We want to calculate an expression for:
(a + i b + j c + k d)n
where n is an integer.
When we calculated the powers of complex numbers we used the binomial theorem:
| (a + i b)n= |
|
|
(-i)k a n-k bk |
However there are a couple of differences in this case: these are 4 terms instead of 2 (so we would have to use a quadnomial instead of a binomial) but more importantly the terms don't all commute so we must use all permutations rather than integer multiplications of combinations which the binomial theorem gives.
see combinatorics
Square
So lets try the simplest case where: (a + i b + j c + k d)²
Taking all the permutations gives:
| permutation | dimension | |
|---|---|---|
| a a | 1 | commutes |
| a b | i | commutes |
| a c | j | commutes |
| a d | k | commutes |
| b a | i | commutes |
| b b | -1 | commutes |
| b c | ij = k | anti-commutes |
| b d | ik -j | anti-commutes |
| c a | j | commutes |
| c b | ji =-k | anti-commutes |
| c c | -1 | commutes |
| c d | jk = i | anti-commutes |
| d a | k | commutes |
| d b | ki = j | commutes |
| d c | kj = -i | anti-commutes |
| d d | -1 | anti-commutes |
The anti-commuting values cancel out so taking the other terms gives:
(a + i b + j c + k d)² = a²-b²-c²-d² + i 2ab + j 2ac + k 2ad
Going on to higher powers gives:
| n | (a + i b + j c + k d)n | w | x | y | z |
|---|---|---|---|---|---|
| 1 | (a + i b + j c + k d)1 | a | b | c | d |
| 2 | (a + i b + j c + k d)2 | a²-b²-c²-d² | 2ab | 2ac | 2ad |
| 3 | (a + i b + j c + k d)3 | a3-3b²a-3c²a -3d²a | 3a²b-b3-3c²a-3d²a | 3a²c-c3-3b²c-3d²c | 3a²d-d3-3b²d-3c²d |
| 4 | (a + i b + j c + k d)4 |
