## College Algebra (10th Edition)

$\displaystyle \frac{6}{5}+\frac{8}{5}i$
Recall, $i=\sqrt{-1}$, so $i^{2}=-1$. Thus, we multiply by the conjugate of the denominator to obtain: $\displaystyle \frac{10}{3-4i}$ $\displaystyle =\frac{10}{3-4i}*\frac{3+4i}{3+4i}$ $\displaystyle =\frac{10(3+4i)}{(3-4i)(3+4i)}$ $\displaystyle =\frac{30+40i}{9+12i-12i-16i^{2}}$ $\displaystyle =\frac{30+40i}{9-16*-1}$ $\displaystyle =\frac{30+40i}{25}$ $\displaystyle =\frac{30}{25}+\frac{40}{25}j$ $\displaystyle =\frac{6}{5}+\frac{8}{5}i$