THE COMPLEX EXPONENTIAL FUNCTION

1 Math307 THECOMPLEXEXPONENTIALFUNCTION(Thesenotes assumeyouarealreadyfamiliarwiththebasicp ropertiesof complexnumbers.)We make thefollowingde nitionei = cos +isin :(1)Thisformulais calledEuler' orderto justifythisuseof theexponentialnotationappearingin (1), we will rstverifythefollowingformof theLaw ofExponents:ei 1+i 2=ei 1ei 2(2)To prove thiswe rstexpandtheright-handsideof (1)by rstmultiplyingouttheproduct:ei 1ei 2= (cos 1+isin 1)(cos 2+isin 2). Nextwe applyto thisthetrigonometricidentities:cos 1cos 2 sin 1sin 2= cos( 1+ 2)sin 1cos 2+ cos 1sin 2= sin( 1+ 2):Whenallthisis donetheresultisei 1ei 2= cos( 1+ 2) +isin( 1+ 2):Theright handsideof thelastequationis exactlywhatwe wouldgetif we wroteout(1)with replacedby 1+ 2. We have thereforeproved (2).To justifytheuseofe= 2:718: : : :, thebaseof thenaturallogarithm,in (1),we willdi erentiate(1)withrespectto : We shouldgetiei . Treatingilikeany otherconstant, we nddd ei =dd cos +isin = sin +icos.

2 But sin +icos =i(cos +isin ) =iei . Thus,as expected,dd ei =i ei (3)If onedoes notde neei by (1), thenonemust ndsomeothermeanto de neezandthento derive (1)directlyas a nitionofezismadeusingpower serieswithcomplexnumberszbutthisrequrire sa considerableamount of preliminaryworkwithpower a verybriefdiscussionof thisapproach, seepage154in :ei =2=i; e i= 1;ande2 i= + (x; y) andthepolarcoordinates(r; ) of a point isx=rcos ;y=rsin r=px2+y2; = arctanyx(4)1wherearctan(alsocalledtan 1) is oneof the\branches"of theinversetangentfunction.(Thequadrant which holdsthepoint (x; y) determinesthecorrectbranchof tan 1.) Ifz= 0 thenr= 0 and canbe Euler'sformula,we canexpresspolarrepresentationin thefollowingmanner:z=x+iy=r(cos +isin ) =rei ;(5)wherer=jzj=px2+y2and is givenby (4). Theangle is alsocalledanargumentofzandwe write = arg(z).Asnoted,thereis anambiguity in (4)abouttheinversetangent formulafor which can(andmust) be resolvedby lookingat thesignsofxandyin orderto determinein which quadrantei example,ifx= 0, thentheformulafor in (4)makes nosense;butx= 0 simplymeansthatz= 0 +iyliesontheimaginaryaxisso mustbe =2 or 3 =2 dependingonwhetheryis positive ornegative.

3 Again,ifz= 4 + 4i, thenr=p42+ 42= 4p2 and = 3 =4. Therefore 4 + 4i= 4p2e3 i= ,dueto theperiodicity of sin andcos , ifz=rei , thenwealsohavez=rei( +2k ),k= 0; 1; 2; : : :. Thus,in ourlastexample, 4 + 4i=4p2e11 i=4= 4p2e 5 i= anotherexample:thecomplexnumber2 + 8imay alsobe writtenasp68ei ;where = tan 1(4) 1 + 8i=p68ei = tan 14 1:33radians of two 1andz2=r2ei 2, thenz1=z2if andonlyifr1=r2and 1= 2+ 2k ; k= 0; 1; 2; : : :Despitethisthepolarrepresentationis veryusefulwhenit comesto multplicationanddivision:ifz1=r1ei 1andz2=r2ei 2;thenz1z2=r1r2ei( 1+ 2);(6)z1z2=z1z 12=r1r2ei( 1 2)(z26= 0):(7)2 ThisfollowsfromtheLaw of exponents in equation(2)andtherules:jz1z2j=jz1jjz2j=r 1r2;jz 12j= 1=r2; arg(1=z2) = arg(z2) = 2. For example,letz1= 2 +i=p5ei 1; 1= tan 1(12) :464z2= 2 + 4i=p20ei 2; 2= tan 1(4 2) = + Tan 1( 2) 2:034: : :(Note:Tan 1is theprincipalinversetangent. It is thequantity computedonmostscienti ccalculators.)

4 Thenz3=z1z2where:z3=p5p20ei 3= 10ei 3; 3 :464+ 2:034= 2:498: : :Thisgivesz3 10(cos(2:498)+ sin(2:498) 7:995+i6:001.(Theexactvalueisz3= 8 + 6i.) We leave it to thereaderto ndz1=z2in thisexampleusing(7).(Theexactvalueis i=2 usingthealgebraicmethod.)Applying(6)toz1 =z2= 4 + 4i= 4p2e3 i=4, gives(4 + 4i)2= 4p2e3 i=4 2= 32e3 i=2= 32i:Indeedforany positive (ornegative) integerit is quitestraightforwardto show thatIfz=rei 6= 0;thenzn=rnein :Thisformulamakes it quiteeasytosolve equationssuch asz3= . Thenfortheequationz3= 1, we haver3e3i = 1 = ,r3= 1 andr= 1, becauseris supposedto be a positive realnumber,and3 = 0 + 2k ,k= 0; 1; 2; : : :. It followsthat = 2k =3; k= 0; 1; : : :. Thereareonlythreedistinctnumbersof theforme2k i=3, namely:1 =e0; e2 i=3, ande4 i= gureillustratesthedistinctsolutionsto anotherequation:z3=8i. Thesolutions(calledthecube rootsof 8i= 8e i=2) are:z1= 2ei =6,z2= 2e5i =6,andz3= 2e9i =6= 2e i=6 1:7 +iz2= 2e5 i=6z3= 2i"""""bbbbbFigure2.)

5 Thethreecube rootsof = circlewithradius2 andcenter0. Thendistinctnthrootsof any complexnumberw6= 0 areequallyspacedona circleofradiusjwj1=ncenteredat 0. Oneneedonlylocateoneof gettheothern 1 roots,onerotatatesthe rstonen 1 times,each timethroughanangle2 =n, markingthepoints as distinctpoint correspondstoa distinctroot of theequationzn=w. Afternrotationsonegoes right roundthecircleandarrives at ei n=ein we obtainDeMoivre'sformula:(cos +isin )n= cosn +isinn :Expandingon theleftandequatingrealandimaginaryparts, leadsto trigonometricidentitieswhich canbe usedto expresscosn andsinn as a sumof termsof theform(cos )j(sin )k. For examplewithn= 2 onegets:(cos +isin )2= cos2 sin2 +i2 sin cos = cos2 +isin2 :Hencecos2 = cos2 sin2 andsin2 = 2 sin cos . Forn= 3, letus setC= cos andS= sin . Then(C+iS)3=C3+ 3iC2S 3CS2 iS3, socos3 = Ref(C+iS)3g=C3 3CS2= 4 cos3 3 cos :(becauseS2= 1 C2)).Onecanderive a similaridentity forsin3.

6 1 Theexponential of any nitionofex+iyis givenby theformulaex+iy=exeiy(8)Each termontheright-handsideof (8)alreadyhasa wellde isleftas anexerciseto show thatddte(a+bi)t= (a+bi)e(a+bi)t(9)forany complexconstanta+ 3iandz2= 2 2i.(a)Plotthepointsz1+z2; z1 z2, andz2.(b)Computejz1+z2jandjz1 z2j.(c)Expressz1andz2in 6ei =3andz2= 2e i =6. Plotz1z2, andz1= (a)Findandplotallcomplexnumberswhich satisfyz3= 8.(b)Findallcomplexnumbersz=rei , which satisfyz2=p2ei = (9). Notethate(a+bi)t=eat eibt. Usetheproductdi erentiateeibtby meansof (3) somealgebrato doto gettheformontheright of (9). forsin3 usingn= 3 in DeMoivre' in a way thatinvolves onlysin andsin3 if rstrealuseof cubicequationcanbe tranformedinto theform:x3= 3px+ 2q;wherepandqareconstants by replacingxwithax+bandmultiplyingthecubic bya constant. Thegraphof theright sideis a straight linewhich mustcrossthegraphofx3andthereforetheremu stbe a (real)solutionto :x= q+pq2 p3 1=3+ q pq2 p3 1=3:Try ndingthesolutiontox3= 6x+ 6 usingthisformula(p= 2 andq= 3).

7 Hereis wherecomplexnumbersarise:To solvex3= 15x+ 4,p= 5 andq= 2,so we obtain:x= (2 + 11i)1=3+ (2 11i)1=3:Eventhoughthislookslike a complexnumber,it actuallyis a realnumber:thesecondtermis thecomplexconjugateof the that(2+i)3= 2+11i,andthus thesolutionisx=