The question is not clearly readable, but I'm assuming the expression which makes more sense, so you can have a clue
Answer:
[tex]i^{97}-i=0[/tex]
Step-by-step explanation:
Powers of The Imaginary Unit
The imaginary unit i is defined as
[tex]i=\sqrt{-1}[/tex]
The first powers of i are
[tex]i^0=1[/tex]
[tex]i^1=i[/tex]
[tex]i^2=(\sqrt{-1})^2=-1[/tex]
[tex]i^3=i.i^2=-i[/tex]
[tex]i^4=i^2.i^2=1[/tex]
And so on the cycle repeats every four numbers. To find the value of [tex]i^{96}[/tex] we can find the remainder of 96/4=0. So [tex]i^{96}=i^0=1[/tex]
The given expression is
[tex]i^{97}-i[/tex]
Factoring
[tex]i(i^{96}-1)[/tex]
Since [tex]i^{96}=1[/tex]
[tex]=i(1-1)=0[/tex]
The required value is 0
