Roots of Complex Number/Corollary

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Theorem

Let $z := \polar {r, \theta}$ be a complex number expressed in polar form, such that $z \ne 0$.

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


Let $w$ be one of the complex $n$th roots of $z$.


Then the $n$th roots of $z$ are given by:

$z^{1 / n} = \set {w \epsilon^k: k \in \set {1, 2, \ldots, n - 1} }$

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


Proof

By definition of primitive complex $n$th root of unity:

$\omega = e^{2 m i \pi k}$

for some $m \in \Z: 1 \le m < n$.

Thus:

\(\ds \paren {w \omega^k}^n\) \(=\) \(\ds w^n \paren {e^{2 m i \pi k / n} }^n\)
\(\ds \) \(=\) \(\ds z e^{2 m i \pi k}\)
\(\ds \) \(=\) \(\ds z \paren {e^{2 i \pi} }^{m k}\)
\(\ds \) \(=\) \(\ds z \times 1^{m k}\)
\(\ds \) \(=\) \(\ds z\)

This demonstrates that $w \omega^k$ is one of the complex $n$th roots of $z$.


All of the complex $n$th roots of unity are represented by powers of $\omega$.

Thus it follows from Roots of Complex Number that:

$z^{1 / n} = \set {w \omega^k: k \in \set {1, 2, \ldots, n - 1} }$

are the $n$ complex $n$th roots of $z$.

$\blacksquare$


Examples

Complex Cube Roots

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

Let $z \ne 0$.

Let $w$ be one of the (complex) cube roots of $z$.

Then the complete set of (complex) cube roots of $z$ is:

$\set {w, w \omega, w \omega^2}$

where:

$\omega = e^{2 \pi / 3} = -\dfrac 1 2 + \dfrac {i \sqrt 3} 2$


Fourth Roots of $2 - 2 i$

The complex $4$th roots of $2 - 2 i$ are given by:

$\paren {2 - 2 i}^{1/4} = \set {b, bi, -b, -bi}$

where:

$b = \sqrt [8] 8 \paren {\cos \dfrac \pi {16} + i \sin \dfrac \pi {16} }$


Sources