Question 27911
For the most part, you can treat i's just like x's (it's only when we get to i^2 that we have to do anything differently).

So we start by using FOIL to get:
{{{(5+3i)(2+i)=5*2+5*i+3i*2+3i*i=10+11i+3i^2}}}
The difference now comes in that i^2 term at the end. Since i is the imaginary number defined by {{{i=sqrt(-1)}}} then {{{i^2=-1}}} so whenever you see an i^2 term, replace it with a negative 1:
{{{10+11i+3(-1)}}}
{{{10+11i-3}}}
{{{7+11i}}}