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Suppose $p(z)$ is a polynomial with real coefficients. Prove that
(a) $\overline{p(z)}=p(\bar{z})$;
(b) $p(z)=0$ if and only if $p(\bar{z})=0$;
(c) the roots of $p(z)=0$ appear in conjugate pairs, i.e., if $z_{0}$ is a root of $p(z)=0$, so is $\bar{z}_{0}$.
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Proof. Let $p(z)=a_{0}+a_{1} z+\ldots+a_{n} z^{n}$ for $a_{0}, a_{1}, \ldots, a_{n} \in \mathbb{R} .$ Then
\begin{aligned} \overline{p(z)} &=\overline{a_{0}+a_{1} z+\cdots+a_{n} z^{n}} \\ &=\overline{a_{0}}+\overline{a_{1} z}+\cdots+\overline{a_{n} z^{n}} \\ &=\overline{a_{0}}+\left(\overline{a_{1}}\right) \bar{z}+\cdots+\left(\overline{a_{n}}\right) \overline{z^{n}} \\ &=a_{0}+a_{1} \bar{z}+\cdots+a_{n} \bar{z}^{n}=p(\bar{z}) . \end{aligned}
If $p(z)=0$, then $\overline{p(z)}=0$ and hence $p(\bar{z})=\overline{p(z)}=0 ;$ on the other hand, if $p(\bar{z})=0$, then $\overline{p(z)}=p(\bar{z})=0$ and hence $p(z)=0$ By the above, $p\left(z_{0}\right)=0$ if and only if $p\left(\bar{z}_{0}\right)=0 .$ Therefore, $z_{0}$ is a root of $p(z)=0$ if and only if $\bar{z}_{0}$ is.
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