Question

(5+12i)/(8-6i)+(1+2i)^2/(2+i)

Answer 1

$\frac{5+12i}{8-6i}=\frac{(5+12i)(8+6i)}{(8-6i)(8+6i)}=\frac{40+30i+96i+72i^2}{8^2-(6i)^2}=\frac{40+126i-72}{64-36i^2}=\\\\=\frac{-32+126i}{64+36}=\frac{-32+126i}{100} \\\\\frac{(1+2i)^2}{2+i}=\frac{1+4i+4i^2}{2+i}=\frac{1+4i-4}{2+i}=\frac{(-3+4i)(2-i)}{(2+i)(2-i)}=\frac{-6+3i+8i+4}{4-i^2}=\\\\=\frac{-2+11i}{4+1}=\frac{-2+11i}{5}\\\\\frac{5+12i}{8-6i}+\frac{(1+2i)^2}{2+i}=\frac{-32+126i}{100}+\frac{-2+11i}{5}=\frac{-32+126i-40+220i}{100}=\\\\=\frac{-72+346i}{100}=-0,72+3,46i$