Alternative form of Eisenstein integers I just recently got into number theory and algebra so my knowledge is very limited. I stumbled upon the Eisenstein integers $\mathbb{Z}[\omega]$. They are of the form $\varepsilon = a + b \omega$, where $\omega = e^{2 \pi i/3} = -1/2 + i \sqrt{3}/2$.
Now my problem:

I want to show that you can write every Eisenstein integer in the form of $\varepsilon = \frac{a + bi \sqrt{3}}{2}$ where $a \equiv b \bmod 2.$

For now I got: $\omega = -1/2 + i \sqrt{3}/2 \Rightarrow \varepsilon = a + b \omega = a - b/2 + i \sqrt{3}b/2$ then $a = b + 2k \Rightarrow b = a - 2k \Rightarrow \varepsilon = a - a/2 + k + i \sqrt{3}b/2 = \frac{a + bi\sqrt{3}}{2} + k$.
Now I still have to deal with the superfluous $k \in \mathbb{Z}$. How is the argumentation to get rid of it?

Your end number is $$\frac{(a+2k)+bi\sqrt{3}}{2}$$ which is of the right form.
Another way to argue is that if $x+y\sqrt{-3}$ is an integer where $x,y\in \mathbb{Q}$ then $x-y\sqrt{-3}$ is also an integer and thus 
$2x$ and $x^2+3y^2$ are integers. This implies that $3(2y)^2$ is an integer and thus $2y$ is an integer since $\sqrt{3}$ is irrational.
So we have $x=\frac{a}{2}$ and $y=\frac{b}{2}$ and since $4|a^2+3b^2$ we see that $a$ and $b$ have the same parity. 
