-z_{k})^{2}dfrac{cos z -1}{(e^{z}-1)^{2}}$, but I can't see how to rearrange that and find the limit.
Is there a better way to do this?
complex-analysis limits analysis residue-calculus holomorphic-functions
asked Nov 24 at 16:50
pejet
11
add a comment |
I've found the poles of $frac{cos z -1}{(e^z-1)^2}$ to be double poles at each $z_k = 2kpi i$, where $kinmathbb{Z}$ and $kneq 0$. (At $k=0$ this is a removable singularity instead.)
I have no idea how to find out the residues at each $z_k$ - I tried using the formula $limlimits_{zrightarrow z_{0}}dfrac{d}{dz}(z-z_{k})^{2}dfrac{cos z -1}{(e^{z}-1)^{2}}$, but I can't see how to rearrange that and find the limit.
Is there a better way to do this?
complex-analysis limits analysis residue-calculus holomorphic-functions
asked Nov 24 at 16:50
pejet
11
add a comment |
I've found the poles of $frac{cos z -1}{(e^z-1)^2}$ to be double poles at each $z_k = 2kpi i$, where $kinmathbb{Z}$ and $kneq 0$. (At $k=0$ this is a removable singularity instead.)
I have no idea how to find out the residues at each $z_k$ - I tried using the formula $limlimits_{zrightarrow z_{0}}dfrac{d}{dz}(z-z_{k})^{2}dfrac{cos z -1}{(e^{z}-1)^{2}}$, but I can't see how to rearrange that and find the limit.
Is there a better way to do this?
complex-analysis limits analysis residue-calculus holomorphic-functions
asked Nov 24 at 16:50
pejet
11
I've found the poles of $frac{cos z -1}{(e^z-1)^2}$ to be double poles at each $z_k = 2kpi i$, where $kinmathbb{Z}$ and $kneq 0$. (At $k=0$ this is a removable singularity instead.)
I have no idea how to find out the residues at each $z_k$ - I tried using the formula $limlimits_{zrightarrow z_{0}}dfrac{d}{dz}(z-z_{k})^{2}dfrac{cos z -1}{(e^{z}-1)^{2}}$, but I can't see how to rearrange that and find the limit.
Is there a better way to do this?
complex-analysis limits analysis residue-calculus holomorphic-functions
complex-analysis limits analysis residue-calculus holomorphic-functions
asked Nov 24 at 16:50
pejet
11
asked Nov 24 at 16:50
pejet
11
asked Nov 24 at 16:50
pejet
11
asked Nov 24 at 16:50
pejet
11
asked Nov 24 at 16:50
pejet
11
11
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
Note thatbegin{align}cos(z)-1&=cosbigl((z-2kpi i)+2kpi i)-1\&=cos(z-2kpi i)cosh(2kpi)-1-sin(z-2kpi i)sinh(2kpi)i\&=cosh(2kpi)-1-isinh(2kpi)(z-2kpi i)+cdotsend{align}and thatbegin{align}(e^z-1)^2&=(e^{z-2kpi i}-1)^2\&=(z-2pi i)^2+(z-2pi i)^3+cdotsend{align}So, if$$frac{cos(z)-1}{(e^z-1)^2}=frac{a_{-2}}{(z-2kpi i)^2}+frac{a_{-1}}{z-kpi i}+a_0+cdots,$$then you have$$cos(z)-1=cosh(2kpi)-1-isinh(2kpi)(z-2kpi i)+cdots$$and alsobegin{align}cos(z)-1&=cosh(2kpi)-1-isinh(2kpi)(z-2kpi i)+cdots\&=(e^z-1)^2frac{cos(z)-1}{(e^z-1)^2}\&=bigl((z-2pi i)^2+(z-2pi i)^3+cdotsbigr)left(frac{a_{-2}}{(z-2kpi i)^2}+frac{a_{-1}}{z-kpi i}+a_0+cdotsright)\&=bigl(1+(z-2kpi i)bigr)bigl(a_{-2}+a_{-1}(z-2kpi i)+a_0(z-2kpi i)^2+cdotsbigr)\&=a_{-2}+(a_{-2}+a_{-1})(z-2kpi i)+cdotsend{align}So, $a_{-2}=cos(2kpi)-1$, and the residue, which is $a_{-1}$ can now be obtained from the equality$$a_{-2}+a_{-1}=-isinh(2kpi).$$In other words, the residue is equal to $1-cosh(2kpi)-isinh(2kpi)$.
answered Nov 24 at 17:27
José Carlos Santos
149k22119221
add a comment |
You have already found that $z=0$ is a removable singularity and the poles are $z_k=2kπi$ with $kinmathbb{Z}, kneq 0$.
$(e^z-1)^2$ is analytic on $mathbb{C}$, so it can be expanded in a Taylor series (converging to it at all $zinmathbb{C}$) about $z=z_k$:
begin{align}(e^z-1)^2&=(z-z_k)^2+(z-z_k)^3+cdots\&=(z-z_k)^2left[1+(z-z_k)+cdotsright]\&=(z-z_k)^2 g_k(z),end{align}
with $g_k(z_k)=g'_k(z_k)=1.$
The residue of the function at $z=z_k$ is:
begin{align}&limlimits_{zrightarrow z_k}frac{d}{dz}frac{cos z -1}{g_k(z)}\&=limlimits_{zrightarrow z_k}frac{(-sin z)g_k(z)+(1-cos z)g'_k(z)}{[g_k(z)]^2}
\&=1-cos (2kpi i)-sin (2kpi i)
\&=1-cosh (2kpi)-isinh (2kpi).end{align}
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3011780%2ffinding-the-residues-of-frac-cos-z-1ez-12%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Note thatbegin{align}cos(z)-1&=cosbigl((z-2kpi i)+2kpi i)-1\&=cos(z-2kpi i)cosh(2kpi)-1-sin(z-2kpi i)sinh(2kpi)i\&=cosh(2kpi)-1-isinh(2kpi)(z-2kpi i)+cdotsend{align}and thatbegin{align}(e^z-1)^2&=(e^{z-2kpi i}-1)^2\&=(z-2pi i)^2+(z-2pi i)^3+cdotsend{align}So, if$$frac{cos(z)-1}{(e^z-1)^2}=frac{a_{-2}}{(z-2kpi i)^2}+frac{a_{-1}}{z-kpi i}+a_0+cdots,$$then you have$$cos(z)-1=cosh(2kpi)-1-isinh(2kpi)(z-2kpi i)+cdots$$and alsobegin{align}cos(z)-1&=cosh(2kpi)-1-isinh(2kpi)(z-2kpi i)+cdots\&=(e^z-1)^2frac{cos(z)-1}{(e^z-1)^2}\&=bigl((z-2pi i)^2+(z-2pi i)^3+cdotsbigr)left(frac{a_{-2}}{(z-2kpi i)^2}+frac{a_{-1}}{z-kpi i}+a_0+cdotsright)\&=bigl(1+(z-2kpi i)bigr)bigl(a_{-2}+a_{-1}(z-2kpi i)+a_0(z-2kpi i)^2+cdotsbigr)\&=a_{-2}+(a_{-2}+a_{-1})(z-2kpi i)+cdotsend{align}So, $a_{-2}=cos(2kpi)-1$, and the residue, which is $a_{-1}$ can now be obtained from the equality$$a_{-2}+a_{-1}=-isinh(2kpi).$$In other words, the residue is equal to $1-cosh(2kpi)-isinh(2kpi)$.
answered Nov 24 at 17:27
José Carlos Santos
149k22119221
add a comment |
Note thatbegin{align}cos(z)-1&=cosbigl((z-2kpi i)+2kpi i)-1\&=cos(z-2kpi i)cosh(2kpi)-1-sin(z-2kpi i)sinh(2kpi)i\&=cosh(2kpi)-1-isinh(2kpi)(z-2kpi i)+cdotsend{align}and thatbegin{align}(e^z-1)^2&=(e^{z-2kpi i}-1)^2\&=(z-2pi i)^2+(z-2pi i)^3+cdotsend{align}So, if$$frac{cos(z)-1}{(e^z-1)^2}=frac{a_{-2}}{(z-2kpi i)^2}+frac{a_{-1}}{z-kpi i}+a_0+cdots,$$then you have$$cos(z)-1=cosh(2kpi)-1-isinh(2kpi)(z-2kpi i)+cdots$$and alsobegin{align}cos(z)-1&=cosh(2kpi)-1-isinh(2kpi)(z-2kpi i)+cdots\&=(e^z-1)^2frac{cos(z)-1}{(e^z-1)^2}\&=bigl((z-2pi i)^2+(z-2pi i)^3+cdotsbigr)left(frac{a_{-2}}{(z-2kpi i)^2}+frac{a_{-1}}{z-kpi i}+a_0+cdotsright)\&=bigl(1+(z-2kpi i)bigr)bigl(a_{-2}+a_{-1}(z-2kpi i)+a_0(z-2kpi i)^2+cdotsbigr)\&=a_{-2}+(a_{-2}+a_{-1})(z-2kpi i)+cdotsend{align}So, $a_{-2}=cos(2kpi)-1$, and the residue, which is $a_{-1}$ can now be obtained from the equality$$a_{-2}+a_{-1}=-isinh(2kpi).$$In other words, the residue is equal to $1-cosh(2kpi)-isinh(2kpi)$.
answered Nov 24 at 17:27
José Carlos Santos
149k22119221
add a comment |
Note thatbegin{align}cos(z)-1&=cosbigl((z-2kpi i)+2kpi i)-1\&=cos(z-2kpi i)cosh(2kpi)-1-sin(z-2kpi i)sinh(2kpi)i\&=cosh(2kpi)-1-isinh(2kpi)(z-2kpi i)+cdotsend{align}and thatbegin{align}(e^z-1)^2&=(e^{z-2kpi i}-1)^2\&=(z-2pi i)^2+(z-2pi i)^3+cdotsend{align}So, if$$frac{cos(z)-1}{(e^z-1)^2}=frac{a_{-2}}{(z-2kpi i)^2}+frac{a_{-1}}{z-kpi i}+a_0+cdots,$$then you have$$cos(z)-1=cosh(2kpi)-1-isinh(2kpi)(z-2kpi i)+cdots$$and alsobegin{align}cos(z)-1&=cosh(2kpi)-1-isinh(2kpi)(z-2kpi i)+cdots\&=(e^z-1)^2frac{cos(z)-1}{(e^z-1)^2}\&=bigl((z-2pi i)^2+(z-2pi i)^3+cdotsbigr)left(frac{a_{-2}}{(z-2kpi i)^2}+frac{a_{-1}}{z-kpi i}+a_0+cdotsright)\&=bigl(1+(z-2kpi i)bigr)bigl(a_{-2}+a_{-1}(z-2kpi i)+a_0(z-2kpi i)^2+cdotsbigr)\&=a_{-2}+(a_{-2}+a_{-1})(z-2kpi i)+cdotsend{align}So, $a_{-2}=cos(2kpi)-1$, and the residue, which is $a_{-1}$ can now be obtained from the equality$$a_{-2}+a_{-1}=-isinh(2kpi).$$In other words, the residue is equal to $1-cosh(2kpi)-isinh(2kpi)$.
answered Nov 24 at 17:27
José Carlos Santos
149k22119221
Note thatbegin{align}cos(z)-1&=cosbigl((z-2kpi i)+2kpi i)-1\&=cos(z-2kpi i)cosh(2kpi)-1-sin(z-2kpi i)sinh(2kpi)i\&=cosh(2kpi)-1-isinh(2kpi)(z-2kpi i)+cdotsend{align}and thatbegin{align}(e^z-1)^2&=(e^{z-2kpi i}-1)^2\&=(z-2pi i)^2+(z-2pi i)^3+cdotsend{align}So, if$$frac{cos(z)-1}{(e^z-1)^2}=frac{a_{-2}}{(z-2kpi i)^2}+frac{a_{-1}}{z-kpi i}+a_0+cdots,$$then you have$$cos(z)-1=cosh(2kpi)-1-isinh(2kpi)(z-2kpi i)+cdots$$and alsobegin{align}cos(z)-1&=cosh(2kpi)-1-isinh(2kpi)(z-2kpi i)+cdots\&=(e^z-1)^2frac{cos(z)-1}{(e^z-1)^2}\&=bigl((z-2pi i)^2+(z-2pi i)^3+cdotsbigr)left(frac{a_{-2}}{(z-2kpi i)^2}+frac{a_{-1}}{z-kpi i}+a_0+cdotsright)\&=bigl(1+(z-2kpi i)bigr)bigl(a_{-2}+a_{-1}(z-2kpi i)+a_0(z-2kpi i)^2+cdotsbigr)\&=a_{-2}+(a_{-2}+a_{-1})(z-2kpi i)+cdotsend{align}So, $a_{-2}=cos(2kpi)-1$, and the residue, which is $a_{-1}$ can now be obtained from the equality$$a_{-2}+a_{-1}=-isinh(2kpi).$$In other words, the residue is equal to $1-cosh(2kpi)-isinh(2kpi)$.
answered Nov 24 at 17:27
José Carlos Santos
149k22119221
edited Nov 26 at 8:54
answered Nov 24 at 17:27
José Carlos Santos
149k22119221
answered Nov 24 at 17:27
José Carlos Santos
149k22119221
answered Nov 24 at 17:27
José Carlos Santos
149k22119221
149k22119221
add a comment |
add a comment |
You have already found that $z=0$ is a removable singularity and the poles are $z_k=2kπi$ with $kinmathbb{Z}, kneq 0$.
$(e^z-1)^2$ is analytic on $mathbb{C}$, so it can be expanded in a Taylor series (converging to it at all $zinmathbb{C}$) about $z=z_k$:
begin{align}(e^z-1)^2&=(z-z_k)^2+(z-z_k)^3+cdots\&=(z-z_k)^2left[1+(z-z_k)+cdotsright]\&=(z-z_k)^2 g_k(z),end{align}
with $g_k(z_k)=g'_k(z_k)=1.$
The residue of the function at $z=z_k$ is:
begin{align}&limlimits_{zrightarrow z_k}frac{d}{dz}frac{cos z -1}{g_k(z)}\&=limlimits_{zrightarrow z_k}frac{(-sin z)g_k(z)+(1-cos z)g'_k(z)}{[g_k(z)]^2}
\&=1-cos (2kpi i)-sin (2kpi i)
\&=1-cosh (2kpi)-isinh (2kpi).end{align}
add a comment |
You have already found that $z=0$ is a removable singularity and the poles are $z_k=2kπi$ with $kinmathbb{Z}, kneq 0$.
$(e^z-1)^2$ is analytic on $mathbb{C}$, so it can be expanded in a Taylor series (converging to it at all $zinmathbb{C}$) about $z=z_k$:
begin{align}(e^z-1)^2&=(z-z_k)^2+(z-z_k)^3+cdots\&=(z-z_k)^2left[1+(z-z_k)+cdotsright]\&=(z-z_k)^2 g_k(z),end{align}
with $g_k(z_k)=g'_k(z_k)=1.$
The residue of the function at $z=z_k$ is:
begin{align}&limlimits_{zrightarrow z_k}frac{d}{dz}frac{cos z -1}{g_k(z)}\&=limlimits_{zrightarrow z_k}frac{(-sin z)g_k(z)+(1-cos z)g'_k(z)}{[g_k(z)]^2}
\&=1-cos (2kpi i)-sin (2kpi i)
\&=1-cosh (2kpi)-isinh (2kpi).end{align}
add a comment |
You have already found that $z=0$ is a removable singularity and the poles are $z_k=2kπi$ with $kinmathbb{Z}, kneq 0$.
$(e^z-1)^2$ is analytic on $mathbb{C}$, so it can be expanded in a Taylor series (converging to it at all $zinmathbb{C}$) about $z=z_k$:
begin{align}(e^z-1)^2&=(z-z_k)^2+(z-z_k)^3+cdots\&=(z-z_k)^2left[1+(z-z_k)+cdotsright]\&=(z-z_k)^2 g_k(z),end{align}
with $g_k(z_k)=g'_k(z_k)=1.$
The residue of the function at $z=z_k$ is:
begin{align}&limlimits_{zrightarrow z_k}frac{d}{dz}frac{cos z -1}{g_k(z)}\&=limlimits_{zrightarrow z_k}frac{(-sin z)g_k(z)+(1-cos z)g'_k(z)}{[g_k(z)]^2}
\&=1-cos (2kpi i)-sin (2kpi i)
\&=1-cosh (2kpi)-isinh (2kpi).end{align}
You have already found that $z=0$ is a removable singularity and the poles are $z_k=2kπi$ with $kinmathbb{Z}, kneq 0$.
$(e^z-1)^2$ is analytic on $mathbb{C}$, so it can be expanded in a Taylor series (converging to it at all $zinmathbb{C}$) about $z=z_k$:
begin{align}(e^z-1)^2&=(z-z_k)^2+(z-z_k)^3+cdots\&=(z-z_k)^2left[1+(z-z_k)+cdotsright]\&=(z-z_k)^2 g_k(z),end{align}
with $g_k(z_k)=g'_k(z_k)=1.$
The residue of the function at $z=z_k$ is:
begin{align}&limlimits_{zrightarrow z_k}frac{d}{dz}frac{cos z -1}{g_k(z)}\&=limlimits_{zrightarrow z_k}frac{(-sin z)g_k(z)+(1-cos z)g'_k(z)}{[g_k(z)]^2}
\&=1-cos (2kpi i)-sin (2kpi i)
\&=1-cosh (2kpi)-isinh (2kpi).end{align}
edited Dec 2 at 16:20
answered Dec 2 at 16:11
user621367
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3011780%2ffinding-the-residues-of-frac-cos-z-1ez-12%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
