Question

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=

Answer 1

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