## Calculus 8th Edition

$-\frac{5}{17}+\frac{14}{17}i$
$\displaystyle \frac{3+2i}{1-4i}$ We multiply the numerator and denominator by the conjugate of the denominator: $=\frac{3+2i}{1-4i}\cdot\frac{1+4i}{1+4i}$ We use the fact that $i^2=-1$ (because $i=\sqrt{-1}$) $=\frac{3+12i+2i+8*-1}{1^{2}+4^{2}}$ $=\frac{-5+14i}{17}$ $=-\frac{5}{17}+\frac{14}{17}i$