If AB = i and BC = 0, then B would be in the same 2D space as C, but one of them would be “above” the other in 3D space (which doesn’t exist in this context, just as sqrt(-1) doesn’t exist in the traditional sense).
So this triangle represents a 2D object that is “standing up” on the page.
It makes sense if you represent complex numbers as (a, b) pairs, where a is the real part and b is the imaginary part (just like the popular a + bi representation that can be expanded to a * (1, 0) + b * (0, 1)). AB’s length is (1,0), AC’s length is (0,1), and BC’s length will also be a complex number.
Yes. Also if you think of i as a 90° rotation (with a length of the scalar coefficient infront of i, in this case 1) . Thus one rotates you outwards away from the 2D plane, and two of those gets you back to the 2D plane, just going the other direction.
If AB = i and BC = 0, then B would be in the same 2D space as C, but one of them would be “above” the other in 3D space (which doesn’t exist in this context, just as sqrt(-1) doesn’t exist in the traditional sense).
So this triangle represents a 2D object that is “standing up” on the page.
It makes sense if you represent complex numbers as
(a, b)
pairs, wherea
is the real part andb
is the imaginary part (just like the populara + bi
representation that can be expanded toa * (1, 0) + b * (0, 1)
). AB’s length is(1, 0)
, AC’s length is(0, 1)
, and BC’s length will also be a complex number.I think.
Yes. Also if you think of i as a 90° rotation (with a length of the scalar coefficient infront of i, in this case 1) . Thus one rotates you outwards away from the 2D plane, and two of those gets you back to the 2D plane, just going the other direction.