Roots of Complex Number/Corollary/Examples/Cube Roots of 8i

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Example of Roots of Complex Number: Corollary

The complex $4$th roots of $8 i$ are given by:

$\paren {2 - 2 i}^{1/4} = \set {2 i, \sqrt 3 + i, -\sqrt 3 + i}$


Proof

Let $z = 8 i$.

Then:

$z = 8 \exp \paren {\dfrac {i \pi} 2}$


Let $b$ be defined as:

\(\ds b\) \(=\) \(\ds \sqrt [3] 8 \map \exp {\dfrac 1 3 \dfrac {i \pi} 2}\)
\(\ds \) \(=\) \(\ds 2 \map \exp {\dfrac {i \pi} 6}\)


Then we have that the complex cube roots of unity are:

$1, \exp {\dfrac {2 i \pi} 3}, \exp {\dfrac {-2 i \pi} 3}$


Thus from Roots of Complex Number: Corollary:

\(\ds b\) \(=\) \(\ds \sqrt [3] 8 \map \exp {\dfrac 1 3 \dfrac {i \pi} 2}\)
\(\ds \) \(=\) \(\ds 2 \map \exp {\dfrac {i \pi} 6}\)
\(\ds \) \(=\) \(\ds 2 \paren {\cos \dfrac \pi 6 + i \sin \dfrac \pi 6}\)
\(\ds \) \(=\) \(\ds 2 \paren {\dfrac {\sqrt 3} 2 + i \frac 1 2}\) Cosine of $\dfrac \pi 6$, Sine of $\dfrac \pi 6$
\(\ds \) \(=\) \(\ds \sqrt 3 + i\)


\(\ds b \exp {\dfrac {2 i \pi} 3}\) \(=\) \(\ds 2 \map \exp {\dfrac {i \pi} 6} \exp {\dfrac {2 i \pi} 3}\)
\(\ds \) \(=\) \(\ds 2 \map \exp {\dfrac {i \pi} 6 + \dfrac {2 i \pi} 3}\)
\(\ds \) \(=\) \(\ds 2 \map \exp {\dfrac {5 i \pi} 6}\)
\(\ds \) \(=\) \(\ds 2 \paren {\cos \dfrac {5 \pi} 6 - i \sin \dfrac {5 \pi} 6}\)
\(\ds \) \(=\) \(\ds 2 \paren {-\dfrac {\sqrt 3} 2 + i \frac 1 2}\) Cosine of $\dfrac {5 \pi} 6$, Sine of $\dfrac {5 \pi} 6$
\(\ds \) \(=\) \(\ds -\sqrt 3 + i\)


\(\ds b \exp {\dfrac {-2 i \pi} 3}\) \(=\) \(\ds 2 \map \exp {\dfrac {i \pi} 6} \exp {\dfrac {-2 i \pi} 3}\)
\(\ds \) \(=\) \(\ds 2 \map \exp {\dfrac {i \pi} 6 - \dfrac {2 i \pi} 3}\)
\(\ds \) \(=\) \(\ds 2 \map \exp {\dfrac {-i \pi} 2}\)
\(\ds \) \(=\) \(\ds -2 i\)

$\blacksquare$


Sources

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