Roots of Complex Number/Examples/6th Roots of 64
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Example of Roots of Complex Number
The complex $6$th roots of $64$ are given by:
- $\paren {64}^{1/6} = \set {2 \, \map \cis {60 k} \degrees}$
for $k = 0, 1, 2, 3, 4, 5$.
That is:
\(\ds k = 0: \ \ \) | \(\ds z = z_1\) | \(=\) | \(\ds 2\) | |||||||||||
\(\ds k = 1: \ \ \) | \(\ds z = z_2\) | \(=\) | \(\ds 2 \cis 60 \degrees = 1 + \sqrt 3 i\) | |||||||||||
\(\ds k = 2: \ \ \) | \(\ds z = z_3\) | \(=\) | \(\ds 2 \cis 120 \degrees = -1 + \sqrt 3 i\) | |||||||||||
\(\ds k = 3: \ \ \) | \(\ds z = z_4\) | \(=\) | \(\ds -2\) | |||||||||||
\(\ds k = 4: \ \ \) | \(\ds z = z_5\) | \(=\) | \(\ds 2 \cis 240 \degrees = -1 - \sqrt 3 i\) | |||||||||||
\(\ds k = 5: \ \ \) | \(\ds z = z_6\) | \(=\) | \(\ds 2 \cis 300 \degrees = 1 - \sqrt 3 i\) |
Proof
Let $z^6 = 64$.
We have that:
- $z^6 = 64 \, \map \cis {0 + 2 k \pi}$
Let $z = r \cis \theta$.
Then:
\(\ds z^6\) | \(=\) | \(\ds r^6 \cis 6 \theta\) | De Moivre's Theorem | |||||||||||
\(\ds \) | \(=\) | \(\ds 64 \, \map \cis {0 + 2 k \pi}\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds r^6\) | \(=\) | \(\ds 64\) | |||||||||||
\(\ds 6 \theta\) | \(=\) | \(\ds 2 k \pi\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds r\) | \(=\) | \(\ds 64^{1/6}\) | |||||||||||
\(\ds \) | \(=\) | \(\ds 2\) | ||||||||||||
\(\ds \theta\) | \(=\) | \(\ds \dfrac {k \pi} 3\) | for $k = 0, 1, 2, 3, 4, 5$ | |||||||||||
\(\ds \theta\) | \(=\) | \(\ds k \times 60 \degrees\) | for $k = 0, 1, 2, 3, 4, 5$ |
When $k = 3$ we have:
- $z_4 = 2 \cis 180 \degrees = - 2$
$\blacksquare$
Sources
- 1981: Murray R. Spiegel: Theory and Problems of Complex Variables (SI ed.) ... (previous) ... (next): $1$: Complex Numbers: Supplementary Problems: Roots of Complex Numbers: $95 \ \text{(e)}$