How do you simplify #i^-12#?

1 Answer
Lotusbluete
Nov 10, 2015

#1#

Explanation:

You can use regular power rules:
#a^(-n) = 1/a^n#
#a^(n+m) = a^n * a^m#
#(a^n)^m = a^(n*m)#

And in addition, whenever you have the term #i^2#, always use #i^2 = -1#.

So, let me break it down in very small steps:

#i^(-12) = 1/i^12 = 1/i^(2*6) = 1/(i^2)^6 = 1 / (-1)^6 = 1 / (-1)^(2*3) = 1 / ((-1)^2)^3 = 1 / 1^3 = 1#

Alternatively, you can use #i^2 = -1 => i^4 = 1# and get there a bit quicker:
#i^(-12) = 1/i^12 = 1/i^(4 *3) = 1/(i^4)^3 = 1/1^3 = 1#

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