Question 30908
(1+3i)/(1-2i)
In imaginary numbers, there's a rule that you can never have an i in the denominator.  To get rid of that, you multiply by the inverse of (1-2i) which is (1+2i) in both the denominator and numerator.  You must use FOIL to multiply the numbers. 
(1+3i)/(1-2i)* (1+2i)/(1+2i).  Multiply numerators together and denominators. 
(1+3i)(1+2i)/(1-2i)(1+2i)
(1 + 2i + 3i + 6i^2)/(1 -2i + 2i - 4i^2) 
(1 + 5i + 6(-1))/(1 -4(-1))
(1-6 + 5i)/(1 + 4)
(-5 + 5i)/(5) =now reduce
5(-1 + i)/5
-1 + i