How do you simplify #i^100#?

Question

18657 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-25T17:51:17

#i^100=1#

#i^100=(i^2)^50#

From the fact that #i^2=-1,# we get

#(-1)^50=1# as #-1# raised to any even power is #1.#

Alternatively, we can rewrite in trigonometric form and then in the form #re^(itheta)#:

#i=cos(pi/2)+isin(pi/2)#

#=e^(ipi/2)#

Raise the exponential to the power of #100:#

#(e^(ipi/2))^100=e^(50pi)#

#=cos(50pi)+isin(50pi)#

#=cos2pi+isin2pi#

#cos2pi=1, sin2pi=0#

so we get

#=1#

Answer 2 · 2018-04-25T17:55:06

#i^100=1#

#i^100=(i^2)^50=(-1)^50=1#

#(-a)^n=a^n#, where n is an even number.