A Return to Your Roots

The expansion of $\large P(x)$ would be

$P(x)=x^{8}+2x^{7}+3x^{6}+4x^{5}+4x^{4}+4x^{3}+3x^{2}+2x+1$

This can be factored into

$P(x)=(x^{5}+x^{4}+x^{3}+x^{2}+x+1)(x^{3}+x^{2}+x+1)$

We know that $(x^{5}+x^{4}+x^{3}+x^{2}+x+1)(x-1)=x^{6}-1$ and $(x^{3}+x^{2}+x+1)(x-1)=x^{4}-1$

so we can rewrite $\large P(x)$ as:

$P(x)=\frac{x^{6}-1}{x-1} \times \frac{x^{4}-1}{x-1}$ .

for $\large P(x)$ to equal $0$ , $x^{6}-1=0$ or $x^{4}-1=0$ , where $x$ is not equal to $1$ .

This means we have all of the $6^{th}$ and $4^{th}$ roots of unity, exlcuding $1$ itself.

We can rewrite $1$ as

$\cos (0+2\pi k) +i \sin (0+2\pi k)$

where $k$ is an integer.

De Moivre's Theorem states that

$(\cos (\theta) +i \sin (\theta))^{n}=(\cos (n \theta) +i \sin (n \theta))$ .

Using this, we can conclude that the sixth roots of unity are

$1^{\frac{1}{6}}=\cos (\frac{1}{6}(0+2\pi k)) +i \sin (\frac{1}{6}(0+2\pi k))$

which simplifies to

$\cos (\frac{1}{3}\pi k) +i \sin (\frac{1}{3}\pi k))$ .

Plugging in some values for $k$ will give us our distinct sixth roots of unity are

$1, -1, \frac{1}{2}+\frac{\sqrt{3}}{2}i,\frac{1}{2}-\frac{\sqrt{3}}{2}i,-\frac{1}{2}+\frac{\sqrt{3}}{2}i,-\frac{1}{2}-\frac{\sqrt{3}}{2}i$ .

We repeat the same process for our fourth roots, and get

$1^{\frac{1}{4}}= \cos (\frac{\pi}{2}k)+i \sin (\frac{\pi}{2}k)$

which, when plugging in some values for $k$ gets our distinct fourth roots of unity as

$1,-1,i,-i$ .

We can ignore $1$ , since it is not in our domain, and $-1$ is included twice, once as a fourth root and once as a sixth root. This means that all of our distinct roots are

$-1, i, -i, \frac{1}{2}+\frac{\sqrt{3}}{2}i,\frac{1}{2}-\frac{\sqrt{3}}{2}i,-\frac{1}{2}+\frac{\sqrt{3}}{2}i,-\frac{1}{2}-\frac{\sqrt{3}}{2}i$ .

We were asked to find the sum of the magnitudes of the imaginary part of all of our distinct roots. This gives us

$2+4(\frac{\sqrt{3}}{2})$

which simplifies to

$2+2\sqrt{3}$ .

Since that value satisfies the constraints of $m,p,q$ , our answer is

$\large 2+2+3=\boxed{7}$

write as $x^8+2x^7+3x^6+4x^5+5x^4+4x^3+3x^2+2x+1-x^4$ by the link provided, we can easily deduce this is equal to $(x^4+x^3+x^2+x+1)^2-x^4$ apply basic algebra identity $(x^4+x^3+x+1)(x^4+x^3+x^2+x^2+x+1)$ $(x^3(x+1)+(x+1))(x^2(x^2+x+1)+(x^2+x+1))$ $(x^3+1)(x+1)(x^2+1)(x^2+x+1)$ we have that $(x^3+1)=0\Longrightarrow x=-1,\dfrac{1}{2}\pm i\dfrac{\sqrt{3}}{2}$ $x^2+1=0\Longrightarrow x=\pm i$ $x^2+x+1=0\Longrightarrow x=-\dfrac{1}{2}\pm i\dfrac{\sqrt{3}}{2}$ $x+1=0\Longrightarrow x=-1$ the sum of the modulus to im part is $\dfrac{\sqrt{3}}{2}+\dfrac{\sqrt{3}}{2}+1+1+\dfrac{\sqrt{3}}{2}+\dfrac{\sqrt{3}}{2}$ $=2+2\sqrt{3}$