And welcome. My name is Anthony Varela, and today we're going to learn about complex numbers in electrical engineering. So we're going to be using complex numbers in problems that involve this formula for electrical circuits, and specifically, we're going to be multiplying and dividing two complex numbers.
So first, let's review what a complex number is, as well as complex conjugates. So a complex number has a real part and an imaginary part with this imaginary unit, i, which is the square root of negative 1. Now, this is imaginary because no real number squared results in a negative number. They're always positive. So the square root of a negative number is imaginary.
I'd also like to talk about complex conjugates, which are used when we divide two complex numbers. And basically, we reverse the sign between a and bi. So for example, 7 plus 3i and 7 minus 3i are complex conjugates. And we use these when dividing complex numbers so that our denominator doesn't have any imaginary parts to it at all, and we'll see that in just a couple of minutes.
Now, when we use complex conjugates, we multiply it by another complex number. So we use FOIL to do this multiplication. So let's go through an example using non-complex numbers to review the process for FOIL. Well, FOIL stands for First Outside Inside Last. So we multiply the first terms, the outside terms, the inside terms, and then the last terms. And then we can combine like terms, so we have in this case x squared minus x minus 6 is this binomial multiplication FOILed out.
Well, let's get to our examples involving complex numbers, and we're going to be using this formula V equals I times R. So V stands for voltage. It's measured in volts, I is the current in amps, and R is the resistance in ohms. Now, just a couple of notes before we move on is commonly engineers will use the letter j to represent i, and this is because this equation has a capital I, and we don't want to confuse the imaginary unit with current. So if you see j, just know that j equals i, which is the square root of negative 1.
You might also see engineers write or use j in a different way. So they might write a complex number as 3 plus j6. This is the same as 3 plus 6j. This is the same as 3 plus 6i, so a couple of different ways to write the same thing. Well, we're going to go through two examples using this formula. Our first one is that an electrical circuit has a current of 2 plus j3 amps, and the resistance of 5 minus j2 ohms. What is the voltage of the circuit?
Well, we're given information about current and resistance, and if you multiply those two together, we'll get our voltage. So here, we have to multiply two complex numbers. Now, if you like, you can write this using i in a way you may be more familiar with. So we're going to use FOIL to multiply these two complex numbers.
So first, we get 2 times 5, then we get 2 times negative 2i, then we get 3i times 5, and finally, 3i times negative 2i. So we can combine our negative 4i and our positive 15i, but now let's think about our minus 6i squared. Well, since i is the square root of negative 1, i squared equals negative 1. So to rewrite negative 6i squared, you can eliminate i squared but change the sign, so we have plus 6.
Well, notice that's a real number. We can add it to 10. So our complex number here is 16 plus 11i, and if you prefer, you can write this as an engineer might 16 plus j11 volts.
Our second example reads, an electrical circuit has a voltage of 13 minus j34 volts and a current of 7 minus j2 amps. What is the circuit's resistance? So here were given information about the volts, we're given information about the current, but not the resistance. So plugging what we know into V equals I times R, we see that we need to isolate R, so we can divide both sides of our equation by that current, and now we have R equals the voltage divided by the current.
Now, we could write this then as 13 minus 34i divided by 7 minus 2i, different way to write the same thing. So now you see that we have to divide two complex numbers. Now, when dividing two complex numbers, we use the complex conjugate. So I'm going to multiply this by a fraction, and I'm going to have the denominator's conjugate here, so 7 plus 2i, and to keep things equal, I have to make that my numerator as well.
So really, what we're doing when we're dividing two complex numbers is we're multiplying two complex numbers twice. So let's focus on our numerators, so FOILing out our two numerators, 13 times 7, then we have 13 times 2i, and next we have negative 34i times 7, and finally, negative 34i times 2i.
So once again, we're going to combine the two terms in the middle. So we have 91 minus 212i minus 68i squared. Once again, that nice trick to simplify minus 68i squared is to eliminate the i squared but change the sign out in front. That can be added to 91 because that's a real number. So we have 159 minus 212i. That is our simplified numerator.
We have to FOIL two more complex numbers, this time our denominators. This involves our complex conjugate, so pay attention to how this works. This is pretty neat. 7 times 7 is 49. Then we have 7 times 2i, then we have negative 2i time 7, and finally, negative 2i times positive 2i.
Notice that when we combine 14i and negative 14i, that results in 0, so that actually goes away. We just have 49 minus 4i squared. But we can write negative 4i squared as a positive 4. So we actually don't have any imaginary numbers in our denominator at all. This is just 53.
So now what we'd like to do to finish this is to divide both 159 by 53 and negative 212i by 53. Now, this doesn't always result in a nice, clean number, but in this case, it does. 159 divided by 53 is 3, and negative 212 divided by 53 is 4, so we have minus 4i, and that is then our resistance. So as an engineer, you might write this as 3 minus j4 ohms.
So let's review then our lesson on complex numbers in electrical engineering. Well, complex number has a real part and an imaginary part with an imaginary unit square, root of negative 1. We used complex conjugates when dividing, and that's just changing the sign between the real part, a, and then bi. And we used FOIL then to multiply two complex numbers, and we actually used it in division as well, and our two problems were related to voltage equals current times resistance.
So thanks for watching this tutorial on complex numbers in electrical engineering. Hope to see you next time.
