Before I get started o this let me first make it clear that this documet is ot iteded to teach you everythig there is to kow about complex umbers. That is a subject that ca (ad does) take a whole course to cover. The purpose of this documet is to give you a brief overview of complex umbers, otatio associated with complex umbers, ad some of the basic operatios ivolvig complex umbers. This documet has bee writte with the assumptio that you ve see complex umbers at some poit i the past, kow (or at least kew at some poit i time) that complex umbers ca be solutios to quadratic equatios, kow (or recall) i =, ad that you ve see how to do basic arithmetic with complex umbers. If you do t remember how to do arithmetic I will show a example or two to remid you how to do arithmetic, but I m goig to assume that you do t eed more tha that as a remider. For most studets the assumptios I ve made above about their exposure to complex umbers is the extet of their exposure. Problems ted to arise however because most istructors seem to assume that either studets will see beyod this exposure i some later class or have already see beyod this i some earlier class. Studets are the all of a sudde expected to kow more tha basic arithmetic of complex umbers but ofte have t actually see it aywhere ad have to quickly pick it up o their ow i order to survive i the class. That is the purpose of this documet. We will go beyod the basics that most studets have see at some poit ad show you some of the otatio ad operatios ivolvig complex umbers that may studets do t ever see oce they lear how to deal with complex umbers as solutios to quadratic equatios. We ll also be seeig a slightly differet way of lookig at some of the basics that you probably did t see whe you were first itroduced to complex umbers ad provig some of the basic facts. The first sectio is a more mathematical defiitio of complex umbers ad is ot really required for uderstadig the remaider of the documet. It is preseted solely for those who might be iterested. The secod sectio (arithmetic) is assumed to be mostly a review for those readig this documet ad ca be read if you eed a quick refresher o how to do basic arithmetic with complex umbers. Also icluded i this sectio is a more precise defiitio of subtractio ad divisio tha is ormally give whe a perso is first itroduced to complex umbers. Agai, uderstadig these defiitios is ot required for the remaider of the documet they are oly preseted so you ca say you ve see it. The remaiig sectios are the real poit of this documet ad ivolve the topics that are typically ot taught whe studets are first exposed to complex umbers. So, with that out of the way, let s get started
The Defiitio As I ve already stated, I am assumig that you have see complex umbers to this poit ad that you re aware that i = ad so i =. This is a idea that most people first see i a algebra class (or wherever they first saw complex umbers) ad i = is defied so that we ca deal with square roots of egative umbers as follows, ( )( ) 00 = 00 = 00 = 00 i = 0i What I d like to do is give a more mathematical defiitio of a complex umbers ad show that i = (ad hece i = ) ca be thought of as a cosequece of this defiitio. We ll also take a look at how we defie arithmetic for complex umbers. What we re goig to do here is goig to seem a little backwards from what you ve probably already see but is i fact a more accurate ad mathematical defiitio of complex umbers. Also ote that this sectio is ot really required to uderstad the remaiig portios of this documet. It is here solely to show you a differet way to defie complex umbers. So, let s give the defiitio of a complex umber. Give two real umbers a ad b we will defie the complex umber as, = a+ bi () Note that at this poit we ve ot actually defied just what i is at this poit. The umber a is called the real part of ad the umber b is called the imagiary part of ad are ofte deoted as, Re = a Im = b () There are a couple of special cases that we eed to look at before proceedig. First, let s take a look at a complex umber that has a ero real part, = 0 + bi = bi I these cases, we call the complex umber a pure imagiary umber. Next, let s take a look at a complex umber that has a ero imagiary part, = a+ 0i = a I this case we ca see that the complex umber is i fact a real umber. Because of this we ca thik of the real umbers as beig a subset of the complex umbers. We ext eed to defie how we do additio ad multiplicatio with complex umbers. Give two complex umbers = a+ bi ad = c+ di we defie additio ad multiplicatio as follows,
( ) ( ) ( ) ( ) + = a+ c + b+ d i (3) = ac bd + ad + cb i (4) Now, if you ve see complex umbers prior to this poit you will probably recall that these are the formulas that were give for additio ad multiplicatio of complex umbers at that poit. However, the multiplicatio formula that you were give at that poit i time required the use of i = to completely derive ad for this sectio we do t yet kow that is true. I fact, as oted previously i = will be a cosequece of this defiitio as we ll see shortly. Above we oted that we ca thik of the real umbers as a subset of the complex umbers. Note that the formulas for additio ad multiplicatio of complex umbers give the stadard real umber formulas as well. For istace give the two complex umbers, = a+ 0i = c+ 0i the formulas yield the correct formulas for real umbers as see below. ( ) ( 0 0) ( )( ) + = a+ c + + i = a+ c ( 0 0 ) ( ( 0) ( 0) ) = ac + a + c i= ac The last thig to do i this sectio is to show that i = is a cosequece of the defiitio of multiplicatio. However, before we do that we eed to ackowledge that powers of complex umbers work just as they do for real umbers. I other words, if is a positive iteger we will defie expoetiatio as, = times So, let s start by lookig at i, use the defiitio of expoetiatio ad the use the defiitio of multiplicatio o that. Doig this gives, i = i i = 0+ i 0+ i ( )( ) ( )( ) ( )( ) ( 0 0 ) (( 0)( ) ( 0)( ) ) = + + = So, by defiig multiplicatio as we ve doe above we get that i = as a cosequece of the defiitio istead of just statig that this is a true fact. If we ow take to be the stadard square root, i.e. what did we square to get the quatity uder the radical, we ca see that i =. i
Arithmetic Before proceedig i this sectio let me first say that I m assumig that you ve see arithmetic with complex umbers at some poit before ad most of what is i this sectio is goig to be a review for you. I am also goig to be itroducig subtractio ad divisio i a way that you probably have t see prior to this poit, but the results will be the same ad are t importat for the remaiig sectios of this documet. I the previous sectio we defied additio ad multiplicatio of complex umbers ad showed that i = is a cosequece of how we defied multiplicatio. However, i practice, we geerally do t multiply complex umbers usig the defiitio. I practice we ted to just multiply two complex umbers much like they were polyomials ad the make use of the fact that we ow kow that i =. Just so we ca say that we ve worked a example let s do a quick additio ad multiplicatio of complex umbers. Example Compute each of the followig. (a) ( 58 i) + ( 7i) 6+ 3i 0+ 8i (b) ( )( (c) ( )( ) ) 4+ i 4 i Solutio As oted above, I m assumig that this is a review for you ad so wo t be goig ito great detail here. 58 i + 7i = 58 i+ 7i = 60 8i (a) ( ) ( ) (b) ( i)( i) i i i i ( ) (c) ( 4 + i) ( 4 i) = 6 8i+ 8i 4i = 6 + 4 = 0 6 + 3 0 + 8 = 60 + 48 + 30 + 4 = 60 + 78 + 4 = 36 + 78i It is importat to recall that sometimes whe addig or multiplyig two complex umbers the result might be a real umber as show i the third part of the previous example! The third part of the previous example also gives a ice property about complex umbers. ( )( ) a bi a bi a b + = + (5) We ll be usig this fact with divisio ad lookig at it i slightly more detail i the ext sectio. Let s ow take a look at the subtractio ad divisio of two complex umbers. Hopefully, you recall that if we have two complex umbers, = a+ bi ad = c+ di the you subtract them as, = a+ bi c+ di = a c + b d i (6) ( ) ( ) ( ) ( )
Ad that divisio of two complex umbers, a+ bi = (7) c + di ca be thought of as simply a process for elimiatig the i from the deomiator ad writig the result as a ew complex umber u+ vi. Let s take a quick look at a example of both to remid us how they work. Example Compute each of the followig. (a) ( 58 i) ( 7i) (b) 6 + 3 i 0 + 8i 5i (c) 7i Solutio (a) There really is t too much to do here so here is the work, ( 58 i) ( 7i) = 58 i + 7i = 56 + 6i (b) Recall that with divisio we just eed to elimiate the i from the deomiator ad usig () we kow how to do that. All we eed to do is multiply the umerator ad deomiator by 0 8i ad we will elimiate the i from the deomiator. 6+ 3i ( 6+ 3i) ( 0 8i) = 0 + 8i 0 + 8i 0 8i ( ) ( ) 60 48i+ 30i 4i = 00 + 64 84 8i = 64 84 8 9 = i = i 64 64 4 8 (c) We ll do this oe a little quicker. 5i 5i ( + 7i) 35+ 5i 7 = = = + i 7i 7i + 7i + 49 0 0 ( ) ( ) Now, for the most part this is all that you eed to kow about subtractio ad multiplicatio of complex umbers for this rest of this documet. However, let s take a look at a more precise ad mathematical defiitio of both of these. If you are t iterested i this the you ca skip this ad still be able to uderstad the remaider of this documet. The remaider of this documet ivolves topics that are typically first taught i a Abstract/Moder Algebra class. Sice we are goig to be applyig them to the field of
complex variables we wo t be goig ito great detail about the cocepts. Also ote that we re goig to be skippig some of the ideas ad glossig over some of the details that do t really come ito play i complex umbers. This will especially be true with the defiitios of iverses. The defiitios I ll be givig below are correct for complex umbers, but i a more geeral settig are ot quite correct. You do t eed to worry about this i geeral to uderstad what were goig to be doig below. I just wated to make it clear that I m skippig some of the more geeral defiitios for easier to work with defiitios that are valid i complex umbers. Okay, ow that I ve got the warigs/otes out of the way let s get started o the actual topic Techically, the oly arithmetic operatios that are defied o complex umbers are additio ad multiplicatio. This meas that both subtractio ad divisio will, i some way, eed to be defied i terms of these two operatios. We ll start with subtractio sice it is (hopefully) a little easier to see. We first eed to defie somethig called a additive iverse. A additive iverse is some elemet typically deoted by so that + = (8) ( ) 0 Now, i the geeral field of abstract algebra, is just the otatio for the additive iverse ad i may cases is NOT give by = ( )! Luckily for us however, with complex variables that is exactly how the additive iverse is defied ad so for a give complex umber = a+ bi the additive iverse,, is give by, = ( ) = a bi It is easy to see that this does meet the defiitio of the additive iverse ad so that wo t be show. With this defiitio we ca ow officially defie the subtractio of two complex umbers. Give two complex umbers = a+ bi ad = c+ di we defie the subtractio of them as, = + (9) ( ) Or, i other words, whe subtractig from we are really just addig the additive iverse of (which is deoted by ) to. If we further use the defiitio of the additive iverses for complex umbers we ca arrive at the formula give above for subtractio. = + = a+ bi + c di = a c + b d i ( ) ( ) ( ) ( ) ( ) So, that was t too bad I hope. Most of the problems that studets have with these kids of topics is that they eed to forget some otatio ad ideas that they are very used to workig with. Or, to put it aother way, you ve always bee taught that is just a
shorthad otatio for ( ), but i the geeral topic of abstract algebra this does ot ecessarily have to be the case. It s just that i all of the examples where you are liable to ru ito the otatio i real life, whatever that meas, we really do mea =. ( ) Okay, ow that we have subtractio out of the way, let s move o to divisio. As with subtractio we first eed to defie a iverse. This time we ll eed a multiplicative iverse. A multiplicative iverse for a o-ero complex umber is a elemet deoted by such that = Now, agai, be careful ot to make the assumptio that the expoet of - o the otatio is i fact a expoet. It is t! It is just a otatio that is used to deote the multiplicative iverse. With real (o-ero) umbers this turs out to be a real expoet ad we do have that 4 4 for istace. However, with complex umbers this will ot be the case! I fact, let s see just what the multiplicative iverse for a complex umber is. Let s start out with the complex umber = a+ bi ad let s call its multiplicative iverse = u+ vi. Now, we kow that we must have = so, let s actual do the multiplicatio. = ( a + bi)( u + vi) = ( au bv) + ( av + bu) i = = This tells us that we have to have the followig, au bv = av + bu = 0 Solvig this system of two equatios for the two ukows u ad v (remember a ad b are kow quatities from the origial complex umber) gives, a b u = v= a + b a + b Therefore, the multiplicative iverse of the complex umber is, a b = i (0) a + b a + b As you ca see, i this case, the expoet of - is ot i fact a expoet! Agai, you really eed to forget some otatio that you ve become familiar with i other math courses.
So, ow that we have the defiitio of the multiplicative iverse we ca fially defie divisio of two complex umbers. Suppose that we have two complex umbers ad the the divisio of these two is defied to be, = () I other words, divisio is defied to be the multiplicatio of the umerator ad the multiplicative iverse of the deomiator. Note as well that this actually does match with the process that we used above. Let s take aother look at oe of the examples that we looked at earlier oly this time let s do it usig multiplicative iverses. So, let s start out with the followig divisio. 6 + 3 i = ( 6+ 3i)( 0+ 8i) 0 + 8i We ow eed the multiplicative iverse of the deomiator ad usig (6) this is, ( ) 0 8 0 8i 0 + 8i = i = 0 + 8 0 + 8 64 Now, we ca do the multiplicatio, 6+ 3i 0 8i 60 48i+ 30i 4i 9 = ( 6+ 3i)( 0+ 8i) = ( 6+ 3i) = = i 0 + 8i 64 64 4 8 Notice that the secod to last step is idetical to oe of the steps we had i the origial workig of this problem ad, of course, the aswer is the same. As a fial topic let s ote that if we do t wat to remember the formula for the multiplicative iverse we ca get it by usig the process we used i the origial multiplicatio. I other words, to get the multiplicative iverse we ca do the followig 0 8i 0 8i ( 0 + 8i) = = 0 + 8i 0 8i 0 + 8 ( ) ( ) As you ca see this is essetially the process we used i doig the divisio iitially.
Cojugate ad Modulus I the previous sectio we looked at algebraic operatios o complex umbers. There are a couple of other operatios that we should take a look at sice they ted to show up o occasio. We ll also take a look at quite a few ice facts about these operatios. Complex Cojugate The first oe we ll look at is the complex cojugate, (or just the cojugate). Give the complex umber = a+ bi the complex cojugate is deoted by ad is defied to be, = a bi () I other words, we just switch the sig o the imagiary part of the umber. Here are some basic facts about cojugates. = () ± = ± (3) = (4) = (5) The first oe just says that if we cojugate twice we get back to what we started with origially ad hopefully this makes some sese. The remaiig three just say we ca break up sum, differeces, products ad quotiets ito the idividual pieces ad the cojugate. So, just so we ca say that we worked a umber example or two let s do a couple of examples illustratig the above facts. Example Compute each of the followig. (a) for = 3 5i (b) for = 5 + i ad = 8+ 3i (c) for = 5 + i ad = 8+ 3i Solutio There really is t much to do with these other tha to so the work so, (a) = 3+ 5i = 3+ 5i = 3 5i = Sure eough we ca see that after cojugatig twice we get back to our origial umber. = 3 i = 3 i = 3 + i (b) = 5+ 8+ 3 = 5 8 3 = 3+i We ca see that results from (b) ad (c) are the same as the fact implied they would be. (c) i ( i) i ( i)
There is aother ice fact that uses cojugates that we should probably take a look at. However, istead of just givig the fact away let s derive it. We ll start with a complex umber = a+ bi ad the perform each of the followig operatios. + = a+ bi+ ( a bi) = a+ bi ( a bi) = a = bi Now, recallig that Re = a ad Im = b we see that we have, Re + Im = = (6) Modulus The other operatio we wat to take a look at i this sectio is the modulus of a complex umber. Give a complex umber = a+ bi the modulus is deoted by ad is defied by = a + b (7) Notice that the modulus of a complex umber is always a real umber ad i fact it will ever be egative sice square roots always retur a positive umber or ero depedig o what is uder the radical. Notice that if is a real umber (i.e. = a+ 0i) the, = a = a where the o the is the modulus of the complex umber ad the o the a is the absolute value of a real umber (recall that i geeral for ay real umber a we have a = a ). So, from this we ca see that for real umbers the modulus ad absolute value are essetially the same thig. We ca get a ice fact about the relatioship betwee the modulus of a complex umbers ad its real ad imagiary parts. To see this let s square both sides of (7) ad use the fact that Re = a ad Im = b. Doig this we arrive at ( ) = a + b = Re + ( Im ) Sice all three of these terms are positive we ca drop the Im part o the left which gives the followig iequality, = ( Re ) + ( Im ) ( Re ) If we the square root both sides of this we get, Re where the o the is the modulus of the complex umber ad the o the Re are absolute value bars. Fially, for ay real umber a we also kow that a a (absolute value ) ad so we get, Re Re (8)
We ca use a similar argumet to arrive at, Im Im (9) There is a very ice relatioship betwee the modulus of a complex umber ad it s cojugate. Let s start with a complex umber = a+ bi ad take a look at the followig product. = ( a+ bi)( a bi) = a + b From this product we ca see that = (0) This is a ice ad coveiet fact o occasio. Notice as well that i computig the modulus the sig o the real ad imagiary part of the complex umber wo t affect the value of the modulus ad so we ca also see that, = () ad = () We ca also ow formalie the process for divisio from the previous sectio ow that we have the modulus ad cojugate otatios. I order to get the i out of the deomiator of the quotiet we really multiplied the umerator ad deomiator by the cojugate of the deomiator. The usig (0) we ca simplify the otatio a little. Doig all this gives the followig formula for divisio, = = Here s a quick example illustratig this, Example Evaluate 6 + 3 i. 0 + 8i Solutio I this case we have = 6+ 3i ad = 0 + 8i. The computig the various parts of the formula gives, = 0 8i = 0 + 8 = 64 The quotiet is the, 6 + 3i ( 6+ 3i)( 0 8i) 60 48i+ 30i 4i 9 = = = i 0 + 8i 64 64 4 8 Here are some more ice facts about the modulus of a complex umber. If = 0 the = 0 (3)
= (4) = (5) Property (3) should make some sese to you. If the modulus is ero the a + b = 0, but the oly way this ca be ero is if both a ad b are ero. To verify (4) cosider the followig, ( )( ) ( )( ) = = = = So, from this we ca see that usig property (0) usig property (4) rearragig terms usig property (0) agai (twice) = Fially, recall that we kow that the modulus is always positive so take the square root of both sides to arrive at = Property (5) ca be verified usig a similar argumet. Triagle Iequality ad Variats Properties (4) ad (5) relate the modulus of a product/quotiet of two complex umbers to the product/quotiet of the modulus of the idividual umbers. We ow eed to take a look at a similar relatioship for sums of complex umbers. This relatioship is called the triagle iequality ad is, + + (6) We ll also be able to use this to get a relatioship for the differece of complex umbers. The triagle iequality is actually fairly simple to prove so let s do that. We'll start with the left side squared ad use (0) ad (3) to rewrite it a little. + = + + = + + ( )( ) ( )( ) Now multiply out the right side to get, + = + + + (7) Next otice that,
= = ad so usig (6), (8) ad () we ca write middle two terms of the right side of (7) as + = + = Re = = ( ) Also use (0) o the first ad fourth term i (7) to write them as, = = With the rewrite o the middle two terms we ca ow write (7) as + = + + + = + + + + + ( ) = + So, puttig all this together gives, ( ) + + Now, recallig that the modulus is always positive we ca square root both sides ad we ll arrive at the triagle iequality. + + There are several variatios of the triagle iequality that ca all be easily derived. Let s first start by assumig that. This is ot required for the derivatio, but will help to get a more geeral versio of what we re goig to derive here. So, let s start with ad do some work o it. = + + + = + + Usig triagle iequality Now, rewrite thigs a little ad we get, + (8) If we ow assume that time switch ad ad we get, 0 we ca go through a similar process as above except this ( ) + = (9) 0 Now, recallig the defiitio of absolute value we ca combie (8) ad (9) ito the followig variatio of the triagle iequality.
+ (0) Also, if we replace with i (6) ad (0) we arrive at two more variatios of the triagle iequality. + () () O occasio you ll see () called the reverse triagle iequality.
Polar & Expoetial Form Most people are familiar with complex umbers i the form = a+ bi, however there are some alterate forms that are useful at times. I this sectio we ll look at both of those as well as a couple of ice facts that arise from them. Geometric Iterpretatio Before we get ito the alterate forms we should first take a very brief look at a atural geometric iterpretatio to a complex umbers sice this will lead us ito our first alterate form. Cosider the complex umber = a+ bi. We ca thik of this complex umber as either the poit ( ab, i the stadard Cartesia coordiate system or as the vector that starts at ) the origi ad eds at the poit ( ab, ). A example of this is show i the figure below. I this iterpretatio we call the x-axis the real axis ad the y-axis the imagiary axis. We ofte call the xy-plae i this iterpretatio the complex plae. Note as well that we ca ow get a geometric iterpretatio of the modulus. From the image above we ca see that = a + b is othig more tha the legth of the vector that we re usig to represet the complex umber = a+ bi. This iterpretatio also tells us that the iequality < meas that is closer to the origi (i the complex plae) tha is. Polar Form Let s ow take a look at the first alterate form for a complex umber. If we thik of the o-ero complex umber a bi as the poit ab, i the xy-plae we also kow that = + ( ) we ca represet this poit by the polar coordiates ( r, θ ), where r is the distace of the poit from the origi ad θ is the agle, i radias, from the positive x-axis to the ray coectig the origi to the poit. 006 Paul Dawkis 5 http://tutorial.math.lamar.edu/terms.aspx
Whe workig with complex umbers we assume that r is positive ad that θ ca be ay of the possible (both positive ad egative) agles that ed at the ray. Note that this meas that there are literally a ifiite umber of choices for θ. We excluded = 0 sice θ is ot defied for the poit (0,0). We will therefore oly cosider the polar form of o-ero complex umbers. We have the followig coversio formulas for covertig the polar coordiates ( r, θ ) ito the correspodig Cartesia coordiates of the poit, ( ab, ). a= rcosθ b= rsiθ If we substitute these ito the complex umber, = a+ bi ad factor a r out we arrive at the polar form of ( cosθ siθ ) = r + i () Note as well that we also have the followig formula from polar coordiates relatig r to a ad b. r = a + b but, the right side is othig more tha the defiitio of the modulus ad so we see that, r = () So, sometimes the polar form will be writte as, = cosθ + isiθ (3) ( ) The agle θ is called the argumet of ad is deoted by, θ = arg The argumet of ca be ay of the ifiite possible values of θ each of which ca be foud by solvig b taθ = (4) a 006 Paul Dawkis 6 http://tutorial.math.lamar.edu/terms.aspx
ad makig sure that θ is i the correct quadrat. Note as well that ay two values of the argumet will differ from each other by a iteger multiple of π. This makes sese whe you cosider the followig. For a give complex umber pick ay of the possible values of the argumet, say θ. If you ow icrease the value of θ, which is really just icreasig the agle that the poit makes with the positive x-axis, you are rotatig the poit about the origi i a couterclockwise maer. Sice it takes π radias to make oe complete revolutio you will be back at your iitial startig poit whe you reach θ + π ad so have a ew value of the argumet. See the figure below. If you keep icreasig the agle you will agai be back at the startig poit whe you reach θ + 4π, which is agai a ew value of the argumet. Cotiuig i this fashio we ca see that every time we reach a ew value of the argumet we will simply be addig multiples of π oto the origial value of the argumet. Likewise, if you start at θ ad decrease the agle you will be rotatig the poit about the origi i a clockwise maer ad will retur to your origial startig poit whe you reach θ π. Cotiuig i this fashio ad we ca agai see that each ew value of the argumet will be foud by subtractig a multiple of π from the origial value of the argumet. So we ca see that if θ ad θ are two values of arg the for some iteger k we will have, θ θ = πk (5) 006 Paul Dawkis 7 http://tutorial.math.lamar.edu/terms.aspx
Note that we ve also show here that r( cosθ isiθ ) = + is a parametric represetatio for a circle of radius r cetered at the origi ad that it will trace out a complete circle i a couter-clockwise directio as the agle icreases from θ to θ + π. The priciple value of the argumet (sometimes called the priciple argumet) is the uique value of the argumet that is i the rage π < arg π ad is deoted by Arg. Note that the iequalities at either ed of the rage tells that a egative real umber will have a priciple value of the argumet of Arg = π. Recallig that we oted above that ay two values of the argumet will differ from each other by a multiple of π leads us to the followig fact. arg = Arg + π = 0, ±, ±, (6) We should probably do a couple of quick umerical examples at this poit before we move o to look the secod alterate form of a complex umber. Example Write dow the polar form of each of the followig complex umbers. (a) = + i 3 (b) = 9 (c) = i Solutio (a) Let s first get r. r = = + 3 = Now let s fid the argumet of. This ca be ay agle that satisfies (4), but it s usually easiest to fid the priciple value so we ll fid that oe. The priciple value of the argumet will be the value of θ that is i the rage π < θ π, satisfies, 3 taθ = θ = ta ( 3) ad is i the secod quadrat sice that is the locatio the complex umber i the complex plae. If you re usig a calculator to fid the value of this iverse taget make sure that you π π uderstad that your calculator will oly retur values i the rage < θ < ad so you may get the icorrect value. Recall that if your calculator returs a value of θ the the secod value that will also satisfy the equatio will be θ = θ+ π. So, if you re usig a calculator be careful. You will eed to compute both ad the determie which falls ito the correct quadrat to match the complex umber we have because oly oe of them will be i the correct quadrat. I our case the two values are, 006 Paul Dawkis 8 http://tutorial.math.lamar.edu/terms.aspx
θ = π θ π π = + π = 3 3 3 The first oe is i quadrat four ad the secod oe is i quadrat two ad so is the oe that we re after. Therefore, the priciple value of the argumet is, π Arg = 3 ad all possible values of the argumet are the π arg = + π = 0, ±, ±, 3 Now, let s actually do what we were origially asked to do. Here is the polar form of = + i 3. π π = cos + isi 3 3 Now, for the sake of completeess we should ackowledge that there are may more equally valid polar forms for this complex umber. To get ay of the other forms we just eed to compute a differet value of the argumet by pickig. Here are a couple of other possible polar forms. 8π 8π = cos + isi 3 3 = 6π 6π = cos + isi = 3 3 3 (b) I this case we ve already oted that the priciple value of a egative real umber is π so we do t eed to compute that. For completeess sake here are all possible values of the argumet of ay egative umber. arg = π + π= π + = 0, ±, ±, ( ) Now, r is, r = = 8+ 0 = 9 The polar form (usig the priciple value) is, = 9cosπ + isiπ ( ( ) ( )) Note that if we d had a positive real umber the priciple value would be Arg = 0 (c) This aother special case much like real umbers. If we were to use (4) to fid the argumet we would ru ito problems sice the imagiary part is ero ad this would give divisio by ero. However, all we eed to do to get the argumet is thik about where this complex umber is i the complex plae. I the complex plae purely imagiary umbers are either o the positive y-axis or the egative y-axis depedig o the sig of the imagiary part. 006 Paul Dawkis 9 http://tutorial.math.lamar.edu/terms.aspx
For our case the imagiary part is positive ad so this complex umber will be o the positive y-axis. Therefore, the priciple value ad the geeral argumet for this complex umber is, π π Arg = arg = + π= π + = 0, ±, ±, Also, i this case r = ad so the polar form (agai usig the priciple value) is, π π = cos + isi Expoetial Form Now that we ve discussed the polar form of a complex umber we ca itroduce the secod alterate form of a complex umber. First, we ll eed Euler s formula, iθ e = cosθ + i siθ (7) With Euler s formula we ca rewrite the polar form of a complex umber ito its expoetial form as follows. i = re θ where θ = arg ad so we ca see that, much like the polar form, there are a ifiite umber of possible expoetial forms for a give complex umber. Also, because ay two argumets for a give complex umber differ by a iteger multiple of π we will sometimes write the expoetial form as, i( θ+ π) = re = 0, ±, ±, where θ is ay value of the argumet although it is more ofte tha ot the priciple value of the argumet. To get the value of r we ca either use (3) to write the expoetial form or we ca take a more direct approach. Let s take the direct approach. Take the modulus of both sides ad the do a little simplificatio as follows, iθ iθ = re = r e = r cosθ + isiθ = r + 0 cos θ + si θ = r ad so we see that r =. Note as well that because we ca cosider r( cosθ isiθ ) = + as a parametric represetatio of a circle of radius r ad the expoetial form of a complex umber is i really aother way of writig the polar form we ca also cosider = re θ a parametric represetatio of a circle of radius r. Now that we ve got the expoetial form of a complex umber out of the way we ca use this alog with basic expoet properties to derive some ice facts about complex umbers ad their argumets. 006 Paul Dawkis 0 http://tutorial.math.lamar.edu/terms.aspx
First, let s start with the o-ero complex umber = e. I the arithmetic sectio we gave a fairly complex formula for the multiplicative iverse, however, with the expoetial form of the complex umber we ca get a much icer formula for the multiplicative iverse. i r θ ( i ( ) ) ( i θ θ ) i θ θ e e e e = r = r = r = r i Note that sice r is a o-ero real umber we kow that r =. So, puttig this r together the expoetial form of the multiplicative iverse is, i( θ ) = e (8) r ad the polar form of the multiplicative iverse is, ( ( θ ) i ( θ ) ) = cos si r + (9) We ca also get some ice formulas for the product or quotiet of complex umbers. iθ iθ Give two complex umbers = re ad = re, where θ is ay value of arg ad θ is ay value of arg, we have iθ iθ i( θ+ θ ) ( )( ) = re re = rre (0) iθ e = iθ = r r e r r i( e θ θ ) () The polar forms for both of these are, ( cos( θ θ ) si ( θ )) = rr + + i +θ () ( cos( θ θ) i si ( θ θ) ) r = + (3) r We ca also use (0) ad () to get some ice facts about the argumets of a product ad a quotiet of complex umbers. Sice θ is ay value of arg ad θ is ay value of arg we ca see that, arg = arg + arg (4) ( ) = arg arg arg (5) 006 Paul Dawkis http://tutorial.math.lamar.edu/terms.aspx
Note that (4) ad (5) may or may ot work if you use the priciple value of the argumet, Arg. For example, cosider = i ad =. I this case we have = i ad the priciple value of the argumet for each is, π π Arg () i = Arg ( ) = π Arg ( i) = However, 3π π Arg () i + Arg ( ) = ad so (4) does t hold if we use the priciple value of the argumet. Note however, if we use, π arg () i = arg ( ) = π the, 3π arg() i + arg ( ) = is valid sice 3 π is a possible argumet for i, it just is t the priciple value of the argumet. As a iterestig side ote, (5) actually does work for this example if we use the priciple argumets. That wo t always happe, but it does i this case so be careful! We will close this sectio with a ice fact about the equality of two complex umbers that we will make heavy use of i the ext sectio. Suppose that we have two complex iθ iθ umbers give by their expoetial forms, = re ad = re. Also suppose that we kow that. I this case we have, = iθ iθ e = re r The we will have if ad oly if, r = r ad θ = θ + πk for some iteger k ie.. k = 0, ±, ±, = ( ) (6) Note that the phrase if ad oly if is a facy mathematical phrase that meas that if is true the so is (6) ad likewise, if (6) is true the we ll have. = = This may seem like a silly fact, but we are goig to use this i the ext sectio to help us fid the powers ad roots of complex umbers. 006 Paul Dawkis http://tutorial.math.lamar.edu/terms.aspx
Powers ad Roots I this sectio we re goig to take a look at a really ice way of quickly computig iteger powers ad roots of complex umbers. We ll start with iteger powers of the, i r θ = e sice they are easy eough. If is a iteger i ( θ ) = r = r i θ e e (7) There really is t too much to do with powers other tha workig a quick example. Example Compute ( 3+ 3i ). 5 Solutio Of course we could just do this by multiplyig the umber out, but this would be time cosumig ad proe to mistakes. Istead we ca covert to expoetial form ad the use () to quickly get the aswer. Here is the expoetial form of 3+ 3i. 3 π r = 9+ 9 = 3 taθ = Arg = 3 4 3+ 3i = 3 e i Note that we used the priciple value of the argumet for the expoetial form, although we did t have to. Now, use () to quickly do the computatio. ( 3+ 3i ) 5 = ( 3 ) 5 e 5π i 4 π 4 5π 5π = 97 cos + i si 4 4 = 97 i = 97 97i So, there really is t too much to iteger powers of a complex umber. Note that if r = the we have, θ ( ) i i θ = e = e ad if we take the last two terms ad covert to polar form we arrive at a formula that is called de Moivre s formula. 006 Paul Dawkis 3 http://tutorial.math.lamar.edu/terms.aspx
( θ θ) ( θ) ( θ) cos + isi = cos + isi = 0, ±, ±, We ow eed to move oto computig roots of complex umbers. We ll start this off simple by fidig the th roots of uity. The th roots of uity for =,3, are the distict solutios to the equatio, = Clearly (hopefully) = is oe of the solutios. We wat to determie if there are ay other solutios. To do this we will use the fact from the previous sectios that states that if ad oly if = r = r ad θ = θ + πk for some iteger k ( ie.. k = 0, ±, ±, ) So, let s start by covertig both sides of the equatio to complex form ad the computig the power o the left side. Doig this gives, iθ ( 0) ( 0) ( ) i i θ i re = e r e = e So, accordig to the fact these will be equal provided, r = θ = 0+ πk k = 0, ±, ±, Now, r is a positive iteger (by assumptio of the expoetial/polar form) ad so solvig gives, π k r = θ = k = 0, ±, ±, The solutios to the equatio are the, π k = exp i k = 0, ±, ±, Recall from our discussio o the polar form (ad hece the expoetial form) that these poits will lie o the circle of radius r. So, our poits will lie o the uit circle ad they will be equally spaced o the uit circle at every π radias. Note this also tells us that there distict roots correspodig to k = 0,,,, sice we will get back to where we started oce we reach k = Therefore there are th roots of uity ad they are give by, πk πk πk exp i = cos + isi k = 0,,,, (8) There is a simpler otatio that is ofte used to deote th roots of uity. First defie, π ω = exp i (9) 006 Paul Dawkis 4 http://tutorial.math.lamar.edu/terms.aspx
the the th roots of uity are, k k π πk ω = exp i = exp i k = 0,,, Or, more simply the th roots of uity are,, ω, ω,, ω (0) where ω is defied i (3). Example Compute the th roots of uity for =, 3, ad 4. Solutio We ll start with =. This gives, π ω = exp i = e iπ = ad ω = e iπ ( π ) i si ( π ) = cos + = So, for = we have -, ad as the th roots of uity. This should be too surprisig as all we were doig was solvig the equatio = ad we all kow that - ad are the two solutios. While the result for = may ot be that surprisig that for = 3 may be somewhat surprisig. I this case we are really solvig 3 = ad i the world of real umbers we kow that the solutio to this is =. However, from the work above we kow that there are 3 th roots of uity i this case. The problem here is that the remaiig two are complex solutios ad so are usually ot thought about whe solvig for real solutio to this equatio which is geerally what we wated up to this poit. So, let s go ahead ad fid the th roots of uity for = 3. π ω3 = exp i 3 This gives, 006 Paul Dawkis 5 http://tutorial.math.lamar.edu/terms.aspx
π 4π = ω3 = exp i ω3 = exp i 3 3 π π 4π 4π = cos + isi = cos + isi 3 3 3 3 3 3 = + i = i I ll leave it to you to check that if you cube the last two values you will i fact get. Fially, let s go through = 4. We ll do this oe much quicker tha the previous cases. π π ω4 = exp i = exp i 4 This gives, π 3 3π = ω4 = exp i ω4 = exp( i π) ω4 = exp i = i = = i Now, let s move o to more geeral roots. First let s get some otatio out of the way. We ll defie 0 to be ay umber that will satisfy the equatio = 0 () To fid the values of 0 we ll eed to solve this equatio ad we ca do that i the same way that we foud the th roots of uity. So, if r 0 = 0 ad θ 0 = arg 0 (ote θ 0 ca be ay value of the argumet, but we usually use the priciple value) we have, iθ iθ0 i θ iθ0 ( re ) = r0e r e = r0e So, this tells us that, θ0 π k r = r0 θ = + k = 0, ±, ±, The distict solutios to (5) are the, θ0 π k a k = r0 exp i + k = 0,,,, () So, we ca see that just as there were th roots of uity there are also th roots of. 0 Fially, we ca agai simplify the otatio up a little. If a is ay of the th roots of the all the roots ca be writte as, a, aω, aω,,, aω where ω is defied i (3). 0 006 Paul Dawkis 6 http://tutorial.math.lamar.edu/terms.aspx
Example 3 Compute all values of the followig. (a) ( i ) (b) ( ) 3 3 i Solutio (a) The first thig to do is write dow the expoetial form of the complex umber we re takig the root of. i= exp i π π So, if we use θ 0 = we ca use (6) to write dow the roots. π ak = exp i + π k k 0, 4 = Pluggig i for k gives, π 5π a0 = exp i a = exp i 4 4 π π 5π 5π = cos + isi = cos + isi 4 4 4 4 = + i = i I ll leave it to you to check that if you square both of these will get i. (b) Here s the expoetial form of the umber, π 3 i = exp i 6 Usig (6) the roots are, 3 π πk ak = exp i + k = 0,, 8 3 Pluggig i for k gives, 3 π 3 π π a0 = exp i cos si.4078 0.878 8 = + i 8 8 = + i 3 π 3 π π a = exp i = cos isi 0.4309.8394 8 + 8 8 = + i 3 3π 3 3π 3π a = exp i = cos + isi 0.80986 0.9656i 8 = 8 8 As with the previous part I ll leave it to you to check that if you cube each of these you will get 3 i. 006 Paul Dawkis 7 http://tutorial.math.lamar.edu/terms.aspx
006 Paul Dawkis 8 http://tutorial.math.lamar.edu/terms.aspx