1) let $Z_0$ be a solution of $Z^{13}-13Z^{7}+7Z^{3}-3Z+1=0$, Is it true that $Z_0$'s conjugate is also a solution?
2) let $Z_0$ be a solution of $Z^{2}+iZ+2=0$, Is it true that $Z_0$'s conjugate is also a solution?
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More generally, if $f\in\mathbb{R}[x]$ is a polynomial and $f(a)=0$ for some $a\in\mathbb{C}$, then it is true that $f(\bar a)=0$ as well. You have to use that complex conjugation is an $\mathbb{R}$-linear operation (which is easy to check, really). Then, $$ f(\bar a) = \sum_{i=0}^n b_i \overline{a}^i = \overline{\sum_{i=0}^n b_i a^i} = \overline{f(a)}= 0.$$ |
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The main (first) question has been answered. We deal with the second. Let $z_0$ be a solution of the equation $z^2+iz+2=0$. Then $\overline{z_0}$ is a solution of the equation $z^2-iz+2=0$. This equation is different from the equation $z^2+iz+2=0$. So certainly we cannot conclude that $\overline{z_0}$ is a solution of the equation $z^2+iz+2=0$. But this does not show that $\overline{z_0}$ is not a solution of the equation $z^2+iz+2=0$. In fact, it isn't. There are various ways to see this. We use an idea which is more general than necessary. Note that the sum of the two roots of $z^2+iz+2$ is the negative of the coefficient of $z$. So the sum of the roots is $-i$. But $z_0+\overline{z_0}$ is real, so if $z_0$ is one root, $\overline{z_0}$ cannot be the other one. |
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The two solutions of your second equation are $i$ and $-2i$, and these are not conjugates. |
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