Roots of Complex Number/Examples/5th Roots of -32

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Example of Roots of Complex Number

Let $z^5 = -32$.

Then:

$z = 2 \paren {\map \cos {\dfrac {\pi + 2 k \pi} 5} + i \, \map \sin {\dfrac {\pi + 2 k \pi} 5} }$

for $k = 0, 1, 2, 3, 4$.


That is:

\(\ds k = 0: \ \ \) \(\ds z = z_1\) \(=\) \(\ds 2 \paren {\cos \dfrac \pi 5 + i \sin \dfrac \pi 5}\)
\(\ds k = 1: \ \ \) \(\ds z = z_2\) \(=\) \(\ds 2 \paren {\cos \dfrac {3 \pi} 5 + i \sin \dfrac {3 \pi} 5}\)
\(\ds k = 2: \ \ \) \(\ds z = z_3\) \(=\) \(\ds 2 \paren {\cos \dfrac {5 \pi} 5 + i \sin \dfrac {5 \pi} 5} = -2\)
\(\ds k = 3: \ \ \) \(\ds z = z_4\) \(=\) \(\ds 2 \paren {\cos \dfrac {7 \pi} 5 + i \sin \dfrac {7 \pi} 5}\)
\(\ds k = 4: \ \ \) \(\ds z = z_5\) \(=\) \(\ds 2 \paren {\cos \dfrac {9 \pi} 5 + i \sin \dfrac {9 \pi} 5}\)


Proof

Complex 5th Roots of -32.png

In polar form:

$-32 = \polar {32, \pi + 2 k \pi}$


Let $z = r \cis \theta$.

Then:

\(\ds z^5\) \(=\) \(\ds r^5 \cis 5 \theta\) De Moivre's Theorem
\(\ds \) \(=\) \(\ds 32 \, \map \cis {\pi + 2 k \pi}\)
\(\ds \leadsto \ \ \) \(\ds r^5\) \(=\) \(\ds 32\)
\(\ds 5 \theta\) \(=\) \(\ds \pi + 2 k \pi\)
\(\ds \leadsto \ \ \) \(\ds r\) \(=\) \(\ds 2\)
\(\ds \theta\) \(=\) \(\ds \dfrac {\pi + 2 k \pi} 5\)

$\blacksquare$


Sources

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