Product of the 5th roots of $(1+\sqrt{2}i)^3$
The following is a question that I'm not currently seeing the solution to:
Find the product of the 5th roots of $(1+\sqrt{2}i)^3.$
I considered literally finding the roots by converting either $1+\sqrt{2}i$ or $(1+\sqrt{2}i)^{-2}$ to mod-arg form, but neither argument of
$$1+\sqrt{2}i$$
nor of
$$(1+\sqrt{2}i)^{-2}=-\frac{1}{9}-\frac{2\sqrt{2}}{9}i$$
end up being 'nice', leading me to imagine I'm going down the wrong route. In the worked solutions book this question is skipped, and the answers show a simple
$$(1+\sqrt{2}i)^3 = -5+\sqrt{2}i$$
which doesn't even seem to answer what the question asked. I must be missing something very obvious?
$\endgroup$ 12 Answers
$\begingroup$Let $b$ be a fifth root of $a\ne0$. Then $b^5=a$, and the other fifth roots of $a$ are $b\zeta$, $b\zeta^2$, $b\zeta^3$, $b\zeta^4$ with $\zeta=\exp(2\pi i/5)$. The product of all the fifth roots of $a$ is $b^5\zeta^{10}=a$.
$\endgroup$ 0 $\begingroup$The meaning of the statement is:
Find the product of the 5th roots of $(1+\sqrt{2}i)^3$, that is, find the product of all $x$ satisfying the equation: $$x^5=(1+\sqrt 2 i)^3 \tag 1$$
Hence, using Vieta’s formulae, we can find that product of all $x$ satisfying $(1)$ is $$P = (1+\sqrt 2i)^3 = -5+\sqrt 2i$$
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