Answer
$i$
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 $i$.
$i^{33}=(i^{4})^{8}\times i=(1)^{8}\times i=1\times i=i$
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