If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content
Current time:0:00Total duration:13:04

Video transcript

what I want to do in this video is make sure we're comfortable with ways to represent and visualize complex 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 number let's say that Z is equal to a plus bi we call it complex because it has a real part it has a real part and it has an imaginary 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'll 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 the imaginary part of Z you input some complex number it'll output the imaginary part the imaginary part or it'll it'll say how much you're scaling up I and in this case it would be be it would be B this is a real number but this tells us how much the eye 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 Argand diagram 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 it has a real and 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 zero and that has a tip at the coordinate a comma B so this right here 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 U and if you're familiar with polar coordinates you're probably thinking hey I don't have to represent this complex number just as kind of coordinates just as an a plus bi maybe maybe I can represent this as some angle here as some angle here let's call that angle Phi and some distance here let's call that R which is kind of the magnitude of this vector and you could if you if you gave some angle in some distance that would also specify this point in a 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 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 let's say 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 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 so Toa tangent deals with opposite over adjacent so the tangent the tangent of this angle which we call the argument of the complex number the the argument is going to be equal to the opposite side over the adjacent side it is equal to B over a and so if we wanted to solve for this argument we would say that the argument is equal to the arc tan or the inverse tangent of B over a now if we wanted to represent let's say we had we had the complex number let's say that we were given the argument and the let's say that we were given let's say that we're given the modulus and the argument let's say we're given that how do we go the other way right now if we have the A's and the B's the real in the 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 give it an angle and the hypotenuse so adjacent over hypotenuse is equal to cosine so you would have cosine of the argument is equal to R is equal to D is equal to the adjacent over the hypotenuse is equal to a over R multiplied 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 over R is equal to B over the magnitude multiplied 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 the complex number Z is going to be equal to its real part which is R cosine of Phi R cosine of Phi plus plus the imaginary part times I plus R let me do that same green plus R sine of Phi times I times I now this might pop out at you is something that's pretty interesting if you've ever seen Euler's formula let's factor out this are over here so this is going to be equal to factor out an R R times cosine of Phi cosine of Phi plus I'll put the I up front I sine ax Phi I I sine ax Phi now what is this and if you've seen the videos 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 in all of mathematics it still gives me chills this is Euler's formula this is or this by Euler's formula is the same thing this is the same thing and we show it by looking at the Taylor series representations of e to the 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 e to the I Phi so Z is going to be equal to R is going to be equal to R times e to the I Phi e to the I Phi so there's two ways to write a complex number you can write it like this where you have the real and the 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 we're going to see that this can 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 with an actual example so let's say that I had I don't know let's say that I had z1 is equal to let's say that z1 is equal to square root of let's say it's square root of 3 over 2 square root of 3 over 2 plus plus I and so we want to figure out we want to figure out the we want to figure out its magnitude and we want to figure out its argument so let's do that so the magnitude the magnitude of Z 1 is going to be equal to the square root of this squared so this is going to be equal to this is going to be equal to 3 over 4 3/4 plus one squared plus one squared right should say plus four fourths plus four fourths so this is going to be equal to square root of 7 over 4 which is equal to the square root of 7 over two and now let's figure out its argument so let's draw if I were to draw this on an Argand diagram on an Argand diagram it would look like this so I'm 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 know we want to make our first our first example a simple one so let's make the square root of 3 over 2 plus 1/2 plus 1/2 I so let's figure out the magnitude the magnitude here z1 is equal to the square root of square root of 3 over 2 squared is equal to 3 over 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 that is my imaginary axis this is my real axis my real axis and so this complex number is square root of 3 over 2 square root of 3 is like 1.7 so if we have like 1 it'll be like so this is 1 it'll be like right right over here someplace right over here this is square root of 3 over 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 how do we figure out 5 over here we know that this side over here is square root of 3 over I'll let me be careful we know that side over there is 1/2 that's the imaginary part and we know the base is square root of 3 over 2 so a bunch of ways we can do this one you could just look at 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 over 2 and then you could take the inverse tan of both sides so this would be the same thing as 5 being equal to the inverse tangent or the arctangent of if you multiply the numerator in the denominator by 2 this is 1 over the square root of 3 you could do it like that you could 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 arc sine of 1/2 and you could put that in your calculator or you could recognize this is a 30-60-90 triangle this base right here is square root of 3 over 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 Phi is equal to 30 degrees which is the same thing 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 1 times e to the PI over 6 i e to the PI over 6 I and we're done