Complex Numbers Prmer Before I get started o ths let me frst make t clear that ths documet s ot teded to teach you everythg there s to kow about complex umbers. That s a subject that ca (ad does) take a whole class to cover. The purpose of ths documet s to gve you a bref overvew of complex umbers, otato assocated wth complex umbers, ad some of the basc operatos volvg complex umbers. Ths documet has bee wrtte wth the assumpto that you ve see complex umbers at some pot the past, kow (or at least kew at some pot tme) that complex umbers ca be solutos to quadratc equatos, kow (or recall) =, ad that you ve see how to do basc arthmetc wth complex umbers. If you do t remember how to do arthmetc I wll show a example or two to remd you how to do arthmetc, but I m gog to assume that you do t eed more tha that as a remder. For most studets the assumptos I ve made above about exposure to complex umbers s the extet of ther exposure. Problems ted to arse however because most structors seem to assume that ether studets wll see beyod ths some later class or have already see beyod ths some earler class. Studets are the all of a sudde expected to kow more tha basc arthmetc of complex umbers but ofte have t actually see t aywhere ad have to quckly pck t up. That s the purpose of ths documet. We wll go beyod the bascs that most studets have see at some pot ad show you some of the otato ad operatos volvg complex umbers that may studets do t ever see oce they lear how to deal wth complex umbers as solutos to quadratc equatos. We ll also be seeg a slghtly dfferet way of lookg at some of the bascs that you probably dd t see whe you were frst troduced to complex umbers ad provg some of the basc facts. The frst secto s a more mathematcal defto of complex umbers ad s ot really requred for uderstadg the remader of the documet. It s preseted solely for those who mght be terested. The secod secto (arthmetc) s assumed to be mostly a revew for those readg ths documet ad ca be read f you eed a quck refresher o how to do basc arthmetc wth complex umbers. Also cluded ths secto s a more precse defto of subtracto ad dvso tha s ormally gve whe a perso s frst troduced to complex umbers. Aga, uderstadg these deftos s ot requred for the remader of the documet t s oly preseted so you ca say you ve see t. The remag sectos are the real pot of ths documet ad volve the topcs that are typcally ot taught whe studets are frst exposed to complex umbers. So, wth that out of the way, let s get started 005 Paul Dawks
The Defto As I ve already stated, I am assumg that you have see complex umbers to ths pot ad that you re aware that = ad so =. Ths s a dea that most people frst see a algebra class ad = s defed so that we ca deal wth square roots of egatve umbers as follows, ( )( ) 00 = 00 = 00 = 00 = 0 What I d lke to do s gve a more mathematcal defto of a complex umbers ad show that = s fact a cosequece of ths defto ad how we defe the arthmetc for complex umbers. What I m gog to do here s gog to seem a lttle backwards from what you ve probably already see but s fact a more accurate ad mathematcal defto of complex umbers. Also ote that ths secto s ot really requred to uderstad the remag portos of ths documet. It s here solely to show you a dfferet way to defe complex umbers. So, let s gve the defto of a complex umber. Gve two real umbers a ad b we wll defe the complex umber as, = a+ b () Note that at ths pot we ve ot actually defed just what s at ths pot. The umber a s called the real part of ad the umber b s called the magary part of ad are ofte deoted as, Re = a Im = b () There are a couple of specal cases that we eed to look at before proceedg. Frst, let s take a look at a complex umber that has a ero real part, = 0 + b = b I these cases, we call the complex umber a pure magary umber. Next, let s take a look at a complex umber that has a ero magary part, = a+ 0 = a I ths case we ca see that the complex umber s fact a real umber. Because of ths we ca thk of the real umbers as beg a subset of the complex umbers. We ext eed to defe how we do addto ad multplcato wth complex umbers. Gve two complex umbers = a+ b ad = c+ d we defe addto ad multplcato as follows, 005 Paul Dawks
( ) ( ) ( ) ( ) + = a+ c + b+ d (3) = ac bd + ad + cb (4) Now, f you ve see complex umbers pror to ths pot you wll probably recall that these are the formulas that were gve for addto ad multplcato of complex umbers at that pot. However, the multplcato formula that you were gve at that pot tme requred the use of = ad for ths secto we do t yet kow that s true. I fact, as oted prevously = s fact a cosequece of ths defto as we ll see shortly. Above we oted that we ca thk of the real umbers as a subset of the complex umbers. Note that the formulas for addto ad multplcato gve the stadard real umber formulas as well. For stace gve the two complex umbers, = a+ 0 = c+ 0 the formulas yeld, + = a+ c + 0+ 0 = a+ c ( ) ( ) ( )( ) ( 0 0 ) ( ( 0) ( 0) ) = ac + a + c = ac The last thg to do ths secto s to show that = s a cosequece of the defto of multplcato. 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, f s a postve teger we wll defe expoetato as, = tmes So, let s start by lookg at, use the defto of expoetato ad the use the defto of multplcato o that. Dog ths gves, = = 0+ 0+ ( )( ) ( )( ) ( )( ) ( 0 0 ) (( 0)( ) ( 0)( ) ) = + + = So, by defg multplcato as we ve doe above we get that = as a cosequece of the defto. Note that we stll ca t techcally clam that = yet sce we have ot yet offcally defed roots for complex umbers. However, t does tur out that ths wll fact be the case. 005 Paul Dawks 3
Arthmetc Before proceedg ths secto let me frst say that I m assumg that you ve see arthmetc wth complex umbers at some pot before ad most of what s ths secto s gog to be a revew for you. I am also gog to be troducg subtracto ad dvso a way that you probably have t see pror to ths pot, but the results wll be the same. I the prevous secto we defed addto ad multplcato of complex umbers ad showed that = s a cosequece of how we defed multplcato. However, practce, we geerally do t multply complex umbers usg the defto. I practce we ted to just multply two complex umbers much lke they were polyomals ad the make use of the fact that =. Just so we ca say that we ve worked a example let s do a quck addto ad multplcato of complex umbers. Example Compute each of the followg. (a) ( 58 ) + ( 7) 6+ 3 0+ 8 (b) ( )( ) (c) ( 4+ )( 4 ) Soluto As oted above, I m assumg that ths s a revew for you ad so wo t be gog to great detal here. 58 + 7 = 58 + 7 = 60 8 (a) ( ) ( ) (b) ( )( ) ( ) (c) ( 4 + ) ( 4 ) = 6 8+ 8 4 = 6 + 4 = 0 6 + 3 0 + 8 = 60 + 48 + 30 + 4 = 60 + 78 + 4 = 36 + 78 It s mportat to recall that sometmes whe addg or multplyg two complex umbers the result mght be a real umber! The thrd example above also gves a ce property about complex umbers. Namely, ( )( ) a b a b a b + = + () We ll be usg ths fact wth dvso ad lookg at t slghtly more detal the ext secto. Let s ow take a look at the subtracto ad dvso of two complex umbers. Hopefully, you recall that f we have two complex umbers, = a+ b ad = c+ d the you subtract them as, 005 Paul Dawks 4
( ) ( ) ( ) ( ) = a+ b c+ d = a c + b d () Ad that dvso of two complex umbers, a+ b = (3) c + d ca be thought of as smply a process for elmatg the from the deomator ad wrtg the result as a ew complex umber u+ v. Let s take a quck look at a example of both to remd us how they work. Example Compute each of the followg. (a) ( 58 ) ( 7) (b) 6 + 3 0 + 8 5 (c) 7 Soluto (a) There really s t too much to do here so here s the work, 58 7 = 58 + 7 = 56 + 6 ( ) ( ) (b) Recall that wth dvso we just eed to elmate the from the deomator ad usg () we kow how to do that. All we eed to do s multply the umerator ad deomator by 0 8 ad we wll elmate the from the deomator. 6+ 3 ( 6+ 3) ( 0 8) = 0 + 8 0 + 8 0 8 ( ) ( ) 60 48+ 30 4 = 00 + 64 84 8 = 64 84 8 9 = = 64 64 4 8 (c) We ll do ths oe a lttle qucker. 5 5 ( + 7) 35+ 5 7 = = = + 7 7 + 7 + 49 0 0 ( ) ( ) Now, for the most part ths s all that you eed to kow about subtracto ad multplcato of complex umbers for ths rest of ths documet. However, let s take a look at a more precse ad mathematcal defto of both of these. If you are t terested ths the you ca skp ths ad stll be able to uderstad the remader of ths documet. 005 Paul Dawks 5
The remader of ths documet volves topcs that are typcally frst taught a Abstract/Moder Algebra class. Sce we are gog to be applyg them to the feld of complex varables we wo t be gog to great detal about the cocepts. Also ote, that we re gog to be skppg some of the deas ad glossg over some of the detals that do t really come to play complex umbers. Ths wll especally be true wth the deftos of verses. The deftos I ll be gvg below are correct for complex umbers, but a more geeral settg are ot qute correct. You do t eed to worry about ths geeral to uderstad what were gog to be dog below. I just wated to make t clear that I m skppg some of the more geeral deftos for easer to work wth deftos that are vald complex umbers. Okay, ow that I ve got the wargs/otes out of the way let s get started o the actual topc Techcally, the oly arthmetc operatos that are defed o complex umbers are addto ad multplcato. Ths meas that both subtracto ad dvso wll, some way, eed to be defed terms of these two operatos. We ll start wth subtracto sce t s (hopefully) a lttle easer to see. We frst eed to defe somethg called a addtve verse. A addtve verse s some elemet typcally deoted by so that + = (4) ( ) 0 Now, the geeral feld of abstract algebra, s just the otato for the addtve =! Luckly for us however, wth verse ad may cases s NOT gve by ( ) complex varables that s exactly how the addtve verse s defed ad so for a gve complex umber = a+ b the addtve verse,, s gve by, = = a b ( ) It s easy to see that ths does meet the defto of the addtve verse ad so that wo t be show. Wth ths defto we ca ow offcally defe the subtracto of two complex umbers. Gve two complex umbers = a+ b ad = c+ d we defe the subtracto of them as, = + (5) ( ) Or, other words, whe subtractg from we are really just addg the addtve verse of (whch s deoted by ) to. If we further use the defto of the addtve verses for complex umbers we ca arrve at the formula gve above for subtracto. = + = a+ b + c d = a c + b d ( ) ( ) ( ) ( ) ( ) 005 Paul Dawks 6
So, that was t too bad I hope. Most of the problems that studets have wth these kds of topcs s havg to forget some otato ad deas that they are very used to workg =, but the wth. Or, to put t aother way, you ve always bee taught that ( ) geeral topc of abstract algebra ths does ot ecessarly have to be the case. It s just that all of the examples where you are lable to ru to the otato real lfe, whatever that meas, we really do have =. Okay, ow that we have subtracto out of the way, let s move o to dvso. As wth subtracto we frst eed to defe a verse. Ths tme we ll eed a multplcatve verse. A multplcatve verse for a o-ero complex umber s a elemet deoted by such that = Now, aga, be careful ot to make the assumpto that the expoet of - o the otato s fact a expoet. It s t! It s just a otato that s used to deote the multplcatve verse. Wth real (o-ero) umbers ths turs out to be a real expoet ad we do have that 4 4 for stace. However, wth complex umbers ths wll ot be the case! I fact, let s see just what the multplcatve verse for a complex umber s. Let s start out wth the complex umber = a+ b ad let s call ts multplcatve verse = u+ v. Now, we kow that we must have = so, let s actual do the multplcato. = ( a + b)( u + v) = ( au bv) + ( av + bu) = ( ) = Ths tells us that we have to have the followg, au bv = av + bu = 0 Solvg ths system of two equatos for the two ukows u ad v (remember a ad b are kow quattes from the orgal complex umber) gves, a b u = v = a + b a + b Therefore, the multplcatve verse of the complex umber s, a b = (6) a + b a + b As you ca see, ths case, the expoet of - s ot fact a expoet! 005 Paul Dawks 7
So, ow that we have the defto of the multplcatve verse we ca fally defe dvso of two complex umbers. Suppose that we have two complex umbers ad the the dvso of these two s defed to be, = (7) I other words, dvso s defed to be the multplcato of the umerator ad the multplcatve verse of the deomator. Note as well that ths actually does match wth the process that we used above. Let s take aother look at oe of the examples that we looked at earler oly ths tme let s do t usg multplcatve verses. So, let s start out wth the followg dvso. 6 + 3 = ( 6+ 3)( 0+ 8) 0 + 8 We ow eed the multplcatve verse of the deomator ad ths s, ( ) 0 8 0 8 0 + 8 = = 0 + 8 0 + 8 64 Now, we ca do the multplcato, 6+ 3 0 8 60 48+ 30 4 9 = ( 6 + 3)( 0 + 8) = ( 6 + 3) = = 0 + 8 64 64 4 8 Notce that the secod to last step s detcal to oe of the steps we had the orgal workg of ths problem ad, of course, the aswer s the same. As a fal topc let s ote that f we do t wat to remember the formula for the multplcatve verse we ca get t by usg the process we used the orgal multplcato. I other words, to get the multplcatve verse we ca do the followg 0 8 0 8 ( 0 + 8) = = 0 + 8 0 8 0 + 8 ( ) ( ) As you ca see ths s essetally the process we used dog the dvso tally. Cojugate ad Modulus I the prevous secto we looked at algebrac operatos o complex umbers. There are a couple of other operatos that we should take a look at sce they ted to show up o occaso. We ll also take a look at qute a few ce facts about these operatos. Complex Cojugate The frst oe we ll look at s the complex cojugate, (or just the cojugate). Gve the complex umber = a+ b the complex cojugate s deoted by ad s defed to be, 005 Paul Dawks 8
= a b () I other words, we just swtch the sg o the magary part of the umber. Here are some basc facts about cojugates. = () ± = ± (3) = (4) = (5) The frst oe just says that f we cojugate twce we get back to what we started wth orgally ad hopefully ths makes some sese. The remag three just say we ca break up sum, dffereces, products ad quotets to the dvdual peces ad the cojugate. So, just so we ca say that we worked a umber example or two let s do a couple of examples llustratg the above facts. Example 3 Compute each of the followg. (a) for = 3 5 (b) for = 5 + ad = 8+ 3 (c) for = 5 + ad = 8+ 3 Soluto There really s t much to do wth these other tha to so the work so, (a) = 3+ 5 = 3+ 5= 3 5= Sure eough we ca see that after cojugatg twce we get back to our orgal umber. = 3 = 3 = 3+ (b) (c) ( ) ( ) = 5 + 8 + 3 = 5 8 3 = 3+ We ca see that results from (b) ad (c) are the same as the fact mpled they would be. There s aother ce fact that uses cojugates that we should probably take a look at. However, stead of just gvg the fact away let s derve t. We ll start wth a complex umber = a+ b ad the perform each of the followg operatos. 005 Paul Dawks 9
( ) ( ) + = a + b + a b = a + b a b = a = b Now, recallg that Re = a ad Im = b we see that we have, Re + Im = = (6) Modulus The other operato we wat to take a look at ths secto s the modulus of a complex umber. Gve a complex umber = a+ b the modulus s deoted by ad s defed by = a + b (7) Notce that the modulus of a complex umber s always a real umber ad fact t wll ever be egatve sce square roots always retur a postve umber or ero depedg o what s uder the radcal. Notce that f s a real umber (.e. = a+ 0) the, = a = a where the o the s the modulus of the complex umber ad the o the a s the absolute value of a real umber (recall that geeral for ay real umber a we have a = a ). So, from ths we ca see that for real umbers the modulus ad absolute value are essetally the same thg. We ca get a ce fact about the relatoshp betwee the modulus of a complex umbers ad ts real ad magary parts. To see ths let s square both sdes of (7) ad use the fact that Re = a ad Im = b. Dog ths we arrve at = a + b = ( Re ) + ( Im ) Sce all three of these terms are postve we ca drop the Im part o the left whch gves the followg equalty, = ( Re ) + ( Im ) ( Re ) If we the square root both sdes of ths we get, Re where the o the s the modulus of the complex umber ad the o the Re are absolute value bars. Fally, for ay real umber a we also kow that a a (absolute value ) ad so we get, Re Re (8) We ca use a smlar argumet to arrve at, Im Im (9) 005 Paul Dawks 0
There s a very ce relatoshp betwee the modulus of a complex umber ad t s cojugate. Let s start wth a complex umber = a+ b ad take a look at the followg product. = ( a+ b)( a b) = a + b From ths product we ca see that Ths s a ce ad coveet fact o occaso. = (0) Notce as well that computg the modulus the sg o the real ad magary 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 formale the process for dfferetato from the prevous secto ow that we have the modulus ad cojugate otatos. I order to get the out of the deomator of the quotet we really multpled the umerator ad deomator by the cojugate of the deomator. The usg (0) we ca smplfy the otato a lttle. Dog all ths gves the followg formula for dervatves, = = Here s a quck example llustratg ths, Example 4 Evaluate 6 + 3. 0 + 8 Soluto I ths case we have = 6+ 3 ad = 0 + 8. The computg the varous parts of the formula gves, = 0 8 = 0 + 8 = 64 The quotet s the, 6 + 3 ( 6 + 3)( 0 8) 60 48+ 30 4 9 = = = 0 + 8 64 64 4 8 Here are some more ce facts about the modulus of a complex umber. If = 0 the = 0 (3) = (4) 005 Paul Dawks
= (5) Property (3) should make some sese to you. If the modulus s ero the a + b = 0, but the oly way ths ca be ero s f both a ad b are ero. To verfy (4) cosder the followg, ( )( ) ( )( ) = = = = So, from ths we ca see that usg property (0) usg property (4) rearragg terms usg property (0) aga (twce) = Fally, recall that we kow that the modulus s always postve so take the square root of both sdes to arrve at = Property (5) ca be verfed usg a smlar argumet. Tragle Iequalty ad Varats Propertes (4) ad (5) relate the modulus of a product/quotet of two complex umbers to the product/quotet of the modulus of the dvdual umbers. We ow eed to take a look at a smlar relatoshp for sums of complex umbers. Ths relatoshp s called the tragle equalty ad s, + + (6) We ll also be able to use ths to get a relatoshp for the dfferece of complex umbers. The tragle equalty s actually farly smple to prove so let s do that. We'll start wth the left sde squared ad use (0) ad (3) to rewrte t a lttle. + = ( + )( + ) = ( + )( + ) Now multply out the rght sde to get, + = + + + (7) Next otce that, = = ad so usg (6), (8) ad () we ca wrte mddle two terms of the rght sde of (7) as + = + = Re = = ( ) 005 Paul Dawks
Also use (0) o the frst ad fourth term (7) to wrte them as, = = Wth the rewrte o the mddle two terms we ca ow wrte (7) as So, puttg all ths together gves, + = + + + = + + + + + ( ) = + ( ) + + Now, recallg that the modulus s always postve we ca square root both sdes ad we ll arrve at the tragle equalty. + + There are several varatos of the tragle equalty that ca all be easly derved. Let s frst start by assumg that. Ths s ot requred for the dervato, but wll help to get a more geeral verso of what we re gog to derve here. So, let s start wth ad do some work o t. = + + + = + + Usg tragle equalty Now, rewrte thgs a lttle ad we get, + (8) 0 If we ow assume that we ca go through a smlar process as above except ths tme swtch ad ad we get, + = (9) ( ) 0 Now, recallg the defto of absolute value we ca combe (8) ad (9) to the followg varato of the tragle equalty. + (0) Also, f we replace wth (6) ad (0) we arrve at two more varatos of the tragle equalty. 005 Paul Dawks 3
+ () () O occaso you ll see () called the reverse tragle equalty. Polar & Expoetal Form Most people are famlar wth complex umbers the form = a+ b, however there are some alterate forms that are useful at tmes. I ths secto we ll look at both of those as well as a couple of ce facts that arse from them. Geometrc Iterpretato Before we get to the alterate forms we should frst take a very bref look at a atural geometrc terpretato to a complex umbers sce ths wll lead us to our frst alterate form. Cosder the complex umber = a+ b. We ca thk of ths complex umber as ether the pot ( ab, ) the stadard Cartesa coordate system or as the vector that starts at the org ad eds at the pot ( ab, ). A example of ths s show the fgure below. I ths terpretato we call the x-axs the real axs ad the y-axs the magary axs. We ofte call the xy-plae ths terpretato the complex plae. Note as well that we ca ow get a geometrc terpretato of the modulus. From the mage above we ca see that = a + b s othg more tha the legth of the vector that we re usg to represet the complex umber = a+ b. Ths terpretato also tells us that the equalty < meas that s closer to the org ( the complex plae) tha s. 005 Paul Dawks 4
Polar Form Let s ow take a look at the frst alterate form for a complex umber. If we thk of the o-ero complex umber a b ab, the xy-plae we also kow that = + as the pot ( ) we ca represet ths pot by the polar coordates ( r, θ ), where r s the dstace of the pot from the org ad θ s the agle, radas, from the postve x-axs to the ray coectg the org to the pot. Whe workg wth complex umbers we assume that r s postve ad that θ ca be ay of the possble (both postve ad egatve) agles that ed at the ray. Note that ths meas that there are lterally a fte umber of choces for θ. We excluded = 0 sce θ s ot defed for the pot (0,0). We wll therefore oly cosder the polar form of o-ero complex umbers. We have the followg coverso formulas for covertg the polar coordates ( r, θ ) to the correspodg Cartesa coordates of the pot, ( ab, ). a= rcosθ b= rsθ If we substtute these to = a+ b ad factor a r out we arrve at the polar form of the complex umber, = r cosθ + sθ () ( ) Note as well that we also have the followg formula from polar coordates relatg r to a ad b. r = a + b but, the rght sde s othg more tha the defto of the modulus ad so we see that, r = () ad sometmes the polar form wll be wrtte as, = cosθ + sθ (3) ( ) 005 Paul Dawks 5
The agle θ s called the argumet of ad s deoted by, θ = arg The argumet of ca be ay of the fte possble values of θ each of whch ca be foud by solvg b taθ = (4) a ad makg sure that θ s the correct quadrat. Note as well that ay two values of the argumet wll dffer from each other by a teger multple of π. Ths makes sese whe you cosder the followg. For a gve complex umber pck ay of the possble values of the argumet, say θ. If you ow crease the value of θ, whch s really just creasg the agle that the pot makes wth the postve x-axs, you are rotatg the pot about the org a couterclockwse maer. Sce t takes π radas to make oe complete revoluto you wll be back at your tal startg pot whe you reach θ + π ad so have a ew value of the argumet. See the fgure below. If you keep creasg the agle you wll aga be back at the startg pot whe you reach θ + 4π, whch s aga a ew value of the argumet. Cotug ths fasho we ca see that every tme we reach a ew value of the argumet we wll smply be addg multples of π oto the orgal value of the argumet. Lkewse, f you start at θ ad decrease the agle you wll be rotatg the pot about the org a clockwse maer ad wll retur to your orgal startg pot whe you reach θ π. Cotug ths fasho ad we ca aga see that each ew value of the argumet wll be foud by subtractg a multple of π from the orgal value of the argumet. 005 Paul Dawks 6
So we ca see that f θ ad θ are two values of arg the for some teger k we wll have, θ θ = πk (5) Note that we ve also show here that r( cosθ sθ ) = + s a parametrc represetato for a crcle of radus r cetered at the org ad that t wll trace out a complete crcle a couter-clockwse drecto as the agle creases from θ to θ + π. The prcple value of the argumet (sometmes called the prcple argumet) s the uque value of the argumet that s the rage π < arg π ad s deoted by Arg. Note that the equaltes at ether ed of the rage tells that a egatve real umber wll have a prcple value of the argumet of Arg = π. Recallg that we oted above that ay two values of the argumet wll dffer from each other by a multple of π leads us to the followg fact. arg = Arg + π = 0, ±, ±, (6) We should probably do a couple of quck umercal examples at ths pot before we move o to look the secod alterate form of a complex umber. Example 5 Wrte dow the polar form of each of the followg complex umbers. (a) = + 3 (b) = 9 (c) = Soluto (a) Let s frst get r. r = = + 3 = Now let s fd the argumet of. Ths ca be ay agle that satsfes (4), but t s usually easest to fd the prcple value so we ll fd that oe. The prcple value of the argumet wll be the value of θ that s the rage π < θ π, satsfes, 3 taθ = θ = ta ( 3) ad s the secod quadrat sce that s the locato the complex umber the complex plae. If you re usg a calculator to fd the value of ths verse taget make sure that you π π uderstad that your calculator wll oly retur values the rage < θ < ad so you may get the correct value. Recall that f your calculator returs a value of θ the the secod value that wll also satsfy the equato wll be θ = θ+ π. So, f you re 005 Paul Dawks 7
usg a calculator be careful. You wll eed to compute both ad the determe whch falls to the correct quadrat to match the complex umber we have because oly oe of them wll be the correct quadrat. I our case the two values are, θ = π θ π π = + π = 3 3 3 The frst oe s quadrat four ad the secod oe s quadrat two ad so s the oe that we re after. Therefore, the prcple value of the argumet s, π Arg = 3 ad all possble values of the argumet are the π arg = + π = 0, ±, ±, 3 Now, let s actually do what we were orgally asked to do. Here s the polar form of = + 3. π π = cos + s 3 3 Now, for the sake of completeess we should ackowledge that there are may more equally vald polar forms for ths complex umber. To get ay of the other forms we just eed to compute a dfferet value of the argumet by pckg. Here are a couple of other possble polar forms. 8π 8π = cos + s 3 3 = 6π 6π = cos + s = 3 3 3 (b) I ths case we ve already oted that the prcple value of a egatve real umber s π so we do t eed to compute that. For completeess sake here s all possble values of the argumet of ay egatve umber. arg = π + π= π + = 0, ±, ±, ( ) Now, r s, r = = 8+ 0 = 9 The polar form (usg the prcple value) s, = 9cosπ + sπ ( ( ) ( )) Note that f we d had a postve real umber the prcple value would be Arg = 0 (c) Ths aother specal case much lke real umbers. If we were to use (4) to fd the 005 Paul Dawks 8
argumet we would ru to problems sce the magary part s ero ad ths would gve dvso by ero. However, all we eed to do to get the argumet s thk about where ths complex umber s the complex plae. I the complex plae purely magary umbers are ether o the postve y-axs or the egatve y-axs depedg o the sg of the magary part. For our case the magary part s postve ad so ths complex umber wll be o the postve y-axs. Therefore, the prcple value ad the geeral argumet for ths complex umber s, π π Arg = arg = + π= π + = 0, ±, ±, Also, ths case r = ad so the polar form (aga usg the prcple value) s, π π = cos + s Expoetal Form Now that we ve dscussed the polar form of a complex umber we ca troduce the secod alterate form of a complex umber. Frst, we ll eed Euler s formula, θ e = cosθ + sθ (7) Wth Euler s formula we ca rewrte the polar form of a complex umber to ts expoetal form as follows. = re θ where θ = arg ad so we ca see that, much lke the polar form, there are a fte umber of possble expoetal forms for a gve complex umber. Also, because ay two argumets for a gve complex umber dffer by a teger multple of π we wll sometmes wrte the expoetal form as, ( θ+ π) = re = 0, ±, ±, where θ s ay value of the argumet although t s more ofte tha ot the prcple value of the argumet. To get the value of r we ca ether use (3) to wrte the expoetal form or we ca take a more drect approach. Let s take the drect approach. Take the modulus of both sdes ad the do a lttle smplfcato as follows, ad so we see that r θ θ = = = cosθ + sθ = + 0 cos θ + s θ = re r e r r r =. Note as well that because we ca cosder r( cosθ sθ ) = + as a parametrc represetato of a crcle of radus r ad the expoetal form of a complex umber s really aother way of wrtg the polar form we ca also cosder = re θ a parametrc represetato of a crcle of radus r. 005 Paul Dawks 9
Now that we ve got the expoetal form of a complex umber out of the way we ca use ths alog wth basc expoet propertes to derve some ce facts about complex umbers ad ther argumets. Frst, let s start wth the o-ero complex umber = e. I the arthmetc secto we gave a farly complex formula for the multplcatve verse, however, wth the expoetal form of the complex umber we ca get a much cer formula for the multplcatve verse. r θ ( ( ) ) ( θ θ ) θ θ e e e e = r = r = r = r Note that sce r s a o-ero real umber we kow that r =. So, puttg ths r together the expoetal form of the multplcatve verse s, ( θ ) = e (8) r ad the polar form of the multplcatve verse s, ( ( θ ) ( θ )) = cos s r + (9) We ca also get some ce formulas for the product or quotet of complex umbers. θ Gve two complex umbers = re ad = re, where θ s ay value of arg ad θ s ay value of arg, we have θ θ θ ( θ+ θ ) ( )( ) = re re = rre (0) θ e = = θ r r e r r ( e θ θ ) () The polar forms for both of these are, ( cos( θ θ ) s ( θ )) = rr + + +θ () ( cos( θ θ ) s ( θ θ ) ) r = + (3) r We ca also use (0) ad () to get some ce facts about the argumets of a product ad a quotet of complex umbers. Sce θ s ay value of arg ad θ s ay value of arg we ca see that, arg = arg + arg (4) ( ) = arg arg arg (5) 005 Paul Dawks 0
Note that (4) ad (5) may or may ot work f you use the prcple value of the argumet, Arg. For example, cosder = ad =. I ths case we have = ad the prcple value of the argumet for each s, π π Arg () = Arg ( ) = π Arg ( ) = However, 3π π Arg () + Arg ( ) = ad so (4) does t hold f we use the prcple value of the argumet. Note however, f we use, π arg () = arg ( ) = π the, 3π arg () + arg ( ) = s vald sce 3 π s a possble argumet for, t just s t the prcple value of the argumet. As a terestg sde ote, (5) actually does work for ths example f we use the prcple argumets. That wo t always happe, but t does ths case so be careful! We wll close ths secto wth a ce fact about the equalty of two complex umbers that we wll make heavy use of the ext secto. Suppose that we have two complex θ θ umbers gve by ther expoetal forms, = re ad = re. Also suppose that we kow that. I ths case we have, = θ θ e = re r Ths wll be true f ad oly f, r = r ad θ = θ + πk for some teger k e.. k = 0, ±, ±, ( ) (6) Note that the phrase f ad oly f s a facy mathematcal phrase that meas that f s true the so s (6). Lkewse, f (6) s true the we ll have. = = Ths may seem lke a slly fact, but we are gog to use ths the ext secto to help us fd the powers ad roots of complex umbers. Powers ad Roots 005 Paul Dawks
I ths secto we re gog to take a look at a really ce way of quckly computg teger powers ad roots of complex umbers. r θ We ll start wth teger powers of = e sce they are easy eough. If s a teger the, ( θ ) = r = r r θ e e () There really s t too much to do wth powers other tha workg a quck example. Example 6 Compute ( ). 3+ 3 5 Soluto Of course we could just do ths by multplyg the umber out, but ths would be tme cosumg ad proe to mstakes. Istead we ca covert to expoetal form ad the use () to quckly get the aswer. Here s the expoetal form of 3+ 3. 3 π r = 9+ 9 = 3 taθ = Arg = 3 4 3+ 3 = 3 e Note that we used the prcple value of the argumet for the expoetal form, although we dd t have to. Now, use () to quckly do the computato. ( 3+ 3 ) 5 = ( 3 ) 5 e 5π 4 π 4 5π 5π = 97 cos + s 4 4 = 97 = 97 97 So, there really s t too much to teger powers of a complex umber. Note that f r = the we have, θ ( ) θ = e = e ad f we take the last two terms ad covert to polar form we arrve at a formula that s called de Movre s formula. ( θ θ) ( θ) ( θ) cos + s = cos + s = 0, ±, ±, 005 Paul Dawks
We ow eed to move oto computg roots of complex umbers. We ll start ths off smple by fdg the th roots of uty. The th roots of uty for =,3, are the dstct solutos to the equato, = Clearly (hopefully) = s oe of the solutos. We wat to determe f there are ay other solutos. To do ths we wll use the fact from the prevous sectos that states that f ad oly f = r = r ad θ = θ + πk for some teger k ( e.. k = 0, ±, ±, ) So, let s start by covertg both sdes of the equato to complex form ad the computg the power o the left sde. Dog ths gves, θ ( 0) ( 0) ( ) θ re = e r e = e So, accordg to the fact these wll be equal provded, r = θ = 0+ πk k = 0, ±, ±, Now, r s a postve teger (by assumpto of the expoetal/polar form) ad so solvg gves, π k r = θ = k = 0, ±, ±, The solutos to the equato are the, π k = exp k = 0, ±, ±, Recall from our dscusso o the polar form (ad hece the expoetal form) that these pots wll le o the crcle of radus r. So, our pots wll le o the ut crcle ad they wll be equally spaced o the utl crcle at every π radas. Note ths also tells us that there dstct roots correspodg to k = 0,,,, sce we wll get back to where we started oce we reach k = Therefore there are th roots of uty ad they are gve by, πk πk πk exp = cos + s k = 0,,,, () There s a smpler otato that s ofte used to deote th roots of uty. Frst defe, π ω = exp (3) the the th roots of uty are, 005 Paul Dawks 3
k k π πk ω = exp = exp k = 0,,, Or, more smply the th roots of uty are,, ω, ω,, (4) where ω s defed (3). ω Example 7 Compute the th roots of uty for =, 3, ad 4. Soluto We ll start wth =. Ths gves, π ω = exp = e π = ad ω = e π ( π ) s ( π ) = cos + = So, for = we have -, ad as the th roots of uty. Ths should be too surprsg as all we were dog was solvg the equato = ad we all kow that - ad are the two solutos. Whle the result for = may ot be that surprsg that for = 3 may be somewhat surprsg. I ths case we are really solvg 3 = ad the world of real umbers we kow that the soluto to ths s =. However, from the work above we kow that there are 3 th roots of uty ths case. The problem here s that the remag two are complex solutos ad so are usually ot thought about whe solvg for real soluto to ths equato whch s usually what we were after up to ths pot. So, let s go ahead ad fd the th roots of uty for = 3. π ω3 = exp 3 Ths gves, 005 Paul Dawks 4
π 4π = ω3 = exp ω3 = exp 3 3 π π 4π 4π = cos + s = cos + s 3 3 3 3 3 3 = + = I ll leave t to you to check that f you cube the last two values you wll fact get. Fally, let s go through = 4. We ll do ths oe much qucker tha the prevous case. π π ω4 = exp = exp 4 Ths gves, π 3 3π = ω4 = exp ω4 = exp( π) ω4 = exp = = = Now, let s move o to more geeral roots. Frst let s get some otato out of the way. We ll defe 0 to be ay umber that wll satsfy the equato = 0 (5) To fd the values of 0 we ll eed to solve ths equato ad we ca do that the same way that we foud the th roots of uty. So, f r 0 = 0 ad θ 0 = arg 0 (ote θ 0 ca be ay value of the argumet, but we usually use the prcple value) we have, θ θ0 θ θ0 ( re ) = r0e r e = r0e So, ths tells us that, θ0 π k r = r0 θ = + k = 0, ±, ±, The dstct solutos to (5) are the, θ0 π k a k = r0 exp + k = 0,,,, (6) So, we ca see that just as there were th roots of uty there are also th roots of. 0 Fally, we ca aga smplfy the otato up a lttle. If a s ay of the th roots of the all the roots ca be wrtte as, a, aω, aω,,, aω where ω s defed (3). 0 Example 8 Compute all values of the followg. (a) ( ) 005 Paul Dawks 5
(b) ( ) 3 3 Soluto (a) The frst thg to do s wrte dow the expoetal form of the complex umber we re takg the root of. = exp π π So, f we use θ 0 = we ca use (6) to wrte dow the roots. π ak = exp + π k k 0, 4 = Pluggg for k gves, π 5π a0 = exp a = exp 4 4 π π 5π 5π = cos + s = cos + s 4 4 4 4 = + = I ll leave t to you to check that f you square both of these wll get. (b) Here s the expoetal form of the umber, π 3 = exp 6 Usg (6) the roots are, 3 π πk ak = exp + k = 0,, 8 3 Pluggg for k gves, 3 π 3 π π a0 = exp = cos + s =.4078 + 0.878 8 8 8 π π π = exp = cos + s = 0.4309 +.8394 8 8 8 3 3 a 3 3π 3 3π 3π a = exp = cos + s 0.80986 0.9656 8 = 8 8 As wth the prevous part I ll leave t to you to check that f you cube each of these you wll get 3. 005 Paul Dawks 6