MHB Proving x^8+1=P_0^3(x^2-2xcos(2k+1)π/8+1)

[Suvadip]

I have to prove
$$x^8+1= P_0^3(x^2-2xcos\frac{(2k+1)\pi}{8}+1)$$ where $$P_0^3$$ means product from $$k=0$$ to $$k=3$$.I tried it but got $$x^8+1= P_0^3(x^2-a_k^2)$$ where $$a_k=cos\frac{(2k+1)\pi}{8}+isin\frac{(2k+1)\pi}{8}$$. How to arrive at the correct answer

 

[polygamma]

Re: complex number

The roots of $x^8+1=0$ are $x = e^{i/8(\pi +2 \pi n)}$ for $n=0,1, \ldots$, and $7$.

That is, the roots are $ e^{i \pi/8},e^{3 \pi i /8},e^{5 \pi i/8},e^{7 \pi i/8},e^{9 \pi i/8},e^{11 \pi i/8},e^{13 \pi i/8}$, and $e^{15 \pi i/8}$.

Then

$$x^8+1 = (x- e^{i \pi/8})(x-e^{15 \pi i/8})(x-e^{3 \pi i/8})(x-e^{13 \pi i/8})(x- e^{5 \pi i/8})(x-e^{11 \pi i/8})(x-e^{7 \pi i/8})(x-e^{9 \pi i/8})$$

$$ = (x- e^{i \pi/8})(x-e^{- \pi i/8})(x-e^{3 \pi i/8})(x-e^{-3 \pi i/8})(x- e^{5 \pi i/8})(x-e^{-5 \pi i/8})(x-e^{7 \pi i/8})(x-e^{-7 \pi i/8}) $$

$$ = \Big( x^{2} - 2 x \cos(\frac{\pi}{8}) + 1 \Big) \Big( x^{2} - 2 x \cos(\frac{3 \pi}{8}) + 1\Big) \Big(x^{2} - 2 x \cos(\frac{5 \pi}{8}) + 1 \Big)\Big(x^{2} - 2 x \cos(\frac{7\pi}{8}) + 1 \Big)$$

 

[Deveno]

Re: complex number

This is just RandomVariable's solution, but re-arranged a bit.

We'll start here:

Let $w = \cos(\frac{\pi}{8}) + i\sin(\frac{\pi}{8})$

If you know DeMoivre's formula, you can see that:

$w^8 = -1$, so $w$ is one root of $x^8 + 1$.

With any real polynomial, if a complex number $z$ is a root, so is $\overline{z}$.

This means that $\overline{w}^8 = -1$ as well.

It should also be clear that:

$(w^n)^8 = w^{8n} = (w^8)^n = (-1)^n = -1$ for $n = 1,3,5,7$.

Thus, likewise:

$((\overline{w})^n)^8 = 1$ for $n = 1,3,5,7$.

so this gives us 8 factors.

Let's look at the 4 quadratics:

$(x - w^n)(x - \overline{w}^n)$ for $n = 1,3,5,7$.

Expanding this out, this becomes:

$(x - \cos(\frac{n\pi}{8}) - i\sin(\frac{n\pi}{8}))(x - \cos(\frac{n\pi}{8}) + i\sin(\frac{n\pi}{8}))$

$= x^2 -2\cos(\frac{n\pi}{8})x + \cos^2(\frac{n\pi}{8}) + \sin^2(\frac{n\pi}{8})$

$= x^2 - 2\cos(\frac{n\pi}{8})x + 1$

Since we are only considering odd $n$, we can use the index $k = \frac{n-1}{2}$, in which case we get:

$n = 2k+1$ for $k = 0,1,2,3$ and our quadratics become:

$x^2 - 2\cos(\frac{(2k+1)\pi}{8})x + 1$

so that:

[math]x^8 + 1 = \prod_{k = 0}^3 [x^2 - 2\cos\left(\frac{(2k+1)\pi}{8}\right)x + 1][/math]

In other words, it's not the product of $(x^2 - a_k^2)$ like you had, but instead the product of:

$(x - a_k)(x - \overline{a_k})$