# Symmetries in the equations and graphs of complex-valued polynomials

I tried to visualize the complex roots of a polynomial with real coefficients $$a_i \in \mathbb{R}$$:

$$f(z) = z^n + a_{n-1}z^{n-1} + \dots + a_1z + a_0$$

following some obvious thoughts:

1. For any complex function $$f(z) = 0$$ just means $$\operatorname{Re}f(z) = 0$$ and $$\operatorname{Im}f(z) = 0$$.

2. $$\operatorname{Re}f(z)$$ and $$\operatorname{Im}f(z)$$ can both be considered as functions $$F(u,v)$$, $$G(u,v)$$ of two real variables (with $$z = u + iv$$), thus defining two two-dimensional surfaces "over" the plane $$\mathbb{R}^2$$.

3. $$F(u,v) = 0$$ and $$G(u,v) = 0$$ define the intersections of these surfaces with the plane $$\mathbb{R}^2$$, which for polynomials are two different one-dimensional objects, i.e. unions of some curves and possibly isolated points ("generalized curves").

4. The roots of $$f$$, i.e. the numbers with $$f(z) = 0$$, are exactly the intersections of these intersections: the points (= numbers) with $$\operatorname{Re}f(z) = F(u,v) = 0$$ and $$\operatorname{Im}f(z) = G(u,v) = 0.$$

For polynomials the roots are isolated points: at least 1 and at most $$n$$ of them (for $$n$$ the degree of $$f$$). That's the essence of the fundamental theorem of algebra.

For $$f(z) = z^3 + a_2z^2 + a_1z + a_0$$ and $$a_0 = a_2 = 1$$ and $$a_1 = 1,2,3,4$$ the curves $$F(u,v)=0$$ (red) and $$G(u,v)=0$$ (blue) look like this:

Note, that the generalized curve $$\operatorname{Re}f(z) = F(u,v) = 0$$ (red) plays the role of the graph of $$f(x)$$ for real arguments, while the generalized curve $$\operatorname{Im}f(z) = G(u,v) = 0$$ (blue) plays the role of the real axis $$y = 0$$: the (real resp. complex) roots are their intersections. (Note, that the real axis $$v=0$$ is contained in $$G(u,v) = 0$$ for all polynomials.)

Determining $$F(u,v)$$ and $$G(u,v)$$ for degree $$n=2,3,4$$ yields

n=2: $$f(z) = z^2 + a_1z +a_0$$
$$F(u,v) = a_1u + (u^2-v^2) + a_0$$
$$G(u,v) = a_1v + (uv +uv)$$
$$= v(a_1 + 2u)$$

n=3: $$f(z) = z^3 + a_2z^2 + a_1z +c$$
$$F(u,v) = a_1u + a_2(u^2-v^2) + (u^3 -3uv^2) + a_0$$
$$G(u,v) = a_1v + a_2(uv + vu) -(v^3 -3u^2v)$$
$$= v(a_1 + 2a_2u - (v^2 - 3u^2))$$

n=4: $$f(z) = z^4 + a_3z^3 + a_2z^2 + a_1z +c$$
$$F(u,v) = a_1u + a_2(u^2-v^2) + a_3(u^3 -3uv^2) + (u^4 + v^4) + a_0$$
$$G(u,v) = a_1v + a_2(uv + vu) -a_3(v^3 -3u^2v) + 4(u^3v - uv^3)$$
$$= v(a_1 + 2a_2u - a_3(v^2 - 3u^2) + 4(u^3 - uv^2))$$

which shows some "formulaic symmetry" between $$F(u,v)$$ and $$G(u,v)$$. It also reveals that

• $$G(u,0) = 0$$

• $$F(u,v) = F(u,-v)$$

• $$G(u,v) = G(u,-v)$$

which is the reason that complex roots always come in conjugate pairs.

My questions are:

What are the general formulas for $$F(u,v)$$ and $$G(u,v)$$ for arbitrary degree $$n$$? (I didn't manage to write it down.)

Does possibly this relation between $$F(u,v)$$ and $$G(u,v)$$ lie at the heart of the fundamental theorem of algebra: Because $$F(u,v)$$ and $$G(u,v)$$ are related like this, each polynomial has 1 to $$n$$ roots?

How are these symmetries related to the symmetries between the roots of $$f(z)$$ investigated in Galois theory and to the fact that the coefficients $$a_i$$ are symmetric functions of the roots $$z_i$$?

For the sake of comparison this is how $$F(u,v) = 0$$ and $$G(u,v)=0$$ look like for $$f(z) = (z-a)(z-b)(z-c)$$ with $$a = 1, c = 2$$ and $$b = 1,2,3,4$$

• A random comment: the fundamental theorem of algebra also deals with polynomials with complex coefficients, which do not have the symmetries $G(u, v) = G(u, -v)$ and so on. – Joppy Oct 5 '18 at 8:53
• Ok then, but there's a restricted version for real coefficients which might be proved independently. – Hans-Peter Stricker Oct 5 '18 at 8:54
• @HansStricker, the FTA for real polynomials is equivalent to the FTA for complex polynomials because $g\bar g$ is a real polynomial with the same zeros as $g$. – lhf Oct 5 '18 at 13:11
• Does this make Joppy's comment obsolete? – Hans-Peter Stricker Oct 5 '18 at 13:30