Roots of Complex Number/Examples/Cube Roots of 2 + 2 root 3 i

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Example of Roots of Complex Number

The complex cube roots of $2 + 2 \sqrt 3 i$ are given by:

$\paren {2 + 2 \sqrt 3 i}^{1/3} = \set {\sqrt [3] 4 \, \map \cis {20 + 120 k} \degrees}$

for $k = 0, 1, 2$.


That is:

\(\ds k = 0: \ \ \) \(\ds z = z_1\) \(=\) \(\ds \sqrt [3] 4 \cis 20 \degrees\)
\(\ds k = 1: \ \ \) \(\ds z = z_2\) \(=\) \(\ds \sqrt [3] 4 \cis 140 \degrees\)
\(\ds k = 2: \ \ \) \(\ds z = z_3\) \(=\) \(\ds \sqrt [3] 4 \cis 260 \degrees\)


Proof

Complex Cube Roots of 2 + 2 root 3 i.png


Let $z^3 = 2 + 2 \sqrt 3 i$.

We have that:

$z^3 = 4 \, \map \cis {\dfrac \pi 3 + 2 k \pi} = 4 \, \map \cis {60 \degrees + k \times 360 \degrees}$


Let $z = r \cis \theta$.

Then:

\(\ds z^3\) \(=\) \(\ds r^3 \cis 3 \theta\) De Moivre's Theorem
\(\ds \) \(=\) \(\ds 4 \, \map \cis {\dfrac \pi 3 + 2 k \pi}\)
\(\ds \leadsto \ \ \) \(\ds r^3\) \(=\) \(\ds 4\)
\(\ds 3 \theta\) \(=\) \(\ds \dfrac \pi 3 + 2 k \pi\)
\(\ds \leadsto \ \ \) \(\ds r\) \(=\) \(\ds 4^{1/3}\)
\(\ds \) \(=\) \(\ds \sqrt [3] 4\)
\(\ds \theta\) \(=\) \(\ds \dfrac \pi 9 + \dfrac {2 k \pi} 3\) for $k = 0, 1, 2$
\(\ds \) \(=\) \(\ds 20 \degrees + 120 k \degrees\) for $k = 0, 1, 2$

$\blacksquare$


Sources

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