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:

enter image description here enter image description here

enter image description here enter image description here interactive version

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 $b = 1,2,3,4$

enter image description here enter image description here

enter image description here enter image description here interactive version

  • $\begingroup$ 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. $\endgroup$ – Joppy Oct 5 '18 at 8:53
  • $\begingroup$ Ok then, but there's a restricted version for real coefficients which might be proved independently. $\endgroup$ – Hans-Peter Stricker Oct 5 '18 at 8:54
  • $\begingroup$ @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 $g$. $\endgroup$ – lhf Oct 5 '18 at 13:11
  • $\begingroup$ Does this make Joppy's comment obsolete? $\endgroup$ – Hans-Peter Stricker Oct 5 '18 at 13:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.