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Let $w\neq 1$ 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 $w=\cos(\frac{2\pi}{13})+i\sin(\frac{2\pi}{13})$ 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?

Andrei
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Rituraj Tripathy
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  • Welcome to the site. Please typeset your equations with Mathjax for better presentation. – Shubham Johri Jan 03 '19 at 15:46
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    These are Gaussian periods: https://en.wikipedia.org/wiki/Gaussian_period – Angina Seng Jan 03 '19 at 15:46
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    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 – K Split X Jan 03 '19 at 15:55
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    "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? – fleablood Jan 03 '19 at 16:04
  • 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. – rschwieb Jan 03 '19 at 16:19

2 Answers2

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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 $$w^n+w^{-n}=2\cos\frac{2\pi n}{13}$$

Edit: After some manipulations, and using $w^{n+13}=w^n$, I've got $$ab=3(w+w^2+w^3+w^4+w^5+w^6+w^7+w^8+w^9+w^{10}+w^{11}+w^{12})=\\-3+3(1+w+w^2+w^3+w^4+w^5+w^6+w^7+w^8+w^9+w^{10}+w^{11}+w^{12})=-3$$

Andrei
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  • I got to this point. I can not solve the trigonometric part. – Rituraj Tripathy Jan 03 '19 at 16:27
  • I went back to $w$ and wrote explicitly the product – Andrei Jan 03 '19 at 17:03
  • I think Step 3 should give $a+b=-1$. Note that $\frac{w^{13}-1}{w-1}=1+w+w^2+\dots+w^{12} = a+b+1$. – gandalf61 Jan 03 '19 at 17:09
  • Fixed that. I forgot to add and subtract one. I've just added. – Andrei Jan 03 '19 at 17:14
  • Andrei, it is not entirely necessary to relate $a$ and $b.$ See page 16 in https://books.google.com/books?id=wt7lgfeYqMQC&pg=PR1&lpg=PR1&dq=reuschle++tafeln+complexer+primzahlen&source=bl&ots=VGZFPrfUBn&sig=MlQ667PqXaQ9rAvLWkG3_F1rwsk&hl=en&sa=X&ved=0ahUKEwiIwtSvm9TQAhUJ-2MKHXJIA_kQ6AEIODAE#v=onepage&q=reuschle%20%20tafeln%20complexer%20primzahlen&f=false The method is due to Gauss, modern discussion (but just a few examples) in http://zakuski.utsa.edu/~jagy/cox_galois_Gaussian_periods.pdf – Will Jagy Jan 03 '19 at 18:52
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$$ a^2 + a = \frac{ w^{16} + 2w^{15} + w^{14} + 2w^{13} + 3w^{12} + 3w^{11} + 3w^{10} + 3w^9 + 6w^8 + 3w^7 + 3w^6 + 3w^5 + 3w^4 + 2w^3 + w^2 + 2w + 1}{w^8} $$

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 $$ a^2 + a - 3 = 0 $$

Will Jagy
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  • Elegant solution, but it might be at a higher level than what the question wanted. It would have probably had different tags. – Andrei Jan 03 '19 at 19:12