In my paper above I stated specifically: "(B & y) could be ANY combination of 'real numbers'."
Therefore when y= 1, we arrive at a real number:
Consider: B^y = a + bi.......eq.909 Since B= a real number, the RHS of the following equation can be correctly stated at: B^1= a + 0i B^1= a Now lets' assume: B= 2 y=1 subbing into eq. 909: 2^1= 2 + 0i
So, our coordinates are: a= 2 b= 0 y= 1
The graph point described above does NOT fall on the 'complex plane' because it represents a number from the 'real number set'. My system would certainly be 'out-of-order' if I contended such a point ever fell upon the complex plane!