Proving $Res(f'/f,z)$ is an integer for f analytic on a domain $Omega$ and $z in Omega$
$begingroup$
Here is what I have so far,
since f is analytic at z we know it is infinitely differentiable at z, so $f'(z) not = 0$ and so we know that $Res(f'/f,z) = frac{f'(z)}{f'(z)} = 1$ and any pole of higher order will be a multiple of this result so it will still be an integer. Is there a mistake in this reasoning, it seems a bit too simple but I think I covered all the possibilities for singularities of f'/f
complex-analysis residue-calculus
asked Dec 12 '18 at 0:29
Richard VillalobosRichard Villalobos
1687
$endgroup$
add a comment |
$begingroup$
Here is what I have so far,
since f is analytic at z we know it is infinitely differentiable at z, so $f'(z) not = 0$ and so we know that $Res(f'/f,z) = frac{f'(z)}{f'(z)} = 1$ and any pole of higher order will be a multiple of this result so it will still be an integer. Is there a mistake in this reasoning, it seems a bit too simple but I think I covered all the possibilities for singularities of f'/f
complex-analysis residue-calculus
asked Dec 12 '18 at 0:29
Richard VillalobosRichard Villalobos
1687
$endgroup$
2
$begingroup$
No, try with $f(z) = z^2$ and the residue at $0$. The correct argument is that $f(z) = (z-a)^n h(z)$ with $h(a) ne 0$ thus $f'(z)/f(z) = ...$
$endgroup$
– reuns
Dec 12 '18 at 0:45
add a comment |
$begingroup$
Here is what I have so far,
since f is analytic at z we know it is infinitely differentiable at z, so $f'(z) not = 0$ and so we know that $Res(f'/f,z) = frac{f'(z)}{f'(z)} = 1$ and any pole of higher order will be a multiple of this result so it will still be an integer. Is there a mistake in this reasoning, it seems a bit too simple but I think I covered all the possibilities for singularities of f'/f
complex-analysis residue-calculus
asked Dec 12 '18 at 0:29
Richard VillalobosRichard Villalobos
1687
$endgroup$
Here is what I have so far,
since f is analytic at z we know it is infinitely differentiable at z, so $f'(z) not = 0$ and so we know that $Res(f'/f,z) = frac{f'(z)}{f'(z)} = 1$ and any pole of higher order will be a multiple of this result so it will still be an integer. Is there a mistake in this reasoning, it seems a bit too simple but I think I covered all the possibilities for singularities of f'/f
complex-analysis residue-calculus
complex-analysis residue-calculus
asked Dec 12 '18 at 0:29
Richard VillalobosRichard Villalobos
1687
asked Dec 12 '18 at 0:29
Richard VillalobosRichard Villalobos
1687
asked Dec 12 '18 at 0:29
Richard VillalobosRichard Villalobos
1687
asked Dec 12 '18 at 0:29
Richard VillalobosRichard Villalobos
1687
asked Dec 12 '18 at 0:29
Richard VillalobosRichard Villalobos
1687
1687
2
$begingroup$
No, try with $f(z) = z^2$ and the residue at $0$. The correct argument is that $f(z) = (z-a)^n h(z)$ with $h(a) ne 0$ thus $f'(z)/f(z) = ...$
$endgroup$
– reuns
Dec 12 '18 at 0:45
add a comment |
2
$begingroup$
No, try with $f(z) = z^2$ and the residue at $0$. The correct argument is that $f(z) = (z-a)^n h(z)$ with $h(a) ne 0$ thus $f'(z)/f(z) = ...$
$endgroup$
– reuns
Dec 12 '18 at 0:45
2
2
$begingroup$
No, try with $f(z) = z^2$ and the residue at $0$. The correct argument is that $f(z) = (z-a)^n h(z)$ with $h(a) ne 0$ thus $f'(z)/f(z) = ...$
$endgroup$
– reuns
Dec 12 '18 at 0:45
$begingroup$
No, try with $f(z) = z^2$ and the residue at $0$. The correct argument is that $f(z) = (z-a)^n h(z)$ with $h(a) ne 0$ thus $f'(z)/f(z) = ...$
$endgroup$
– reuns
Dec 12 '18 at 0:45
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The statement $Resleft(frac{f'}{f}, zright) = = frac{f'}{f'}$ isn't quite correct. In general, if $g$ has an order $k$ pole at $z_0$ and $h$ has an order $k+1$ pole at $z_0$, then begin{align*}
Resleft(frac{g}{h},z_0right) &= (k+1)frac{g^{(k)}(z_0)}{h^{(k+1)}(z_0)}
end{align*}
If $f$ has an order $k+1$ pole at $z_0$, then $f'$ has an order $k$ pole at $z_0$. Therefore begin{align*}
Resleft(frac{f'}{f},z_0right) = (k+1)frac{(f')^{(k)}(z_0)}{f^{(k+1)}(z_0)} = (k+1)frac{f^{(k+1)}(z_0)}{f^{(k+1)}(z_0)}= k+1
end{align*}
answered Dec 13 '18 at 2:47
J. PistachioJ. Pistachio
488212
$endgroup$
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%2f3036063%2fproving-resf-f-z-is-an-integer-for-f-analytic-on-a-domain-omega-and-z%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The statement $Resleft(frac{f'}{f}, zright) = = frac{f'}{f'}$ isn't quite correct. In general, if $g$ has an order $k$ pole at $z_0$ and $h$ has an order $k+1$ pole at $z_0$, then begin{align*}
Resleft(frac{g}{h},z_0right) &= (k+1)frac{g^{(k)}(z_0)}{h^{(k+1)}(z_0)}
end{align*}
If $f$ has an order $k+1$ pole at $z_0$, then $f'$ has an order $k$ pole at $z_0$. Therefore begin{align*}
Resleft(frac{f'}{f},z_0right) = (k+1)frac{(f')^{(k)}(z_0)}{f^{(k+1)}(z_0)} = (k+1)frac{f^{(k+1)}(z_0)}{f^{(k+1)}(z_0)}= k+1
end{align*}
answered Dec 13 '18 at 2:47
J. PistachioJ. Pistachio
488212
$endgroup$
add a comment |
$begingroup$
The statement $Resleft(frac{f'}{f}, zright) = = frac{f'}{f'}$ isn't quite correct. In general, if $g$ has an order $k$ pole at $z_0$ and $h$ has an order $k+1$ pole at $z_0$, then begin{align*}
Resleft(frac{g}{h},z_0right) &= (k+1)frac{g^{(k)}(z_0)}{h^{(k+1)}(z_0)}
end{align*}
If $f$ has an order $k+1$ pole at $z_0$, then $f'$ has an order $k$ pole at $z_0$. Therefore begin{align*}
Resleft(frac{f'}{f},z_0right) = (k+1)frac{(f')^{(k)}(z_0)}{f^{(k+1)}(z_0)} = (k+1)frac{f^{(k+1)}(z_0)}{f^{(k+1)}(z_0)}= k+1
end{align*}
answered Dec 13 '18 at 2:47
J. PistachioJ. Pistachio
488212
$endgroup$
add a comment |
$begingroup$
The statement $Resleft(frac{f'}{f}, zright) = = frac{f'}{f'}$ isn't quite correct. In general, if $g$ has an order $k$ pole at $z_0$ and $h$ has an order $k+1$ pole at $z_0$, then begin{align*}
Resleft(frac{g}{h},z_0right) &= (k+1)frac{g^{(k)}(z_0)}{h^{(k+1)}(z_0)}
end{align*}
If $f$ has an order $k+1$ pole at $z_0$, then $f'$ has an order $k$ pole at $z_0$. Therefore begin{align*}
Resleft(frac{f'}{f},z_0right) = (k+1)frac{(f')^{(k)}(z_0)}{f^{(k+1)}(z_0)} = (k+1)frac{f^{(k+1)}(z_0)}{f^{(k+1)}(z_0)}= k+1
end{align*}
answered Dec 13 '18 at 2:47
J. PistachioJ. Pistachio
488212
$endgroup$
The statement $Resleft(frac{f'}{f}, zright) = = frac{f'}{f'}$ isn't quite correct. In general, if $g$ has an order $k$ pole at $z_0$ and $h$ has an order $k+1$ pole at $z_0$, then begin{align*}
Resleft(frac{g}{h},z_0right) &= (k+1)frac{g^{(k)}(z_0)}{h^{(k+1)}(z_0)}
end{align*}
If $f$ has an order $k+1$ pole at $z_0$, then $f'$ has an order $k$ pole at $z_0$. Therefore begin{align*}
Resleft(frac{f'}{f},z_0right) = (k+1)frac{(f')^{(k)}(z_0)}{f^{(k+1)}(z_0)} = (k+1)frac{f^{(k+1)}(z_0)}{f^{(k+1)}(z_0)}= k+1
end{align*}
answered Dec 13 '18 at 2:47
J. PistachioJ. Pistachio
488212
answered Dec 13 '18 at 2:47
J. PistachioJ. Pistachio
488212
answered Dec 13 '18 at 2:47
J. PistachioJ. Pistachio
488212
answered Dec 13 '18 at 2:47
J. PistachioJ. Pistachio
488212
488212
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.
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%2f3036063%2fproving-resf-f-z-is-an-integer-for-f-analytic-on-a-domain-omega-and-z%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
2
$begingroup$
No, try with $f(z) = z^2$ and the residue at $0$. The correct argument is that $f(z) = (z-a)^n h(z)$ with $h(a) ne 0$ thus $f'(z)/f(z) = ...$
$endgroup$
– reuns
Dec 12 '18 at 0:45