1) (1-i)(1+i)^2= 2) (1-i)^3/(1+i)^5= 3) (1-i)^3/(1+i)^7= 4) (1-2i)^2-(1+i)^3/(3-2i)^3-(2-i)^2=

[latex]1) (1-i)(1+i)^2=(1-i)(1+i)(1+i)=(1^2-i^2)(1+i)=\\=2(1+i) \\2) \frac{(1-i)^3}{(1+i)^5}=\frac{(1-i)^3(1-i)^5}{(1+i)^5(1-i)^5}=\frac{(1-i)^8}{((1+i)(1-i))^5}=\frac{((1-i)^2)^4}{(1-i^2)^5}=\\=\frac{(1-2i-1)^4}{2^5}=\frac{16i^4}{32}=0.5[/latex] [latex]3) \frac{(1-i)^3}{(1+i)^7}=\frac{(1-i)^3(1-i)^7}{(1+i)^7(1-i)^7}=\frac{(1-i)^{10}}{((1+i)(1-i))^7}=\frac{((1-i)^2)^5}{(1-i^2)^7}=\\=\frac{(1-2i-1)^5}{2^7}=\frac{2^5}{2^7}=0.25[/latex] [latex]4) \frac{(1-2i)^2-(1+i)^3}{(3-2i)^3-(2-i)^2}=\frac{1-4i-4-(1-3+3i-i)}{27-12-54i+8i-(4-4i-1)}=\frac{-3-4i+2-2i}{15-46i-3+4i}=\\=\frac{-6i-1}{-42i-12}=\frac{6i+1}{6(7i+1)}=\frac{1}{6}*\frac{(6i+1)(7i-1)}{(7i+1)(7i-1)}=\frac{1}{6}*\frac{-42+i-1}{-49-1}=\\=\frac{43-i}{300}[/latex]