Question 252022
(2-4i)/2 = 2(1-2i)/2 = 1-2i



(2+4i)/(-5+2i) = (2+4i)(-5-2i)/(-5+2i)(-5-2i) 
               = (-10-4i+20i+8)/[(-5)^2-(2i)^2]
               = (-2+14i)/(25+4)
               = 2(-1+7i)/29


(1-i)/(-4-6i) = (1-i)(-4+6i)/(-4-6i)(-4+6i)
              = (-4+6i+4i+6)/[(-4)^2-(6i)^2]
              = (2+10i)/(16+36)
              = 2(1+5i)/52
              = (1+5i)/26


(3+2i)/(-2+i) = (3+2i)(-2-i)/(-2+i)(-2-i)
              = (-6-3i-4i+2)/ [(-2)^2-(i)^2]
              = (-4-7i)/(4+1)
              = -(4+7i)/5