Algebraic geometry interpretation of deducing $\Re(z)= -1$ from $(z+2)^6=z^6$ This question asks

[I]f $z$ is a solution of the equation 
  $$(z+2)^6=z^6$$
  then we must have $\Re (z)=-1$.

How can I translate the geometry of that question into algebraic geometry, and break it down into a polynomial equation system? (Or, if I am mistaken and this is not possible, why doesn't it work?)
The reason which makes me think this works is that :


*

*$Re(z) = k$ for a complex $z$ is a line in the plane. A line in the plane is also a linear equation ( multivariate polynomial of order 1  ). Using convention $z = a+bi$ $$1\cdot a + 0\cdot b  +1=0$$

*the fact that a complex polynomial we can split into a separate polynomial for real and imaginary parts.
 A: This can be formulated as a system of polynomial equations, but this doesn't seem helpful. 
First of all expanding $(z+2)^6=z^6$ yields
$$12z^5+60z^4+160z^3+240z^2+192z+64=0,$$
and next writing $z=x+yi$ with $x,y\in\Bbb{R}$ yields the equation
\begin{eqnarray*}
0&=&12(x^5+5x^4yi-10x^3y^2-10x^2y^3i+5xy^4+y^5i)\\
&+&60(x^4+4x^3yi-6x^2y^2-4xy^3i+y^4)\\
&+&160(x^3+3x^2yi-3xy^2-y^3i)\\
&+&240(x^2+2xyi-y^2)\\
&+&192(x+yi)\\
&+&64.
\end{eqnarray*}
Collecting terms yields the two equations
\begin{eqnarray*}
(12x^5-120x^3y^2+60xy^4)+(60x^4-360x^2y^2+60y^4)+(160x^3-480xy^2)+(240x^2-240y^2)+192x+64=0,\\
(60x^4y-120x^2y^3+12y^5)+(240x^3y-240xy^3)+(480x^2y-160y^3)+480xy+192y=0.
\end{eqnarray*}
This shows the solution set is the intersection of a quintic and a quartic surface. The corresponding geometric picture isn't particularly illuminating, but as noted in the comments, the quartic polynomial is a multiple of $y$. Setting $y=0$ takes us back to our original quintic, but now in the real variable $x$ in stead of the complex variable $z$.

What makes the answer work, is the observation that any solution $z\in\Bbb{C}$ must satisfy $|z+2|=|z|$. This is equivalent to solving
$$(z+2)(\overline{z+2})=z\overline{z},$$
and here writing $z=x+yi$ with $x,y\in\Bbb{R}$ yields the equation
$$x^2+4x+4+y^2=x^2+y^2,$$
from which it easily follows that $x=-1$. From here it is not hard to find the appropriate values of $y$.
