Question 1180448
 Determine without converting the form and leave answer in the form of a+bi
{{{((2-i)(3+4i))/(i+3)}}}
same as
{{{((2-i)(3+4i))/(3+i)}}}
FOIL the numerator
{{{(6+8i-3i-4(i^2))/(3+i)}}} = {{{(6+5i-4(-1))/(3+i)}}} = {{{(6+5i+4)/(3+i)}}} = {{{(10+5i)/(3+i)}}}
multiply by the conjugate of the denominator over itself
{{{(10+5i)/(3+i)}}} *{{{(3-i)/(3-i)}}} = {{{((10+5i)(3-i))/((3+i)(3-i))}}} = {{{(30-10i+15i-5(i^2))/(9-i^2)}}} = {{{(30+5i-5(-1))/(9-(-1))}}} = {{{(30+5i+5)/10}}} =
{{{(35+5i)/10}}}
cancel out 5
{{{(7+i)/2}}} or (3.5 + .5i)