When we compute $ \sqrt{i} $, we start like this:
$ \sqrt{i} = a+bi $ , when $a \in \mathbb R, b \in \mathbb R $
Then we square both sides, and find out the values of $a$ and $b$.
But the procedure above implies that $ \sqrt{i} \in \mathbb C $.
While $ \sqrt{r} \notin \mathbb R $ for some $ r \in \mathbb R $, why can we be sure that $ \sqrt{i} \in \mathbb C $ ?
On
Either there's a $z\in\Bbb C$ with $z^2=i$ or there isn't. Even before you check to see whether $a^2-b^2=0,\,2ab=1$ is solvable over $\Bbb R$, we know from e.g. the fundamental theorem of algebra that a solution will exist. The real difficulty in writing $\sqrt{i}$ is that there are two roots $\pm z$, so we have to decide which one deserves that symbol.
Formally, $\sqrt{x} \in Y$ means that equation $y^2 = x$ has solution in $Y$. One way to prove $\sqrt{i} \in \mathbb C$ is to solve your particular equations (that has two solutions $\pm\frac{\sqrt{2}}{2} (1 + i)$). Another is to apply fundamental theorem of algebra - it implies that equation $x^2 - i = 0$ has solutions in $\mathbb C$.