Mixing
Existence of an entire function
$begingroup$
my Complex Analysis final is in a couple of days and i'm struggling with this question -
"Is there an entire function $f$ that satisfies $|f(z)| = |z| + 1$ for every $z$ in the complex plane for which $|z| ge 2017$?"
I deducted if such function exists, it must be a polynomial. I tried tinkering around with $1/f$ and $f(1/z)$ but didn't really find anything useful, also tried using Rouché's theorem but I can't seem to prove/disprove the existence of a function.
I'd love a hint! :)
complex-analysis entire-functions
asked Jan 26 at 6:48
Guy SchwartzbergGuy Schwartzberg
1247
$endgroup$
add a comment |
$begingroup$
my Complex Analysis final is in a couple of days and i'm struggling with this question -
"Is there an entire function $f$ that satisfies $|f(z)| = |z| + 1$ for every $z$ in the complex plane for which $|z| ge 2017$?"
I deducted if such function exists, it must be a polynomial. I tried tinkering around with $1/f$ and $f(1/z)$ but didn't really find anything useful, also tried using Rouché's theorem but I can't seem to prove/disprove the existence of a function.
I'd love a hint! :)
complex-analysis entire-functions
asked Jan 26 at 6:48
Guy SchwartzbergGuy Schwartzberg
1247
$endgroup$
$begingroup$
Is this the actual question, or are there further conditions? There are trivial solutions such as $f$ being the zero function.
$endgroup$
– Lord Shark the Unknown
Jan 26 at 6:54
$begingroup$
My mistake, I wrote it wrong. So sorry, I edited it. It should have been equals instead of smaller than.
$endgroup$
– Guy Schwartzberg
Jan 26 at 7:00
$begingroup$
Can you prove that $f$ must be a degree one polynomial?
$endgroup$
– Lord Shark the Unknown
Jan 26 at 7:01
$begingroup$
That's what I tried to do, but for that I have to prove that f has only one root inside the circle of radius 2017, right? Also, I am not sure, how does that prove such $f$ doesn't exist?
$endgroup$
– Guy Schwartzberg
Jan 26 at 7:05
$begingroup$
Once you know that $f$ is a polynomial, its degree can extracted from the knowledge on its growth speed as $|z|toinfty$.
$endgroup$
– Sangchul Lee
Jan 26 at 7:23
add a comment |
$begingroup$
my Complex Analysis final is in a couple of days and i'm struggling with this question -
"Is there an entire function $f$ that satisfies $|f(z)| = |z| + 1$ for every $z$ in the complex plane for which $|z| ge 2017$?"
I deducted if such function exists, it must be a polynomial. I tried tinkering around with $1/f$ and $f(1/z)$ but didn't really find anything useful, also tried using Rouché's theorem but I can't seem to prove/disprove the existence of a function.
I'd love a hint! :)
complex-analysis entire-functions
asked Jan 26 at 6:48
Guy SchwartzbergGuy Schwartzberg
1247
$endgroup$
my Complex Analysis final is in a couple of days and i'm struggling with this question -
"Is there an entire function $f$ that satisfies $|f(z)| = |z| + 1$ for every $z$ in the complex plane for which $|z| ge 2017$?"
I deducted if such function exists, it must be a polynomial. I tried tinkering around with $1/f$ and $f(1/z)$ but didn't really find anything useful, also tried using Rouché's theorem but I can't seem to prove/disprove the existence of a function.
I'd love a hint! :)
complex-analysis entire-functions
complex-analysis entire-functions
asked Jan 26 at 6:48
Guy SchwartzbergGuy Schwartzberg
1247
asked Jan 26 at 6:48
Guy SchwartzbergGuy Schwartzberg
1247
edited Jan 26 at 6:59
Guy Schwartzberg
asked Jan 26 at 6:48
Guy SchwartzbergGuy Schwartzberg
1247
asked Jan 26 at 6:48
Guy SchwartzbergGuy Schwartzberg
1247
asked Jan 26 at 6:48
Guy SchwartzbergGuy Schwartzberg
1247
1247
$begingroup$
Is this the actual question, or are there further conditions? There are trivial solutions such as $f$ being the zero function.
$endgroup$
– Lord Shark the Unknown
Jan 26 at 6:54
$begingroup$
My mistake, I wrote it wrong. So sorry, I edited it. It should have been equals instead of smaller than.
$endgroup$
– Guy Schwartzberg
Jan 26 at 7:00
$begingroup$
Can you prove that $f$ must be a degree one polynomial?
$endgroup$
– Lord Shark the Unknown
Jan 26 at 7:01
$begingroup$
That's what I tried to do, but for that I have to prove that f has only one root inside the circle of radius 2017, right? Also, I am not sure, how does that prove such $f$ doesn't exist?
$endgroup$
– Guy Schwartzberg
Jan 26 at 7:05
$begingroup$
Once you know that $f$ is a polynomial, its degree can extracted from the knowledge on its growth speed as $|z|toinfty$.
$endgroup$
– Sangchul Lee
Jan 26 at 7:23
add a comment |
$begingroup$
Is this the actual question, or are there further conditions? There are trivial solutions such as $f$ being the zero function.
$endgroup$
– Lord Shark the Unknown
Jan 26 at 6:54
$begingroup$
My mistake, I wrote it wrong. So sorry, I edited it. It should have been equals instead of smaller than.
$endgroup$
– Guy Schwartzberg
Jan 26 at 7:00
$begingroup$
Can you prove that $f$ must be a degree one polynomial?
$endgroup$
– Lord Shark the Unknown
Jan 26 at 7:01
$begingroup$
That's what I tried to do, but for that I have to prove that f has only one root inside the circle of radius 2017, right? Also, I am not sure, how does that prove such $f$ doesn't exist?
$endgroup$
– Guy Schwartzberg
Jan 26 at 7:05
$begingroup$
Once you know that $f$ is a polynomial, its degree can extracted from the knowledge on its growth speed as $|z|toinfty$.
$endgroup$
– Sangchul Lee
Jan 26 at 7:23
$begingroup$
Is this the actual question, or are there further conditions? There are trivial solutions such as $f$ being the zero function.
$endgroup$
– Lord Shark the Unknown
Jan 26 at 6:54
$begingroup$
Is this the actual question, or are there further conditions? There are trivial solutions such as $f$ being the zero function.
$endgroup$
– Lord Shark the Unknown
Jan 26 at 6:54
$begingroup$
My mistake, I wrote it wrong. So sorry, I edited it. It should have been equals instead of smaller than.
$endgroup$
– Guy Schwartzberg
Jan 26 at 7:00
$begingroup$
My mistake, I wrote it wrong. So sorry, I edited it. It should have been equals instead of smaller than.
$endgroup$
– Guy Schwartzberg
Jan 26 at 7:00
$begingroup$
Can you prove that $f$ must be a degree one polynomial?
$endgroup$
– Lord Shark the Unknown
Jan 26 at 7:01
$begingroup$
Can you prove that $f$ must be a degree one polynomial?
$endgroup$
– Lord Shark the Unknown
Jan 26 at 7:01
$begingroup$
That's what I tried to do, but for that I have to prove that f has only one root inside the circle of radius 2017, right? Also, I am not sure, how does that prove such $f$ doesn't exist?
$endgroup$
– Guy Schwartzberg
Jan 26 at 7:05
$begingroup$
That's what I tried to do, but for that I have to prove that f has only one root inside the circle of radius 2017, right? Also, I am not sure, how does that prove such $f$ doesn't exist?
$endgroup$
– Guy Schwartzberg
Jan 26 at 7:05
$begingroup$
Once you know that $f$ is a polynomial, its degree can extracted from the knowledge on its growth speed as $|z|toinfty$.
$endgroup$
– Sangchul Lee
Jan 26 at 7:23
$begingroup$
Once you know that $f$ is a polynomial, its degree can extracted from the knowledge on its growth speed as $|z|toinfty$.
$endgroup$
– Sangchul Lee
Jan 26 at 7:23
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You already know that $f$ is a polynomial function. Let $n$ be its degree. Then$$lim_{lvert zrverttoinfty}frac{lvert f(z)rvert}{lvert zrvert^n}=1.$$But then it follows from your hypothesis that $n=1$ and that $f(z)=az+b$ with $lvert arvert=1$. However there are no such numbers $a$ and $b$ so that $lvert az+brvert=lvert zrvert+1$ when $lvert zrvert$ is large enough.
answered Jan 26 at 7:27
José Carlos SantosJosé Carlos Santos
169k23132237
$endgroup$
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%2f3087964%2fexistence-of-an-entire-function%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
$begingroup$
You already know that $f$ is a polynomial function. Let $n$ be its degree. Then$$lim_{lvert zrverttoinfty}frac{lvert f(z)rvert}{lvert zrvert^n}=1.$$But then it follows from your hypothesis that $n=1$ and that $f(z)=az+b$ with $lvert arvert=1$. However there are no such numbers $a$ and $b$ so that $lvert az+brvert=lvert zrvert+1$ when $lvert zrvert$ is large enough.
answered Jan 26 at 7:27
José Carlos SantosJosé Carlos Santos
169k23132237
$endgroup$
add a comment |
$begingroup$
You already know that $f$ is a polynomial function. Let $n$ be its degree. Then$$lim_{lvert zrverttoinfty}frac{lvert f(z)rvert}{lvert zrvert^n}=1.$$But then it follows from your hypothesis that $n=1$ and that $f(z)=az+b$ with $lvert arvert=1$. However there are no such numbers $a$ and $b$ so that $lvert az+brvert=lvert zrvert+1$ when $lvert zrvert$ is large enough.
answered Jan 26 at 7:27
José Carlos SantosJosé Carlos Santos
169k23132237
$endgroup$
add a comment |
$begingroup$
You already know that $f$ is a polynomial function. Let $n$ be its degree. Then$$lim_{lvert zrverttoinfty}frac{lvert f(z)rvert}{lvert zrvert^n}=1.$$But then it follows from your hypothesis that $n=1$ and that $f(z)=az+b$ with $lvert arvert=1$. However there are no such numbers $a$ and $b$ so that $lvert az+brvert=lvert zrvert+1$ when $lvert zrvert$ is large enough.
answered Jan 26 at 7:27
José Carlos SantosJosé Carlos Santos
169k23132237
$endgroup$
You already know that $f$ is a polynomial function. Let $n$ be its degree. Then$$lim_{lvert zrverttoinfty}frac{lvert f(z)rvert}{lvert zrvert^n}=1.$$But then it follows from your hypothesis that $n=1$ and that $f(z)=az+b$ with $lvert arvert=1$. However there are no such numbers $a$ and $b$ so that $lvert az+brvert=lvert zrvert+1$ when $lvert zrvert$ is large enough.
answered Jan 26 at 7:27
José Carlos SantosJosé Carlos Santos
169k23132237
answered Jan 26 at 7:27
José Carlos SantosJosé Carlos Santos
169k23132237
answered Jan 26 at 7:27
José Carlos SantosJosé Carlos Santos
169k23132237
answered Jan 26 at 7:27
José Carlos SantosJosé Carlos Santos
169k23132237
169k23132237
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%2f3087964%2fexistence-of-an-entire-function%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
$begingroup$
Is this the actual question, or are there further conditions? There are trivial solutions such as $f$ being the zero function.
$endgroup$
– Lord Shark the Unknown
Jan 26 at 6:54
$begingroup$
My mistake, I wrote it wrong. So sorry, I edited it. It should have been equals instead of smaller than.
$endgroup$
– Guy Schwartzberg
Jan 26 at 7:00
$begingroup$
Can you prove that $f$ must be a degree one polynomial?
$endgroup$
– Lord Shark the Unknown
Jan 26 at 7:01
$begingroup$
That's what I tried to do, but for that I have to prove that f has only one root inside the circle of radius 2017, right? Also, I am not sure, how does that prove such $f$ doesn't exist?
$endgroup$
– Guy Schwartzberg
Jan 26 at 7:05
$begingroup$
Once you know that $f$ is a polynomial, its degree can extracted from the knowledge on its growth speed as $|z|toinfty$.
$endgroup$
– Sangchul Lee
Jan 26 at 7:23