Question 666452
1) (-3-3sqrt(3))^2 = (-3)^2*(1+sqrt(3))^2 = 9*(1+2*sqrt(3) +3) = 18*(2+sqrt(3))
2) 5/i = 5*i/i*i = 5i/(-1) = -5i
3) 1/(2i-3) = (2i+3)/((2i-3)*(2i+3)) = (2i+3)/(-4+2i-2i-9) = (2i+3)/(-13) = -3/13 - (2/13)i
4) 3+(2i)/(2-i) = 3+(2i)(2+i)/(2-i)(2+i) = 3 + 2i(2+i)/5 = 3+(4i-2)/5 = (13+4i)/5 = 13/5 + (4/5)i
5) (3+2i)/4 = 3/4 + i/2
6) 1/(3+sqrt(-2)) = 1/(3+i*sqrt(2)) = (3-i*sqrt(2))/((3+i*sqrt(2))*(3-i*sqrt(2))) = (3-i*sqrt(2))/(9+2) = 3/11 - i*(sqrt(2))/11