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Show that the number $z=\sqrt[3]{4}-2i$ is algebraic, that is satisfied a polynomial equation with integer coefficients.

I thought I could use the Fundamental theorem of Algebra, but it seems to be false.

Is anyone is able to give me a hint how to solve this problem? According to Wolfram Alpha, the answer is $x^6+12 x^4-8 x^3+48 x^2+96 x+80 = 0$

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    $\begingroup$ Take $i$ to one side, cube it to eliminate radicals, then again take $i$ to one side and square it. That should eliminate everything. $\endgroup$
    – Shailesh
    Jan 16, 2016 at 1:31
  • $\begingroup$ Both $\mathbb Q(\sqrt[3]4,i)/\mathbb Q(i)$ and $\mathbb Q(i)/\mathbb Q$ are algebraic field extension, and the composition of algebraic field extensions is still algebraic. $\endgroup$
    – Vim
    Jan 16, 2016 at 2:46
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    $\begingroup$ Notice that $\root 3 \of 4$ is in the ring of algebraic integers of $\textbf{Q}(\root 3 \of 2)$, which is a domain of cubic integers (that is, degree $3$). Specifically, $\root 3 \of 4 = (\root 3 \of 2)^2$. But this is a domain of real numbers only. $2i$, on the other hand, comes from $\textbf{Z}[i]$, a ring of degree $2$ containing complex numbers. thus it makes sense that 6 is the $\endgroup$
    – David R.
    Sep 4, 2018 at 21:00
  • $\begingroup$ @DavidR. It makes sense that 6 is the highest exponent in the polynomial Wolfram Alpha gave? I'm only asking because you seem to have posted your comment without completing what you were trying to say. $\endgroup$ Sep 5, 2018 at 0:56

3 Answers 3

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Here's a hint. Start with $$x = \sqrt[3]{4}-2i$$ rearrange and cube both sides $$ (x+2i)^3 = 4 $$ From there you can expand the polynomial and do some similar steps to eliminate the $i$'s.

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$$(z+2i)^{3}=4$$ $$i(6z^{2}-8)=4+12z-z^{3}$$ $$[i(6z^{2}-8)]^{2}=-36z^{4}+96z^{2}-64=[4+12z-z^{3}]^{2}=16+144z^{2}+z^{6}+2(48z-4z^{3}-12z^{4})$$ We have $$z^{6}+(36-24)z^{4}-8z^{3}+(144-96)z^{2}+96z+16+64=0$$ So $$z^{6}+12z^{4}-8z^{3}+48z^{2}+96z+80=0$$

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If $a$ and $b$ are algebraic, then $a+b$ is algebraic.

We have

$a^3-4 = 0$

and

$b^2 +4 = 0$.

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