Roots of Complex Number/Examples/Cube Roots of -1+i

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Example of Roots of Complex Number: Corollary

The complex cube roots of $-1 + i$ are given by:

$\paren {-1 + i}^{1/3} = \set {2^{1/6} \paren {\cos \dfrac \pi 4 + i \sin \dfrac \pi 4}, 2^{1/6} \paren {\cos \dfrac {11 \pi} {12} + i \sin \dfrac {11 \pi} {12} }, 2^{1/6} \paren {\cos \dfrac {19 \pi} {12} + i \sin \dfrac {19 \pi} {12} } }$


Proof

Complex Cube Roots of -1+i.png


Let $z^3 = -1 + i$.

We have that:

$z^3 = \sqrt 2 \paren {\map \cos {\dfrac {3 \pi} 4 + 2 k \pi} + i \, \map \sin {\dfrac {3 \pi} 4 + 2 k \pi} }$


Let $z = r \cis \theta$.

Then:

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

$\blacksquare$


Sources

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