## Intermediate Algebra for College Students (7th Edition)

$-i$
Solve $i^{23}$ Since, $i^2=-1$ and $i^2 \cdot i^2=(-1)(-1)=1$ and $i^3=i^2 \cdot i=-i$ Thus, $i^{23}=(i^2)^{11} \cdot i=(-1)^{11} \cdot i$ Hence, $i^{23}=-i$