If $i$ is a root, then $-i$ is also a root? I have the following question:
Prove that i is a roof of the equation $g(z) = 0$, where $g(z) = z^3 - 3z^2 + z - 3$. Find the other roots of this equation.
The answer says without any working out:
Since i is a root, then -i is a root. (The answers are $[z - i], [z + i]$, and $[z-3]$)
Why is this?
I know that I can factor out [z-3] and then continue from there, however the answer doesn't do that. I also plotted a graph:


However, that doesn't show the roots at i and -i.
Basically, can I assume that for any polynomial with i as a root, -i is also a root? And why?
Thank you.

You know that $$i^3-3i^2+i-3=0.$$
By taking the conjugate of this identity, 
$$(i^3-3i^2+i-3)^*=0^*$$
so that 
$${i^*}^3-3{i^*}^2+{i^*}-3=0.$$
And $i^*=-i$ is also a root.
This works with any complex root of a real polynomial: $p(z)=0\implies p^*(z)=p(z^*)=0$.
