hello my name is jared skeins and welcome to the zoom room today we want to cover pure mass 3 complex numbers part 3 modulus and argument which is basically talking about the trigonometric form and the exponential form of a complex number so first i'm going to give you this formula it's called demavra's theorem and it's going to be up to you how much if at all you use it in some of the examples that i have for you we don't actually use this particular formula but it is an extremely uh formula for you to be aware of that it is at your disposable disposal and there might be certain times where you choose to use this method as opposed to a different method so after we go through some of the questions at the end of the video i might give you a short demonstration illustrating how this formula can be useful as you compare the different platforms that is the algebraic platform versus a trigonometric platform versus an exponential platform so i'll try to give an illustration and whether or not you actually choose to use it in a particular cambridge question will be up to you so what we want to do now is just basically review and introduce this trigonometric aspect of complex numbers really it's just like when you learn trigonometry uh in the beginning you have your unit circle and in your unit circle you have this right triangle put so instead of focusing on the cartesian coordinate system where you have x and then y value to get to a particular point so instead of having the x y now you're going to have the r theta so another way of doing it is r theta kind of like this so there are two different ways of getting to the same location one is with your cartesian coordinate where you go over on the x and up on the y and then the other one is called polar coordinates polar coordinates have concentric circles and so you start facing zero you rotate a particular angle and then when you're facing that angle you go forwards a certain number of spaces so the how far you go is the r and the theta is your angle of rotation from zero and then you learn because it's a right triangle and you use your sohcahtoa you know that uh sine is the opposite over the hypotenuse and cosine is the adjacent over the hypotenuse so your adjacent side is the x-axis and so x equals r times cosine of theta and your y-axis is our sine of theta so what we do is we when we introduce complex numbers now it's fall instead of the x-axis and y-axis it's the real axis and the imaginary axis so your x-axis is the real axis your y-axis is the imaginary axis but it ends up working basically the same way and we will use the designation r cis theta cis again is the abbreviation of r cosine theta plus r times i sine theta okay so that's the abbreviation so the i actually goes with the sine theta and your r cosine theta is the real component okay so what we're going to do here in the first example let me take you to a question here so you can see it doesn't really say much it just says find the exact modulus and argument of you so here's our modulus argument again when you see those words you want to think the trigonometric platform and so we go back to our board here and the u i've already written up here it was actually from the previous part of the question our u is a negative root 3 plus i and now we want to find the modulus and argument in other words we want to convert it from the algebraic platform that is the a plus bi and convert it into the trigonometric platform the rsis theta so remember these are like sides of your triangle so minus root 3 plus i is going to put you in the second quadrant you do need to pay attention to the quadrant the rest of it is quite easy but for the angle it is important to know what quadrant you're supposed to be in so back on the x-axis up on the y-axis that second quadrant so i'll just write that right here so that we don't forget second quadrant now to find the r the r is just the hypotenuse of this right triangle so we can just use the pythagorean theorem so for our our our for the r value is going to be the square root of negative root 3 squared plus 1 squared it's an understood one on your eye and so you have negative root 3 and 1 and you square it and you end up with 3 plus one which is root four which comes out evenly as a two then we also want to find the angle part so to find the angle which is the argument the modulus is the magnitude or the distance that you go forward that's two the argument part is the angle that's the argument of the trig function the angle and so to find the angle notice if we have the y and the x we can use tangent from our sohcahtoa tangent is opposite over adjacent so tangent of the angle is going to equal the opposite which is 1 over the adjacent negative root 3. so theta equals the inverse tangent of 1 over negative root 3 and when you solve this on your calculator don't forget to be in radian mode uh radians is kind of the understood domain that we're working with with complex numbers and so our theta when you do that in radians comes out to a negative pi 6 now before we write this out we need to check what quadrant we're in remember we're in second quadrant tangent is negative in the second and fourth quadrant remember you're all students take calculus so all students take calculus tangent is positive in the first and third negative in the second and fourth so when we say a negative pi 6 that's going this way from zero it's going in the negative direction that's a fourth quadrant angle we want the pi 6 up here in the second quadrant so we're going to take pi minus pi 6 so that's 6 pi 6 minus 1 pi 6 that's going to give us 5 pi 6. so our modulus argument form is going to be 2 cis 5 pi 6. so there we've converted a complex number that's in algebraic form and we've put it into our modulus argument form okay let me set up another question okay for this next problem we're actually working in the opposite direction says of course throughout this question the use of a calculator is not permitted so remember you're supposed to be able to work with certain trigonometric features without using your calculator complex number with modulus 1 and argument pi thirds is denoted by w express w in the form of x plus i y where x and y are real and exact it's only one point so you need to know what you're doing right away so that you don't take too long on this particular question type and basically they're giving you the modulus and argument so you actually have one cis pi thirds the modulus is the r the argument is the angle and they want us to convert it into uh basically uh algebra format accepted using the vector notation instead of the a plus bi but it's still in your algebraic format and so all we have to do is actually just follow this through what it says one cis pi thirds means that we do one cosine of pi thirds and we do one sine of pi thirds now remember pi thirds is 60 degrees i i think better in degrees than i do radians so 180 divided by 360. we're not allowed to use our calculator so we're going to pull up our special triangle here here's our 30 16 90 right triangle and remember your sides are one two and root three it's one of the two special triangles that you should have memorized so if we're doing cosine of 60 that's adjacent over hypotenuse so this equals one half and if we do sine of 60 it's opposite over hypotenuse that's root three over two we also want to pay attention to the quadrant so pi thirds is in the first quadrant meaning both of these are going to be positive because it's over and up when it's in the first quadrant 60 degrees is in the first quadrant so now we can write this as w equals one-half plus root 3 over 2 i and there we have our complex number now written in the uh vector format instead of in the trigonometric format so not too hard pretty easy let's try something different okay finally we get something a little bit more challenging than simply converting from one to the other so let's take a look at this one says again not allowed to use your calculator the complex number root 3 plus i is denoted by u express u in the form r e to the i theta this is our exponential form and of course our domains between negative pi and pi we talked about that in the first video and then hence or otherwise state the exact values of the modulus and argument of u to the fourth power notice the word hence we're going to use that to our advantage so i wrote here on the board our u in algebraic form u equals root 3 plus i and they want us to come up with the exponential form and they also want us to find u to the fourth power but it used the word hence so first thing we're going to do is we're going to convert it to the trigonometric form first so again we need to find the r the r equals the square root of root three squared plus one squared equals root three plus one equals root four equals two so there's our r and we also need to find the theta so the theta is going to be the inverse tangent so i just went ahead and moved the tangent over inverse tangent of opposite over adjacent so 1 over root 3 and we end up with pi 6 and radians again since you're not allowed to use your calculator you're you're gonna want to use your special triangle just like we did on the previous one it's it's still uh uh except this time we're going the other way so when you look at your special triangle i guess i'll draw that again really quickly so that you can here's 30 60 91 2 root 3. this is a 1 over root 3 1 over root 3. um where now we have six i think in the previous one we were dealing with pi thirds this one is pi six uh but we have one over root three opposite over adjacent is coming from this angle right here opposite is one adjacent is root three that's this so we know it's the 30 degrees convert your 30 degrees into radians so 180 divided by 6 is 30. so you have pi 6 is this okay so use your special triangles you're not supposed to use your calculator so now we end up with our trigonometric form which is 2 cis pi 6. okay we're in the first quadrant over and up so we're okay on that to the right and up so that's first quadrant so this is the trigonometric form now to move it into the exponential form is very easy because it already has r and theta in it so all you have to do is rewrite it as 2 e to the pi 6 i okay 2 e to the theta i i think cambridge says i theta it's multiplication order doesn't matter if they're i like to put i at the end so 2 e to the pi 6 i or you see your r and your theta so it converts very easily you just need to know the form so that's the first part of your answer right here is the exponential form then it says find the modulus and argument of u to the fourth power you can actually do this several ways and this is where we're going to actually choose to use the mavr's theorem you could go back and do the fourth power of root three plus i but that would take a while to do you'd have to do root 3 plus i times root 3 plus i plus root 3 plus i times root 3 plus i times another root 3 plus i you'd have to have four of those and it take a while to multiply all that out algebraically this is where the mavr's theorem is so wonderful is when it comes to multiplication it's so much easier to work in the trigonometric or exponential form so you can do this actually one of two ways we'll look at both of these you can do the demovrous theorem which deals with the trigonometric so we're going to put this here to the fourth power because this is our u we already calculated our u now in trigonometric form raise it to the fourth power we're going to use de mauvre's theorem and all that you need to do is take two to the fourth power so two squared is four and another two squared is four four times four is sixteen so we get sixteen cis and then you take 4 times the pi 6 so 4 pi over 6 reduces to 2 pi over 3. so here is the modulus and here is the argument the theta the modulus and the argument so that's one way of doing it is doing it that way the other way that you can do it is you can do it using exponential rules so this time i'll put this in blue raise it to the fourth power remember exponents distribute to multiply so you still get two to the fourth power which is 16 e and when you have an exponent raised to an exponent you just multiply and so you still get 2 pi thirds over i so you get the same result you still can see your modulus right here and you can still see your angle right there the argument two root three two pi thirds and it's just a matter of which platform you're going to work in again the algebraic platform would take a long time because you have to manually multiply all those out or if you're really good with pascal's triangle uh you could try to do this with pascal's triangle but i wouldn't really recommend it because these other ways are so much easier to use the mavre's theorem is the trigonometric application and then in the exponential you're just following your exponent rules that you learned back in junior high school so either way you get your modulus your argument and you have your exponential form that they wanted you to end up in okay let's look at another problem here okay got an interesting question coming up here here we have find the exact value of a for which the argument of the conjugate of u is equal to pi thirds remember u asterisk or u star is the conjugate of u so find the exact value of a for which the argument of the conjugate of u is equal to pi thirds now we already solved part i pre in a previous video so i have already put that on the board for us so what we have here is actually the argument in the previous part i which we solved in one of the previous videos either part one or part two we had u equals six over a squared plus four plus three a over a squared plus four now the conjugate or the u asterisk is six over a squared plus four minus three a over a squared plus four i okay so that's our conjugate and now what we need to do is find the value of a for which the argument of this is equal to pi thirds so let's unpack that going backwards what is the argument of you uh asterix well the argument would be your tan theta equals the opposite that is the negative three a over a squared plus four over the adjacent opposite over adjacent six over a squared plus four of course this is a complex fraction it's going to invert and multiply so the a squared plus four will end up on top and the six will end up on the bottom and you can see that the a squared plus 4 cancels out you can also see that the negative 3 and 6 cancel so what we get is tan theta equals negative a over two negative a over two so tan theta equals a negative a over two and we want to solve for theta so theta equals the inverse tangent of a negative a over 2. now we want to solve for a given that the argument is pi third so if the argument is pi thirds then that is the angle so pi thirds equals the inverse tangent of a negative a over two so now we're going to work backwards this way so tan of pi thirds equals a negative a over two well what is a tangent of pi thirds so let's go back to our triangle maybe i should have drew it over here and just left it over here so here's our 30 60 90 1 2 root 3 and we have pi thirds remember pi thirds 180 divided by 360. so tangent is opposite over adjacent it's just root three or root three over one so root three equals a negative a over two so when you solve for a a equals a negative two root three is the value of a a negative two root three multiply by two divide by the negative a equals a negative 2 root 3. that is our value of a in here for which the argument of u is pi thirds okay so again focused on what argument means so to find the argument you use your tangent then once you find that the it's actually already giving you the answer the answer is pi thirds and then you've noticed that you've already reduced down to where you have a in a manageable solvable situation where you can get a by itself okay now i'll set up for one more it won't be a cambridge question but it will be another illustration of how you can use demovra's theorem so that maybe gives you a idea we already looked at it in the exponent with that fourth power came in really handy and i'd like to also show you one more application of it just in case you run into it okay i just want to do a simple illustration to show you the flexibility of these platforms and how you can move from one to the other and maybe it will help you understand also a little bit better how this democracy theorem can help you last time we looked at a fourth power but what if we have a root like a fractional power like a square root or cube root so i'll just run you through this simple illustration hopefully it will help you make some connections between the different platforms you're working with and also which tool will be more beneficial in which platform so let's just take a simple concept like x cubed equals eight of course we know that if we take the cube root of eight we get two but actually there are three cube roots of eight and we always think of two because it is the real cube root there are two imaginary cube roots that are um conjugate pairs so how can we come up with those well here's a little trick that you can use these roots have to occur in our complex number field here where we have our wheel and our imaginary so if these three roots occur in this field there's 360 degrees so we can take 360 degrees divided by 3. that means there's 120 degrees between each root we already know that 2 is right here on the real number line and so 120 degrees would be over here and we also know if you go 120 degrees more it's going to be down here so here you have zero there you have 120 here you have 240 and then you're back to 360. so there's 120 degrees between each of our roots and our groups are length of 2. now if we take these and make a right triangle with it and you know that this is 180 180 minus 20 gives us a 60 degree and we know then that this is 30 and we have a 1 2 root 3 and the same thing coming down here we have one two and three all we have to do now is put this into the proper quadrant this one is second quadrant so not only do we have two but x also equals a negative one because it's going backwards on the x-axis and up plus root three i and negative one minus root three i there are your three roots of eight x cubed equals eight gives you two negative one plus root three i negative one minus root three i won't show you the proof but i'll let you try you want to cube obviously if you cube 2 you're going to get 8. you can also cube this negative 1 plus root 3i multiply it out and you'll be surprised that you will get 8 and you can take negative 1 minus root 3i and cube it remember you need to do negative 1 minus root 3i times negative 1 minus root 3i times negative 1 minus root 3i and again you will get 8. i'll let you try that what i want to show you is the flexibility of these other platforms so in the algebraic platform it was very easy to get the real value but the imaginary values we kind of had to work with it and think about it a little bit in order to get those but what if we go into the trigonometric platform so if we take this into the trigonometric platform this would be eight cis zero degrees x equal or x cubed equals eight fifths zero eight on the real number line and there's no bi a plus bi if you think of a plus bi eight is the a there's no bi there's no imaginary part so you're facing the zero degree because there's no imaginary part so then if we cube root that x equals eight cis zero degrees to the one third now we're talking about a fractional exponent doing roots instead of exponents and your formula still works the same way so we end up with eight to the one third power so x equals two eight to the one third power cis zero times one third n times theta zero times one third is zero so the first root is at two zero notice that's exactly where we end up then all you have to do is add your 120 degrees so you have two cis 120 and you also have 2 cis 240 and of course if you use your calculator to convert that doing 2 times cosine of the angle you will get the x component and if you take 2 times for example cosine of 120 is a negative one-half negative half times two is your negative one okay so if you take your two cosine of the angle you'll get the a value and if you take two times sine of the angle you will get the y value or the b the b value in a plus b i format so this was very easy to come up with the three we didn't have to do this extra thinking all we had to do was know how many degrees are between each root and we were able to come up with it quite handily what if we are in the uh exponent form so let's take this down here and go over here and we have x equals uh eight e to the zero i we have zero degrees right so here's our r here's our theta well what's zero times i well zero times i is zero and what is anything to the zero power one one times eight is eight so this we want to be x cubed equals eight e to the zero i that matches our r sub theta our e to the theta i and it matches our eight which is in the algebraic form up here so algebraic trigonometric exponential so e to the zero is one one times eight is eight now we wanna cube root that so x equals 8 e to the 0 i one third again cube rooting it showing fractional exponent instead of a radical and again you distribute it very similar to this so you end up with x equals two a to the third power two e to the zero i and then you end up with 2 e to the 120 i and also 2 e 2 4 i now i use degrees you can plug radians into those actually would be better i just showed it to you in degrees because i think faster in degrees technically you should put those in radians but that's easy enough to do the zero is the same thing whether it's zero degrees or in zero radians so i i showed it to you in degrees because the point is to show you the flexibility if you are in the exponential platform you're simply using exponent rules if you are in the trigonometric platform you're going to use the barbarous theorem and if you're in the algebraic platform where you kind of have to work things out manually you can use pascal's triangle if you are familiar with how to do that and or you can use some a little bit of geometry trigonometry going on in here to try to figure it out so that's how they all tie together i hope that kind of helps you see some flexibility that you have depending on the cambridge question that you get what's important is all of your prerequisite skills are at your disposal and you want to use those all you're doing is then applying it to complex numbers i hope that helps and thank you for joining me hope to see you again next time for part four where we look at the graphing part of complex numbers on the argan diagram thanks again and see you next time