Mixing
a problem on complex numbers
$begingroup$
Let $wneq 1$ and $w^{13} = 1$.
If $a = w+ w^3 + w^4 + w^{-4} + w^{-3} + w^{-1}$ and $b = w^2+ w^5 + w^6 + w^{-6} + w^{-5} + w^{-2}$, then the quadratic equation whose roots are $a$ and $b$ is ... ?
I got $w=cos(frac{2pi}{13})+isin(frac{2pi}{13})$
And then I found $a$ and $b$ in trigonometric form. But when I multiplied them to get the product of roots it gets very difficult. How to solve it?
complex-numbers
edited Jan 3 at 15:57
Andrei
11.7k21026
asked Jan 3 at 15:43
Rituraj TripathyRituraj Tripathy
112
$endgroup$
add a comment |
$begingroup$
Let $wneq 1$ and $w^{13} = 1$.
If $a = w+ w^3 + w^4 + w^{-4} + w^{-3} + w^{-1}$ and $b = w^2+ w^5 + w^6 + w^{-6} + w^{-5} + w^{-2}$, then the quadratic equation whose roots are $a$ and $b$ is ... ?
I got $w=cos(frac{2pi}{13})+isin(frac{2pi}{13})$
And then I found $a$ and $b$ in trigonometric form. But when I multiplied them to get the product of roots it gets very difficult. How to solve it?
complex-numbers
edited Jan 3 at 15:57
Andrei
11.7k21026
asked Jan 3 at 15:43
Rituraj TripathyRituraj Tripathy
112
$endgroup$
$begingroup$
Welcome to the site. Please typeset your equations with Mathjax for better presentation.
$endgroup$
– Shubham Johri
Jan 3 at 15:46
1
$begingroup$
These are Gaussian periods: en.wikipedia.org/wiki/Gaussian_period
$endgroup$
– Lord Shark the Unknown
Jan 3 at 15:46
1
$begingroup$
Also, you are more likely to get views if you actually title your "question" properly. Something like "Solving quadratic roots of complex numbers" is much better then your current title
$endgroup$
– K Split X
Jan 3 at 15:55
2
$begingroup$
"And then I found a and b in trigonometric form. But when I multiplied them to get the product of roots it gets very difficult." If you found a and b in trig form how can multiplying them together be difficult?
$endgroup$
– fleablood
Jan 3 at 16:04
$begingroup$
Ditto what K Split X said. Many people will simply ignore questions that are titled something like "question about subject-tag". I know I certainly am not going to read the problem statement. Keep the tags in the tags section, and put your question in the title, or some adequate abbreviation of it.
$endgroup$
– rschwieb
Jan 3 at 16:19
add a comment |
$begingroup$
Let $wneq 1$ and $w^{13} = 1$.
If $a = w+ w^3 + w^4 + w^{-4} + w^{-3} + w^{-1}$ and $b = w^2+ w^5 + w^6 + w^{-6} + w^{-5} + w^{-2}$, then the quadratic equation whose roots are $a$ and $b$ is ... ?
I got $w=cos(frac{2pi}{13})+isin(frac{2pi}{13})$
And then I found $a$ and $b$ in trigonometric form. But when I multiplied them to get the product of roots it gets very difficult. How to solve it?
complex-numbers
edited Jan 3 at 15:57
Andrei
11.7k21026
asked Jan 3 at 15:43
Rituraj TripathyRituraj Tripathy
112
$endgroup$
Let $wneq 1$ and $w^{13} = 1$.
If $a = w+ w^3 + w^4 + w^{-4} + w^{-3} + w^{-1}$ and $b = w^2+ w^5 + w^6 + w^{-6} + w^{-5} + w^{-2}$, then the quadratic equation whose roots are $a$ and $b$ is ... ?
I got $w=cos(frac{2pi}{13})+isin(frac{2pi}{13})$
And then I found $a$ and $b$ in trigonometric form. But when I multiplied them to get the product of roots it gets very difficult. How to solve it?
complex-numbers
complex-numbers
edited Jan 3 at 15:57
Andrei
11.7k21026
asked Jan 3 at 15:43
Rituraj TripathyRituraj Tripathy
112
edited Jan 3 at 15:57
Andrei
11.7k21026
asked Jan 3 at 15:43
Rituraj TripathyRituraj Tripathy
112
edited Jan 3 at 15:57
Andrei
11.7k21026
edited Jan 3 at 15:57
Andrei
11.7k21026
edited Jan 3 at 15:57
Andrei
11.7k21026
11.7k21026
asked Jan 3 at 15:43
Rituraj TripathyRituraj Tripathy
112
asked Jan 3 at 15:43
Rituraj TripathyRituraj Tripathy
112
asked Jan 3 at 15:43
Rituraj TripathyRituraj Tripathy
112
112
$begingroup$
Welcome to the site. Please typeset your equations with Mathjax for better presentation.
$endgroup$
– Shubham Johri
Jan 3 at 15:46
1
$begingroup$
These are Gaussian periods: en.wikipedia.org/wiki/Gaussian_period
$endgroup$
– Lord Shark the Unknown
Jan 3 at 15:46
1
$begingroup$
Also, you are more likely to get views if you actually title your "question" properly. Something like "Solving quadratic roots of complex numbers" is much better then your current title
$endgroup$
– K Split X
Jan 3 at 15:55
2
$begingroup$
"And then I found a and b in trigonometric form. But when I multiplied them to get the product of roots it gets very difficult." If you found a and b in trig form how can multiplying them together be difficult?
$endgroup$
– fleablood
Jan 3 at 16:04
$begingroup$
Ditto what K Split X said. Many people will simply ignore questions that are titled something like "question about subject-tag". I know I certainly am not going to read the problem statement. Keep the tags in the tags section, and put your question in the title, or some adequate abbreviation of it.
$endgroup$
– rschwieb
Jan 3 at 16:19
add a comment |
$begingroup$
Welcome to the site. Please typeset your equations with Mathjax for better presentation.
$endgroup$
– Shubham Johri
Jan 3 at 15:46
1
$begingroup$
These are Gaussian periods: en.wikipedia.org/wiki/Gaussian_period
$endgroup$
– Lord Shark the Unknown
Jan 3 at 15:46
1
$begingroup$
Also, you are more likely to get views if you actually title your "question" properly. Something like "Solving quadratic roots of complex numbers" is much better then your current title
$endgroup$
– K Split X
Jan 3 at 15:55
2
$begingroup$
"And then I found a and b in trigonometric form. But when I multiplied them to get the product of roots it gets very difficult." If you found a and b in trig form how can multiplying them together be difficult?
$endgroup$
– fleablood
Jan 3 at 16:04
$begingroup$
Ditto what K Split X said. Many people will simply ignore questions that are titled something like "question about subject-tag". I know I certainly am not going to read the problem statement. Keep the tags in the tags section, and put your question in the title, or some adequate abbreviation of it.
$endgroup$
– rschwieb
Jan 3 at 16:19
$begingroup$
Welcome to the site. Please typeset your equations with Mathjax for better presentation.
$endgroup$
– Shubham Johri
Jan 3 at 15:46
$begingroup$
Welcome to the site. Please typeset your equations with Mathjax for better presentation.
$endgroup$
– Shubham Johri
Jan 3 at 15:46
1
1
$begingroup$
These are Gaussian periods: en.wikipedia.org/wiki/Gaussian_period
$endgroup$
– Lord Shark the Unknown
Jan 3 at 15:46
$begingroup$
These are Gaussian periods: en.wikipedia.org/wiki/Gaussian_period
$endgroup$
– Lord Shark the Unknown
Jan 3 at 15:46
1
1
$begingroup$
Also, you are more likely to get views if you actually title your "question" properly. Something like "Solving quadratic roots of complex numbers" is much better then your current title
$endgroup$
– K Split X
Jan 3 at 15:55
$begingroup$
Also, you are more likely to get views if you actually title your "question" properly. Something like "Solving quadratic roots of complex numbers" is much better then your current title
$endgroup$
– K Split X
Jan 3 at 15:55
2
2
$begingroup$
"And then I found a and b in trigonometric form. But when I multiplied them to get the product of roots it gets very difficult." If you found a and b in trig form how can multiplying them together be difficult?
$endgroup$
– fleablood
Jan 3 at 16:04
$begingroup$
"And then I found a and b in trigonometric form. But when I multiplied them to get the product of roots it gets very difficult." If you found a and b in trig form how can multiplying them together be difficult?
$endgroup$
– fleablood
Jan 3 at 16:04
$begingroup$
Ditto what K Split X said. Many people will simply ignore questions that are titled something like "question about subject-tag". I know I certainly am not going to read the problem statement. Keep the tags in the tags section, and put your question in the title, or some adequate abbreviation of it.
$endgroup$
– rschwieb
Jan 3 at 16:19
$begingroup$
Ditto what K Split X said. Many people will simply ignore questions that are titled something like "question about subject-tag". I know I certainly am not going to read the problem statement. Keep the tags in the tags section, and put your question in the title, or some adequate abbreviation of it.
$endgroup$
– rschwieb
Jan 3 at 16:19
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Step 1: the equation you want is $(z-a)(z-b)=0$. Expand the product and you get $$z^2-(a+b)z+ab=0$$
Step 2: Use $w^{13}=1$, so $w^{-1}=w^{13}w^{-1}=w^{12}$ similarly, for all negative powers $$w^{-n}=w^{13-n}$$
Step 3: $$a+b=w+w^3+w^4+w^9+w^{10}+w^{12}+w^2+w^5+w^6+w^7+w^8+w^{11}=\=frac{w^{13}-1}{w-1}-1=-1$$
Step 4: To find $ab$, go to the trigonometric representation, and notice $$w^n+w^{-n}=2cosfrac{2pi n}{13}$$
Edit: After some manipulations, and using $w^{n+13}=w^n$, I've got $$ab=3(w+w^2+w^3+w^4+w^5+w^6+w^7+w^8+w^9+w^{10}+w^{11}+w^{12})=\-3+3(1+w+w^2+w^3+w^4+w^5+w^6+w^7+w^8+w^9+w^{10}+w^{11}+w^{12})=-3$$
answered Jan 3 at 16:15
AndreiAndrei
11.7k21026
$endgroup$
$begingroup$
I got to this point. I can not solve the trigonometric part.
$endgroup$
– Rituraj Tripathy
Jan 3 at 16:27
$begingroup$
I went back to $w$ and wrote explicitly the product
$endgroup$
– Andrei
Jan 3 at 17:03
$begingroup$
I think Step 3 should give $a+b=-1$. Note that $frac{w^{13}-1}{w-1}=1+w+w^2+dots+w^{12} = a+b+1$.
$endgroup$
– gandalf61
Jan 3 at 17:09
$begingroup$
Fixed that. I forgot to add and subtract one. I've just added.
$endgroup$
– Andrei
Jan 3 at 17:14
$begingroup$
Andrei, it is not entirely necessary to relate $a$ and $b.$ See page 16 in books.google.com/… The method is due to Gauss, modern discussion (but just a few examples) in zakuski.utsa.edu/~jagy/cox_galois_Gaussian_periods.pdf
$endgroup$
– Will Jagy
Jan 3 at 18:52
add a comment |
$begingroup$
$$ a^2 + a = frac{ w^{16} + 2w^{15} + w^{14} + 2w^{13} + 3w^{12} + 3w^{11} + 3w^{10} + 3w^9 + 6w^8 + 3w^7 + 3w^6 + 3w^5 + 3w^4 + 2w^3 + w^2 + 2w + 1}{w^8} $$
This is not impressive without
$$ w^{16} + 2w^{15} + w^{14} + 2w^{13} = w^3+2w^2+w+2. $$ Therefore
$$ a^2 + a = frac{ 3w^{12} + 3w^{11} + 3w^{10} + 3w^9 + 6w^8 + 3w^7 + 3w^6 + 3w^5 + 3w^4 + 3w^3 + 3w^2 + 3w + 3}{w^8} $$
The one coefficient out of line is $6 w^8 / w^8,$ so we need to subtract 3
$$ a^2 + a -3 = frac{ 3w^{12} + 3w^{11} + 3w^{10} + 3w^9 + 3w^8 + 3w^7 + 3w^6 + 3w^5 + 3w^4 + 3w^3 + 3w^2 + 3w + 3}{w^8} $$
$$ a^2 + a -3 = frac{ 3 left( w^{12} + w^{11} + w^{10} + w^9 + w^8 + w^7 + w^6 + w^5 + w^4 + w^3 + w^2 + w + 1 right)}{w^8} $$
and
$$ a^2 + a - 3 = 0 $$
answered Jan 3 at 18:43
Will JagyWill Jagy
102k5101199
$endgroup$
$begingroup$
Elegant solution, but it might be at a higher level than what the question wanted. It would have probably had different tags.
$endgroup$
– Andrei
Jan 3 at 19:12
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%2f3060694%2fa-problem-on-complex-numbers%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
$begingroup$
Step 1: the equation you want is $(z-a)(z-b)=0$. Expand the product and you get $$z^2-(a+b)z+ab=0$$
Step 2: Use $w^{13}=1$, so $w^{-1}=w^{13}w^{-1}=w^{12}$ similarly, for all negative powers $$w^{-n}=w^{13-n}$$
Step 3: $$a+b=w+w^3+w^4+w^9+w^{10}+w^{12}+w^2+w^5+w^6+w^7+w^8+w^{11}=\=frac{w^{13}-1}{w-1}-1=-1$$
Step 4: To find $ab$, go to the trigonometric representation, and notice $$w^n+w^{-n}=2cosfrac{2pi n}{13}$$
Edit: After some manipulations, and using $w^{n+13}=w^n$, I've got $$ab=3(w+w^2+w^3+w^4+w^5+w^6+w^7+w^8+w^9+w^{10}+w^{11}+w^{12})=\-3+3(1+w+w^2+w^3+w^4+w^5+w^6+w^7+w^8+w^9+w^{10}+w^{11}+w^{12})=-3$$
answered Jan 3 at 16:15
AndreiAndrei
11.7k21026
$endgroup$
$begingroup$
I got to this point. I can not solve the trigonometric part.
$endgroup$
– Rituraj Tripathy
Jan 3 at 16:27
$begingroup$
I went back to $w$ and wrote explicitly the product
$endgroup$
– Andrei
Jan 3 at 17:03
$begingroup$
I think Step 3 should give $a+b=-1$. Note that $frac{w^{13}-1}{w-1}=1+w+w^2+dots+w^{12} = a+b+1$.
$endgroup$
– gandalf61
Jan 3 at 17:09
$begingroup$
Fixed that. I forgot to add and subtract one. I've just added.
$endgroup$
– Andrei
Jan 3 at 17:14
$begingroup$
Andrei, it is not entirely necessary to relate $a$ and $b.$ See page 16 in books.google.com/… The method is due to Gauss, modern discussion (but just a few examples) in zakuski.utsa.edu/~jagy/cox_galois_Gaussian_periods.pdf
$endgroup$
– Will Jagy
Jan 3 at 18:52
add a comment |
$begingroup$
Step 1: the equation you want is $(z-a)(z-b)=0$. Expand the product and you get $$z^2-(a+b)z+ab=0$$
Step 2: Use $w^{13}=1$, so $w^{-1}=w^{13}w^{-1}=w^{12}$ similarly, for all negative powers $$w^{-n}=w^{13-n}$$
Step 3: $$a+b=w+w^3+w^4+w^9+w^{10}+w^{12}+w^2+w^5+w^6+w^7+w^8+w^{11}=\=frac{w^{13}-1}{w-1}-1=-1$$
Step 4: To find $ab$, go to the trigonometric representation, and notice $$w^n+w^{-n}=2cosfrac{2pi n}{13}$$
Edit: After some manipulations, and using $w^{n+13}=w^n$, I've got $$ab=3(w+w^2+w^3+w^4+w^5+w^6+w^7+w^8+w^9+w^{10}+w^{11}+w^{12})=\-3+3(1+w+w^2+w^3+w^4+w^5+w^6+w^7+w^8+w^9+w^{10}+w^{11}+w^{12})=-3$$
answered Jan 3 at 16:15
AndreiAndrei
11.7k21026
$endgroup$
$begingroup$
I got to this point. I can not solve the trigonometric part.
$endgroup$
– Rituraj Tripathy
Jan 3 at 16:27
$begingroup$
I went back to $w$ and wrote explicitly the product
$endgroup$
– Andrei
Jan 3 at 17:03
$begingroup$
I think Step 3 should give $a+b=-1$. Note that $frac{w^{13}-1}{w-1}=1+w+w^2+dots+w^{12} = a+b+1$.
$endgroup$
– gandalf61
Jan 3 at 17:09
$begingroup$
Fixed that. I forgot to add and subtract one. I've just added.
$endgroup$
– Andrei
Jan 3 at 17:14
$begingroup$
Andrei, it is not entirely necessary to relate $a$ and $b.$ See page 16 in books.google.com/… The method is due to Gauss, modern discussion (but just a few examples) in zakuski.utsa.edu/~jagy/cox_galois_Gaussian_periods.pdf
$endgroup$
– Will Jagy
Jan 3 at 18:52
add a comment |
$begingroup$
Step 1: the equation you want is $(z-a)(z-b)=0$. Expand the product and you get $$z^2-(a+b)z+ab=0$$
Step 2: Use $w^{13}=1$, so $w^{-1}=w^{13}w^{-1}=w^{12}$ similarly, for all negative powers $$w^{-n}=w^{13-n}$$
Step 3: $$a+b=w+w^3+w^4+w^9+w^{10}+w^{12}+w^2+w^5+w^6+w^7+w^8+w^{11}=\=frac{w^{13}-1}{w-1}-1=-1$$
Step 4: To find $ab$, go to the trigonometric representation, and notice $$w^n+w^{-n}=2cosfrac{2pi n}{13}$$
Edit: After some manipulations, and using $w^{n+13}=w^n$, I've got $$ab=3(w+w^2+w^3+w^4+w^5+w^6+w^7+w^8+w^9+w^{10}+w^{11}+w^{12})=\-3+3(1+w+w^2+w^3+w^4+w^5+w^6+w^7+w^8+w^9+w^{10}+w^{11}+w^{12})=-3$$
answered Jan 3 at 16:15
AndreiAndrei
11.7k21026
$endgroup$
Step 1: the equation you want is $(z-a)(z-b)=0$. Expand the product and you get $$z^2-(a+b)z+ab=0$$
Step 2: Use $w^{13}=1$, so $w^{-1}=w^{13}w^{-1}=w^{12}$ similarly, for all negative powers $$w^{-n}=w^{13-n}$$
Step 3: $$a+b=w+w^3+w^4+w^9+w^{10}+w^{12}+w^2+w^5+w^6+w^7+w^8+w^{11}=\=frac{w^{13}-1}{w-1}-1=-1$$
Step 4: To find $ab$, go to the trigonometric representation, and notice $$w^n+w^{-n}=2cosfrac{2pi n}{13}$$
Edit: After some manipulations, and using $w^{n+13}=w^n$, I've got $$ab=3(w+w^2+w^3+w^4+w^5+w^6+w^7+w^8+w^9+w^{10}+w^{11}+w^{12})=\-3+3(1+w+w^2+w^3+w^4+w^5+w^6+w^7+w^8+w^9+w^{10}+w^{11}+w^{12})=-3$$
answered Jan 3 at 16:15
AndreiAndrei
11.7k21026
edited Jan 3 at 17:02
answered Jan 3 at 16:15
AndreiAndrei
11.7k21026
answered Jan 3 at 16:15
AndreiAndrei
11.7k21026
answered Jan 3 at 16:15
AndreiAndrei
11.7k21026
11.7k21026
$begingroup$
I got to this point. I can not solve the trigonometric part.
$endgroup$
– Rituraj Tripathy
Jan 3 at 16:27
$begingroup$
I went back to $w$ and wrote explicitly the product
$endgroup$
– Andrei
Jan 3 at 17:03
$begingroup$
I think Step 3 should give $a+b=-1$. Note that $frac{w^{13}-1}{w-1}=1+w+w^2+dots+w^{12} = a+b+1$.
$endgroup$
– gandalf61
Jan 3 at 17:09
$begingroup$
Fixed that. I forgot to add and subtract one. I've just added.
$endgroup$
– Andrei
Jan 3 at 17:14
$begingroup$
Andrei, it is not entirely necessary to relate $a$ and $b.$ See page 16 in books.google.com/… The method is due to Gauss, modern discussion (but just a few examples) in zakuski.utsa.edu/~jagy/cox_galois_Gaussian_periods.pdf
$endgroup$
– Will Jagy
Jan 3 at 18:52
add a comment |
$begingroup$
I got to this point. I can not solve the trigonometric part.
$endgroup$
– Rituraj Tripathy
Jan 3 at 16:27
$begingroup$
I went back to $w$ and wrote explicitly the product
$endgroup$
– Andrei
Jan 3 at 17:03
$begingroup$
I think Step 3 should give $a+b=-1$. Note that $frac{w^{13}-1}{w-1}=1+w+w^2+dots+w^{12} = a+b+1$.
$endgroup$
– gandalf61
Jan 3 at 17:09
$begingroup$
Fixed that. I forgot to add and subtract one. I've just added.
$endgroup$
– Andrei
Jan 3 at 17:14
$begingroup$
Andrei, it is not entirely necessary to relate $a$ and $b.$ See page 16 in books.google.com/… The method is due to Gauss, modern discussion (but just a few examples) in zakuski.utsa.edu/~jagy/cox_galois_Gaussian_periods.pdf
$endgroup$
– Will Jagy
Jan 3 at 18:52
$begingroup$
I got to this point. I can not solve the trigonometric part.
$endgroup$
– Rituraj Tripathy
Jan 3 at 16:27
$begingroup$
I got to this point. I can not solve the trigonometric part.
$endgroup$
– Rituraj Tripathy
Jan 3 at 16:27
$begingroup$
I went back to $w$ and wrote explicitly the product
$endgroup$
– Andrei
Jan 3 at 17:03
$begingroup$
I went back to $w$ and wrote explicitly the product
$endgroup$
– Andrei
Jan 3 at 17:03
$begingroup$
I think Step 3 should give $a+b=-1$. Note that $frac{w^{13}-1}{w-1}=1+w+w^2+dots+w^{12} = a+b+1$.
$endgroup$
– gandalf61
Jan 3 at 17:09
$begingroup$
I think Step 3 should give $a+b=-1$. Note that $frac{w^{13}-1}{w-1}=1+w+w^2+dots+w^{12} = a+b+1$.
$endgroup$
– gandalf61
Jan 3 at 17:09
$begingroup$
Fixed that. I forgot to add and subtract one. I've just added.
$endgroup$
– Andrei
Jan 3 at 17:14
$begingroup$
Fixed that. I forgot to add and subtract one. I've just added.
$endgroup$
– Andrei
Jan 3 at 17:14
$begingroup$
Andrei, it is not entirely necessary to relate $a$ and $b.$ See page 16 in books.google.com/… The method is due to Gauss, modern discussion (but just a few examples) in zakuski.utsa.edu/~jagy/cox_galois_Gaussian_periods.pdf
$endgroup$
– Will Jagy
Jan 3 at 18:52
$begingroup$
Andrei, it is not entirely necessary to relate $a$ and $b.$ See page 16 in books.google.com/… The method is due to Gauss, modern discussion (but just a few examples) in zakuski.utsa.edu/~jagy/cox_galois_Gaussian_periods.pdf
$endgroup$
– Will Jagy
Jan 3 at 18:52
add a comment |
$begingroup$
$$ a^2 + a = frac{ w^{16} + 2w^{15} + w^{14} + 2w^{13} + 3w^{12} + 3w^{11} + 3w^{10} + 3w^9 + 6w^8 + 3w^7 + 3w^6 + 3w^5 + 3w^4 + 2w^3 + w^2 + 2w + 1}{w^8} $$
This is not impressive without
$$ w^{16} + 2w^{15} + w^{14} + 2w^{13} = w^3+2w^2+w+2. $$ Therefore
$$ a^2 + a = frac{ 3w^{12} + 3w^{11} + 3w^{10} + 3w^9 + 6w^8 + 3w^7 + 3w^6 + 3w^5 + 3w^4 + 3w^3 + 3w^2 + 3w + 3}{w^8} $$
The one coefficient out of line is $6 w^8 / w^8,$ so we need to subtract 3
$$ a^2 + a -3 = frac{ 3w^{12} + 3w^{11} + 3w^{10} + 3w^9 + 3w^8 + 3w^7 + 3w^6 + 3w^5 + 3w^4 + 3w^3 + 3w^2 + 3w + 3}{w^8} $$
$$ a^2 + a -3 = frac{ 3 left( w^{12} + w^{11} + w^{10} + w^9 + w^8 + w^7 + w^6 + w^5 + w^4 + w^3 + w^2 + w + 1 right)}{w^8} $$
and
$$ a^2 + a - 3 = 0 $$
answered Jan 3 at 18:43
Will JagyWill Jagy
102k5101199
$endgroup$
$begingroup$
Elegant solution, but it might be at a higher level than what the question wanted. It would have probably had different tags.
$endgroup$
– Andrei
Jan 3 at 19:12
add a comment |
$begingroup$
$$ a^2 + a = frac{ w^{16} + 2w^{15} + w^{14} + 2w^{13} + 3w^{12} + 3w^{11} + 3w^{10} + 3w^9 + 6w^8 + 3w^7 + 3w^6 + 3w^5 + 3w^4 + 2w^3 + w^2 + 2w + 1}{w^8} $$
This is not impressive without
$$ w^{16} + 2w^{15} + w^{14} + 2w^{13} = w^3+2w^2+w+2. $$ Therefore
$$ a^2 + a = frac{ 3w^{12} + 3w^{11} + 3w^{10} + 3w^9 + 6w^8 + 3w^7 + 3w^6 + 3w^5 + 3w^4 + 3w^3 + 3w^2 + 3w + 3}{w^8} $$
The one coefficient out of line is $6 w^8 / w^8,$ so we need to subtract 3
$$ a^2 + a -3 = frac{ 3w^{12} + 3w^{11} + 3w^{10} + 3w^9 + 3w^8 + 3w^7 + 3w^6 + 3w^5 + 3w^4 + 3w^3 + 3w^2 + 3w + 3}{w^8} $$
$$ a^2 + a -3 = frac{ 3 left( w^{12} + w^{11} + w^{10} + w^9 + w^8 + w^7 + w^6 + w^5 + w^4 + w^3 + w^2 + w + 1 right)}{w^8} $$
and
$$ a^2 + a - 3 = 0 $$
answered Jan 3 at 18:43
Will JagyWill Jagy
102k5101199
$endgroup$
$begingroup$
Elegant solution, but it might be at a higher level than what the question wanted. It would have probably had different tags.
$endgroup$
– Andrei
Jan 3 at 19:12
add a comment |
$begingroup$
$$ a^2 + a = frac{ w^{16} + 2w^{15} + w^{14} + 2w^{13} + 3w^{12} + 3w^{11} + 3w^{10} + 3w^9 + 6w^8 + 3w^7 + 3w^6 + 3w^5 + 3w^4 + 2w^3 + w^2 + 2w + 1}{w^8} $$
This is not impressive without
$$ w^{16} + 2w^{15} + w^{14} + 2w^{13} = w^3+2w^2+w+2. $$ Therefore
$$ a^2 + a = frac{ 3w^{12} + 3w^{11} + 3w^{10} + 3w^9 + 6w^8 + 3w^7 + 3w^6 + 3w^5 + 3w^4 + 3w^3 + 3w^2 + 3w + 3}{w^8} $$
The one coefficient out of line is $6 w^8 / w^8,$ so we need to subtract 3
$$ a^2 + a -3 = frac{ 3w^{12} + 3w^{11} + 3w^{10} + 3w^9 + 3w^8 + 3w^7 + 3w^6 + 3w^5 + 3w^4 + 3w^3 + 3w^2 + 3w + 3}{w^8} $$
$$ a^2 + a -3 = frac{ 3 left( w^{12} + w^{11} + w^{10} + w^9 + w^8 + w^7 + w^6 + w^5 + w^4 + w^3 + w^2 + w + 1 right)}{w^8} $$
and
$$ a^2 + a - 3 = 0 $$
answered Jan 3 at 18:43
Will JagyWill Jagy
102k5101199
$endgroup$
$$ a^2 + a = frac{ w^{16} + 2w^{15} + w^{14} + 2w^{13} + 3w^{12} + 3w^{11} + 3w^{10} + 3w^9 + 6w^8 + 3w^7 + 3w^6 + 3w^5 + 3w^4 + 2w^3 + w^2 + 2w + 1}{w^8} $$
This is not impressive without
$$ w^{16} + 2w^{15} + w^{14} + 2w^{13} = w^3+2w^2+w+2. $$ Therefore
$$ a^2 + a = frac{ 3w^{12} + 3w^{11} + 3w^{10} + 3w^9 + 6w^8 + 3w^7 + 3w^6 + 3w^5 + 3w^4 + 3w^3 + 3w^2 + 3w + 3}{w^8} $$
The one coefficient out of line is $6 w^8 / w^8,$ so we need to subtract 3
$$ a^2 + a -3 = frac{ 3w^{12} + 3w^{11} + 3w^{10} + 3w^9 + 3w^8 + 3w^7 + 3w^6 + 3w^5 + 3w^4 + 3w^3 + 3w^2 + 3w + 3}{w^8} $$
$$ a^2 + a -3 = frac{ 3 left( w^{12} + w^{11} + w^{10} + w^9 + w^8 + w^7 + w^6 + w^5 + w^4 + w^3 + w^2 + w + 1 right)}{w^8} $$
and
$$ a^2 + a - 3 = 0 $$
answered Jan 3 at 18:43
Will JagyWill Jagy
102k5101199
answered Jan 3 at 18:43
Will JagyWill Jagy
102k5101199
answered Jan 3 at 18:43
Will JagyWill Jagy
102k5101199
answered Jan 3 at 18:43
Will JagyWill Jagy
102k5101199
102k5101199
$begingroup$
Elegant solution, but it might be at a higher level than what the question wanted. It would have probably had different tags.
$endgroup$
– Andrei
Jan 3 at 19:12
add a comment |
$begingroup$
Elegant solution, but it might be at a higher level than what the question wanted. It would have probably had different tags.
$endgroup$
– Andrei
Jan 3 at 19:12
$begingroup$
Elegant solution, but it might be at a higher level than what the question wanted. It would have probably had different tags.
$endgroup$
– Andrei
Jan 3 at 19:12
$begingroup$
Elegant solution, but it might be at a higher level than what the question wanted. It would have probably had different tags.
$endgroup$
– Andrei
Jan 3 at 19:12
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%2f3060694%2fa-problem-on-complex-numbers%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
$begingroup$
Welcome to the site. Please typeset your equations with Mathjax for better presentation.
$endgroup$
– Shubham Johri
Jan 3 at 15:46
1
$begingroup$
These are Gaussian periods: en.wikipedia.org/wiki/Gaussian_period
$endgroup$
– Lord Shark the Unknown
Jan 3 at 15:46
1
$begingroup$
Also, you are more likely to get views if you actually title your "question" properly. Something like "Solving quadratic roots of complex numbers" is much better then your current title
$endgroup$
– K Split X
Jan 3 at 15:55
2
$begingroup$
"And then I found a and b in trigonometric form. But when I multiplied them to get the product of roots it gets very difficult." If you found a and b in trig form how can multiplying them together be difficult?
$endgroup$
– fleablood
Jan 3 at 16:04
$begingroup$
Ditto what K Split X said. Many people will simply ignore questions that are titled something like "question about subject-tag". I know I certainly am not going to read the problem statement. Keep the tags in the tags section, and put your question in the title, or some adequate abbreviation of it.
$endgroup$
– rschwieb
Jan 3 at 16:19