Show that, if $ a + bi$ is prime in $\mathbb{Z} [i]$, then $a - bi$ is prime in $\mathbb{Z}[i]$ 
Show that, if $ a + bi$ is prime in $\mathbb{Z} [i]$, then $a - bi $ is prime in $\mathbb{Z}[i]$

Since $\mathbb{Z}[i]$ is $ED$, then if $a+bi$ is irreducible then $a+bi$ is prime.
But now how I can relate $a+bi$ with $a-bi$?
Thanks for help.
 A: If $a-bi$ is not irreducible then take conjugate on both sides to obtain a contradiction.Conjugate of $a+bi$ is $a-bi$. So if $a-ib=(p+iq)(r+is)$ then clearly $a+ib=(p-iq)(r-is)$.
A: In fact the property holds in every quadratic ring $\mathbb Z[\sqrt{d}]$ if one uses the definition of prime elements: say $a+b\sqrt{d}$ is prime, and want to show that $a-b\sqrt{d}$ is also prime. 
Following a nitpick from the comments let's first show that $a-b\sqrt{d}$ is not invertible: suppose it is, and then there is $z\in\mathbb Z[\sqrt{d}]$ such that $(a-b\sqrt d)z=1$. From here we get $(a+b\sqrt d)\overline{z}=1$, so $a+b\sqrt d$ is invertible, a contradiction.
Now suppose $a-b\sqrt{d}\mid z_1z_2$ where $z_i\in\mathbb Z[\sqrt{d}]$. Then $a+b\sqrt{d}\mid\overline{z}_1\overline{z}_2$ where $\overline{z}_i$ is the conjugate of $z_i$. Since $a+b\sqrt{d}$ is prime it follows $a+b\sqrt{d}\mid\overline{z}_1$ or $a+b\sqrt{d}\mid\overline{z}_2$. Suppose $a+b\sqrt{d}\mid\overline{z}_1$. Then $a-b\sqrt{d}\mid z_1$ and we are done.
