Taking the square root of an imaginary number
Every non-zero complex number $z$ has exactly two complex square roots - this is a consequence of the field of complex numbers being algebraically closed (Wikipedia link). If $$z=re^{i\theta}=r\cos(\theta)+ri\sin(\theta)$$ then the square roots of $z$ are $$\begin{align*} \sqrt{r}e^{i\theta/2}&=\sqrt{r}\cos(\theta/2)+\sqrt{r}\,i\sin(\theta/2)\\ -\sqrt{r}e^{i\theta/2}&=-\sqrt{r}\cos(\theta/2)-\sqrt{r}\,i\sin(\theta/2) \end{align*}$$ In short, there are no complex numbers whose square roots are not already present in the complex numbers themselves, so there is nothing more to "add in".
Maybe you want to solve $z^2=i$. It is easy to verify that $z=\pm\frac{\sqrt2}{2}(1+i)$ satisfy the equation. Generally, every polynomial with complex coefficient has a root in $\mathbb C$, or, equivalently, complex field has no algebraic field extension.
The square root function is not as nicely behaved on the complex numbers, $\Bbb C$ as it is on the (nonnegative) real numbers, $[0, \infty)$. It's true that we can find exactly two solutions to the equation $w^2 = z$ for any nonzero $z \in \Bbb C$, but unlike in the usual real setting, we cannot make choice of $w$ that continuously depends on $z$, or more precisely, there is no continuous function $g: \Bbb C \to \Bbb C$ such that $g(z)^2 = z$ (for all $z \in \Bbb C$).
Computing directly gives that $$(\pm e^{\pi i / 4})^2 = \left[\pm \frac{1}{\sqrt{2}} (1 + i)\right]^2 = i,$$ so for any imaginary number $iy,$ we have that $$(\pm \sqrt{y} e^{\pi i / 4})^2 = iy,$$ that is $\pm \sqrt{y} e^{\pi i / 4}$ are precisely the square roots of $iy$. (Here, if $y < 0$, we can take $\sqrt{y}$ to mean either $i \sqrt{-y}$ or its negative.)
More generally, we can write any complex number in polar form as $$z = r e^{i \theta}$$ (here, $r \geq 0$ and $\theta \in \Bbb R$), and its square roots are precisely $$\pm \sqrt{r} e^{i \theta / 2}.$$
The fact that imaginary (and more generally, complex) numbers have square roots again in the complex numbers (and in particular, that we needn't expand our set of numbers as we did when taking the square root of a general real number) is a consequence of the Fundamental Theorem of Algebra, which says that any polynomial $p(z)$ (in our case, $z^2 - i y$) has precisely $\deg p$ roots in $\Bbb C$. More suggestively, this means that $\Bbb C$ is algebraically closed.
Remark All this said, one can naturally embed $\Bbb C$ into larger algebraic objects (say, $\Bbb R$-algebras), in which the notion of square root still makes sense (but subject to the caveat that it is generally less well-behaved). Perhaps the prettiest example is the quaternions, usually denoted $\Bbb H$, which we may identify (as a vector space) with $\Bbb R^4$ (or $\Bbb C^2$) with basis elements $1, i, j, k$ satisfying the product formulas $i^2 = j^2 = k^2 = -1$, $ijk = -1$. This is a division ring, but, as $ij = k \neq -k = ji$, it is noncommutative and hence, unlike $\Bbb R$ and $\Bbb C$, not a field. The only square roots of $1$ in $\Bbb C$ are $\pm 1$, but computing directly shows that $$(a i + b j + c k)^2 = -1$$ for any $(a, b, c)$ such that $a^2 + b^2 + c^2 = 1$, that is, there is an entire $2$-sphere of (and in particular, infinitely many) square roots of $-1$ in $\Bbb H$.