Question 1207482
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The number of "i" is very special.


Consider the sequence i, i^2, i^3, i^4, i^5, i^6, i^7, i^8, . . . 


This sequence is      i, -1,  -i,   1,   i,   -1,  -i,  1,  . . . 


Thus the sequence is cyclic and the numbers in it repeat with the period of 4.


Therefore, if you need to find  i^n,  you should find the remainder  m = (n mod 4)  
of the index n modulo 4.

Then  i^n = i^m, which is one of four numbers i, -1, -i or 1.



For example, for i^14,  the index is  n= 14;  the remainder  (14 mod 4) is 2;
                        therefore,  i^14 = i^2 = -1.


Next example, for i^73, the index is  n= 73;  the remainder  (73 mod 4) is 1;
                        therefore,  i^73 = i^1 = i.


Thus, ALL calculations of this kind are quick, simple and easy. Making them is a  fun.



Regarding your question (B),  i^8 = 1;  i^4 = 1;  so, the value of (B) is  2*1*(2+1) = 2*3 = 6.
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Solved.