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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.

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  • $\begingroup$ 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
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    $\begingroup$ @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
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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 $z_i$.

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