Question #4c230

Question

705 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-01-23T21:25:22

0

I'm guessing this is #i^235+i^29#.

First we need to know:

#i^1=i#, #i^2=-1#, #i^3=-i#, and #i^4=1#.

#i^235=i^(232+3) = i^232*i^3 = (i^4)^58*i^3 = (1)^58*(-i)=-i#

#i^29= i^(28+1)=i^28*i=(i^4)^7*i=i#

So:

#i^235+i^29 = -i+i=0#