Помогите решить 9) и 10)тема: комплексные числа

Решите задачу:

$9) \frac{1-3i}{-2+i}+ \frac{1+4i}{-1+3i} =\\ =\frac{(1-3i)(-2-i)}{5}+ \frac{(1+4i)(-1-3i)}{10} =\\ =\frac{-2+6i-i+3i^2}{5}+ \frac{-1-4i-3i-12i^2}{10} =\\ =\frac{-10+10i}{10}+ \frac{11-7i}{10} =\\ = \frac{1+3i}{10}$

$10) \frac{a+bi}{a-bi} - \frac{a-bi}{a+bi} =\\ = \frac{a^2-b^2+2abi-a^2+b^2+2abi}{a^2+b^2} =\\ = \frac{4abi}{a^2+b^2}$