Power of Complex Number minus 1/Corollary

Theorem

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

Then:

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

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

Proof

\(\ds z^n - 1\)

\(=\)

\(\ds \prod_{k \mathop = 0}^{n - 1} \paren {z - \alpha^k}\)

\(\ds \leadsto \ \ \)

\(\ds \frac {z^n - 1} {z - 1}\)

\(=\)

\(\ds \prod_{k \mathop = 1}^{n - 1} \paren {z - \alpha^k}\)

as $\alpha^k = 1$ when $k = 0$

\(\ds \leadsto \ \ \)

\(\ds \sum_{k \mathop = 0}^{n - 1} z^k\)

\(=\)

\(\ds \prod_{k \mathop = 1}^{n - 1} \paren {z - \alpha^k}\)

Sum of Geometric Sequence

$\blacksquare$

Sources

• 1960: Walter Ledermann: Complex Numbers ... (previous) ... (next): $\S 3$. Roots of Unity: $(3.11)$