8
$\begingroup$

The Complex Conjugate Theorem states that, given the polynomial $p(x)$ with real coefficients $p \in \mathbb{R}[p]$ and one of its roots being $a+bi \in \mathbb{C}$, its complex-conjugate pair $\overline {z}$ must be a root as well. We could also intuitively conclude that only by multiplying $(a+bi)$ with $(a-bi)$ will the imaginary parts be destroyed, allowing $p$ to be a real polynomial.

Following this line of reasoning, it shouldn't be necessary for complex roots of a complex polynomial to come in pairs. However, this premise was used in solving one of my school assignments.

Do complex roots always have to come in pairs, regardless of the field in which the polynomial was defined?

$\endgroup$
2
  • 2
    $\begingroup$ No, not necessarily. You can always factor a polynomial over $\mathbb{C}$ into the product of distinct factors: $p(z)=a(z-z_1)\cdots(z-z_n)$. These roots need not be paired. $\endgroup$
    – MrMazgari
    Nov 22, 2015 at 8:57
  • 9
    $\begingroup$ The main words in the theorem are real coefficients $\endgroup$
    – Shailesh
    Nov 22, 2015 at 8:57

2 Answers 2

Reset to default
23
$\begingroup$

Consider the polynomial $z-i$ . . .

It's good to remember the reason the complex conjugate theorem is true in the first place: the map $z\mapsto \overline{z}$ is an automorphism of the field $\mathbb{C}$, and fixes the subfield $\mathbb{R}$ (this is a fancy way of saying $\overline{r}=r$ if $r$ is real). Thus - defining the "conjugate" $\overline{p}$ of a polynomial (with coefficients in $\mathbb{C}$) to be the polynomial whose coefficients are the conjugates of the corresponding coefficients of $p$ - we have the following:

  • If $p$ is any polynomial and $z$ is a root of $p$, then $\overline{z}$ is a root of $\overline{p}$.

  • If the coefficients of $p$ are from $\mathbb{R}$, then $\overline{p}=p$.

This second fact is no longer true for polynomials with coefficients from outside $\mathbb{R}$!

Note that this generalizes in a natural way to:

Suppose $F$ is any subfield of $\mathbb{C}$, $\varphi$ is a field automorphism of $\mathbb{C}$ which fixes $F$, and $p$ is a polynomial with coefficients from $F$. Then if $z$ is a root of $p$, so is $\varphi(z)$.

This is the (fine, "a") first step towards Galois theory . . .

$\endgroup$
1
  • 6
    $\begingroup$ By the way, lest anyone be too impressed with the "theorem" in my answer: try thinking of any other automorphism of $\mathbb{C}$, besides conjugation! It is consistent with ZF (=set theory without the axiom of choice) that there are in fact no other nontrivial automorphisms of $\mathbb{C}$, and even with the axiom of choice, the automorphisms which do exist are "pathological"; see math.stackexchange.com/questions/412010/…. However, we do care very much about the general question, "Understand the automorphisms of [big] which fix [small]." $\endgroup$
    – Noah Schweber
    Nov 22, 2015 at 9:16
9
$\begingroup$

$(z-i)(z-2i)$ has two complex roots, $i$ and $2i$, but their conjugates are not roots.

$\endgroup$
3
  • $\begingroup$ The field in which the polynomial was given was not specified. Does this mean $p \in \mathbb{R}[p]$ by default? $\endgroup$
    – 0lt
    Nov 22, 2015 at 8:59
  • 2
    $\begingroup$ @banananina I suspect that yes, in the assignment you mention, "coefficients in $\mathbb{R}$" was a default assumption (although one which perhaps ought to have been spelled out). $\endgroup$
    – Noah Schweber
    Nov 22, 2015 at 9:05
  • $\begingroup$ This polynomial doesn't even have real coefficients? $\endgroup$
    – Simply Beautiful Art
    Dec 30, 2015 at 2:20

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.