Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Consider the equation $$31z^{15} - z^{10} + 32 = 0.$$ What would be the sum of all those roots of the equation whose real part is positive? Only trivially trying to solve the equation I find not helpful. Even factorizing we get , by putting $z^5=x $ ,

$31x^3 - x^2 + 32 = (x+1)(31x^2 - 32x + 32) =0 $ which is of no help .

share|cite|improve this question
Why do you think this sum is interesting and/or has a nice formula? – Did Nov 2 '12 at 12:23

$31z^{15}-z^{10}+32=0\implies 31(z^{15}+1)-(z^{10}-1)=0$


$(z^5+1)\{31(z^{10}-z^5+1)-(z^5-1) \}=0 $

$(z^5+1)\{31z^{10}-32z^5+32 \}=0 $

If $z^5+1=0, z^5=-1=e^{(2m+1)\pi i} $ where $m$ is any integer.

So, $z=e^{\frac{(2m+1)\pi i}5}$ where any $5$ in-congruent values of $m\pmod 5$ will give us essentially the same set of $5$ distinct solutions, find the explanation here, the simplest example can be $0,1,2,3,4$.

Now, $e^\frac{(2m+1)\pi i}5=\cos \frac{(2m+1)\pi}5+i\sin \frac{(2m+1)\pi}5$

For the real part to be positive, $-\frac \pi 2<\frac{(2m+1)\pi}5< \frac \pi 2$

$\implies -\frac74<m<\frac 3 4\implies m=-1,0$

So, those roots are $\cos \frac{\pi}5\pm \sin \frac{\pi}5,$ the sum being $2\cos \frac{\pi}5$

If $31z^{10}-32z^5+32=0,z^5=\frac{16\pm12\sqrt 5 i}{31}$

Let $r\cos A= \frac{16}{31}--->(1), r\sin A= \frac{12\sqrt 5 }{31}--->(2)$ where $r>0$

so that $z^5=r(\cos A\pm i\sin A)=re^{\pm iA}=re^{i(2n\pi\pm A)}$ where $n$ is any integer.

$z=r^\frac 15 e^{\frac{i(2n\pi\pm A)}5}$ where $0\le n<5$

Dividing $(2)$ by $(1),\tan A=\frac{12\sqrt 5}{16}=\frac{3\sqrt5}4$

So, $A=\arctan \frac{3\sqrt5}4$ where $A$ lies $\in(0,\frac \pi 2)$ as $\cos A,\sin A>0$

Squaring and adding $(2)$ and $(1), r^2=\left(\frac{16}{31}\right)^2+\left(\frac{12\sqrt 5 }{31}\right)^2$

share|cite|improve this answer

The equation $31x^3-x^2+32=0$, has the roots, $$ -1\qquad \frac{4}{31(4 + i\sqrt{46})}\qquad \frac{4}{31(4 - i\sqrt{46})} $$ so take all them to the $1/5$-th power and sum there real parts.

share|cite|improve this answer
Vieta's formula wold give me only the sum of all the roots , but I want sum of those roots whose real part is positive. – Souvik Dey Nov 2 '12 at 4:56
Would you mind explaining how Vieta's formulas help? – wj32 Nov 2 '12 at 4:56
It can be factored, (z+1)(z^4-z^3+z^2-z+1)(31z^10-32z^5+32)=0, therefore the only real root is z=-1 – boby Nov 2 '12 at 4:58
@boby : Yeah , but the factorization is horrible – Souvik Dey Nov 2 '12 at 5:00
@boby learn some TeX instead of posting awful quality answers! – Norbert Nov 2 '12 at 5:27

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.