Find the continued fraction digits of $\sqrt{3+i} \notin \mathbb{Q}(i)$ We can show that $\sqrt{3+i} \notin \mathbb{Q}(i)$ without too much effort, e.g. by solving $$(x+yi)^2 = (x^2 - y^2) + (2xy)i = 3+i$$  Therefore, we can try to find approximations, using only rational elements of the field.  What are the continued fraction digits
$$ \sqrt{3+i} = [a_0;\overline{a_1, a_2, \dots, a_n}] \in \mathbb{C}$$
with $a_i \in \mathbb{Z}[i]$. I'm quite sure they repeat.  
Even the first digit, I wonder how we can compute it?
$$ \lfloor \sqrt{3+i}\rfloor \in \mathbb{Z}[i] $$
We could solve an equation of some kind perhaps $a^2 + b^2 < 3^2 + 1^2 = 10$.  Not too many solutions to that.

I guess, there's no "positive" $\sqrt{3+i}$ so I guess it's the one with positive real an imaginary part, something like $\mathbb{R}_{\geq}\oplus\mathbb{R}_{\geq}$.
 A: The floor of $\sqrt{3+i}$ isn't well defined (at least, at present, without thinking about it and giving it a definition). For that matter, neither is $\sqrt{3+i}$ (there are two of them). Thus your question as stated can't be answered without some clarification.
That said, let's think about how to define continued fraction expansions with entries in $\Bbb{Z}[i]$ for a general complex number. 
Let $a+bi$ be some complex number. What is the closest Gaussian integer to $a+bi$?
Well, let $n$ be the closest integer to $a$ and let $m$ be the closest integer to $b$.
Then $$|(a+bi)-(n+mi)|=(a-n)^2+(b-m)^2\le \sqrt{\frac{1}{4} +\frac{1}{4}} = \frac{1}{\sqrt{2}}<1.$$
Thus we can call this Gaussian integer (which may not be unique) a floor of $a+bi$. The floors of a complex number have the important property that $|z-\newcommand\floor[1]{\left\lfloor {#1}\right\rfloor}\floor{z}|<1$, so we can then do the usual continued fraction expansion, by recursively defining
$$z_0=z, \quad a_i = \floor{z_i},\quad\text{and }z_{i+1}=\frac{1}{z_i-a_i}.$$
Then we should have
$$z=a_0 + \frac{1}{a_1+\frac{1}{a_2+\cdots}},$$
probably. There's some stuff to check here, with convergence and all that, but it should probably work the same as with the real numbers and the integers.
Note however that the coefficients aren't necessarily unique, and this continued fraction expansion won't necessarily be as nice as it is for the integers.
Apparently I've described the Hurwitz algorithm for complex continued fractions, described in more detail in the linked paper.
A: Continued fraction are defined for $x\in\mathbb R_+$ and can be calculated via the Euclidean algorithm, i.e. integer division.
Complex numbers can not be compared to "less" and "greater", so the operation of integer division is not defined on $\mathbb C.$ This possibility can be defined only for special subsets of $\mathbb C.$
Therefore, in the Hurwitz decomposition algorithm, instead of integer division, rounding is used independently for real and imaginary parts.
Let us find the point $a(x,y)\in \mathbb Z_i,$ the square of which is the nearest to $3+i.$
Attempt of approximation using norm
$$N=|a^2(x,y)-(3+i)|$$
cannot guarantee N < 1 for the considered example.

More perspective looks preliminary square root extraction,
$$r(s,t) = \sqrt{3+i},$$
$$s^2-t^2 + i(2st) = 3+i,\quad s^2+t^2 = \sqrt{10},$$
$$s^2=\dfrac{\sqrt{10}+3}2,\quad t^2=\dfrac{\sqrt{10}+3}2,\quad 2st=1,$$
with the solutions
$$r=\pm(s+it),\quad s=\dfrac{\sqrt{\sqrt{40}+6}}2\approx1.75532,\quad 
t = \dfrac1{2s}=\dfrac{\sqrt{\sqrt{40}-6}}2\approx0.284849.$$
For the "positive" root Wolfram Alpha gives continued fraction
$$r=[2; \{-2 - 2i, 4\}]$$
(using the Hurwitz expansion).
How does it work?
$$a_0 = [\Re r+0.5]+i[\Im r+0.5] = 2,$$
$$r_1 = \dfrac1{r-a_0} = \dfrac{(r-a_0)^*}{|r-a_0|^2}= \dfrac {s-2-i t}{(s-2)^2+t^2},$$
$$r_1 =  \dfrac{\sqrt{\sqrt{40}+6}-4}{2(4+\sqrt{10}-2\sqrt{\sqrt{40}+6})} - \dfrac{\sqrt{\sqrt{40}-6}}{2(4+\sqrt{10}-2\sqrt{\sqrt{40}+6})}i \approx -1.73523 - 2.02008 i,$$
$$r_1=\dfrac1{\sqrt{3+i}-2}=\dfrac{\sqrt{3+i}+2}{-1+i} = -\dfrac{1+i}2(\sqrt{3+i}+2)$$
$$a_1 = [\Re r_1+0.5]+i[\Im r_1+0.5] = -2-2i,$$
$$r_2 = \dfrac1{r_1-a_1} = \dfrac{(r_1-a_1)^*}{|r_1-a_1|^2}= \dfrac {s+2-i(t+2)}{(s+2)^2+(t+2)^2},$$
$$r_2=-\dfrac2{(1+i)(\sqrt{3+i}-2]}= -2\dfrac{\sqrt{3+i}+2}{(i+1)(i-1)} = \sqrt{3+i}+2\approx3.75532+i0.284849$$
(see also Wolfram Alpha).
Easily to see that
$$a_2=a_0+2,$$
and that the denominator $r_2-a_2$ repeated.
