1−1=1−1=1−1=1i=2i=2i+2i3=i⋅(2i+2i3)=2i2+2i3i=2−1+23=1=−11−11−11i12i12i1+2i3i⋅(2i1+2i3)2ii+2i3i21+232…(1)…(2)…(3)…(4)…(5)…(6)…(7)…(8)…(9)
Consider the steps above.
In which step is the mistake committed?
Note: i=−1 in equation (9) and the last one we assume to work out the L.H.S. and the R.H.S. separately.
Source:Spencer,1998