Factorisation of z^n-a

Theorem
Let $n \in \Z_{>0}$ be a (strictly) positive integer.

Let $a \in \C$ be a complex number.

Then:


 * $z^n - a = \ds \prod_{k \mathop = 0}^{n - 1} \paren {z - \alpha^k b}$

where:
 * $\alpha$ is a primitive complex $n$th root of unity
 * $b$ is any complex number such that $b^n = a$.

Proof
From $z^n - a = 0$ we have that:
 * $a = z^n$

Let $b = a^{1 / n}$, hence $b^n = a$, with $a, b \in \C$.

From Roots of Complex Number:

and so each of $b e^{2 i k \pi / n}$ is a root of $z^n - a$.

From First Complex Root of Unity is Primitive:

For each $n \in \Z_{>0}$ there exists at least one primitive complex $n$th root of unity, $\alpha$, hence each of $b \alpha^k$ is a unique root of $z^n - a$.

From the corollary to the Polynomial Factor Theorem:

If $\zeta_1, \zeta_2, \ldots, \zeta_n \in \C$ such that all are different, and $\map P {\zeta_1} = \map P {\zeta_2} = \cdots = \map P {\zeta_n} = 0$, then:
 * $\ds \map P z = k \prod_{j \mathop = 1}^n \paren {z - \zeta_j}$

where $k \in \C$.

$z^n - a$ is a monic polynomial, hence $k = 1$ in the above product.

Choose $\zeta_j = b \alpha^{j - 1}$ and we have the desired result.