10 класс комплексные числа 2 и 3 номер

Решите задачу:

$2)\quad \frac{-4+3i}{-4-3i}- \frac{-4-3i}{-4+3i}=\frac{(-4+3i)^2-(-4-3i)^2}{(-4-3i)(-4+3i)}=\frac{16-24i+9i^2-(16+24i+9i^2)}{16-9i^2} =\\\\= \frac{-48i}{16+9} = -\frac{48}{25} i\\\\3)\quad (2-i)z=3-2i\\\\z= \frac{3-2i}{2-i}= \frac{(3-2i)(2+i)}{(2-i)(2+i)}=\frac{6+3i-4i-2i^2}{4-i^2}= \frac{6-i+2}{4+1}=\frac{8-i}{5}= \frac{8}{5}-\frac{1}{5} i$