Question

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

Answer 1

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