Quote:
Originally posted by o0zi
Checking up the definition of (a+ib)(c+id), it's
(ac-bd) + i (ad+bc)
[snip]
We've only used 1 multiplication, and that's multiplying (ad+bc) by i.
The only further step I could find was:
No, it's not a trick like that. It would still take 4 real multiplications to compute (ac-bd) and (ad+bc). There is a way to do it so you really only have to compute 3 products; and though it takes 1 extra addition and 2 extra subtractions, you would still get savings from not having to do the more expensive multiplication.