Question

What is the value of i^97 - i

Answer 1

$if\ i\ is\ the\ imaginary\ unit\ (i=\sqrt{-1}\Rightarrow i^2=-1):\\\\i^{97}-i=i(i^{96}-1)=i(i^{48\cdot2}-1)=i[(i^2)^{48}-1]=i[(-1)^{48}-1]\\\\=i(1-1)=i\cdot0=0\\\\therefore\ \boxed{i^{97}-i=0}$

Answer 2

this can also be explained as b on edg