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?
 A: Let $p,q$ be integers. Then, as you tried, we have
$$p+q\omega=
p+\dfrac{-q+qi\sqrt 3}{2}=
\dfrac{2p-q+qi\sqrt 3}{2}$$
Let $a=2p-q$ and $b=q$. What can you conclude from here?
A: I think this excessive emphasis on $\omega$ often serves to confuse students. Don't get me wrong, $\omega$ is important. $$\omega = -\frac{1}{2} + \frac{\sqrt{-3}}{2},$$ and $N(\omega) = 1$, which means that $\omega$ is a unit. And furthermore $\omega$ is a complex cubic root of 1.
Let's review the norm function for numbers in $\mathbb{Q}(\sqrt{-3})$: $$N\left(\frac{a}{2} + \frac{b \sqrt{-3}}{2}\right) = \frac{a^2}{4} + \frac{3b^2}{4}.$$ An algebraic number in $\mathbb{Q}(\sqrt{-3})$ is also an algebraic integer if its norm is an integer from $\mathbb{Z}$.
If both $a$ and $b$ are even, then $a^2$ and $3b^2$ are both multiples of 4, so that when you divide them by 4, you still have integers.
If both $a$ and $b$ are odd, then $a^2 \equiv 1 \pmod 4$ and $b^2 \equiv 1 \pmod 4$ as well. But then $3b^2 \equiv 3 \pmod 4$, so that $a^2 + 3b^2 \equiv 0 \pmod 4$. Then, when you divide $a^2 + 3b^2$ by 4, you still have an integer.
Now, how to convert $$m + n \omega = \frac{a}{2} + \frac{b \sqrt{-3}}{2},$$ with $m$ and $n$ both integers from $\mathbb{Z}$?
Well, $a = 2m - n$ and $b = n$.
If both $m$ and $n$ are even, we have no problem, as $a$ and $b$ are as well.
If both $m$ and $n$ are odd, $a$ and $b$ are as well.
If $m$ is odd but $n$ is even, we have $2m$ even and $2m - n$ is still even, so $a$ and $b$ are both even as desired.
If $m$ is even but $n$ is odd, $2m - n$ is odd, so $a$ and $b$ are both odd as desired.

For exercises, I suggest the following: convert the numbers $-\omega$, $-2 + \omega$, $3 - \omega$ and $4 + \omega$ to the $a$ and $b$ form. Also, compute their norms. Bonus: figure out the conversion for $m + n \omega^2$.
A: 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. 
