Roots of Complex Number/Examples/z^6 + 1 = root 3 i

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Theorem

The roots of the equation:

$z^6 + 1 = \sqrt 3 i$

are:

$\set {\sqrt [6] 2 \cis 20 \degrees, \sqrt [6] 2 \cis 80 \degrees, \sqrt [6] 2 \cis 140 \degrees, \sqrt [6] 2 \cis 200 \degrees, \sqrt [6] 2 \cis 260 \degrees, \sqrt [6] 2 \cis 320 \degrees}$


Proof

We have:

$z^6 = -1 + \sqrt 3 i$

and so this is equivalent to finding the complex $6$th roots of $-1 + \sqrt 3 i$:


We have that:

$z^6 = 2 \, \map \cis {\dfrac {2 \pi} 3 + 2 k \pi}$


Let $z = r \cis \theta$.

Then:

\(\ds z^6\) \(=\) \(\ds r^6 \cis 6 \theta\) De Moivre's Theorem
\(\ds \) \(=\) \(\ds 2 \, \map \cis {\dfrac {2 \pi} 3 + 2 k \pi}\)
\(\ds \leadsto \ \ \) \(\ds r^6\) \(=\) \(\ds 2\)
\(\ds 6 \theta\) \(=\) \(\ds \dfrac {2 \pi} 3 + 2 k \pi\)
\(\ds \leadsto \ \ \) \(\ds r\) \(=\) \(\ds \sqrt [6] 2\)
\(\ds \theta\) \(=\) \(\ds \dfrac \dfrac \pi 9 + \dfrac {k \pi} 3\) for $k = 0, 1, 2, 3, 4, 5$
\(\ds \theta\) \(=\) \(\ds 20 \degrees + k \times 60 \degrees\) for $k = 0, 1, 2, 3, 4, 5$

$\blacksquare$


Sources

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