Group structure of the unit circle in $\mathbb{F}_q[\sqrt{-1}]$ Consider $\mathbb{F}_q[\sqrt{-1}]$, where $q = p^n$ where $p$ is a prime, $p \equiv 3 \pmod 4$. Then consider the unit circle $\{a \vert a \overline{a} = 1\}$, where $\overline{a}$ denotes conjugation, i.e. $a + b\sqrt{-1} \to a - b\sqrt{-1}$. Then there is a group structure, as if $a \overline{a} = b \overline{b} =1$, then $ab \overline{ab} =1$ and there are inverses and an identity.
I would like to understand the structure of this group. Based on numerical evidence, I believe it always has $q+1$ elements and is cyclic. 
We can recast the problem as looking for solutions to $a^2 + b^2 = 1$ over $\mathbb{F}_q$, but I don't think that this is helpful. However, it is easy to show using Jacobi sums that there are $p+1$ solutions over $\mathbb{F}_p = \mathbb{Z} / p \mathbb{Z}$, but this doesn't give much insight and I don't see how to generalize it to prime powers.
EDIT: So it is cyclic because it is a subgroup of $F_{q^2}$.
 A: What you say is right when $n$ is odd, wrong when $n$ is even. Let’s look at the nice case first.
When $n$ is odd, $q=p^n\equiv3\pmod4$, and $k=\Bbb F_q$ has no square root of $-1$. Then it’s fine to write formal expressions $a+b\sqrt{-1}$ as you do: you’re considering elements of $K=\Bbb F_{q^2}$, and by asking for $a^2+b^2=1$, you’re asking for the kernel of the norm, $\mathbf N^K_k(w)=ww^q$. It’s a homomorphism between the multiplicative groups of the fields, and its kernel is the set of $(q+1)$-th roots of unity in $K$, $q+1$ of them.
The bad case is that $n$ is even, and you’re in the same situation as for $p\equiv1\pmod4$, with there being an $i$ already in $\Bbb F_q$, since $q=p^{2m}\equiv1\pmod4$. Here, the curve $x^2+y^2=(x+iy)(x-iy)=1$ can be recoordinatized with $X=x+iy$, $Y=x-iy$, giving the new equation $XY=1$, in other words $Y=1/X$. Clearly $X$ can take any nonzero value in $k=\Bbb F_q$, and choice of this value determines $Y$. So there are $q-1$ solutions.
For an example, look at $\Bbb F_9=\Bbb F_3[i]$. Here, the solutions of $x^2+y^2=1$ are $(0,\pm1)$, for two solutions, $(\pm1,0)$, for two, and $(\pm i,\pm i)$ for four, eight in all.
