Roots of Complex Number/Examples/Cube Roots of 2+11i
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Example of Roots of Complex Number
The complex cube roots of $2 + 11 i$ are given by:
- $\paren {2 + 11 i}^{1/3} = \set {2 + i, -1 + \dfrac {\sqrt 3} 2 + i \paren {-\dfrac 1 2 - \sqrt 3 }, -1 - \dfrac {\sqrt 3} 2 + i \paren {-\dfrac 1 2 + \sqrt 3}}$
Proof
Let $z^3 = 2 + 11 i = \paren {p + iq}^3$.
Then:
| \(\ds \paren {p + iq}^3\) | \(=\) | \(\ds 2 + 11 i\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds p^3 + 3 i p^2 q - 3 p q^2 - i q^3\) | \(=\) | \(\ds 2 + 11 i\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds p^3 - 3 p q^2\) | \(=\) | \(\ds 2\) | |||||||||||
| \(\ds 3 p^2 q - q^3\) | \(=\) | \(\ds 11\) |
From this we have:
| \(\ds p^3 - 3 p q^2\) | \(=\) | \(\ds 2\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \frac {p^3} {q^3} - \frac {3 p} q\) | \(=\) | \(\ds \frac 2 {q^3}\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \frac {2 p^3} {q^3} - \frac {6 p} q\) | \(=\) | \(\ds \frac 4 {q^3}\) |
and:
| \(\ds 3 p^2 q - q^3\) | \(=\) | \(\ds 11\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \frac {3 p^2} {q^2} - 1\) | \(=\) | \(\ds \frac {11} {q^3}\) |
Thus:
| \(\ds \frac {2 p^3} {q^3} - \frac {6 p} q\) | \(=\) | \(\ds \frac 4 {q^3}\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \frac {2 p^3} {q^3} - \frac {6 p} q\) | \(=\) | \(\ds \dfrac 4 {11} \paren {\frac {3 p^2} {q^2} - 1 }\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds 2 \paren {p/q}^3 - \dfrac{12} {11} \paren {p/q}^2 - 6 \paren {p/q} + \dfrac 4 {11}\) | \(=\) | \(\ds 0\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds 11 \paren {p/q}^3 - 6 \paren {p/q}^2 - 33 \paren {p/q} + 2\) | \(=\) | \(\ds 0\) | multiplying by $\dfrac {11} 2$ |
Let $w = \dfrac p q$:
| \(\ds 11 w^3 - 6 w^2 - 33 w + 2\) | \(=\) | \(\ds 0\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \paren {w - 2} \paren {11w^2 + 16 w - 1}\) | \(=\) | \(\ds 0\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds w = \dfrac p q\) | \(=\) | \(\ds 2 \textrm { or } \dfrac {-8 \pm 5 \sqrt 3 } {11 }\) | Quadratic Formula on $11w^2 + 16 w - 1$ |
Putting $\dfrac p q = 2$ leads to::
- $p = 2q$
and hence:
| \(\ds 3 p^2 q - q^3\) | \(=\) | \(\ds 11\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds 12 q^3 - q^3\) | \(=\) | \(\ds 11\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds 11 q^3\) | \(=\) | \(\ds 11\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds q\) | \(=\) | \(\ds 1 \textrm { or } -\dfrac 1 2 \pm i \dfrac {\sqrt 3} 2\) | Cube Roots of Unity |
So this gives:
- $z = \begin {cases} 2 + i \\ -1 + \dfrac {\sqrt 3} 2 + i \paren {-\dfrac 1 2 - \sqrt 3 } \\ -1 - \dfrac {\sqrt 3} 2 + i \paren {-\dfrac 1 2 + \sqrt 3} \end{cases}$
$\blacksquare$