2
$\begingroup$

Let $P(z)=a_nz^n+a_{n-1}z^{n-1}+\cdots+a_1z+a_0$ be a polynomial of degree n

Suppose not all of $a_n,a_{n-1},\ldots,a_1,a_0$ are real. Show that $P$ has at least one root whose complex conjugate is not a root.

This problem suggests me to prove the contrapositive. Any advice?

(What I understand for contrapositive: "If the roots of $P$ are complex conjugates, then all of $a_n,a_{n-1},\ldots,a_1,a_0$ are real.")

If so, I believe I should prove that $(z-z_1)(z-\bar {z_1})$ is a polynomial with real coefficients (I know how to do this) and that multiplication of polynomials with real coefficients is a polynomial with real coefficients.

Correct me if I'm wrong.

$\endgroup$
  • $\begingroup$ How about $P(z)=iz-i$? $\endgroup$ – Lord Shark the Unknown Jun 11 '17 at 4:56
  • $\begingroup$ For that matter, let $n=0$ and $P(z)=i$. Then $P$ does not have any roots at all. $\endgroup$ – Chris Culter Jun 11 '17 at 5:10
  • 2
    $\begingroup$ @L. Salvetti: Your proof works if $P$ is monic and has degree at least $1$, or more generally, again assuming degree at least $1$, if at least one coefficient of $P$ is real and nonzero. $\endgroup$ – quasi Jun 11 '17 at 5:11
  • $\begingroup$ @quasi There should be something about "counting multiplicities" as well, otherwise $(z+i)(z-i)^2=z^3 - i z^2 + z - i$ satisfies the rest of conditions, but the only roots are $\pm i\,$. $\endgroup$ – dxiv Jun 11 '17 at 5:19
  • $\begingroup$ @dxiv: Yes, I missed that. So a corrected statement might be something like this: If $P$ has degree at least one, and at least one nonzero real coefficient, then if some of the non-real roots of $P$ are grouped in complex conjugate pairs, at least one non-real root is not paired. Ugh! $\endgroup$ – quasi Jun 11 '17 at 5:33
0
$\begingroup$

As pointed out in the comments you have an issue with multiplying by scalars. A way to remedy this is to assume your polynomial is monic, that is $a_n = 1$. Then the statement would be:

A monic polynomial which has a nonreal coefficient has nonreal roots which are not able to be partitioned into conjugate pairs.

You were correct to point out that the contrapositive is

A monic polynomial whose complex roots can all be partitioned into conjugate pairs has all real coefficients

Given any monic polynomial $p$ we can represent it $p = \prod (x - z_i)$. Then splitting into real roots and complex conjugates we have $p = \prod (x - a_i) \prod (x - z_i) (x - \bar{z_i})$. It is an easy calculation to check that $(x- z_i)(x-\bar{z_i})$ has all real coefficients for any complex $z_i$.

$\endgroup$

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.