# Characterise polynomial roots as intersections of solution sets of its real and imaginary parts

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?

• Gauss, who first proved the fundamental theorem of algebra, used a similar method if I recall correctly. Nov 11, 2019 at 19:20
• See also The Fundamental Theorem of Algebra: A Visual Approach by Velleman.
– lhf
Nov 11, 2019 at 19:22

For large enough $$|z|$$, the highest-degree term dominates. If that term is $$c z^d$$, with $$c = k e^{i\phi}$$, note that $$\text{Re}(c (r e^{i\theta})^d) = 0$$ for $$\cos(\phi + d \theta) = 0$$ and $$\text{Im}(c (r e^{i\theta})^d = 0$$ for $$\sin(\phi + d \theta) = 0$$. Thus for large enough $$|z|$$ we alternate orange and blue curves around the circle of radius $$|z|$$. These must somehow link up inside the circle, but they can't do so without any orange and blue curves crossing, corresponding to roots of the polynomial.
• That tells us that the "generic" case is similar to the first figure in the OPs examples. But it does not allow us to say how many times orange and blue curves must cross inside the circle. This argument, combine with continuity statements, seems to be enough to prove that the $n$-th degree polynomial has at least $n$ roots, but not exactly $n$. Nov 11, 2019 at 19:22
• @MarkFischler Well, you certainly can't have more than $n$ roots, because $\prod_{j=1}^k (z - r_j)$ has degree $k$. Nov 11, 2019 at 20:14