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  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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I see that $i$ is one solution of the equation in 2)... – David Mitra Dec 12 '11 at 18:59
Hint: $z^2 + i z + 2 = (z+2i)(z-i)$. – JavaMan Dec 12 '11 at 19:02
up vote 1 down vote accepted

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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but, is it the same for the second one where one of the numbers is $i$? – Some1 Dec 12 '11 at 18:47
@Some1, why did you accept a solution which, according to your own comment, does not fully answer your question(s)? – Did Dec 12 '11 at 19:49
@Some1: The point to my statement is that $f$ has real coefficients, but the second polynomial does not satisfy this condition. See I.J.Kennedy's answer. – Jesko Hüttenhain Dec 12 '11 at 22:33

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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That, or the fact that if $z$ solves $z^2+iz+2=0$ and $z^2-iz+2=0$, then $2iz=0$. Since $z=0$ does not solve $z^2+iz+2=0$, this is a contradiction. – Did Dec 12 '11 at 19:51

The two solutions of your second equation are $i$ and $-2i$, and these are not conjugates.

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