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R. Brown
Posts:
54
From:
texas
Registered:
12/29/05
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Re: "Specific Complex Numbers"
Posted:
Mar 20, 2007 2:35 PM
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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!
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