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Dividing Complex Numbers  Problem 2
Carl Horowitz
Carl Horowitz
University of Michigan
Runs his own tutoring company
Carl taught upperlevel math in several schools and currently runs his own tutoring company. He bets that no one can beat his love for intensive outdoor activities!
Dividing by complex numbers, so in this particular problem we are looking at a complex number over a complex number. First thing we want to do is simplify everything out so it’s in a form that looks a little bit more familiar to us and by that we have square root of 4 which is just going to be 2i and square root of 9 which is just going to be 3i.
Rewriting our problem we have 2, 1 plus 2i over 4 plus 3i. Now we can’t have square roots in the denominator and i is the square root of 1, so we somehow need to get rid of that, and we have to figure out what we can multiply by in order to get that i to disappear. We want to take a side note for a second.
Common thing is people just want to multiply by i. If we take 4 plus 3i and multiply it by i what we end up with is 4i plus 3i². This 3i², the i disappears so we end up with 4i minus 3, but what we’ve really done is we’ve kept our i and rearranged the order. So nothing’s really changed we haven’t gotten rid of that i all together.
What we have to multiply by is the conjugate which is the exact same numbers but just a different sign in between. What that means in this case is 4 minus 3i. And the reason we do that is that we have now a sum here and a difference here. When we FOIL that out what we end up getting is 16, we have plus 12i and minus 12i which disappear, so our single i term disappears and we have minus 9i². Remember i² is 1. This turns into minus 9 times 1 which turns into plus 9 so our denominator is now 25. So when you multiply by the conjugate all of our i’s disappear.
I just focused on our denominator I sort of left alone our numerator so let’s go back. Remember whenever you multiply by something it has to be 1, so we need a 4 minus 3i in the top as well. We have to FOIL this out and this time we’re not going to be quite as lucky because it’s not the conjugate, we’re going to be left with three terms instead of just the single term.
Let’s go over here and multiply this out. If we FOIL this out, 1 times 4, 4, 1 times 3i turns into plus 3i, 2i times 4 plus 8i and the 2i times 3i turns into 6i². We can combine like terms so this is 4 plus 11i and then i² is 1 this turns into 6 times 1 which is just plus 6. Combining more like terms the 4 and the 6, what we have it 2 plus 11i in the numerator, we still have the denominator which we found over here, the 25. So we put this over 25 and by multiplying by the conjugate we’re able to get the i’s out of the denominator.
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Carl Horowitz
B.S. in Mathematics University of Michigan
He knows how to make difficult math concepts easy for everyone to understand. He speaks at a steady pace and his stepbystep explanations are easy to follow.
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