Question

I^89 simplified. Complex number.

Answer 1

This problem can be simplified by remembering that since $i= \sqrt{-1}$, we have $i^{4}=(-1)^{2}=1$. Then we have $i^{89}=i^{88+1}=i^{88}i^{1}=(i^{4})^{22}i=(1)^{22}i=i$.