# 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:

$\alpha_k$ are the complex $n$th roots of negative unity
$b$ is any complex number such that $b^n = a$.

## Proof

From $z^n + a = 0$ we have that:

$z^n = - a$

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

 $\displaystyle z^{1 / n}$ $=$ $\displaystyle \set {a^{1 / n} e^{i \paren {\theta + 2 k \pi} / n}: k \in \set {0, 1, 2, \ldots, n - 1}, \theta = \arg -a}$ $z^n = -a$ so we need $\theta = \arg -a$ $\displaystyle$ $=$ $\displaystyle \set {a^{1 / n} e^{i \paren {2 k + 1} \pi / n}: k \in \set {0, 1, 2, \ldots, n - 1} }$ $\theta = \pi$ $\displaystyle$ $=$ $\displaystyle \set {b e^{i \paren {2 k + 1} \pi / n}: k \in \set {0, 1, 2, \ldots, n - 1} }$ $b = a^{1 / n}$

and so each of $\alpha_k b = b e^{i \paren {2 k + 1} \pi / n}$ is a unique root of $z^n + a$ with $n \in \Z_{>0}$.

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:

$\displaystyle \map P z = k \prod_{j \mathop = 1}^n \paren {z - \zeta_j}$

where $k \in \C$.

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

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

$\blacksquare$