Simply, what do I get when I take the square root of negative imaginary number? But, it cant be an imaginary number since the answer to the $2^{\mathrm{nd}}$ power must equal the original negative imaginary number. So:
$$ \sqrt{{-(|x|{_i}})} = y $$
where
$$ y{^2} = {-(|x|{_i}}) $$
So, if $ y=\sqrt{-\sqrt{-1}} $, then what is y ?
So, what would it be? (a double imaginary ($x{_{i}}_i$)?)
And, how would your answer to the 2nd power equal the original negative imaginary number?
Best Answer
The concept that you are missing is the Complex plane. Not surprisingly, given that I often seen it introduced a full year after imaginary numbers.
On the complex plane, taking a square root becomes almost entirely geometric. The non geometric part is, you have to take the square root of the distance from your number to 0. For instance $15+20i$ is at a distance of $25$ from 0 (per Pythagorean Theorem) so the distance of the square root will be 5.
To find the angle of the square root—just note the angle of the original number (measuring counter-clockwise from the real axis—a.k.a. the x axis) and cut it in half.
You may notice that there are two ways to cut the angle in half—clockwise, and counter-clockwise. That is correct; every number except 0 has two square roots.
You asked about negative numbers. Let's take $-i$ as an example. Note that the position of $-i$ on the complex plane is one unit "south" of 0, the same place that $(0,-1)$ is on a rectangular coordinate plane. Since the distance from 0 is 1, and the positive square root of 1 is 1, the distance from 0 for our answer will also be 1, making that part easy.
Note that the angle of that point, measured counter-clockwise from the x axis, can be stated as either $270$ or $-90$. Use both. Divide each by two. That gives the two answers.
Now solving with the pythagorean theorem (or using trig, but in this case the angles are just 45 degrees in the 2nd and 4th quadrants, so we don't need trig), we get the two square roots of $-i$, which are $\frac{1-i}{\sqrt2}$ and $\frac{-1+i}{\sqrt2}$.
Try squaring each of those with pencil and paper to convince yourself they are indeed correct.