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Let $F={P\over Q} $ be a fraction with $deg(P)< deg(Q)$ and such that $P$ and $Q$ are polynomials with real coefficients. Suppose $Q$ has the form $Q=(X-a)(X-z)(X-\bar z)$ with $a\in \mathbb R$ and $z$ is a non real root of $Q$ and $\bar z$ is the conjugate of $z$. The partial fraction decomposition of $F$ has the form $$F={s\over X-a }+{r\over X-z}+{t\over X-\bar z}$$ where $s$ is a real and $r, t$ are complex numbers. I would like to say that $r$ and $t$ must be conjugate but i don't see how to prove it. Thank you for your help!

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    $\begingroup$ What happens if you conjugate $F$? $\endgroup$ – Daniel Fischer Dec 3 '15 at 16:16
  • $\begingroup$ I see what you mean, and hence i see that the hypothesis of $P$ and $Q$ being both with real coefficients is necessary (not only $Q$) so that to be able to write $\bar F=F$. $\endgroup$ – palio Dec 3 '15 at 17:12
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Because the sum of the second and third term must be a rational function with real coefficients. Write that sum with the real and imaginary parts of r, t and z.

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