Question

Помогите пожалуйста решить

Answer 1

Ответ:

$\displaystyle 1)\ \ \frac{1-4i}{2+3i}=\frac{(1-4i)(2-3i)}{(2+3i)(2-3i)}=\frac{2-3i-8i+12i^2}{4-9i^2}=\frac{2-11i-12}{4+9}=\\\\\\=\frac{-10-11i}{13}=-\frac{10}{13}-\frac{11}{13}\, i\\\\\\2)\ \ \frac{2+3i}{3+2i}+\frac{1-i}{2\cdot i}=\frac{(2+3i)\cdot 2i+(1-i)(3+2i)}{(3+2i)\cdot 2i}=\\\\\\=\frac{4i+6i^2+3+2i-3i-2i^2}{6i+4i^2}=\frac{3i+3+4i^2}{6i-4}=\frac{3i+3-4}{6i-4}=\frac{3i-1}{6i-4}=\\\\\\=\frac{(3i-1)(6i+4)}{(6i-4)(6i+4)}=\frac{18i^2+12i-6i-4}{36i^2-16}=\frac{-18+6i-4}{-36-16}=\frac{-22+6i}{-52}=$

$=\dfrac{22}{52}-\dfrac{6}{52}\, i=\dfrac{11}{26}-\dfrac{3}{26}\, i$