## College Algebra 7th Edition

(a) $\frac{6}{5}+\frac{8}{5}i$ (b) $2$
(a) We multiply the numerator and denominator by the conjugate: $\frac{4+2i}{2-i}=\frac{4+2i}{2-i}*\frac{2+i}{2+i}=\frac{8+8i+2i^{2}}{4-i^{2}}=\frac{8+8i+2*-1}{4--1}=\frac{8+8i-2}{4+1}=\frac{6+8i}{5}=\frac{6}{5}+\frac{8}{5}i$ (b) $(1-\sqrt{-1})(1+\sqrt{-1})=(1-i)(1+i)=1+i-i-i^{2}=1+i-i--1=1+1=2$