Mixing
Given $x,y,zin mathbb C$, $x+y+z=0$ and $|x|=|y|=|z|$, prove that $x^3=y^3=z^3$.
$begingroup$
Given $z_1,z_2,z_3in mathbb C$ such that $$z_1+z_2+z_3=0 tag {C1}$$ $$|z_1|=|z_2|=|z_3|, tag{C2}$$ prove that $$z_1^3=z_2^3=z_3^3.tag{C3}$$
Source: List of problems for math-contest training.
My attempt: By (C2), let $rho$ be the common modulus for $|z_j|,j=1,2,3.$ Therefore $forall j$,
$$z_j=rho e^{itheta_j} text{and} z_j^3=rho^3 e^{i 3theta_j}$$
And, by (C1) it holds that
$$e^{i theta_1}+e^{i theta_2}+e^{i theta_3}=0.$$
But (C3) implies
$$e^{i 3theta_1}=e^{i 3theta_2}=e^{i 3theta_3}.$$
and that implies $theta_1=theta_2=
theta_3$ (for $theta_j in [0,2pi[$). But that appears to be inconsistent with (C1).
Is there a problem with the problem statement? hints and answers are welcomed.
algebra-precalculus complex-numbers contest-math
edited Jan 6 at 19:12
user376343
3,3933826
asked Dec 22 '18 at 13:44
bluemasterbluemaster
1,638519
$endgroup$
add a comment |
$begingroup$
Given $z_1,z_2,z_3in mathbb C$ such that $$z_1+z_2+z_3=0 tag {C1}$$ $$|z_1|=|z_2|=|z_3|, tag{C2}$$ prove that $$z_1^3=z_2^3=z_3^3.tag{C3}$$
Source: List of problems for math-contest training.
My attempt: By (C2), let $rho$ be the common modulus for $|z_j|,j=1,2,3.$ Therefore $forall j$,
$$z_j=rho e^{itheta_j} text{and} z_j^3=rho^3 e^{i 3theta_j}$$
And, by (C1) it holds that
$$e^{i theta_1}+e^{i theta_2}+e^{i theta_3}=0.$$
But (C3) implies
$$e^{i 3theta_1}=e^{i 3theta_2}=e^{i 3theta_3}.$$
and that implies $theta_1=theta_2=
theta_3$ (for $theta_j in [0,2pi[$). But that appears to be inconsistent with (C1).
Is there a problem with the problem statement? hints and answers are welcomed.
algebra-precalculus complex-numbers contest-math
edited Jan 6 at 19:12
user376343
3,3933826
asked Dec 22 '18 at 13:44
bluemasterbluemaster
1,638519
$endgroup$
2
$begingroup$
Possible duplicate of showing certain vertices form an equilateral triangle
$endgroup$
– Martin R
Dec 22 '18 at 13:52
$begingroup$
math.stackexchange.com/questions/1397066/…
$endgroup$
– lab bhattacharjee
Dec 22 '18 at 14:10
add a comment |
$begingroup$
Given $z_1,z_2,z_3in mathbb C$ such that $$z_1+z_2+z_3=0 tag {C1}$$ $$|z_1|=|z_2|=|z_3|, tag{C2}$$ prove that $$z_1^3=z_2^3=z_3^3.tag{C3}$$
Source: List of problems for math-contest training.
My attempt: By (C2), let $rho$ be the common modulus for $|z_j|,j=1,2,3.$ Therefore $forall j$,
$$z_j=rho e^{itheta_j} text{and} z_j^3=rho^3 e^{i 3theta_j}$$
And, by (C1) it holds that
$$e^{i theta_1}+e^{i theta_2}+e^{i theta_3}=0.$$
But (C3) implies
$$e^{i 3theta_1}=e^{i 3theta_2}=e^{i 3theta_3}.$$
and that implies $theta_1=theta_2=
theta_3$ (for $theta_j in [0,2pi[$). But that appears to be inconsistent with (C1).
Is there a problem with the problem statement? hints and answers are welcomed.
algebra-precalculus complex-numbers contest-math
edited Jan 6 at 19:12
user376343
3,3933826
asked Dec 22 '18 at 13:44
bluemasterbluemaster
1,638519
$endgroup$
Given $z_1,z_2,z_3in mathbb C$ such that $$z_1+z_2+z_3=0 tag {C1}$$ $$|z_1|=|z_2|=|z_3|, tag{C2}$$ prove that $$z_1^3=z_2^3=z_3^3.tag{C3}$$
Source: List of problems for math-contest training.
My attempt: By (C2), let $rho$ be the common modulus for $|z_j|,j=1,2,3.$ Therefore $forall j$,
$$z_j=rho e^{itheta_j} text{and} z_j^3=rho^3 e^{i 3theta_j}$$
And, by (C1) it holds that
$$e^{i theta_1}+e^{i theta_2}+e^{i theta_3}=0.$$
But (C3) implies
$$e^{i 3theta_1}=e^{i 3theta_2}=e^{i 3theta_3}.$$
and that implies $theta_1=theta_2=
theta_3$ (for $theta_j in [0,2pi[$). But that appears to be inconsistent with (C1).
Is there a problem with the problem statement? hints and answers are welcomed.
algebra-precalculus complex-numbers contest-math
algebra-precalculus complex-numbers contest-math
edited Jan 6 at 19:12
user376343
3,3933826
asked Dec 22 '18 at 13:44
bluemasterbluemaster
1,638519
edited Jan 6 at 19:12
user376343
3,3933826
asked Dec 22 '18 at 13:44
bluemasterbluemaster
1,638519
edited Jan 6 at 19:12
user376343
3,3933826
edited Jan 6 at 19:12
user376343
3,3933826
edited Jan 6 at 19:12
user376343
3,3933826
3,3933826
asked Dec 22 '18 at 13:44
bluemasterbluemaster
1,638519
asked Dec 22 '18 at 13:44
bluemasterbluemaster
1,638519
asked Dec 22 '18 at 13:44
bluemasterbluemaster
1,638519
1,638519
2
$begingroup$
Possible duplicate of showing certain vertices form an equilateral triangle
$endgroup$
– Martin R
Dec 22 '18 at 13:52
$begingroup$
math.stackexchange.com/questions/1397066/…
$endgroup$
– lab bhattacharjee
Dec 22 '18 at 14:10
add a comment |
2
$begingroup$
Possible duplicate of showing certain vertices form an equilateral triangle
$endgroup$
– Martin R
Dec 22 '18 at 13:52
$begingroup$
math.stackexchange.com/questions/1397066/…
$endgroup$
– lab bhattacharjee
Dec 22 '18 at 14:10
2
2
$begingroup$
Possible duplicate of showing certain vertices form an equilateral triangle
$endgroup$
– Martin R
Dec 22 '18 at 13:52
$begingroup$
Possible duplicate of showing certain vertices form an equilateral triangle
$endgroup$
– Martin R
Dec 22 '18 at 13:52
$begingroup$
math.stackexchange.com/questions/1397066/…
$endgroup$
– lab bhattacharjee
Dec 22 '18 at 14:10
$begingroup$
math.stackexchange.com/questions/1397066/…
$endgroup$
– lab bhattacharjee
Dec 22 '18 at 14:10
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
(C3) doesn't imply that $theta_1=theta_2=theta_3$.
Take for example $theta_1 = 0, theta_2 = frac{2pi}{3}, theta_3 = frac{4pi}{3}$
Then $$e^{i3theta_1} = e^0 = 1$$ $$e^{i3theta_2} = e^{2pi i} = 1$$ $$e^{i3theta_3} = e^{4pi i} = 1$$
But clearly $theta_1not = theta_2, theta_1 not = theta_3,theta_2not = theta_3$.
answered Dec 22 '18 at 13:48
YankoYanko
6,5471528
$endgroup$
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%2f3049467%2fgiven-x-y-z-in-mathbb-c-xyz-0-and-x-y-z-prove-that-x3-y3-z3%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
$begingroup$
(C3) doesn't imply that $theta_1=theta_2=theta_3$.
Take for example $theta_1 = 0, theta_2 = frac{2pi}{3}, theta_3 = frac{4pi}{3}$
Then $$e^{i3theta_1} = e^0 = 1$$ $$e^{i3theta_2} = e^{2pi i} = 1$$ $$e^{i3theta_3} = e^{4pi i} = 1$$
But clearly $theta_1not = theta_2, theta_1 not = theta_3,theta_2not = theta_3$.
answered Dec 22 '18 at 13:48
YankoYanko
6,5471528
$endgroup$
add a comment |
$begingroup$
(C3) doesn't imply that $theta_1=theta_2=theta_3$.
Take for example $theta_1 = 0, theta_2 = frac{2pi}{3}, theta_3 = frac{4pi}{3}$
Then $$e^{i3theta_1} = e^0 = 1$$ $$e^{i3theta_2} = e^{2pi i} = 1$$ $$e^{i3theta_3} = e^{4pi i} = 1$$
But clearly $theta_1not = theta_2, theta_1 not = theta_3,theta_2not = theta_3$.
answered Dec 22 '18 at 13:48
YankoYanko
6,5471528
$endgroup$
add a comment |
$begingroup$
(C3) doesn't imply that $theta_1=theta_2=theta_3$.
Take for example $theta_1 = 0, theta_2 = frac{2pi}{3}, theta_3 = frac{4pi}{3}$
Then $$e^{i3theta_1} = e^0 = 1$$ $$e^{i3theta_2} = e^{2pi i} = 1$$ $$e^{i3theta_3} = e^{4pi i} = 1$$
But clearly $theta_1not = theta_2, theta_1 not = theta_3,theta_2not = theta_3$.
answered Dec 22 '18 at 13:48
YankoYanko
6,5471528
$endgroup$
(C3) doesn't imply that $theta_1=theta_2=theta_3$.
Take for example $theta_1 = 0, theta_2 = frac{2pi}{3}, theta_3 = frac{4pi}{3}$
Then $$e^{i3theta_1} = e^0 = 1$$ $$e^{i3theta_2} = e^{2pi i} = 1$$ $$e^{i3theta_3} = e^{4pi i} = 1$$
But clearly $theta_1not = theta_2, theta_1 not = theta_3,theta_2not = theta_3$.
answered Dec 22 '18 at 13:48
YankoYanko
6,5471528
answered Dec 22 '18 at 13:48
YankoYanko
6,5471528
answered Dec 22 '18 at 13:48
YankoYanko
6,5471528
answered Dec 22 '18 at 13:48
YankoYanko
6,5471528
6,5471528
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%2f3049467%2fgiven-x-y-z-in-mathbb-c-xyz-0-and-x-y-z-prove-that-x3-y3-z3%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
$begingroup$
Possible duplicate of showing certain vertices form an equilateral triangle
$endgroup$
– Martin R
Dec 22 '18 at 13:52
$begingroup$
math.stackexchange.com/questions/1397066/…
$endgroup$
– lab bhattacharjee
Dec 22 '18 at 14:10