Polynomial equation with complex numbers I'm studying linear algebra and there are a few problems from the book in the section on complex numbers I'm having trouble with. I've solved most of the questions in this chapter - I need a little help with these last five. Here's one of them - there are two statements and I have to answer true or false for each one.


*

*If $z_0$ is a solution for the equation $z^{13}-13z^7+7z^3-3z+1=0$ then $\overline{z}_0$ is also a solution for it.

*if $z_0$ is a solution for the equation $z^2+iz+2=0$ then $\overline{z}_0$ is also a solution for it.
Can anyone help me out?

For 1 just apply conjugation.
For 2, one has to argue differently: suppose $z_0$ and $\overline{z_0}$ are both solutions; then
$$
\begin{cases}
z_0^2+iz_0+2=0\\
\overline{z_0}^2+i\overline{z_0}+2=0
\end{cases}
$$
Subtract the two equations, getting
$$
(z_0-\overline{z_0})(z_0+\overline{z_0})+i(z_0-\overline{z_0})=0
$$
which implies
$$
z_0-\overline{z_0}=0
\qquad\text{or}\qquad
z_0+\overline{z_0}+i=0
$$
The first case gives $z_0=\overline{z_0}$, but this is impossible, because it would entail $z_0$ is real but then $z_0^2+2=-iz_0$ is a contradiction: the left-hand side is real, the right-hand side is purely imaginary.
The second case gives, by conjugation, also
$$
\overline{z_0}+z_0-i=0
$$
and so $i=-i$, which is again a contradiction.
Note that an equation with complex coefficient can have both $z_0$ and $\overline{z_0}$ as roots, provided this is not required for all roots: consider
$$
(z-i)(z+i)(z-2i)=0
$$
that has roots $i$, $-i$ and $2i$, where the first two roots are conjugate of each other.

If all complex (and non real) roots come in pair, that is, for all roots $z_0$ also $\overline{z_0}$ is a root, then the equation has real coefficients, as soon as one of them is real.
