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.
 A: Any polynomial with real coefficients with root $r = a+ib$ also has root $\bar{r} = a-ib$. This is known as the conjugate root theorem.
A: 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: $p(z)=0\implies p^*(z)=p(z^*)=0$.
A: Other answers have pointed you to the conjugate root theorem. It's a very good theorem to know and answers your question immediately. In case you didn't know it, we can approach this specific case as follows:
Divide your polynomial into odd and even degrees: $$g(z) = (x^3 + x) + (-3x^2 -3)$$
Let $g_o(z) = x^3 + x$ and $g_e(z) = -3x^2-3$. Note that $g_o(i)$ is completely imaginary and $g_e(i)$ is real, so if $i$ is a root of $g$, $i$ is a root of both $g_o$ and $g_e$.
Since $(-i)^n = i^n$ if $n$ is even and $(-i)^n = -(i^n)$ if $n$ is odd, we have $g_o(-i) = -g_o(i) = 0$ and $g_e(-i) = g_e(i) = 0$. Thus $g(-i) = g_o(-i) + g_e(-i) = 0$, i.e., $-i$ is a root of $g$
A: There is a rule that if one root is imaginary or irrational then the other root is its conjugate. It's known as Conjugate root theorem. For example if $3+4i $ is one root then $3-4i $ is a root as well. Same goes for $i$ and $- i$. This only applies to polynomials with real coefficients.
