## Trigonometry 7th Edition

$-i$
Knowing that $i^{2}=-1$, we also know that $i^{4}=(i^{2})^{2}=(-1)^{2}=1$. We can use the fact that $i^{4}=1$ to evaluate higher powers of $i$. $i^{35}=i^{32}\times i^{2}\times i=(i^{4})^{8}\times i^{2}\times i=(1)^{8}\times -1\times i=1\times-1\times i=-i$