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

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 $x^6+12 x^4-8 x^3+48 x^2+96 x+80 = 0$

• Take $i$ to one side, cube it to eliminate radicals, then again take $i$ to one side and square it. That should eliminate everything. Jan 16, 2016 at 1:31
• 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.
– Vim
Jan 16, 2016 at 2:46
• 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 Sep 4, 2018 at 21:00
• @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. Sep 5, 2018 at 0:56

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.

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

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

We have

$a^3-4 = 0$

and

$b^2 +4 = 0$.