Question

write (2-i)(4-3i)^2/3+2i in the form a+bi where a,b,er.

Answer 1

$\cfrac{(2-i)(4-3i)^2}{3+2i} = \cfrac{(2-i)(16-24i-9)}{3+2i} =\cfrac{(2-i)(7-24i)}{3+2i} = \\ =\cfrac{14-48i-7i-24}{3+2i} =\cfrac{-10-55i}{3+2i} \\ =\cfrac{-10\cdot3 -55\cdot2}{9+4}+\cfrac{-10\cdot2-55\cdot3}{9+4}i= -\cfrac{140}{13}-\cfrac{145}{13}i$