## Precalculus: Mathematics for Calculus, 7th Edition

$\dfrac{5-i}{3+4i}=\dfrac{11}{25}-\dfrac{23}{25}i$
$\dfrac{5-i}{3+4i}$ Multiply the fraction by $\dfrac{3-4i}{3-4i}$: $\Big(\dfrac{5-i}{3+4i}\Big)\Big(\dfrac{3-4i}{3-4i}\Big)=\dfrac{15-20i-3i+4i^{2}}{3^{2}-(4i)^{2}}=\dfrac{15-23i+4i^{2}}{9-16i^{2}}=...$ Substitute $i^{2}$ with $-1$: $...=\dfrac{15-23i+4(-1)}{9-16(-1)}=\dfrac{15-23i-4}{9+16}=\dfrac{11-23i}{25}=...$ $...=\dfrac{11}{25}-\dfrac{23}{25}i$