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

1 Answer
Oct 21, 2015

Answer:

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

Explanation:

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#