Multiplying Complex Numbers - Problem 2 3,437 views
Multiplying complex numbers or numbers involving i. When we multiply numbers that involve i we pretty much use it as we would any other number, we’re just going to have to distribute it through and the only thing we really have to be careful of is we ever see an i² we know that equals negative 1.
Whenever we’re dealing with the square roots of a negative number we have to simplify it to have i instead of multiplying. For these two examples we actually don’t have to deal with that problem so we can just deal with them as we would any other problem and just distribute these through.
Anytime we have something like this we just have to distribute our coefficient in, so that’s just going to distribute this 3i in leaving us with 21i and then minus 12i². Remembering that i² is -1, this is -1 we actually end up with -12 times -1 which is 12, and then plus our 21i.
In general we write the real number, the number without i first, but it doesn’t matter order can be changed. Just in terms of general math we tend to prefer that real component first.
When we’re dealing with multiplication of two binomials, both involving i. If these were xs instead of i’s we would just FOIL it out, same exact idea when we have i’s. We’re just going to FOIL this out, 2 times 3 is 6, 2 times -i is -2i, 5i times 3 is 15i and then we have 5i and -i which is -5i². From here we can combine like terms, so we have 6, minus 2i and plus 15i turns into 13i and minus 5i².
Just like we did up here, i² is equal to -1 so this last term -5, times -1 turns into a plus 5. We can combine that with the 6 over here leaving us with 11 plus 13i. Remember we’re dealing with multiplying complex numbers it’s really the exact same things as multiplying anything else just remember if you ever see i² switch it to -1 and simplify.