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

Question

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 :

1. $Re(z) = k$ 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 $z = a+bi$ $$1\cdot a + 0\cdot b +1=0$$

2. the fact that a complex polynomial we can split into a separate polynomial for real and imaginary parts.

Answer 1

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 $z$.

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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 $y$.