Question

remind me:

Is there an easier way to solve this i^132

Answer 1

take the integer remainder when you divide \(132\) by \(4\)

Answer 2

is i a complex number ? u mean to say that

Answer 3

since 4 divides 132 evenly, the remainder is zero
that means
\[i^{132}=e^0=1\]

Answer 4

yes and so if say for another example was i^75 . . . all i have to do is to divide by 4?

Answer 5

when i say "divide' i mean take the whole number remainder
in the case of \(75\) it is \(3\) because 4 goes in to 72 evenly, leaving a remainder of 3

Answer 6

that makes
\[i^{75}=i^3=-i\]

Answer 7

i^4n =1
i^4n+1 =i
i^4n +2 =-1
i^4n+3= -i
(where n belongs to natural numbers)
Just remember this

Answer 8

that patter is
\[i^1=i\\
i^2=-1\\
i^3=-i\\
i^4=1\] and so on

Answer 9

ok i understand the dividing part but why isnt it 1. How was it -i

Answer 10

as in why its i^4=1.

Answer 11

What is i
is is basically \[\sqrt{1}\]
square of that becomes -1
and square of -1
becomes 1
so i^4 =1

Answer 12

These values that i @satellite73 wrote are proved accordingly

Answer 13

alright thanks got it