Modulus (absolute value) and argument (angle) of complex numbers
What I want to do in this video is make sure we're comfortable with ways to represent and visualize complex numbers. So you're probably familiar with the idea. A complex number, let's call it z-- and z is the variable we do tend to use for complex numbers-- let's say that z is equal to a plus bi. We call it complex because it has a real part and it has an imaginary part. And just so you're used to the notation, sometimes you'll see someone write the real part, give me the real part of z. This is a function, that you input a complex number, and it will output the real part, and in this case, the real part is equal to a. And you could have another function called the imaginary part of z. You input some complex number it'll output the imaginary part, or it'll say how much are scaling up i, and in this case, it would be b. This is a real number, but this tells us how much the i is scaled up in the complex number z right over there. Now, one way to visualize complex numbers, and this is actually a very helpful way of visualizing it when we start thinking about the roots of numbers, especially the complex roots, is using something called an Argand diagram. So this is this. And so it looks a lot like the coordinate axes and it is a coordinate axes. But instead of having an x and y-axis it has a real and an imaginary axis. So in the example of z being a plus bi, we would plot it really as a position vector, where you have the real part on the horizontal axis. So let's say this is a and then the imaginary part along the vertical axis, or the imaginary axis. So let's say that this is b. And so we would represent, in an Argand diagram, the vector z as a position vector that starts at 0 and that has a tip at the coordinate a comma b. So this right here is our complex number. This right here is a representation in our Argand diagram of the complex number a plus bi, or of z. Now when you draw it this way, when you draw it as a position vector, and if you're familiar with polar coordinates, you're probably thinking, hey, I don't have to represent this complex number just as coordinates, just as an a plus bi. Maybe I could represent this as some angle here, let's call that angle phi, and some the distance here, let's call that r, which is kind of the magnitude of this vector. And you could. If you gave some angle and some distance, that would also specify this point in the complex plane. And this is actually called the argument of the complex number and this right here is called the magnitude, or sometimes the modulus, or the absolute value of the complex number. So let's think about it a little bit. Let's think about how we would actually calculate these values. So r, which is the modulus, or the magnitude. It's denoted by the magnitude or the absolute value of z1. What's this going to be. Well, we have a right triangle here. This side is b, length b. The base right here has length a. So to calculate r, we can just use the Pythagorean Theorem. r squared is going to be equal to a squared plus b squared. Or r is going to be equal to the square root of a squared plus b squared. If we want to figure out the argument, this is going to be equal to what? So let's think about this a little bit. We have b and a. So what trig function deals with the opposite side of an angle and the adjacent side? So let me write all of, let me write the famous sohcahtoa up here. "Soh-cah-toa." Tangent deals with opposite over adjacent. So the tangent of this angle, which we called the argument of the complex number, the tangent of the argument is going to be equal to the opposite side over the adjacent side. It is equal to b/a. And so if we wanted to solve for this argument, we would say that the argument is equal to the arctan, or the inverse tangent, of b/a. Now, if we wanted to represent, let's say that we were given the modulus and the argument. Let's say we were given that. How do we go the other way? Right now if we have the a's and the b's, the real complex part, I just showed you how to get the magnitude and how to get the angle, or the argument. But if you're given this. How do you go the other way? Well here, if you're trying to figure out a, given r and theta-- so you're trying to figure out an adjacent side given angle and the hypotenuse. So adjacent over hypotenuse is equal to cosine. So you would have cosine of the argument is equal to the adjacent over the hypotenuse. It is equal to a/r. Multiply both sides by r, you get r cosine of phi is equal to a. Do something very similar for b. If we use sine, opposite over hypotenuse. Sine of the argument is equal to b/r. It is equal to b over the magnitude. Multiply both sides by r, you get r sine of phi is equal to b. So how would we write this complex number. So this complex number z is going to be equal to it's real part, which is r cosine of phi plus the imaginary part times i. Plus r-- let me do that same green-- plus r sine of phi times i. Now this might pop out at you as something that's pretty interesting, if you ever seen Euler's formula. Let's factor out this r over here. So this is going to be equal to-- factor out an r-- r times cosine of phi plus-- I'll put the i out front-- i sine of phi. Now what is this? And if you've seen the video, I do it in the Taylor series, a series of videos in the calculus playlist. And it's really one of the most profound results and all of mathematics, it still gives me chills. This is Euler's formula. Or this, by Euler's formula, is the same thing. And we show it by looking at the Taylor series representations of e to the x. And the Taylor series representations of cosine of x and sine of x. But this is, if we're dealing with radians, e to the i phi. So z is going to be equal to r times e to the i phi. So there's two ways to write a complex number. You could write it like this, where you have the real and imaginary part, that's maybe what we're used to. Or we can write it in exponential form, where you have the modulus, or the magnitude, being multiplied by a complex exponential. And we're going to see that this going to be super useful, especially when we're trying to find roots. Now just to make this tangible, let's actually do this with an actual example. So let's say that I had, I don't know, let's say that I had to z1 is equal to square root of, let's say it's square root of 3/2 plus i. And so we want to figure out its magnitude, and we want to figure out its argument. So let's do that. So the magnitude of z1 is going to be equal to the square root of this squared. So this is going to be equal to 3/4 plus 1 squared-- or I should say plus 4/4. So this is going to be equal to square root of 7/4, which is equal to the square root of 7/2. And now let's figure out its argument. So if I were to draw this on an Argand diagram, it would look like this. It's going to be in the first quadrant, so that's all I have to worry about. So let me draw it. Let me draw it like this. And so we have a situation. So it's going to be square root of 3, actually, let me change this up a little bit, just so the numbers get a little bit cleaner. Sorry about this. Let me make it a little bit, slightly cleaner. So just so that we have a slightly cleaner result, because we want to make our first example a simple one. So let's make this square root of 3/2 plus 1/2i.. So let's figure out the magnitude, the magnitude here is z1 is equal to the square root of, square root of 3/2 squared, is equal to 3/4 plus 1/2 squared is equal to 1/4,. This makes things a lot nicer. This is equal to the square root of 1, which is 1. And now let's think about it, let's draw it on an Argand diagram to visualize the argument. So this is my imaginary axis. This is my real axis. And so this complex number is square root of 3/2. The square root of 3 is like 1.7. So if we have like 1, it'll be like right over here, someplace right over here. This is square root of 3/2, the real part. The imaginary part is 1/2. So if this is 1, this is 1/2, the imaginary part is right over here, 1/2. And we actually also know its length, its length, or its magnitude is 1. So how do we figure out phi over here? We know that this side over here is square root of 3 over-- oh let me be careful-- we know that side over there is 1/2. That's the imaginary part. And we know the base is the square root of 3/2. So a bunch of ways we can do this. One, you could just do the tangent, because that involves the opposite over the adjacent. You could say that the tangent of phi is equal to the opposite, is equal to 1/2 over the square root of 3/2. And then you can take the inverse tan of both sides. So this would be the same thing as phi being equal to the inverse tangent, or the arctangent of-- if you multiply the numerator and the denominator by 2, this is 1 over the square root of 3. You could do it like that. You can also say that phi is equal to the inverse sine of, so the sine of phi is going to be equal to the opposite over the hypotenuse. So sine of phi is equal to 1/2 over 1, or phi is equal to the arcsine of 1/2. And you could put that into your calculator. Or you could recognize this is a 30-60-90 triangle. This base right here, square root of 3/2, this is 1/2, this is 1. So this angle right here is going to be 30 degrees. And that's just from pattern matching from a 30-60-90 triangle. You could look at these and also get something similar. Now I want to put this in radian form, because whenever I use the exponential form you want it to be in radians. So phi is equal to 30 degrees, which is the same thing as pi over 6. So if I wanted to represent z1 in exponential form, it would be the exact same thing as r, or its magnitude, which is 1-- I'll put the 1 out there even though you really don't have to-- 1 times e to the pi over 6i. And we're done.