What does ##(6+4i)/(3-i)## equal?

I found: ##7/5+9/5i##

In this case you need to first change the denominator into a pure real number; to do that you need to multiply and divide by the complex conjugate of the denominator, i.e.: ##(6+4i)/(3-i)*color(red)((3+i)/(3+i))=## ##((6+4i)(3+i))/(9+1)=## where you used the fact that ##i^2=-1## and: ##(18+6i+12i-4)/10=14/10+18/10i=7/5+9/5i##