1. ## Root of Unity

I am currently doing complex root of unity questions from terry lee's 4u book and while going through some exercises i keep noticing the formula x^2-S(x)+P keeps popping up. I understand that is the sum and product of roots, but when do we use this formula in questions?

2. ## Re: Root of Unity

I don't think I've ever used that for root of unity do you have an example?

3. ## Re: Root of Unity

Originally Posted by Sokc
I am currently doing complex root of unity questions from terry lee's 4u book and while going through some exercises i keep noticing the formula x^2-S(x)+P keeps popping up. I understand that is the sum and product of roots, but when do we use this formula in questions?

Hmm..I think I've seen a couple of questions like that but I might be wrong. If I recall correctly, I mostly used that formula to factorise expressions such as $z^6-1$ into a product of linear or quadratic factors. It's also commonly used for proving some random trigonometric identity but I can't think of any examples off the top of my head.

For example (I probably made some errors along the way here since I wasn't that great at 4U),

$By finding the sixth roots of -1 prove that$

$\: z^6+1 = (z^2+1) (z^2-\sqrt{3}z+1) (z^2+\sqrt{3}z+1)$

$By using your usual method to find roots of unity you would get$
$z = i, -i, cis$ $(\frac{\pi}{6})$ $cis$ $(\frac{-\pi}{6})$ $cis$ $(\frac{5\pi}{6})$ $cis$ $(-\frac{5\pi}{6})$

$Then by using your usual linear factorisation of polynomials via their roots you would get$

$\: z^6+1 = (z-i) (z+i) (z- cis$ $(\frac{\pi}{6}))$ $(z-cis$ $(\frac{-\pi}{6}))$ $(z-cis$ $(\frac{5\pi}{6}))$ $(z-cis$ $(\frac{-5\pi}{6}))$

$Now by expanding the pairs and simplifying you would get$

$(z^2+1) (z^2 -2cos$ $(\frac{\pi}{6})$ $z +1)$ $(z^2 -2cos$ $(\frac{5\pi}{6})$ $z +1)$

$Notice how the last two brackets above are in the form of the formula you mentioned, that is$

$z^2 -Sum of Roots + Product of Roots.$

$I think when you add stuff like the cis and its conjugate you$ $get the double of the real part which is essentially the 2 cos$ $\frac{\pi}{6}$

$and when you multiply (product) the the conjugate of two cis's you get 1 provided that that there is no coefficient in front of it$

$Therefore, from the equation above you just evaluate it and you should get$

$\: z^6+1 = (z^2+1) (z^2-\sqrt{3}z+1) (z^2+\sqrt{3}z+1)$

Idk if that made any sense but hope that helps. This was the first time I used LaTeX so the input is killing me. Let me know if you need more help.

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