MATH 120A COMPLEX VARIABLES NOTES: REVISED …

1 MATH 120a complex variables notes : REVISEDD ecember 3, 2003 BRUCE K. 120A Lecture notes , Fall : REVISED : december 3, (9/26/03) our purposes the definition of COMPLEX VARIABLES is thecalculus ofanalytic functions, where a functionF(x, y)=(u(x, y),v(x, y))fromR2to itself is analytic iffit satisfies the Cauchy Riemann equations:ux= vyandvx= , the associated calculus has some verybeautiful and useful properties which will be explained in this class. The followingfact makes the subject useful in of the common elementary functions, likexn,ex,sinx,tanx,lnx,etc. haveunique extensions to analytic functions. Moreover, the solutions tomany ordinary differential equations extend to analytic functions.

2 So the study ofanalytic functions aids in understandingthese class of real valued Sections ( COMPLEX Numbers).LetC=R2equipped with multiplicationrule( )(a, b)(c, d) (ac bd, bc+ad)and the usual rule for vector addition. As is standard we will write0=(0,0),1=(1,0)andi=(0,1)so that every elementzofCmay be written asz=x1+yiwhich in the future will be written simply asz=x+ +iy,letRez=xandImz= +ibandw=c+id,the multiplication rule in Eq. ( ) becomes( )(a+ib)(c+id) (ac bd)+i(bc+ad)andinparticular12=1andi2= COMPLEX numbersCwith the above multiplication rule sat-isfies the usual definitions of afield. For examplewz=zwandz(w1+w2)=zw1+zw2,etc.

3 Moreover ifz6=0,zhas a multiplicative inverse given by( )z 1=aa2+b2 iba2+ the most painful thing to check directly is the associative law, namely( )u(vw)=(uv) can be checked later in polar form +ib6=0,we wish tofindw=c+idsuch thatzw=1and this happens by Eq. ( ) iffac bd=1and( )bc+ad=0.( )Now takinga( ) +b( )implies a2+b2 c=aand soc=aa2+b2and taking b( ) +a( )implies a2+b2 d= band hencec= ba2+b2as 120a complex variables notes : REVISED december 3, (Not Done in Class).Here is a way to understand some of the basicproperties ofCusing our knowledge of linear algebra. LetMzdenote multiplicationbyz=a+ibthen ifw=c+idwe haveMzw= ac bdbc+ad = a bba cd so thatMz= a bba =aI+bJwhereJ:= 0 110.

4 With this notationwe haveMzMw=Mzwand sinceIandJcommute it follows thatzw= , since matrix multiplication is associative so is COMPLEX multiplication, Eq. ( ) holds. Also notice thatMzis invertible iffdetMz=a2+b2=|z|26=0in which caseM 1z=1|z|2 ab ba =M z/|z|2as we have already seen will write1/zforz 1andw/zto meanz 1 (Conjugation and Modulous).Ifz=a+ibwitha, b Rlet z=a iband|z|2 z z=a2+ that( )Rez=12(z+ z)andImz=12i(z z).Proposition conjugation and the modulus operators satisfy,(1) z=z,(2)zw= z wand z+ w=z+w.(3)| z|=|z|(4)|zw|=|z||w|and in particular|zn|=|z|nfor alln N.(5)|Rez| |z|and|Imz| |z|(6)|z+w| |z|+|w|.

5 (7)z=0iff|z|=0.(8)Ifz6=0thenz 1:= z|z|2(also written as1z)is the inverse ofz.(9) z 1 =|z| 1and more generally|zn|=|z|nfor alln and 3. are geometrically Sayz=a+ibandw=c+id,then z wisthesameaszwwithbreplaced by banddreplaced by d,and looking at Eq. ( ) we see that z w=(ac bd) i(bc+ad)= |zw|2=zw z w=z zw w=|z|2|w|2as real numbers and hence|zw|=|z||w|.5. Geometrically obvious or also follows from|z|=q|Rez|2+|Imz| This is the triangle inequality which may be understood geometrically or bythe computation|z+w|2=(z+w)(z+w)=|z|2+|w|2+w z+ wz=|z|2+|w|2+w z+w z=|z|2+|w|2+2Re(w z) |z|2+|w|2+2|z||w|=(|z|+|w|) Follows from Eq.

6 ( ).9. z 1 = z|z|2 = 1|z|2 | z|=1|z|.MATH 120a complex variables notes : REVISED december 3, 200352.(9/30/03) over Eq. ( ) and properties 8. and 9. in Proposition COMPLEX numberu, v, w, z Cwithv6=06=z,we have1u1v=1uv, 1v 1=(uv) 1uvwz=uwvzanduv+wz=uz+ thefirst item, it suffices to check that(uv) u 1v 1 =u 1uvv 1=1 1= rest follow usinguvwz=uv 1wz 1=uwv 1z 1=uw(vz) 1= +wz=zzuv+vvwz=zuzv+vwvz=(vz) 1(zu+vw)=uz+ Sections 36-37, p. we supposew(t)=c(t)+id(t)and define w(t)= c(t)+i d(t)andZ w(t)dt:=Z c(t)dt+iZ d(t)dtExample /20 et+isint dt=e12 1+ (t)=a(t)+ib(t)andw(t)=c(t)+id(t)and =u+iv Cthen(1)ddt(w(t)+z(t)) = w(t)+ z(t)(2)ddt[w(t)z(t)] =w z+ wz(3)R [w(t)+ z(t)]dt=R w(t)dt+ R z(t)dt(4)R w(t)dt=w( ) w( )In particular if w=0thenwis constant.

7 (5)Z w(t)z(t)dt= Z w(t) z(t)dt+w(t)z(t)| .(6) Z w(t)dt Z |w(t)| and 4. are [wz]=ddt(ac bd)+iddt(bc+ad)=( ac bd)+i( bc+ ad)+(a c b d)+i(b c+a d)= wz+w The only interesting thing to check is thatZ z(t)dt= Z z(t) we simply write out the real and imaginary parts:Z z(t)dt=Z (u+iv)(a(t)+ib(t))dt=Z (ua(t) vb(t)+i[ub(t)+va(t)])dt=Z (ua(t) vb(t))dt+iZ [ub(t)+va(t)]dtwhileZ z(t)dt=(u+iv)Z [a(t)+ib(t)]dt=(u+iv) Z a(t)dt+iZ b(t)dt!=Z (ua(t) vb(t))dt+iZ [ub(t)+va(t)] Alternative:Just check it for =i,this is the only new thing (t)z(t)| =Z ddt[w(t)z(t)]dt=Z w(t)z(t)dt+Z w(t) z(t) Let 0and Rbe chosen so thatZ w(t)dt= ei ,then Z w(t)dt = =e i Z w(t)dt=Z e i w(t)dt=Z Re e i w(t) dt Z Re e i w(t) dt Z e i w(t) dt=Z |w(t)| 120a complex variables notes : REVISED december 3, a natural exten-sion of the functionexto a functionezwithz idea is that sinceddtetx=xetxwithe0x=1we might try to defineezso that( )ddtetz=zetzwithe0z= there is a functionezsuch that Eq.

8 ( ) holds, then itsatisfies:(1)e z=1ezand(2)ew+z= By the product rule,ddt e tzetz = ze tzetz+e tzzetz=0and therefore,e tzetz=e 0ze0z= Again by the product rule,ddthe t(w+z)etwetzi=0and soe t(w+z)etwetz=e t(w+z)etwetz|t=0= showse (w+z)ewez=1and then using Item 1. proves Item to Proposition , tofind the desired functionezit suffices let us writeeit=x(t)+iy(t)then by assumptionddteit=ieitwithei0=1implies x+i y=i(x+iy)= y+ixwithx(0) = 1andy(0) = 0or equivalently that x= y, y=xwithx(0) = 1andy(0) = equation implies x(t)= y(t)= x(t)withx(0) = 1and x(0) = 0which has the unique solutionx(t)=costin which casey(t)= ddtcost= leads to the following (Euler s Formula).

9 For Rletei := cos +isin and forz=x+iylet( )ez=exeiy=ex(cosy+isiny).Quickly reviewezand its properties, in particular Euler s functionezdefined by Eq. ( ) satisfies Eq. ( ) and hencethe results of Proposition Also notice thatez=e is proved on p. 112 of the book and the proof goes as follows,ddtetz=ddt etxeity =xetxeity+etxiyeity=zetxeity= last equality follows fromez=ex(cosy+isiny)=ex(cosy+isiny)=ex( cosy isiny)=ex(cos ( y)+isin ( y)) =e (Addition formulas).For , Rwe havecos ( + )=cos cos sin sin sin ( + )=cos sin +cos sin . follow by comparing the real and imaginary parts of the identityei ei =ei( + )=cos( + )+isin ( + )whileei ei =(cos +isin ) (cos +isin )=cos cos sin sin +i(cos sin +cos sin ).

10 MATH 120a complex variables notes : REVISED december 3, 200393.(10/1/03)Exercise , b R,showZT0eateibtdt=ZT0e(a+ib)tdt=ZT01a+i bddte(a+ib)tdt=1a+ib eaTeibT 1 .By comparing the real and imaginary parts of both sides of this integralfind explicitformulas for the two real integralsZT0eatcos (bt)dtandZT0eatsin (bt) Form of COMPLEX Numbers: Sections 6 :Give the geometric interpretation of each of the following properties.(1)z=rei =|z|ei .(2) z=|z|e i andz 1= z/|z|2=|z| 1e i (3) Ifw=|w|ei thenzw=|z||w|ei( + )andz/w=zw 1=|z|ei |w| 1e i =|z||w| 1ei( ).In particularzn=|z|nein forn let =Arg(z)if < andz=|z|ei while wedefinearg (z)= R:z=|z|ei.