here's another approach (yes, I know the title is a pun, that's intentional and I'm ignoring it):
You want an x such that x^2 = j; but then x^4 = -1 and x^8 = 1. Therefore the number you want is an eighth root of 1.
There are 8 such (complex) numbers, all of the form e^(2*pi*n/8) where n goes from 0 to 7. They are:
1
sqrt(1/2) + sqrt(1/2)j
j
-sqrt(1/2)+sqrt(1/2)j
-1
-sqrt(1/2)-sqrt(1/2)j
-j
sqrt(1/2)-sqrt(1/2)j
Of these, you want the ones with x^2 = j.
The first and fifth (1, -1) have squares = 1.
The second and sixth have squares = j.
The third and seventh (j, -j) have squares = -1.
The fourth and eighth have squares = -j
You want, therefore, the second and sixth. |
