## Precalculus: Mathematics for Calculus, 7th Edition

$a)$ $3+i$ $b)$ $8-i$
$a)$ $(2-3i)+(1+4i)$ Evaluate the sum: $(2-3i)+(1+4i)=(2+1)+(-3+4)i=3+i$ $b)$ $(2+i)(3-2i)$ Evaluate the product: $(2+i)(3-2i)=6-4i+3i-2i^{2}=...$ Substitute $i^{2}$ by $-1$ and simplify: $...=6-4i+3i-2(-1)=6-i+2=8-i$