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R. Brown
Posts:
54
From:
texas
Registered:
12/29/05


Re: "Specific Complex Numbers"
Posted:
Mar 20, 2007 2:35 PM


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 'outoforder' if I contended such a point ever fell upon the complex plane!



