Roots of Complex Number/Examples/Cube Roots of 8

From ProofWiki
Jump to navigation Jump to search

Example of Roots of Complex Number

The complex cube roots of $8$ are given by:

$\paren 8^{1/3} = \set {2 \, \map \cis {120 k} \degrees}$

for $k = 0, 1, 2$.


That is:

\(\ds k = 0: \ \ \) \(\ds z = z_1\) \(=\) \(\ds 2\)
\(\ds k = 1: \ \ \) \(\ds z = z_2\) \(=\) \(\ds 2 \cis 120 \degrees = -1 + \sqrt 3 i\)
\(\ds k = 2: \ \ \) \(\ds z = z_3\) \(=\) \(\ds 2 \cis 240 \degrees = -1 - \sqrt 3 i\)


Proof

Complex Cube Roots of 8.png


Let $z^3 = 8$.

We have that:

$z^3 = 8 \, \map \cis {0 + 2 k \pi}$


Let $z = r \cis \theta$.

Then:

\(\ds z^3\) \(=\) \(\ds r^3 \cis 3 \theta\) De Moivre's Theorem
\(\ds \) \(=\) \(\ds 8 \, \map \cis {0 + 2 k \pi}\)
\(\ds \leadsto \ \ \) \(\ds r^3\) \(=\) \(\ds 8\)
\(\ds 3 \theta\) \(=\) \(\ds 2 k \pi\)
\(\ds \leadsto \ \ \) \(\ds r\) \(=\) \(\ds 8^{1/3}\)
\(\ds \) \(=\) \(\ds 2\)
\(\ds \theta\) \(=\) \(\ds \dfrac {2 k \pi} 3\) for $k = 0, 1, 2$
\(\ds \) \(=\) \(\ds k \times 120 \degrees\) for $k = 0, 1, 2$

$\blacksquare$


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Roots_of_Complex_Number/Examples/Cube_Roots_of_8&oldid=395785"