Can we find the cubic roots of $-46+9i$ given that $(2+3i)^3=-46+9i $ is known?

Question

Is it possible to find the cubic roots of $-46+9i$ not using De Moivre's formula if we know the principal root $2+3i$?

Any hints or complete solutions are welcome!

Answer 1

If you multiply $2+3i$ by any third root of $1$, you will find another solution to the original equation.

If you're lucky, you already know by other means that the third roots of $1$ are $1$ and $-\frac12\pm\frac{\sqrt3}2i$.

This gives you three different cubic roots of $-46+9i$, which (since $\mathbb C$ is a field) must be all of them.

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Alternatively you could do it the slow way: You're trying to find roots of $z^3-(-46+9i)$, which is to say, factor it into linear factors. You know that $z-(2+3i)$ is a factor. Use polynomial division to remove it, and apply the quadratic formula to find the roots of the degree-2 quotient.

Answer 2

Sure.

$(ab)^3 = a^3b^3$ so $((2+3i)*z)^3= (2+3i)^3z^3 = (-46+ 9 i)*z^3 = -46+9i$ if $z^3 = 1$.

The three cube roots of $1$ are $1$ and $ -\frac 12 \pm \frac {\sqrt 3}2i$

and so the three cube roots of $-46 + 9i$ are

$2 + 3i$ and

$(2+ 3i)(-\frac 12 \pm \frac {\sqrt 3}2i)= (-1\mp \frac{3\sqrt 3}2) + (-\frac 32 \pm \sqrt 3)i$