Taking the square root of an imaginary number We know that when we take the square root of a negative real number, it's realness "splits open" and an "imaginary" dimension is introduced (characterized by the presence of iota).
The question is, what would happen if take square root of a positive imaginary number? Will the dimension split up again, or will it stay in level-1 imaginary dimension?
 A: 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 $\Bbb H$.
A: 
Basically instead of degrees and radians, on a complex plane we can use multiplication to express rotations. Multiplying by $1$ takes you take to the same place, so it's the equivalent of 360 degrees. Multiplying by $-1$ is the same as 180 degrees (because if you do that two times it's the equivalent of multiplying by $1$). Multiplying by $i$ is the same as a 90 degree rotation and multiplying by  $\sqrt{i}$ is the same as rotating 45 degrees (because if you multiply by $\sqrt{i}$ eight times it's the equivalent of a $360$ rotation). Looking at the value onto the imaginary unit circle of 45 degrees can give you the value  of $\sqrt{i}$. However you also need to consider the negative value is also a solution when dealing with any even roots.
In general multiplication by $\sqrt[n]{i}$ is the same as a $90/n$ degree rotation so you can find any root of  $i$ if you want. 
We use multiplication to represent rotations on a complex plane because we don't graph $x$ vs $y$, rather we graph one number as a whole.
A: Other answers have explained why $\sqrt{i}$ is a complex number, and shown how to compute it using complex exponentials, but you can also compute it directly. 
If you want to find the complex number $a + bi$, where $a$ and $b$ are real numbers, such that $\sqrt{i} = a + bi$, then square both sides and solve $$i = (a + bi)^2.$$ Expanding the right-hand side, we get $$i = a^2 + 2abi - b^2.$$ Now what we can do is equate the real and imaginary parts on both sides. Intuitively, the real part can't become imaginary and the imaginary part can't become real, so they are independent. More rigorously, $\{1, i\}$ is a linearly independent set over the reals (that in fact forms a basis for $\mathbb{C}$ over $\mathbb{R}$). 
We get the system $$\begin{align} a^2 - b^2 &= 0\\ 2ab &= 1.\end{align}$$ Solving bottom equation for a gives $$a=\frac{1}{2b},$$ and plugging this into the top gives $\frac{1}{4b^2} = b^2,$ that is, $$b^4=\frac{1}{4}.$$ Since this is a quartic, it has four solutions, given by $$b=\pm \sqrt[4]{\frac{1}{4}},\pm i\sqrt[4]{\frac{1}{4}}.$$ Since our original $b$ was supposed to be real, we only use the first two roots. Hence $$b=\pm\sqrt[4]{\frac{1}{4}}=\pm\frac{1}{\sqrt{2}}=\pm\frac{\sqrt{2}}{2},$$ and $$a=\frac{1}{2b}=\pm\frac{\sqrt{2}}{2}.$$ So $$\sqrt{i}=\pm\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}i,$$ which you can verify by squaring. 
A: 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".
A: 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.
