simplify complex polynomial $p(t) \in \mathbb{C}[t]$ How to simplify the following polynomial?
$$
\begin{align}

(t - \sqrt{3} \; e^{ \frac{\pi}{3} i }) (t - \sqrt{3} \; e^{ -\frac{\pi}{3} i }) &= t^2 - \sqrt{3} \; e^{ \frac{\pi}{3} i } \; t - \sqrt{3} \; e^{ -\frac{\pi}{3} i } \; t  + 3\\

&= t^2 - \sqrt{3} \; t \; ( e^{ \frac{\pi}{3} i } + e^{ -\frac{\pi}{3} i } ) + 3

\end{align}
$$
I know the result is
$$
t^2 - \sqrt{3} \; t + 3
$$
but I can't see how to simplify the two complex numbers.

Is it a general rule, that if I've got a polynomial with $(t-c)(t-\bar c)$ where $c,\bar c \in \mathbb{C}$ and $\bar c$ is the complex complement of $c$, the resulting polnomial is
$$(t-c)(t-\bar{c})=t^2-ct-\bar{c}t+c\bar{c}=t^2-(c+\bar{c})t+|c|^2,$$

PS.
$\mathbb{C}[t]$ means that the coefficients are in $\mathbb{C}$? Why we usually write $\mathbb{K}[t]$?
 A: We have that
$$e^{ \frac{\pi}{3} i } + e^{ -\frac{\pi}{3} i }=(\cos(\tfrac{\pi}{3})+i\sin(\tfrac{\pi}{3}))+(\cos(-\tfrac{\pi}{3})+i\sin(-\tfrac{\pi}{3}))=$$
$$(\tfrac{1}{2}+i\tfrac{\sqrt{3}}{2})+(\tfrac{1}{2}-i\tfrac{\sqrt{3}}{2})=\tfrac{1}{2}+\tfrac{1}{2}=1$$

Your equation
$$(t-c)(t-\bar{c})=t^2 - |c|t + |c|^2$$
is not quite correct; expanding out the left side,
$$(t-c)(t-\bar{c})=t^2-ct-\bar{c}t+c\bar{c}=t^2-(c+\bar{c})t+|c|^2,$$
but it is not generally true that $c+\bar{c}=c\bar{c}$.

To answer your other question: in general, if $R$ is a ring (an abstract kind of  "number system" where we can add and multiply), then $R[x]$ denotes the collection of polynomials in the variable $x$ having coefficients in $R$, i.e.
$$R[x]=\{a_0+a_1x+\cdots+a_nx^n\mid a_i\in R\}.$$
Thus, $\mathbb{C}[t]$ denotes the polynomials in the variable $t$ with complex coefficients. A good way to think of it is as "adding in" the new element $x$ to the original number system $R$. This is also its meaning when talking about, for example, $\mathbb{Z}[\sqrt{2}]$.
A: $\bf Hint:$ $e^{\theta i}=cos\theta+isin\theta$.
