Roots of Complex Number/Examples/Cube Roots of -11-2i

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Example of Roots of Complex Number: Corollary

The complex cube roots of $-11 - 2 i$ are given by:

$\paren {-11 - 2 i}^{1/3} = \set {1 + 2 i, -\dfrac 1 2 + \sqrt 3 + \paren {-1 - \dfrac {\sqrt 3} 2} i, -\dfrac 1 2 - \sqrt 3 + \paren {-1 + \dfrac {\sqrt 3} 2} i}$


Proof

Let $z^3 = -11 - 2 i = \paren {p + iq}^3$.

Then:

\(\ds \paren {p + iq}^3\) \(=\) \(\ds -11 - 2 i\)
\(\ds \leadsto \ \ \) \(\ds p^3 + 3 i p^2 q - 3 p q^2 - i q^3\) \(=\) \(\ds -11 - 2 i\)
\(\ds \leadsto \ \ \) \(\ds p^3 - 3 p q^2\) \(=\) \(\ds -11\)
\(\ds 3 p^2 q - q^3\) \(=\) \(\ds -2\)


From this we have:

\(\ds p^3 - 3 p q^2\) \(=\) \(\ds -11\)
\(\ds \leadsto \ \ \) \(\ds \frac {p^3} {q^3} - \frac {3 p} q\) \(=\) \(\ds \frac {-11} {q^3}\)
\(\ds \leadsto \ \ \) \(\ds \frac {2 p^3} {q^3} - \frac {6 p} q\) \(=\) \(\ds -11 \frac 2 {q^3}\)

and:

\(\ds 3 p^2 q - q^3\) \(=\) \(\ds -2\)
\(\ds \leadsto \ \ \) \(\ds \frac {3 p^2} {q^2} - 1\) \(=\) \(\ds -\frac 2 {q^3}\)


Thus:

\(\ds \frac {2 p^3} {q^3} - \frac {6 p} q\) \(=\) \(\ds -11 \frac 2 {q^3}\)
\(\ds \leadsto \ \ \) \(\ds \frac {2 p^3} {q^3} - \frac {6 p} q\) \(=\) \(\ds \frac {33 p^2} {q^2} - 11\)
\(\ds \leadsto \ \ \) \(\ds 2 \paren {p/q}^3 - 33 \paren {p/q}^2 - 6 \paren {p/q} + 11\) \(=\) \(\ds 0\)


Let $w = \dfrac p q$:

\(\ds 2 w^3 - 33 w^2 - 6 w + 11\) \(=\) \(\ds 0\)
\(\ds \leadsto \ \ \) \(\ds \paren {2 w - 1} \paren {w^2 - 16 w - 11}\) \(=\) \(\ds 0\)
\(\ds \leadsto \ \ \) \(\ds w = \dfrac p q\) \(=\) \(\ds \dfrac 1 2 \textrm { or } 8 \pm 5 \sqrt 3\) Quadratic Formula on $w^2 - 16 w - 11$


Putting $\dfrac p q = \dfrac 1 2$ leads to::

$2 p = q$

and hence:

\(\ds 3 p^2 q - q^3\) \(=\) \(\ds -2\)
\(\ds \leadsto \ \ \) \(\ds 6 p^3 - 8 p^3\) \(=\) \(\ds -2\)
\(\ds \leadsto \ \ \) \(\ds p^3\) \(=\) \(\ds 1\)
\(\ds \leadsto \ \ \) \(\ds p\) \(=\) \(\ds 1 \textrm { or } -\dfrac 1 2 \pm i \dfrac {\sqrt 3} 2\)


So this gives:

$z = \begin {cases} 1 + 2i \\ -\dfrac 1 2 + \sqrt 3 + i \paren {-1 - \dfrac {\sqrt 3} 2} \\ -\dfrac 1 2 - \sqrt 3 + i \paren {-1 + \dfrac {\sqrt 3} 2} \end{cases}$

$\blacksquare$


Sources

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