Factorisation of z^n+1

Theorem

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

Then:

$z^n + 1 = \ds \prod_{k \mathop = 0}^{n - 1} \paren {z - \exp \dfrac {\paren {2 k + 1} i \pi} n}$

Proof

From Factorisation of $z^n - a$, setting $a = -1$:

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

where:

$\alpha$ is a primitive complex $n$th root of unity

$b$ is any complex number such that $b^n = a$.

From Euler's Identity:

$-1 = e^{i \pi}$

From Exponential of Product:

$\paren {\exp \dfrac {i \pi} n}^n = e^{i \pi}$

and so:

$b = \exp \dfrac {i \pi} n$

We also have by definition of the first complex $n$th root of unity, and from First Complex Root of Unity is Primitive:

$\alpha = \exp \dfrac {2 i \pi} n$

Hence:

\(\ds z^n + 1\)

\(=\)

\(\ds \prod_{k \mathop = 0}^{n - 1} \paren {z - \paren {\exp \dfrac {2 i \pi} n}^k \exp \dfrac {i \pi} n}\)

\(\ds \)

\(=\)

\(\ds \prod_{k \mathop = 0}^{n - 1} \paren {z - \exp \dfrac {2 k i \pi} n \exp \dfrac {i \pi} n}\)

Exponential of Product

\(\ds \)

\(=\)

\(\ds \prod_{k \mathop = 0}^{n - 1} \paren {z - \exp \dfrac {\paren {2 k + 1} i \pi} n}\)

Exponential of Sum

$\blacksquare$