Yesterday in Further Maths, I got curious about root j (j being root -1), and it turns out that there are two accepted roots of j:
root(1/2) + root(1/2)j
- root(1/2) - root(1/2)j
This much was told to me by my teacher, and he gave me the proof:
root(j) = a + bj
j = a2 + b2j2 + 2abj
j = a2 - b2 + 2abj
re(j) = 0 = a2 - b2
(a + b)(a - b) = 0
Ergo, a = +/- b.
I don't know how he went from there to here, but,
2a2 = 1, or, -2a2 = 1
To work out the known roots, he followed for a = b. He went this far for a = -b, and stopped because it came out with a as an imaginary number, as a2 = -1/2.
This is where I started: if a is imaginary, then -a is imaginary also, so a - aj is still a valid complex number. Therefore,
a = root(-1/2) = j / root(2)
b = -a, so
root(j) = root(-1/2) - root(-1/2)j, or, root(j) = - root(-1/2) + root(-1/2)j
root(j) = j/root(2) + j^2/root(2)
root(j) = (j - 1) / root(2)
j = (j - 1)2 / 2
j = (1 - 2j - 1) / 2
j = 2j / 2 = j
So, the complex numbers root(1/2) - root(1/2)j, and, - root(1/2) + root(1/2)j are valid roots of j.
Now, my maths teacher didn't find these, so I checked a bit on Google, and I can't find anyone who did. I'm forced to conclude one of three things, either that I've made a calculational error in the substitution, my google-fu is weak, or I'm the first to find these two roots (which is unbelievable).
Therefore, please find something I'm doing wrong, or someone who's done this before, because I refuse to accept I'm capable of this discovery. |
