Nth root of Unity

Hi all I am in higher level mathematics and I am taking the IB. We started doing problems associated with nth root of unity. I understand how to find the roots of for example:

$$Z^3 – 1 =0$$

and also understand how the roots will equal 0 when added up. Furthermore my teacher showed me how when:

$Z_2$ (when $k = 1$) $= w$, that means $(Z_2)^2 = w^2 = Z_3$ (when $k = 2$).

He then made us note $Z^3-1 = (Z-1)(Z^2+Z+1)$

And then made us state: if $w$ is the root with the smallest positive argument, that means $Z^2+Z+1=0$ which means $w^2+w+1= 0$ since $w-1$ cannot $= 0$.

The above statement I do not understand, can someone explain this to me since I need it to simplify:


can someone please explain this to me, I will be so grateful thanks so much.

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Note that when $Z^3 – 1$ is factored as $(Z-1)(Z^2+Z+1)$, it reduces the problem of solving $Z^3-1$ into two subproblems: Either $Z-1=0$ or $Z^2+Z+1=0$.

The first of these gives us $Z=1$ as a solution, and the second of these yields, by the quadratic formula, a pair of conjugate complex roots:

$$ Z = \frac{-1 \pm \sqrt{-3}}{2} = -\frac{1}{2} \pm i \frac{\sqrt{3}}{2} $$

These two conjugate complex roots fall in the second and third quadrants. The one in the second quadrant has the smallest positive “argument” (angle with respect to the positive real axis).

You then say you “need this” to simplify $(1+w^2)(1+w)$. That part of the Question is unclear. Do you mean that $w$ is one of the three roots of $Z^3-1=0$? Do you mean that it is specifically the one with “smallest positive argument”?

If you mean the latter, then you can “plug in” the root (found above, with the plus sign for the choice of plus/minus)) and get a complex number. If the former, you can do a bit of algebra, since $(1+w^2)(1+w) = 1 + w + w^2 + w^3$. That is:

$$ 1+w+w^2+w^3 = 0+w^3 = 1 $$

for either of the complex roots of $Z^3 – 1 = 0$.

By the Fundamental Theorem of Algebra, the polynomial $$f(z)=z^3-1$$ has exactly three complex roots, counted by multiplicity. Call them $r_1, r_2, r_3$. Hence $$f(z)=(z-r_1)(z-r_2)(z-r_3)$$

But we can also factor the polynomial as $$f(z)=z^3-1=(z-1)(z^2+z+1)$$
Just multiply it out and you will see that the two expressions agree, there’s a lot of cancellation. Hence we may assume that $r_1$ is the root of $(z-1)$, and $r_2,r_3$ are the roots of $z^2+z+1$. We know then that $r_1=1$, but the roots of $z^2+z+1$ are not so nice. You can find them explicitly with the quadratic formula if you like; one of them will have positive imaginary part and one will have negative imaginary part. Let $r_2$ be the one with positive imaginary part; this one is the non-real root with minimal argument. Since it is a root of $z^2+z+1$, we know that $r_2^2+r_2+1=0$. This is the property you need for the remainder of the problem.