Mixing
Complex analysis: identity verification
$begingroup$
I want to prove if $|z|=S$, then $$S^m=|z|^m=|z^m|$$.
Work done so far:
The first one is quite easy to prove: if $|z|=S$, then raising both sides to the power of $m$ will give us $s^m=|z|^m$.
I am struggling with the second part; any ideas on how to prove it?
Edit:
After a helpful hint this is a proof I came up with:
Let $z=re^{itheta}$, then $z^m=r^me^{mitheta}$. But, $|z|=r$ so $|z^m|=r^m$
Furthermore, since, $|z|=r$, $|z|^m=r^m$
Therefore, $S^m=|z|^m=|z^m|$.
How is this proved?
complex-analysis proof-verification proof-writing
edited Jan 29 at 2:06
J. W. Tanner
4,0611320
asked Jan 28 at 22:54
Bertrand Wittgenstein's GhostBertrand Wittgenstein's Ghost
537217
$endgroup$
add a comment |
$begingroup$
I want to prove if $|z|=S$, then $$S^m=|z|^m=|z^m|$$.
Work done so far:
The first one is quite easy to prove: if $|z|=S$, then raising both sides to the power of $m$ will give us $s^m=|z|^m$.
I am struggling with the second part; any ideas on how to prove it?
Edit:
After a helpful hint this is a proof I came up with:
Let $z=re^{itheta}$, then $z^m=r^me^{mitheta}$. But, $|z|=r$ so $|z^m|=r^m$
Furthermore, since, $|z|=r$, $|z|^m=r^m$
Therefore, $S^m=|z|^m=|z^m|$.
How is this proved?
complex-analysis proof-verification proof-writing
edited Jan 29 at 2:06
J. W. Tanner
4,0611320
asked Jan 28 at 22:54
Bertrand Wittgenstein's GhostBertrand Wittgenstein's Ghost
537217
$endgroup$
2
$begingroup$
write $z=re^{i theta}$ and compare the real and imaginary parts of $|z|^m$ and $|z^m|$
$endgroup$
– Mustafa Said
Jan 28 at 23:00
$begingroup$
@MustafaSaid thanks, i used your hint to write a proof. Please check it.
$endgroup$
– Bertrand Wittgenstein's Ghost
Jan 28 at 23:08
1
$begingroup$
$|z^m|^2=z^m cdot bar z^m=(zcdot bar z)^m=|z|^{2m}$ is another way
$endgroup$
– Matt A Pelto
Jan 28 at 23:11
$begingroup$
@MattAPelto Thanks, that is ingenious!
$endgroup$
– Bertrand Wittgenstein's Ghost
Jan 28 at 23:14
$begingroup$
Let $z=r(cos t+isin t)$ with $rge 0$ and $tin Bbb R$. By deMoivre's Theorem, $z^n=r^n(cos t+isin t)^n=r^n(cos nt+isin nt)$... deMoivres's Theorem is by induction on $n$, using the trig "angle-sum" formulas.
$endgroup$
– DanielWainfleet
Jan 29 at 5:57
add a comment |
$begingroup$
I want to prove if $|z|=S$, then $$S^m=|z|^m=|z^m|$$.
Work done so far:
The first one is quite easy to prove: if $|z|=S$, then raising both sides to the power of $m$ will give us $s^m=|z|^m$.
I am struggling with the second part; any ideas on how to prove it?
Edit:
After a helpful hint this is a proof I came up with:
Let $z=re^{itheta}$, then $z^m=r^me^{mitheta}$. But, $|z|=r$ so $|z^m|=r^m$
Furthermore, since, $|z|=r$, $|z|^m=r^m$
Therefore, $S^m=|z|^m=|z^m|$.
How is this proved?
complex-analysis proof-verification proof-writing
edited Jan 29 at 2:06
J. W. Tanner
4,0611320
asked Jan 28 at 22:54
Bertrand Wittgenstein's GhostBertrand Wittgenstein's Ghost
537217
$endgroup$
I want to prove if $|z|=S$, then $$S^m=|z|^m=|z^m|$$.
Work done so far:
The first one is quite easy to prove: if $|z|=S$, then raising both sides to the power of $m$ will give us $s^m=|z|^m$.
I am struggling with the second part; any ideas on how to prove it?
Edit:
After a helpful hint this is a proof I came up with:
Let $z=re^{itheta}$, then $z^m=r^me^{mitheta}$. But, $|z|=r$ so $|z^m|=r^m$
Furthermore, since, $|z|=r$, $|z|^m=r^m$
Therefore, $S^m=|z|^m=|z^m|$.
How is this proved?
complex-analysis proof-verification proof-writing
complex-analysis proof-verification proof-writing
edited Jan 29 at 2:06
J. W. Tanner
4,0611320
asked Jan 28 at 22:54
Bertrand Wittgenstein's GhostBertrand Wittgenstein's Ghost
537217
edited Jan 29 at 2:06
J. W. Tanner
4,0611320
asked Jan 28 at 22:54
Bertrand Wittgenstein's GhostBertrand Wittgenstein's Ghost
537217
edited Jan 29 at 2:06
J. W. Tanner
4,0611320
edited Jan 29 at 2:06
J. W. Tanner
4,0611320
edited Jan 29 at 2:06
J. W. Tanner
4,0611320
4,0611320
asked Jan 28 at 22:54
Bertrand Wittgenstein's GhostBertrand Wittgenstein's Ghost
537217
asked Jan 28 at 22:54
Bertrand Wittgenstein's GhostBertrand Wittgenstein's Ghost
537217
asked Jan 28 at 22:54
Bertrand Wittgenstein's GhostBertrand Wittgenstein's Ghost
537217
537217
2
$begingroup$
write $z=re^{i theta}$ and compare the real and imaginary parts of $|z|^m$ and $|z^m|$
$endgroup$
– Mustafa Said
Jan 28 at 23:00
$begingroup$
@MustafaSaid thanks, i used your hint to write a proof. Please check it.
$endgroup$
– Bertrand Wittgenstein's Ghost
Jan 28 at 23:08
1
$begingroup$
$|z^m|^2=z^m cdot bar z^m=(zcdot bar z)^m=|z|^{2m}$ is another way
$endgroup$
– Matt A Pelto
Jan 28 at 23:11
$begingroup$
@MattAPelto Thanks, that is ingenious!
$endgroup$
– Bertrand Wittgenstein's Ghost
Jan 28 at 23:14
$begingroup$
Let $z=r(cos t+isin t)$ with $rge 0$ and $tin Bbb R$. By deMoivre's Theorem, $z^n=r^n(cos t+isin t)^n=r^n(cos nt+isin nt)$... deMoivres's Theorem is by induction on $n$, using the trig "angle-sum" formulas.
$endgroup$
– DanielWainfleet
Jan 29 at 5:57
add a comment |
2
$begingroup$
write $z=re^{i theta}$ and compare the real and imaginary parts of $|z|^m$ and $|z^m|$
$endgroup$
– Mustafa Said
Jan 28 at 23:00
$begingroup$
@MustafaSaid thanks, i used your hint to write a proof. Please check it.
$endgroup$
– Bertrand Wittgenstein's Ghost
Jan 28 at 23:08
1
$begingroup$
$|z^m|^2=z^m cdot bar z^m=(zcdot bar z)^m=|z|^{2m}$ is another way
$endgroup$
– Matt A Pelto
Jan 28 at 23:11
$begingroup$
@MattAPelto Thanks, that is ingenious!
$endgroup$
– Bertrand Wittgenstein's Ghost
Jan 28 at 23:14
$begingroup$
Let $z=r(cos t+isin t)$ with $rge 0$ and $tin Bbb R$. By deMoivre's Theorem, $z^n=r^n(cos t+isin t)^n=r^n(cos nt+isin nt)$... deMoivres's Theorem is by induction on $n$, using the trig "angle-sum" formulas.
$endgroup$
– DanielWainfleet
Jan 29 at 5:57
2
2
$begingroup$
write $z=re^{i theta}$ and compare the real and imaginary parts of $|z|^m$ and $|z^m|$
$endgroup$
– Mustafa Said
Jan 28 at 23:00
$begingroup$
write $z=re^{i theta}$ and compare the real and imaginary parts of $|z|^m$ and $|z^m|$
$endgroup$
– Mustafa Said
Jan 28 at 23:00
$begingroup$
@MustafaSaid thanks, i used your hint to write a proof. Please check it.
$endgroup$
– Bertrand Wittgenstein's Ghost
Jan 28 at 23:08
$begingroup$
@MustafaSaid thanks, i used your hint to write a proof. Please check it.
$endgroup$
– Bertrand Wittgenstein's Ghost
Jan 28 at 23:08
1
1
$begingroup$
$|z^m|^2=z^m cdot bar z^m=(zcdot bar z)^m=|z|^{2m}$ is another way
$endgroup$
– Matt A Pelto
Jan 28 at 23:11
$begingroup$
$|z^m|^2=z^m cdot bar z^m=(zcdot bar z)^m=|z|^{2m}$ is another way
$endgroup$
– Matt A Pelto
Jan 28 at 23:11
$begingroup$
@MattAPelto Thanks, that is ingenious!
$endgroup$
– Bertrand Wittgenstein's Ghost
Jan 28 at 23:14
$begingroup$
@MattAPelto Thanks, that is ingenious!
$endgroup$
– Bertrand Wittgenstein's Ghost
Jan 28 at 23:14
$begingroup$
Let $z=r(cos t+isin t)$ with $rge 0$ and $tin Bbb R$. By deMoivre's Theorem, $z^n=r^n(cos t+isin t)^n=r^n(cos nt+isin nt)$... deMoivres's Theorem is by induction on $n$, using the trig "angle-sum" formulas.
$endgroup$
– DanielWainfleet
Jan 29 at 5:57
$begingroup$
Let $z=r(cos t+isin t)$ with $rge 0$ and $tin Bbb R$. By deMoivre's Theorem, $z^n=r^n(cos t+isin t)^n=r^n(cos nt+isin nt)$... deMoivres's Theorem is by induction on $n$, using the trig "angle-sum" formulas.
$endgroup$
– DanielWainfleet
Jan 29 at 5:57
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You already have it there. It'd seem that your problem is recognizing it. My recomendation would be that you write down your proof carefully annotating every derivation step with an arrow ($Rightarrow$). This arrow means if the left-hand-side is true, then we can conclude the right-hand-size, and the reason for that to be true must be easy to understand -otherwise it'd require a prove on it's own.
In the end the full proof can be read as a single, big if-then statement:
$$|z|=sqrt{zbar{z}},Rightarrow,|z|^n,=,sqrt{(zbar{z})^n},Rightarrow,|z|^n,=,|z^n|$$
answered Jan 29 at 2:23
MASLMASL
708313
$endgroup$
$begingroup$
Thanks, appreciate your answer!
$endgroup$
– Bertrand Wittgenstein's Ghost
Jan 29 at 2:29
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%2f3091507%2fcomplex-analysis-identity-verification%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 have it there. It'd seem that your problem is recognizing it. My recomendation would be that you write down your proof carefully annotating every derivation step with an arrow ($Rightarrow$). This arrow means if the left-hand-side is true, then we can conclude the right-hand-size, and the reason for that to be true must be easy to understand -otherwise it'd require a prove on it's own.
In the end the full proof can be read as a single, big if-then statement:
$$|z|=sqrt{zbar{z}},Rightarrow,|z|^n,=,sqrt{(zbar{z})^n},Rightarrow,|z|^n,=,|z^n|$$
answered Jan 29 at 2:23
MASLMASL
708313
$endgroup$
$begingroup$
Thanks, appreciate your answer!
$endgroup$
– Bertrand Wittgenstein's Ghost
Jan 29 at 2:29
add a comment |
$begingroup$
You already have it there. It'd seem that your problem is recognizing it. My recomendation would be that you write down your proof carefully annotating every derivation step with an arrow ($Rightarrow$). This arrow means if the left-hand-side is true, then we can conclude the right-hand-size, and the reason for that to be true must be easy to understand -otherwise it'd require a prove on it's own.
In the end the full proof can be read as a single, big if-then statement:
$$|z|=sqrt{zbar{z}},Rightarrow,|z|^n,=,sqrt{(zbar{z})^n},Rightarrow,|z|^n,=,|z^n|$$
answered Jan 29 at 2:23
MASLMASL
708313
$endgroup$
$begingroup$
Thanks, appreciate your answer!
$endgroup$
– Bertrand Wittgenstein's Ghost
Jan 29 at 2:29
add a comment |
$begingroup$
You already have it there. It'd seem that your problem is recognizing it. My recomendation would be that you write down your proof carefully annotating every derivation step with an arrow ($Rightarrow$). This arrow means if the left-hand-side is true, then we can conclude the right-hand-size, and the reason for that to be true must be easy to understand -otherwise it'd require a prove on it's own.
In the end the full proof can be read as a single, big if-then statement:
$$|z|=sqrt{zbar{z}},Rightarrow,|z|^n,=,sqrt{(zbar{z})^n},Rightarrow,|z|^n,=,|z^n|$$
answered Jan 29 at 2:23
MASLMASL
708313
$endgroup$
You already have it there. It'd seem that your problem is recognizing it. My recomendation would be that you write down your proof carefully annotating every derivation step with an arrow ($Rightarrow$). This arrow means if the left-hand-side is true, then we can conclude the right-hand-size, and the reason for that to be true must be easy to understand -otherwise it'd require a prove on it's own.
In the end the full proof can be read as a single, big if-then statement:
$$|z|=sqrt{zbar{z}},Rightarrow,|z|^n,=,sqrt{(zbar{z})^n},Rightarrow,|z|^n,=,|z^n|$$
answered Jan 29 at 2:23
MASLMASL
708313
edited Jan 29 at 2:31
answered Jan 29 at 2:23
MASLMASL
708313
answered Jan 29 at 2:23
MASLMASL
708313
answered Jan 29 at 2:23
MASLMASL
708313
708313
$begingroup$
Thanks, appreciate your answer!
$endgroup$
– Bertrand Wittgenstein's Ghost
Jan 29 at 2:29
add a comment |
$begingroup$
Thanks, appreciate your answer!
$endgroup$
– Bertrand Wittgenstein's Ghost
Jan 29 at 2:29
$begingroup$
Thanks, appreciate your answer!
$endgroup$
– Bertrand Wittgenstein's Ghost
Jan 29 at 2:29
$begingroup$
Thanks, appreciate your answer!
$endgroup$
– Bertrand Wittgenstein's Ghost
Jan 29 at 2:29
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%2f3091507%2fcomplex-analysis-identity-verification%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
$begingroup$
write $z=re^{i theta}$ and compare the real and imaginary parts of $|z|^m$ and $|z^m|$
$endgroup$
– Mustafa Said
Jan 28 at 23:00
$begingroup$
@MustafaSaid thanks, i used your hint to write a proof. Please check it.
$endgroup$
– Bertrand Wittgenstein's Ghost
Jan 28 at 23:08
1
$begingroup$
$|z^m|^2=z^m cdot bar z^m=(zcdot bar z)^m=|z|^{2m}$ is another way
$endgroup$
– Matt A Pelto
Jan 28 at 23:11
$begingroup$
@MattAPelto Thanks, that is ingenious!
$endgroup$
– Bertrand Wittgenstein's Ghost
Jan 28 at 23:14
$begingroup$
Let $z=r(cos t+isin t)$ with $rge 0$ and $tin Bbb R$. By deMoivre's Theorem, $z^n=r^n(cos t+isin t)^n=r^n(cos nt+isin nt)$... deMoivres's Theorem is by induction on $n$, using the trig "angle-sum" formulas.
$endgroup$
– DanielWainfleet
Jan 29 at 5:57