Roots of Complex Number/Examples/Square of Cube Roots of i

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Example of Roots of Complex Number: Corollary

The squares of the complex cube roots of $i$ are given by:

$\paren i^{2/3} = \set {\map \cis {60 + 120 k} \degrees}$

for $k = 0, 1, 2$.


That is:

\(\ds k = 0: \ \ \) \(\ds z = z_1\) \(=\) \(\ds \cis 60 \degrees = \dfrac 1 2 + \dfrac {\sqrt 3} 2 i\)
\(\ds k = 1: \ \ \) \(\ds z = z_2\) \(=\) \(\ds \cis 180 \degrees = -1\)
\(\ds k = 2: \ \ \) \(\ds z = z_3\) \(=\) \(\ds \cis 300 \degrees = \dfrac 1 2 - \dfrac {\sqrt 3} 2 i\)


Proof

Square of Complex Cube Roots of i.png


Let $z^3 = i$.

We have that:

$z^3 = \map \cis {90 + 2 k \pi}$


Let $z = r \cis \theta$.

Then:

\(\ds z^3\) \(=\) \(\ds r^3 \cis 3 \theta\) De Moivre's Theorem
\(\ds \) \(=\) \(\ds \map \cis {\dfrac \pi 2 + 2 k \pi}\)
\(\ds \leadsto \ \ \) \(\ds r^3\) \(=\) \(\ds 1\)
\(\ds 3 \theta\) \(=\) \(\ds \dfrac \pi 2 + 2 k \pi\)
\(\ds \leadsto \ \ \) \(\ds r\) \(=\) \(\ds 1\)
\(\ds \theta\) \(=\) \(\ds \dfrac \pi 6 + \dfrac {2 k \pi} 3\) for $k = 0, 1, 2$
\(\ds \) \(=\) \(\ds 30 \degrees + k \times 120 \degrees\) for $k = 0, 1, 2$


Hence:

\(\ds z^2\) \(=\) \(\ds r^2 \cis 2 \theta\) De Moivre's Theorem
\(\ds \leadsto \ \ \) \(\ds r\) \(=\) \(\ds 1\)
\(\ds \theta\) \(=\) \(\ds \dfrac {2 \pi} 6 + \dfrac {4 k \pi} 3\) for $k = 0, 1, 2$
\(\ds \) \(=\) \(\ds 60 \degrees + k \times 240 \degrees\) for $k = 0, 1, 2$

$\blacksquare$


Sources