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
- 1981: Murray R. Spiegel: Theory and Problems of Complex Variables (SI ed.) ... (previous) ... (next): $1$: Complex Numbers: Supplementary Problems: Roots of Complex Numbers: $97 \ \text{(b)}$