Roots of Complex Number/Examples/6th Roots of 64

From ProofWiki
Jump to navigation Jump to search

Example of Roots of Complex Number

The complex $6$th roots of $64$ are given by:

$\paren {64}^{1/6} = \set {2 \, \map \cis {60 k} \degrees}$

for $k = 0, 1, 2, 3, 4, 5$.


That is:

\(\ds k = 0: \ \ \) \(\ds z = z_1\) \(=\) \(\ds 2\)
\(\ds k = 1: \ \ \) \(\ds z = z_2\) \(=\) \(\ds 2 \cis 60 \degrees = 1 + \sqrt 3 i\)
\(\ds k = 2: \ \ \) \(\ds z = z_3\) \(=\) \(\ds 2 \cis 120 \degrees = -1 + \sqrt 3 i\)
\(\ds k = 3: \ \ \) \(\ds z = z_4\) \(=\) \(\ds -2\)
\(\ds k = 4: \ \ \) \(\ds z = z_5\) \(=\) \(\ds 2 \cis 240 \degrees = -1 - \sqrt 3 i\)
\(\ds k = 5: \ \ \) \(\ds z = z_6\) \(=\) \(\ds 2 \cis 300 \degrees = 1 - \sqrt 3 i\)


Proof

Complex 6th Roots of 64.png


Let $z^6 = 64$.

We have that:

$z^6 = 64 \, \map \cis {0 + 2 k \pi}$


Let $z = r \cis \theta$.

Then:

\(\ds z^6\) \(=\) \(\ds r^6 \cis 6 \theta\) De Moivre's Theorem
\(\ds \) \(=\) \(\ds 64 \, \map \cis {0 + 2 k \pi}\)
\(\ds \leadsto \ \ \) \(\ds r^6\) \(=\) \(\ds 64\)
\(\ds 6 \theta\) \(=\) \(\ds 2 k \pi\)
\(\ds \leadsto \ \ \) \(\ds r\) \(=\) \(\ds 64^{1/6}\)
\(\ds \) \(=\) \(\ds 2\)
\(\ds \theta\) \(=\) \(\ds \dfrac {k \pi} 3\) for $k = 0, 1, 2, 3, 4, 5$
\(\ds \theta\) \(=\) \(\ds k \times 60 \degrees\) for $k = 0, 1, 2, 3, 4, 5$


When $k = 3$ we have:

$z_4 = 2 \cis 180 \degrees = - 2$

$\blacksquare$


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Roots_of_Complex_Number/Examples/6th_Roots_of_64&oldid=395781"