How do you simplify #i^(-43)+i^(-32)# ?

1 Answer
Apr 11, 2017

#i^(-43)+i^(-32) = i+1#

Explanation:

Look at the first few non-negative powers of #i#:

#i^0 = 1#

#i^1 = i#

#i^2 = -1#

#i^3 = -i#

#i^4 = 1#

Basically this pattern: #1, i, -1, -i# repeats every #4# powers.

In terms of angles, multiplying by #i# is an anticlockwise rotation of #pi/2# in the complex plane. So after #4# rotations we are back facing the same way.

So in general we can write:

#{ (i^(4n) = 1), (i^(4n+1) = i), (i^(4n+2) = -1), (i^(4n+3) = -i) :}#

which holds for any integer #n#.

Now:

#-43 = -44+1 = 4(-11)+1#

#-32 = -32+0 = 4(-8)+0#

So:

#i^(-43)+i^(-32) = i+1#