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Let $j=e^{2i\pi/3}$. How can I prove that the set $A=\mathbb{Z}+j\mathbb{Z}$ is stable under multiplication and conjugaison ? If I understood the question well, I need to prove the following :

a)If $x$ and $y$ are two elements of $A$, then $xy$ is also an element of A.

b)If $x$ is an element of $A$, so is $\bar x$.

For a), I tried basic multiplication stuff but could'nt write the result as $a+bj$ with $a$ and $b$ integers. And similarly for b).

And another problem I would like to solve is to find all inversible elements of $A$ ?

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    We say that $z$ is inversible in $A$ iff there exists $z^' \in A$ such that $zz^'=1$, i.e iff $\frac{1}{z} \in A$.2012-01-11

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a) Suppose $x$ and $y$ are two elements of $A$. Then we may rite $x = a + bj$ and $y = c + dj$, where $a,b,c$ and $d$ are integers. Then $xy = (a+bj)(c+dj) = ac + (ad+bc)j + bdj^2$. Now, we'd like to see if we can write $j^2$ as a linear combination of $j$'s and integers. Luckily, we have the identity $1 + j + j^2 = 0$ (since $j$ is a cube root of unity), giving $j^2 = -1 - j$ and $xy = (ac - bd) + (ad + bc -bd)j$, which is in $A$.

b) As before, write $x = a + bj$. Then $\bar{x} = a + b\bar{j}$, since $a$ and $b$ are real. Now $\bar{j} = e^{-2i\pi /3}$, and we'd like to write this as a linear combination of $j$'s and integers as before. We can use the same identity as we did in a), noticing that $e^{-2i\pi /3} = \frac{1}{j} $. If $1 + j + j^2 = 0$, then $j(j+1) = -1 $ and so $ \frac{1}{j} = - j - 1$. This then gives us that $\bar{x} = (a-b) -bj$, which is in $A$.