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:


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

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

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

From Roots of Complex Number:
 * $z^{1 / n} = \set {a^{1 / n} e^{i \paren {\theta + 2 \pi k} / n}: k \in \set {0, 1, 2, \ldots, n - 1} }$

and so each of $a^{1 / n} e^{i \paren {\theta + 2 \pi k} / n}$ is a root of $z^n - a$.

The result then follows from the corollary to the Polynomial Factor Theorem:

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

where $k \in \C$.