What is #i^97-i#?

Question

(a) #-2i#
(b) #0#
(c) #1-i#
(d) #-1-i#

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Answer 1 · 2016-09-11T17:47:15

#i^97-i=0# that is answer is (b)

As #i=sqrt(-1)#, while #i^1=i#,

#i^2=-1#

and #i^3=i^2xxi=-1xxi=-i#

and #i^4=(i^2)^2=(-1)^2=1#

Similarly we can work out #i^5=i#, #i^6=-1#, #i^7=-i# and #i^8=1#

This goes on in a cycle of #4#,

hence #i^(4n+1)=i#, #i^(4n+2)=-1#, #i^(4n+3)=-i# and #i^(4n)=1#

Hence as #97=4xx24+1#

#i^97=(i^4)^24xxi=1^24xxi=i# and #i^97-i=i-i=0# that is answer is (b).