i^65
2007-09-22 13:49:20 · 2 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
its just a simple recall. Recall that i=sqrt(-1). also, recall that the units digit of a number raised to n will repeat in intervals of 4. To illustrate this better, take for example 3, 3^1=3, 3^2=9, 3^3=27, 3^4=81, 3^5=243, 3^6=729, 3^7=2187, 3^8= 6561 and so on. so, we do this in the case of i:
i^1=i, i^2=iXi=sqrt(-1)sqrt(-1) = -1, i^3=i^2Xi= -1Xi= -i and i^4=i^2Xi^2= -1X-1 = 1.
so, we apply these principles in simplifying i^65. we know that since i^4=1, then i^4 raised to the 14 power is 1
+> (i^4)^14 = i^64 = 1. since i^65 = i^64Xi = i.
hence i^65=i. hope its clear enough. good day!:D
2007-09-22 14:23:27 · answer #1 · answered by mikael 3 · 0⤊ 0⤋
i^65 = i^64*i = (i^4)^16*i
As i^4=1 the result is i
2007-09-22 20:57:35 · answer #2 · answered by santmann2002 7 · 1⤊ 1⤋