## Precalculus (6th Edition)

$-i$
Apply exponent rule: $\displaystyle \quad a^{-n}=\frac{1}{a^{n}}$ $i^{-13}=\displaystyle \frac{1}{i^{13}}$ Calculate $i^{13}$: $i^{13}=i^{12+1}=i^{12}i \qquad$...Apply: $a^{mn}=(a^{m})^{n}$ $=i(i^{2})^{6}\qquad$...Apply: $\quad i^{2}=-1$ $=(-1)^{6}i \qquad$...Apply: $(-a)^{n}=a^{n},$ if $n$ is even $=1i$ $=i$ Thus, $i^{-13}=\displaystyle \frac{1}{i}$ Multiply by the conjugate $\displaystyle \frac{-i}{-i}$ $=\displaystyle \frac{1\cdot(-i)}{i(-i)}$ $=1\cdot(-i)$ $=-i$