Consider a polynomial equation $p(z)\equiv\sum_k \alpha_k z^k=0$ for $\alpha_k\in\mathbb C$.
We can always understand this equation as a system of two polynomial equations, given by real and imaginary parts of $p(z)=0$: $$ p(z)=0\Longleftrightarrow \begin{cases}\operatorname{Re}[p(z)]=0, \\ \operatorname{Im}[p(z)]=0.\end{cases} $$ As a trivial example, the solutions of $z^2-1=0$ are the intersections of the surfaces described by $x^2-y^2-1=0$ and $xy=0$.
We can therefore understand geometrically the solutions of $p(z)=0$ as intersections of two algebraic curves. For example, generating random polynomials of degree $4$ and plotting the curves corresponding to real and imaginary parts of each, we get curves like the following ones:
Code used to generate plots:
With[{exprs = Total[
RandomComplex[{-1 - I, 1 + I}, 5] z^Range[0, 4]
] /. {z -> x + I y} // Expand // ReIm // FullSimplify[#, {x, y} \[Element] Reals] &
},
ContourPlot[Evaluate@Thread[exprs == 0], {x, -5, 5}, {y, -5, 5},
PlotPoints -> 50, MaxRecursion -> 4, ImageSize -> 200
]
]
Where in each figure the blue line is the solution set of $\Re(p(z))=0$ and the orange line that of $\Im(p(z))=0$.
From these figures, we can clearly see that there are always $4$ intersections of blue and orange curves, consistently with the fundamental theorem of algebra.
Can anything be said about these curves from a purely geometrical point of view? Or more generally, can we prove the fundamental theorem of algebra by purely geometrical considerations on the types of pairs of algebraic curves that can be produced by a single complex polynomial?