Question

Помогите пожалуйста решить ​

Answer 1

Объяснение:

$\displaystyle\\\frac{(1+i)*(1+i)}{(1+2i)^3}=\frac{(1+i)^2}{(1+2i)^2*(1+2y)}=\frac{1^2+2*1*i+i^2}{(1+2*1*2i+(2i)^2)*(1+2i)} =\\ \\\\=\frac{1+2i+(\sqrt{-1})^2 }{(1+4i-4)*(1+2i)}=\frac{1+2i-1}{(-3+4i)*(1+2i)} =\\\\\\=\frac{2i}{-3-6i+4i-8} =\frac{2i}{-2i-11} =-\frac{2i}{11+2i}=-\frac{2i*(11-2i)}{(11+2i)*(11-2i)}=\\\\\\=-\frac{4+22i}{11^2+4} =\frac{-4-22i}{121+4} =\frac{-4-22i}{125} =-\frac{4}{125}-\frac{22}{125}i=-0,032-0,176i .$