How to show that $f_1(z) = -f_2(z)$, where $f_1$ and $f_2$ are the square root functions with different...
$begingroup$
Let $f_1(z)=sqrt{r}e^{itheta/2}, (r>0, 0<theta<pi)$ and $f_2(z)=sqrt{r}e^{itheta/2}, (r>0, pi<theta<5pi/2)$. How to show $f_1(z)=-f_2(z)$ in the first quadrant?
If we set $z_0=re^{itheta}, (r>0, 0<theta<pi)$, then we can plug it into $f_1(z)$, but what about $f_2(z)$? The angle of $z_0$ is not in $(pi, 5pi/2)$. Is it correct to add $pi$ to the argument of $z_0$, then plug it into $f_2$?
complex-numbers
asked Nov 30 '18 at 17:37
user398843user398843
648216
$endgroup$
add a comment |
$begingroup$
Let $f_1(z)=sqrt{r}e^{itheta/2}, (r>0, 0<theta<pi)$ and $f_2(z)=sqrt{r}e^{itheta/2}, (r>0, pi<theta<5pi/2)$. How to show $f_1(z)=-f_2(z)$ in the first quadrant?
If we set $z_0=re^{itheta}, (r>0, 0<theta<pi)$, then we can plug it into $f_1(z)$, but what about $f_2(z)$? The angle of $z_0$ is not in $(pi, 5pi/2)$. Is it correct to add $pi$ to the argument of $z_0$, then plug it into $f_2$?
complex-numbers
asked Nov 30 '18 at 17:37
user398843user398843
648216
$endgroup$
$begingroup$
This is part of an exercise on the book. The exercise guides us to see that if $f_2$ is an analytic continuation of $f_1$, and $f_3$ is an analytic continuation of $f_2$, then it is not necessarily true that $f_1=f_3$ in the intersection of the domains of $f_1$ and $f_3$. But it seems to me that the intersection of the domains of $f_1$ and $f_3$ is empty.
$endgroup$
– user398843
Nov 30 '18 at 18:12
add a comment |
$begingroup$
Let $f_1(z)=sqrt{r}e^{itheta/2}, (r>0, 0<theta<pi)$ and $f_2(z)=sqrt{r}e^{itheta/2}, (r>0, pi<theta<5pi/2)$. How to show $f_1(z)=-f_2(z)$ in the first quadrant?
If we set $z_0=re^{itheta}, (r>0, 0<theta<pi)$, then we can plug it into $f_1(z)$, but what about $f_2(z)$? The angle of $z_0$ is not in $(pi, 5pi/2)$. Is it correct to add $pi$ to the argument of $z_0$, then plug it into $f_2$?
complex-numbers
asked Nov 30 '18 at 17:37
user398843user398843
648216
$endgroup$
Let $f_1(z)=sqrt{r}e^{itheta/2}, (r>0, 0<theta<pi)$ and $f_2(z)=sqrt{r}e^{itheta/2}, (r>0, pi<theta<5pi/2)$. How to show $f_1(z)=-f_2(z)$ in the first quadrant?
If we set $z_0=re^{itheta}, (r>0, 0<theta<pi)$, then we can plug it into $f_1(z)$, but what about $f_2(z)$? The angle of $z_0$ is not in $(pi, 5pi/2)$. Is it correct to add $pi$ to the argument of $z_0$, then plug it into $f_2$?
complex-numbers
complex-numbers
asked Nov 30 '18 at 17:37
user398843user398843
648216
asked Nov 30 '18 at 17:37
user398843user398843
648216
edited Nov 30 '18 at 18:19
user398843
asked Nov 30 '18 at 17:37
user398843user398843
648216
asked Nov 30 '18 at 17:37
user398843user398843
648216
asked Nov 30 '18 at 17:37
user398843user398843
648216
648216
$begingroup$
This is part of an exercise on the book. The exercise guides us to see that if $f_2$ is an analytic continuation of $f_1$, and $f_3$ is an analytic continuation of $f_2$, then it is not necessarily true that $f_1=f_3$ in the intersection of the domains of $f_1$ and $f_3$. But it seems to me that the intersection of the domains of $f_1$ and $f_3$ is empty.
$endgroup$
– user398843
Nov 30 '18 at 18:12
add a comment |
$begingroup$
This is part of an exercise on the book. The exercise guides us to see that if $f_2$ is an analytic continuation of $f_1$, and $f_3$ is an analytic continuation of $f_2$, then it is not necessarily true that $f_1=f_3$ in the intersection of the domains of $f_1$ and $f_3$. But it seems to me that the intersection of the domains of $f_1$ and $f_3$ is empty.
$endgroup$
– user398843
Nov 30 '18 at 18:12
$begingroup$
This is part of an exercise on the book. The exercise guides us to see that if $f_2$ is an analytic continuation of $f_1$, and $f_3$ is an analytic continuation of $f_2$, then it is not necessarily true that $f_1=f_3$ in the intersection of the domains of $f_1$ and $f_3$. But it seems to me that the intersection of the domains of $f_1$ and $f_3$ is empty.
$endgroup$
– user398843
Nov 30 '18 at 18:12
$begingroup$
This is part of an exercise on the book. The exercise guides us to see that if $f_2$ is an analytic continuation of $f_1$, and $f_3$ is an analytic continuation of $f_2$, then it is not necessarily true that $f_1=f_3$ in the intersection of the domains of $f_1$ and $f_3$. But it seems to me that the intersection of the domains of $f_1$ and $f_3$ is empty.
$endgroup$
– user398843
Nov 30 '18 at 18:12
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
To get $f_2(x)$ let $z_0=re^{i(theta +2pi) }$. Then $f_2(z_0)=sqrt{r}e^{i(theta/2 +pi) }=-sqrt{r}e^{itheta/2}=-f_1(z_0)$. You can do this because $e^{2npi i}=1$
for all $n$, while $e^{pi i}=-1$.
answered Nov 30 '18 at 18:19
herb steinbergherb steinberg
2,7432310
$endgroup$
add a comment |
$begingroup$
Let $z=x+iy (x,yin mathbb R)$. Then $z=x+iy=re^{iTheta+2npi}$ ($pi<Theta le pi$, $nin mathbb Z$). Hence, $z=re^{iTheta}$ is only one of an infinite number of possiblities for the exponential form of $x+iy$.
answered Nov 30 '18 at 18:37
user398843user398843
648216
$endgroup$
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%2f3020384%2fhow-to-show-that-f-1z-f-2z-where-f-1-and-f-2-are-the-square-root-f%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
$begingroup$
To get $f_2(x)$ let $z_0=re^{i(theta +2pi) }$. Then $f_2(z_0)=sqrt{r}e^{i(theta/2 +pi) }=-sqrt{r}e^{itheta/2}=-f_1(z_0)$. You can do this because $e^{2npi i}=1$
for all $n$, while $e^{pi i}=-1$.
answered Nov 30 '18 at 18:19
herb steinbergherb steinberg
2,7432310
$endgroup$
add a comment |
$begingroup$
To get $f_2(x)$ let $z_0=re^{i(theta +2pi) }$. Then $f_2(z_0)=sqrt{r}e^{i(theta/2 +pi) }=-sqrt{r}e^{itheta/2}=-f_1(z_0)$. You can do this because $e^{2npi i}=1$
for all $n$, while $e^{pi i}=-1$.
answered Nov 30 '18 at 18:19
herb steinbergherb steinberg
2,7432310
$endgroup$
add a comment |
$begingroup$
To get $f_2(x)$ let $z_0=re^{i(theta +2pi) }$. Then $f_2(z_0)=sqrt{r}e^{i(theta/2 +pi) }=-sqrt{r}e^{itheta/2}=-f_1(z_0)$. You can do this because $e^{2npi i}=1$
for all $n$, while $e^{pi i}=-1$.
answered Nov 30 '18 at 18:19
herb steinbergherb steinberg
2,7432310
$endgroup$
To get $f_2(x)$ let $z_0=re^{i(theta +2pi) }$. Then $f_2(z_0)=sqrt{r}e^{i(theta/2 +pi) }=-sqrt{r}e^{itheta/2}=-f_1(z_0)$. You can do this because $e^{2npi i}=1$
for all $n$, while $e^{pi i}=-1$.
answered Nov 30 '18 at 18:19
herb steinbergherb steinberg
2,7432310
edited Nov 30 '18 at 18:38
answered Nov 30 '18 at 18:19
herb steinbergherb steinberg
2,7432310
answered Nov 30 '18 at 18:19
herb steinbergherb steinberg
2,7432310
answered Nov 30 '18 at 18:19
herb steinbergherb steinberg
2,7432310
2,7432310
add a comment |
add a comment |
$begingroup$
Let $z=x+iy (x,yin mathbb R)$. Then $z=x+iy=re^{iTheta+2npi}$ ($pi<Theta le pi$, $nin mathbb Z$). Hence, $z=re^{iTheta}$ is only one of an infinite number of possiblities for the exponential form of $x+iy$.
answered Nov 30 '18 at 18:37
user398843user398843
648216
$endgroup$
add a comment |
$begingroup$
Let $z=x+iy (x,yin mathbb R)$. Then $z=x+iy=re^{iTheta+2npi}$ ($pi<Theta le pi$, $nin mathbb Z$). Hence, $z=re^{iTheta}$ is only one of an infinite number of possiblities for the exponential form of $x+iy$.
answered Nov 30 '18 at 18:37
user398843user398843
648216
$endgroup$
add a comment |
$begingroup$
Let $z=x+iy (x,yin mathbb R)$. Then $z=x+iy=re^{iTheta+2npi}$ ($pi<Theta le pi$, $nin mathbb Z$). Hence, $z=re^{iTheta}$ is only one of an infinite number of possiblities for the exponential form of $x+iy$.
answered Nov 30 '18 at 18:37
user398843user398843
648216
$endgroup$
Let $z=x+iy (x,yin mathbb R)$. Then $z=x+iy=re^{iTheta+2npi}$ ($pi<Theta le pi$, $nin mathbb Z$). Hence, $z=re^{iTheta}$ is only one of an infinite number of possiblities for the exponential form of $x+iy$.
answered Nov 30 '18 at 18:37
user398843user398843
648216
answered Nov 30 '18 at 18:37
user398843user398843
648216
answered Nov 30 '18 at 18:37
user398843user398843
648216
answered Nov 30 '18 at 18:37
user398843user398843
648216
648216
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%2f3020384%2fhow-to-show-that-f-1z-f-2z-where-f-1-and-f-2-are-the-square-root-f%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
$begingroup$
This is part of an exercise on the book. The exercise guides us to see that if $f_2$ is an analytic continuation of $f_1$, and $f_3$ is an analytic continuation of $f_2$, then it is not necessarily true that $f_1=f_3$ in the intersection of the domains of $f_1$ and $f_3$. But it seems to me that the intersection of the domains of $f_1$ and $f_3$ is empty.
$endgroup$
– user398843
Nov 30 '18 at 18:12