Conjugate root theorem confusion For z^3 = 1, the conjugate root theorem is applicable here.
However for z^3 = -i, the conjugate root theorem is not applicable here.
I'm confused on why the -i is considered a non-real coefficient? Isn't it a constant, like 1 in the first line of my post?
I know that it can be expressed as i*z^0 but in maths, don't we consider numbers that aren't multiplied to a variable as a constant term?
For example for y = 2x + 1, we would call "1" as a constant term, not x^0 ha a coefficient of 1. 
Thanks
 A: Nope, we do say that a constant term is a coefficient of $z^0$.  When talking about the coefficients of a polynomial, we include the constant term as a coefficient.
You can think about the proof of the conjugate root theorem to understand why this is necessary.  The proof works by showing that if $f(z)$ is a polynomial with real coefficients, then $f(\overline{z})=\overline{f(z)}$ for all $z$.  In particular, if $f(z)=0$, this says $f(\overline{z})=0$ as well.
Now to prove that $f(\overline{z})=\overline{f(z)}$, you expand out $\overline{f(z)}$ and find that the conjugates ignore all the coefficients (since they're real) and leave you with just $f(\overline{z})$.  To get the same constant term as $f(\overline{z})$, you need the constant term to be real too!
Here's an example of how this looks in practice.  Say $f(z)=z^3+i$.  Then $$\overline{f(z)}=\overline{z^3+i}=\overline{z^3}+\overline{i}=\overline{z}^3-i.$$  But that's not the same as $f(\overline{z})=\overline{z}^3+i$, because the constant term got conjugated!  You need the constant term to be real for it to not change.
