Roots of Complex Number/Examples/5th Roots of -16 + 16 root 3 i

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Example of Roots of Complex Number

The complex $5$th roots of $-16 + 16 \sqrt 3 i$ are given by:

$\paren {-16 + 16 \sqrt 3 i}^{1/5} = \set {2 \, \map \cis {24 + 72 k} \degrees}$

for $k = 0, 1, 2, 3, 4$.


That is:

\(\ds k = 0: \ \ \) \(\ds z = z_1\) \(=\) \(\ds \sqrt 2 \cis 24 \degrees\)
\(\ds k = 1: \ \ \) \(\ds z = z_2\) \(=\) \(\ds \sqrt 2 \cis 96 \degrees\)
\(\ds k = 2: \ \ \) \(\ds z = z_3\) \(=\) \(\ds \sqrt 2 \cis 168 \degrees\)
\(\ds k = 3: \ \ \) \(\ds z = z_4\) \(=\) \(\ds \sqrt 2 \cis 240 \degrees\)
\(\ds k = 4: \ \ \) \(\ds z = z_5\) \(=\) \(\ds \sqrt 2 \cis 321 \degrees\)


Proof

Complex 5th Roots of -16 + 16 root 3 i.png


Let $z^5 = -16 + 16 \sqrt 3 i$.

We have that:

$z^5 = 16 \sqrt 2 \, \map \cis {\dfrac {2 \pi} 3 + 2 k \pi} = 16 \sqrt 2 \, \map \cis {120 \degrees + k \times 360 \degrees}$


Let $z = r \cis \theta$.

Then:

\(\ds z^5\) \(=\) \(\ds r^5 \cis 5 \theta\) De Moivre's Theorem
\(\ds \) \(=\) \(\ds 16 \sqrt 2 \, \map \cis {\dfrac {2 \pi} 3 + 2 k \pi}\)
\(\ds \leadsto \ \ \) \(\ds r^5\) \(=\) \(\ds 16 \sqrt 2\)
\(\ds 5 \theta\) \(=\) \(\ds \dfrac {2 \pi} 3 + 2 k \pi\)
\(\ds \leadsto \ \ \) \(\ds r\) \(=\) \(\ds \paren {16 \sqrt 2}^{1/5}\)
\(\ds \) \(=\) \(\ds 2\)
\(\ds \theta\) \(=\) \(\ds \dfrac {2 \pi} {15} + \dfrac {2 k \pi} 5\) for $k = 0, 1, 2, 3, 4$
\(\ds \) \(=\) \(\ds 24 \degrees + 72 k \degrees\) for $k = 0, 1, 2, 3, 4$

$\blacksquare$


Sources

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