# Chapter 8 - Section 8.1 - Complex Numbers - 8.1 Problem Set - Page 426: 53

$i$

#### 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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