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).