Помогите решить 5i/(3+2i) 6-7i/i

Решите задачу:

$\frac{5i}{3+2i} = \frac{5i(3-2i)}{(3+2i)(3-2i)} = \frac{15i-10i^2}{3^2-(2i)^2} = \frac{15i-10*(-1)}{9-4i^2} = \frac{10-15i}{9-4*(-1)} = \frac{10-15i}{9+4} = \\ = \frac{10-15i}{13} = \frac{10}{13} - \frac{15}{13} i$
$\frac{6-7i}{i} = \frac{(6-7i)(-i)}{i*(-i)}= \frac{-6i+7i^2}{-i^2} = \frac{-6i+7*(-1)}{-(-1)} = \frac{-7-6i}{1} =-7-6i$