Answer
-1
Work Step by Step
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 $i$.
$i^{34}=i^{32}\times i^{2}=(i^{4})^{8}\times i^{2}=(1)^{8}\times-1=1\times-1=-1$