Question

2+i/3-i

Answer 1

I am sure you mean
(2+i)/(3-i)

multiply by (3+i)/(3+i)
Since that has a value of 1, we are not changing the value of our original fraction, only its appearance.
(2+i)/(3-i)(3+i)/(3+i)
= (6 + 5i + i^2)/(9-i^2) , recall that i^2 = -1
= (5+5i)/10
= (1+i)/5

Answer 2

yes that is indeed what i meant ...except when you factor
(3+i)(3+i) would it be 9+6i+i^2?
Then what would you do?

Answer 3

Just noticed your reply as I was about to log off ...

Where do you see a multiplication of (x+3)(x+3) ??

Here is what I had:
(2+i)/(3-i)(3+i)/(3+i)
If you follow the order of operation that is

[(2+i)(3+i)] / [(3-i)(3+i}
= as you see above