Derivative does not vanish on the segment of reflection.
$begingroup$
Suppose $I = (a,b)$ lies on the boundary of $U$, where $U$ is a domain in the upper half of the complex plane $mathbb{H}$. Assume $f$ is analytic on $U$ and continuous on $I$.
Furthermore, assume $f(U cup I) subset bar{mathbb{H}}$, and $f$ is not constant, and $f$ is real valued on $I$.
Prove that $f'(z) neq 0$ for $z in I$.
My thoughts:
This problem looks like a set-up for the applying the reflection principle to obtain $tilde{f}$ which is analytic on $tilde{U}$, but use this properly.
Also, note that if $f = u + iv$, then $u_x(z) geq 0$ for $z in I$ since the image of $f$ lies in $bar{mathbb{H}}.$
Any ideas on how to show that the derivative does not vanish on $I$?
complex-analysis
asked Dec 17 '18 at 5:56
Karagounis ZKaragounis Z
363
$endgroup$
add a comment |
$begingroup$
Suppose $I = (a,b)$ lies on the boundary of $U$, where $U$ is a domain in the upper half of the complex plane $mathbb{H}$. Assume $f$ is analytic on $U$ and continuous on $I$.
Furthermore, assume $f(U cup I) subset bar{mathbb{H}}$, and $f$ is not constant, and $f$ is real valued on $I$.
Prove that $f'(z) neq 0$ for $z in I$.
My thoughts:
This problem looks like a set-up for the applying the reflection principle to obtain $tilde{f}$ which is analytic on $tilde{U}$, but use this properly.
Also, note that if $f = u + iv$, then $u_x(z) geq 0$ for $z in I$ since the image of $f$ lies in $bar{mathbb{H}}.$
Any ideas on how to show that the derivative does not vanish on $I$?
complex-analysis
asked Dec 17 '18 at 5:56
Karagounis ZKaragounis Z
363
$endgroup$
add a comment |
$begingroup$
Suppose $I = (a,b)$ lies on the boundary of $U$, where $U$ is a domain in the upper half of the complex plane $mathbb{H}$. Assume $f$ is analytic on $U$ and continuous on $I$.
Furthermore, assume $f(U cup I) subset bar{mathbb{H}}$, and $f$ is not constant, and $f$ is real valued on $I$.
Prove that $f'(z) neq 0$ for $z in I$.
My thoughts:
This problem looks like a set-up for the applying the reflection principle to obtain $tilde{f}$ which is analytic on $tilde{U}$, but use this properly.
Also, note that if $f = u + iv$, then $u_x(z) geq 0$ for $z in I$ since the image of $f$ lies in $bar{mathbb{H}}.$
Any ideas on how to show that the derivative does not vanish on $I$?
complex-analysis
asked Dec 17 '18 at 5:56
Karagounis ZKaragounis Z
363
$endgroup$
Suppose $I = (a,b)$ lies on the boundary of $U$, where $U$ is a domain in the upper half of the complex plane $mathbb{H}$. Assume $f$ is analytic on $U$ and continuous on $I$.
Furthermore, assume $f(U cup I) subset bar{mathbb{H}}$, and $f$ is not constant, and $f$ is real valued on $I$.
Prove that $f'(z) neq 0$ for $z in I$.
My thoughts:
This problem looks like a set-up for the applying the reflection principle to obtain $tilde{f}$ which is analytic on $tilde{U}$, but use this properly.
Also, note that if $f = u + iv$, then $u_x(z) geq 0$ for $z in I$ since the image of $f$ lies in $bar{mathbb{H}}.$
Any ideas on how to show that the derivative does not vanish on $I$?
complex-analysis
complex-analysis
asked Dec 17 '18 at 5:56
Karagounis ZKaragounis Z
363
asked Dec 17 '18 at 5:56
Karagounis ZKaragounis Z
363
asked Dec 17 '18 at 5:56
Karagounis ZKaragounis Z
363
asked Dec 17 '18 at 5:56
Karagounis ZKaragounis Z
363
asked Dec 17 '18 at 5:56
Karagounis ZKaragounis Z
363
363
add a comment |
add a comment |
0
active
oldest
votes
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%2f3043582%2fderivative-does-not-vanish-on-the-segment-of-reflection%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
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%2f3043582%2fderivative-does-not-vanish-on-the-segment-of-reflection%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
