Roots of Complex Number/Examples/4th Roots of -2 root 3 - 2i

From ProofWiki
Jump to navigation Jump to search

Example of Roots of Complex Number: Corollary

The complex $4$th roots of $-2 \sqrt 3 - 2 i$ are given by:

$\paren {-2 \sqrt 3 - 2 i}^{1/4} = \set {\sqrt 2 \paren {\map \cis {\dfrac {7 \pi} {24} + \dfrac {k \pi} 2} } }$

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


That is:

\(\ds k = 0: \ \ \) \(\ds z = z_1\) \(=\) \(\ds \sqrt 2 \paren {\cos \dfrac {7 \pi} {24} + i \sin \dfrac {7 \pi} {24} }\)
\(\ds k = 1: \ \ \) \(\ds z = z_2\) \(=\) \(\ds \sqrt 2 \paren {\cos \dfrac {19 \pi} {24} + i \sin \dfrac {19 \pi} {24} }\)
\(\ds k = 2: \ \ \) \(\ds z = z_3\) \(=\) \(\ds \sqrt 2 \paren {\cos \dfrac {31 \pi} {24} + i \sin \dfrac {31 \pi} {24} }\)
\(\ds k = 3: \ \ \) \(\ds z = z_4\) \(=\) \(\ds \sqrt 2 \paren {\cos \dfrac {43 \pi} {24} + i \sin \dfrac {43 \pi} {24} }\)


Proof

Complex 4th Roots of -2 root 3 - 2i.png


Let $z^4 = -2 \sqrt 3 - 2 i$.

We have that:

$z^4 = 4 \paren {\map \cis {\dfrac {7 \pi} 6 + 2 k \pi} }$


Let $z = r \cis \theta$.

Then:

\(\ds z^4\) \(=\) \(\ds r^4 \cis 4 \theta\) De Moivre's Theorem
\(\ds \) \(=\) \(\ds 4 \, \map \cis {\dfrac {7 \pi} 6 + 2 k \pi}\)
\(\ds \leadsto \ \ \) \(\ds r^4\) \(=\) \(\ds 4\)
\(\ds 4 \theta\) \(=\) \(\ds \dfrac {7 \pi} 6 + 2 k \pi\)
\(\ds \leadsto \ \ \) \(\ds r\) \(=\) \(\ds 4^{1/4}\)
\(\ds \) \(=\) \(\ds \sqrt 2\)
\(\ds \theta\) \(=\) \(\ds \dfrac {7 \pi} {24} + \dfrac {k \pi} 2\) for $k = 0, 1, 2, 3$

$\blacksquare$


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Roots_of_Complex_Number/Examples/4th_Roots_of_-2_root_3_-_2i&oldid=394779"