TITLE: Taking the square root of an imaginary number
QUESTION [10 upvotes]: 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?

REPLY [4 votes]: 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.