Question 701723
{{{((4-3i))/((7+3i))}}} 
multiply by the conjugate of the denominator over itself
{{{((4-3i))/((7+3i))}}} * {{{((7-3i))/((7-3i))}}} = {{{((28-12i-21i+9(i^2)))/((49-21i+21i-9(i^2)))}}} =
Notice that when you FOIL the denominators, the middle term containing "i" drops out, and i^2 = -1
{{{((28-33i+9(-1)))/((49-9(-1)))}}} = {{{((28-33i-9))/((49+9))}}}  = {{{((19-33i))/((58))}}}