Question

escrevam na forma z=a+bi os números complexos... (3-4i)(4-3i)÷3-2i

Answer 1

$(3-4i)(4-3i):3-2i= \\ \\ (12-9i-16i+12i^2): 3-2i= \\ \\ (12-25i+12(-1)): 3-2i= \\ \\ (12-25i-12): 3-2i= \\ \\ \dfrac{-25i}{3-2i} =\\ \\ \\ \dfrac{-25i}{3-2i} *\dfrac{3+2i}{3+2i} =\dfrac{-25i(3+2i)}{(3-2i)(3+2i)}=\dfrac{-75i-50i^2}{3^2-(2i)^2}=\dfrac{-75i+50}{9+4}= \\ \\ \\ \dfrac{-75i+50}{13}= \boxed{-\dfrac{75}{13}i+\dfrac{50}{13}}$