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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 .

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

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

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

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

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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.

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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
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@boby learn some TeX instead of posting awful quality answers! –  Norbert Nov 2 '12 at 5:27
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