Roots of Complex Number/Examples/5th Roots of -16 + 16 root 3 i
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Example of Roots of Complex Number
The complex $5$th roots of $-16 + 16 \sqrt 3 i$ are given by:
- $\paren {-16 + 16 \sqrt 3 i}^{1/5} = \set {2 \, \map \cis {24 + 72 k} \degrees}$
for $k = 0, 1, 2, 3, 4$.
That is:
\(\ds k = 0: \ \ \) | \(\ds z = z_1\) | \(=\) | \(\ds \sqrt 2 \cis 24 \degrees\) | |||||||||||
\(\ds k = 1: \ \ \) | \(\ds z = z_2\) | \(=\) | \(\ds \sqrt 2 \cis 96 \degrees\) | |||||||||||
\(\ds k = 2: \ \ \) | \(\ds z = z_3\) | \(=\) | \(\ds \sqrt 2 \cis 168 \degrees\) | |||||||||||
\(\ds k = 3: \ \ \) | \(\ds z = z_4\) | \(=\) | \(\ds \sqrt 2 \cis 240 \degrees\) | |||||||||||
\(\ds k = 4: \ \ \) | \(\ds z = z_5\) | \(=\) | \(\ds \sqrt 2 \cis 321 \degrees\) |
Proof
Let $z^5 = -16 + 16 \sqrt 3 i$.
We have that:
- $z^5 = 16 \sqrt 2 \, \map \cis {\dfrac {2 \pi} 3 + 2 k \pi} = 16 \sqrt 2 \, \map \cis {120 \degrees + k \times 360 \degrees}$
Let $z = r \cis \theta$.
Then:
\(\ds z^5\) | \(=\) | \(\ds r^5 \cis 5 \theta\) | De Moivre's Theorem | |||||||||||
\(\ds \) | \(=\) | \(\ds 16 \sqrt 2 \, \map \cis {\dfrac {2 \pi} 3 + 2 k \pi}\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds r^5\) | \(=\) | \(\ds 16 \sqrt 2\) | |||||||||||
\(\ds 5 \theta\) | \(=\) | \(\ds \dfrac {2 \pi} 3 + 2 k \pi\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds r\) | \(=\) | \(\ds \paren {16 \sqrt 2}^{1/5}\) | |||||||||||
\(\ds \) | \(=\) | \(\ds 2\) | ||||||||||||
\(\ds \theta\) | \(=\) | \(\ds \dfrac {2 \pi} {15} + \dfrac {2 k \pi} 5\) | for $k = 0, 1, 2, 3, 4$ | |||||||||||
\(\ds \) | \(=\) | \(\ds 24 \degrees + 72 k \degrees\) | for $k = 0, 1, 2, 3, 4$ |
$\blacksquare$
Sources
- 1981: Murray R. Spiegel: Theory and Problems of Complex Variables (SI ed.) ... (previous) ... (next): $1$: Complex Numbers: Supplementary Problems: Roots of Complex Numbers: $96 \ \text{(c)}$