Elements and structure of units in $\mathbb Q(\sqrt {-7})$ I'm completely stuck on this question:

Find the elements and the structure of the group of units in the ring of algebraic integers of number field $\mathbb Q (\sqrt {-7})$.
Are the group of units in the ring of algebraic integers of $\mathbb Q(\sqrt 5)$ finite? Why?

For the first part, I think it might be something to do with Dirichlet and for the second part, I believe the answer is yes but I don't know how to justify it. Pretty stumped.
Any help would be great
 A: Since $\;K:=\Bbb Q(\sqrt5)\;$ is a totally real number, there are only two embeddings into an algebraic closure of $\;\Bbb Q\;$ , and both real of course. Thus, here we have $\;r_1=2\,,\,r_2=0\implies\;$ the (multiplicative) group of integral units is isomorphic to $\;F\times\Bbb Z\;$ , with $\;F\;$ the finite cyclic groups of roots of unit contained in $\;K\;$, which are then only $\;\pm1\;$.
In $\;L:=\Bbb Q(\sqrt{-7})\;$ we have zero real embeddings and two, conjugate, complex non-real ones, and thus $\;r_1=0,\,r_2=1\implies\;$ the group of integral units has rank equal to zero and is thus the finite, cyclic group of all the roots of unit contained in $\;L\;$.
As before (since, again, $\;-7=1\pmod4\;$ , the ring of integers in $\;L\;$ is 
$$\mathcal O_L=\Bbb Z\left[\frac{1+\sqrt{-7}}2\right]$$
and since the conjugate couple of embedding is determined by $\;\sqrt{-7}\mapsto\pm \sqrt{-7}\;$ , we have that
$$\mathcal N^L_{\Bbb Q}\left(a+\frac12b+\frac{\sqrt{-7}}2b\right)=\left(a+\frac12b+\frac{\sqrt{-7}}2b\right)\left(a+\frac12b-\frac{\sqrt{-7}}2b\right)=$$
$$=a^2+ab+2b^2=\pm1\iff a^2+ab+(2b^2\mp1)=0$$
The above quadratic in $\;a\;$ has a real (in fact, we need integer) solution if
$$\Delta=b^2-8b^2\pm4\ge0\implies\begin{cases} b^2\le\frac47\\{}\\
b^2\le-\frac47\end{cases}\implies |b|\le\frac2{\sqrt7}\iff b=0$$
and thus the only units are $\;a^2=1\iff a=\pm1\;$
