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
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
- 1981: Murray R. Spiegel: Theory and Problems of Complex Variables (SI ed.) ... (previous) ... (next): $1$: Complex Numbers: Supplementary Problems: Roots of Complex Numbers: $95 \ \text{(f)}$