The symbol:
$$\sqrt{i}$$
is not "well-defined": the complex number i has TWO square roots, and there is no clear choice for making one "the positive" one.
Notice Hartlw saying that " has two roots, i and -i" and Deveno saying "there is no clear choice for the positive one". They are both correct. Since the complex numbers do NOT form an "ordered" field, there is no single choice for "positive" and "negative" complex numbers.
Have a look at this discussion.
There is a growing trend away from using $\sqrt{t}$ unless $t\in\mathbb{R}$.
Here is an early theorem in Marsden/Hoffman: If $z\in\mathbb{C}$ then $\exists w\in\mathbb{C}$ such that $w^2=z$ (notice that $-w$ also works).
We would use $-i\left(\frac{1}{i}\right)^{\dfrac{1}{2}}$ or $-i\left(-i\right)^{\dfrac{1}{2}}$
Because $\mathbb{C}$ is not ordered it is difficult to talk about a principal root.
But some authors do it this way.
If $z\in\mathbb{C}$ If $z^{\dfrac{1}{2}}=\sqrt{|z|}$$\exp\left(\dfrac{i \text{Arg}(z)}{2}\right)$
If you use that convention then $(-i)^{\dfrac{1}{2}}=\exp\left(\dfrac{-i\pi}{4}\right)=$$\left(\dfrac{1}{\sqrt{2}}\right) +\left(\dfrac{-i}{\sqrt{2}}\right)$
In actual fact, there are "difficulties" with defining c^{1/n} in the complex numbers. To see why, let's look at the analogous definition in the real numbers:
c^{1/n} = c^{x} = e^{log(c)x} for x = 1/n.
The trouble is: "log" isn't definable on ALL the complex numbers because the analogue of e^{x} isn't 1-1 on the complex numbers anymore (so it doesn't have a UNIQUE inverse: log).
To get around this difficulty, one can "restrict" the domain of the complex exponential to a horizontal "swath" of height 2*pi (this is what the "Arg" function does, but there are actually "many" arg functions we could use, so we have to be CLEAR about "which one" we are using).
The two complex numbers:
$$\frac{\sqrt{2}}{2} + i\frac{\sqrt{2}}{2}$$
$$\frac{\sqrt{2}}{2} - i\frac{\sqrt{2}}{2}$$
BOTH square to i, and there is no clear-cut choice for calling either one THE square root of i.
In a similar vein, there is no reason to say i OR -i is THE square root of -1. In fact, choosing EITHER ONE leads to a version of the complex numbers that we "can't tell apart algebraically". This is because:
(z + w)* = z* + w*
(zw)* = z*w*
So the mapping z-->z* preserves all the algebraic properties of the complex numbers (it is a field automorphism). The choice of i as the "positive direction of the y-axis" is a completely arbitrary one.
Let $a,~b\in\mathbb{C},~~ a\ne 0$ then $a^b=e^{b\log(a)}$
Here is an excellent quote from Jerrold E. Marsden: $a^b$ is singled valued if and only if $b\in\mathbb{Z}$; if $b\in\mathbb{Q}$ where $b=p/q$ in lowest terms then $a^b$ has exactly $q$ distinct values; if $b\in\mathbb{R}\setminus\mathbb{Q}\text{ or if Im}(b)\ne 0$ then $a^b$ has infinitely many values.