Student Solutions Manual and Study Guide to Advanced Engineering Mathematics[Volume 2]

222 Pages·2012·2.99 MB·English

by  Herbert Kreyszig

D
P A R T
Complex
Analysis
Chap. 13 Complex Numbers and Functions.
Complex Differentiation
Complexnumbersappearedinthetextbookbeforeindifferenttopics.SolvinglinearhomogeneousODEs
ledtocharacteristicequations,(3),p.54inSec.2.2,withcomplexnumbersinExample5,p.57,and
CaseIIIofthetableonp.58.SolvingalgebraiceigenvalueproblemsinChap.8ledtocharacteristic
equationsofmatriceswhoseroots,theeigenvalues,couldalsobecomplexasshowninExample4,p.328.
Whereas,inthesetypeofproblems,complexnumbersappearalmostnaturallyascomplexrootsof
polynomials(thesimplestbeingx2C1 D 0),itismuchlessimmediatetoconsidercomplexanalysis—the
systematicstudyofcomplexnumbers,complexfunctions,and“complex”calculus.Indeed,complex
analysiswillbethedirectionofstudyinPartD.Theareahasimportantengineeringapplicationsin
electrostatics,heatflow,andfluidflow.Furthermotivationforthestudyofcomplexanalysisisgivenon
p.607ofthetextbook.
WestartwiththebasicsinChap.13byreviewingcomplexnumbersz D xCyi inSec.13.1and
introducingcomplexintegrationinSec.13.3.Thosefunctionsthataredifferentiableinthecomplex,on
somedomain,arecalledanalyticandwillformthebasisofcomplexanalysis.Notallfunctionsare
analytic.Thisleadstothemostimportanttopicofthischapter,theCauchy–Riemannequations(1),
p.625inSec.13.4,whichallowustotestwhetherafunctionisanalytic.Theyareveryshortbutyouhave
torememberthem!Therestofthechapter(Secs.13.5–13.7)isdevotedtoelementarycomplexfunctions
(exponential,trigonometric,hyperbolic,andlogarithmicfunctions).
Yourknowledgeandunderstandingofrealcalculuswillbeuseful.Conceptsthatyoulearnedinreal
calculuscarryovertocomplexcalculus;however,beawarethattherearedistinctdifferencesbetween
realcalculusandcomplexanalysisthatweclearlymark.Forexample,whereastherealequationex D 1
hasonlyonesolution,itscomplexcounterpartez D 1hasinfinitelymanysolutions.
258 ComplexAnalysis PartD
Sec.13.1 ComplexNumbersandTheirGeometricRepresentation
Muchofthematerialmaybefamiliartoyou,butwestartfromscratchtoassureeveryonestartsatthe
samelevel.Thissectionbeginswiththefourbasicalgebraicoperationsofcomplexnumbers(addition,
subtraction,multiplication,anddivision).Ofthese,theonethatperhapsdiffersmostfromrealnumbersis
division(orformingaquotient).Thusmakesurethatyourememberhowtocalculatethequotientoftwo
complexnumbersasgiveninequation(7),Example2,p.610,andProb.3.In(7)wetakethenumberz
2
fromthedenominatorandformitscomplexconjugatezN andanewquotientzN =zN .Wemultiplythegiven
2 2 2
quotientbythisnewquotientzN =zN (whichisequalto1andthusallowed):
2 2
z z z zN
z D 1 D 1 (cid:1)1 D 1 (cid:1) 2;
z z z zN
2 2 2 2
whichwemultiplyout,recallingthati2 D (cid:0)1[see(5),p.609].Thefinalresultisacomplexnumberina
formthatallowsustoseparateitsreal(Rez/andimaginary(Imz/parts.Alsorememberthat1=i D (cid:0)i
(seeProb.1),asitoccursfrequently.Wecontinuebydefiningthecomplexplaneanduseittograph
complexnumbers(noteFig.318,p.611,andFig.322,p.612).Weuseequation(8),p.612,togofrom
complextoreal.
ProblemSet.13.1.Page612
1. Powersofi.Wecomputethevariouspowersofi bytherulesofaddition,subtraction,
multiplication,anddivisiongivenonpp.609–610ofthetextbook.Wehaveformallythat
i2 D ii
D .0;1/.0;1/ [by(1),p.609]
D .0(cid:1)0(cid:0)1(cid:1)1;0(cid:1)1C1(cid:1)0/ [by(3),p.609]
(I1)
D .0(cid:0)1;0C0/ (arithmetic)
D .(cid:0)1;0/
D (cid:0)1 [by(1)],
wherein(3),thatis,multiplicationofcomplexnumbers,weusedx D 0,x D 0,y D 1,y D 1.
1 2 1 2
(I2) i3 D i2i D .(cid:0)1/(cid:1)i D (cid:0)i:
Hereweused(I1)inthesecondequality.Toget(I3),weapply(I2)twice:
(I3) i4 D i2i2 D .(cid:0)1/(cid:1).(cid:0)1/ D 1:
(I4) i5 D i4i D 1(cid:1)i D i;
andthepatternrepeatsitselfassummarizedinthetablebelow.
Weuse(7),p.610,inthefollowingcalculation:
1 1iN 1.(cid:0)i/ .1C0i/.0(cid:0)i/ 1(cid:1)0C0(cid:1)1 0(cid:1)0(cid:0)1(cid:1)1
(I5) D D D D Ci D 0(cid:0)i D (cid:0)i:
i i iN i .(cid:0)i/ .0Ci/.0(cid:0)i/ 02C12 02C12
Chap.13 ComplexNumbersandFunctions.ComplexDifferentiation 259
By(I5)and(I1)weget
1 1 1
(I6) D (cid:1) D .(cid:0)i/.(cid:0)i/ D .(cid:0)1/i (cid:1).(cid:0)1/i D 1(cid:1)i2 D (cid:0)1;
i2 i i
1 (cid:18)1(cid:19)2(cid:18)1(cid:19)
D D .(cid:0)1/.(cid:0)i/ D i [from(I6)and(I5)];
i3 i i
1 (cid:18)1(cid:19)2(cid:18)1(cid:19)2
D D .(cid:0)1/.(cid:0)1/ D 1,
i4 i i
andthepatternrepeatsitself.Memorizethati2 D (cid:0)1and1=i D (cid:0)i astheywillappearquite
frequently.
i8 i9 : .
i4 i5 i6 i7
Start(cid:0)! i0 i i2 i3
1 i (cid:0)1 (cid:0)i
1=i4 1=i3 1=i2 1=i (cid:0)Start
1=i8 1=i7 1=i6 1=i5
. . 1=i10 1=i9
Sec.13.1. Prob.1. Tableofpowersofi
3. Divisionofcomplexnumbers
a.Thecalculationsof(7),p.610,indetailare
z x Ciy
z D 1 D 1 1 (bydefinitionofz andz )
z x Ciy 1 2
2 2 2
x Ciy x (cid:0)iy
D 1 1 (cid:1) 2 2 (N.B.correspondstomultiplicationby1)
x Ciy x (cid:0)iy
2 2 2 2
.x Ciy /.x (cid:0)iy /
D 1 1 2 2
.x Ciy /.x (cid:0)iy /
2 2 2 2
x x (cid:0)x iy Ciy x (cid:0)iy iy
D 1 2 1 2 1 2 1 2 (multiplyingitout:(3)innotation(4),p.609)
x x (cid:0)x iy Ciy x (cid:0)iy iy
2 2 2 2 2 2 2 2
x x (cid:0)ix y Cix y (cid:0)i2y y
D 1 2 1 2 2 1 1 2 (groupingterms,usingcommutativity)
x2(cid:0)ix y Cix y (cid:0)i2y2
2 2 2 2 2 2
x x (cid:0)ix y Cix y Cy y
D 1 2 1 2 2 1 1 2 (usingi2 D (cid:0)1andsimplifying)
x2Cy2
2 2
x x CCy y x y (cid:0)x y
D 1 2 1 2 Ci 2 1 1 2 (breakingintorealpartandimaginarypart).
x2Cy2 x2Cy2
2 2 2 2
b.Apracticalexampleusing(7)is
26(cid:0)18i .26(cid:0)18i/.6C2i/ 26(cid:1)6C26(cid:1)2i (cid:0)18(cid:1)6i (cid:0)18(cid:1)2i2
D D
6(cid:0)2i .6(cid:0)2i/ .6C2i/ 62C22
156C52i (cid:0)108i C36 192(cid:0)56i
D D D 4:8(cid:0)1:4i:
36C4 40
260 ComplexAnalysis PartD
5. Pureimaginarynumbera.Ifz D xCiy ispureimaginary,thenzN D (cid:0)z:
Proof.Letz D xCiy bepureimaginary.Thenx D 0,bydefinitiononthebottomofp.609.
Hence
(A) z D iy and (B) zN D (cid:0)iy (bydefinition.ofcomplexconjugate,p.612).
Ifwemultipybothsidesof(A)by(cid:0)1,weget
(cid:0)z D (cid:0)iy;
whichisequaltozN,hence
(cid:0)z D zN:
b.IfzN D (cid:0)z thenz D xCiy ispureimaginary.
Proof.Letz D xCiy sothatzN D x(cid:0)iy.WearegiventhatzN D (cid:0)z,so
zN D x(cid:0)iy D (cid:0)z D (cid:0).xCiy/ D (cid:0)x(cid:0)iy:
Bythedefinitionofequality(p.609)weknowthattherealpartsmustbeequalandthatthe
imaginarypartsmustbeequal.Thus
RezN D Re.(cid:0)z/;
x D (cid:0)x;
2x D 0;
x D 0;
and
ImzN D Im.(cid:0)z/;
(cid:0)y D (cid:0)y;
whichistrueforanyy.Thus
z D xCiy D iy:
Butthismeans,bydefinition,thatz ispureimaginary,ashadtobeshown.
11. Complexarithmetic
z (cid:0)z D .(cid:0)2C11i/(cid:0).2(cid:0)i/
1 2
D (cid:0)2C11i (cid:0)2Ci D .(cid:0)2(cid:0)2/C.11C1/i D (cid:0)4C12i
.z (cid:0)z /2 D .(cid:0)4C12i/.(cid:0)4C12i/ D 16(cid:0)48i (cid:0)48i (cid:0)144 D (cid:0)128(cid:0)96i
1 2
.z (cid:0)z /2 128 96 8(cid:1)16 25(cid:1)3
1 2 D (cid:0) (cid:0) i D (cid:0) (cid:0) D (cid:0)8(cid:0)6i:
16 16 16 16 24
Nextconsider
(cid:16)z z (cid:17)2
1 (cid:0) 2 :
4 4
Wehave
z 1 2 11 z 2 1
1 D .(cid:0)2C11i/ D (cid:0) C i; 2 D (cid:0) i:
4 4 4 4 4 4 4
Chap.13 ComplexNumbersandFunctions.ComplexDifferentiation 261
Theirdifferenceis
(cid:18) (cid:19)
z z 2 2 11 1
1 (cid:0) 2 D (cid:0) (cid:0) C C i D (cid:0)1C3i:
4 4 4 4 4 4
Hence
(cid:16)z z (cid:17)2
1 (cid:0) 2 D .(cid:0)1C3i/.(cid:0)1C3i/ D 1(cid:0)3i (cid:0)3i C9i2 D 1(cid:0)6i (cid:0)9 D (cid:0)8(cid:0)6i;
4 4
whichisthesameresultasbefore.
19. Realpartandimaginarypartofz=zN.Forz D xCiy,wehaveby(7),p.610,
z z zNN zz
D D
zN zN zNN zN z
sincetheconjugateoftheconjugateofacomplexnumberisthecomplexnumberitself(whichyou
maywanttoprove!). Then
z z2 .xCiy/2 x2C2ixy (cid:0)y2 x2(cid:0)y2 2xy
D D D D Ci :
zN zNz x2Cy2 x2Cy2 x2Cy2 x2Cy2
Hencewegettheresultasshownonp.A34ofthetextbook:
(cid:18)z(cid:19) x2(cid:0)y2 (cid:18)z(cid:19) 2xy
Re D I Im D :
zN x2Cy2 zN x2Cy2
Sec.13.2 PolarFormofComplexNumbers.PowersandRoots
Polarcoordinates,definedby(1)and(2)onp.613,playamoreimportantroleincomplexanalysisthan
incalculus.Theirstudygivesadeeperunderstandingofmultiplicationanddivisionofcomplexnumbers
(pp.615–616)andabsolutevalues.Moredetailsareasfollows.
Thepolarangle(cid:18) (takencounterclockwise,seeFig.323,p.614)ofacomplexnumberisdetermined
onlyuptointegermultiplesof2(cid:25).Whileoftenthisisnotessential,therearesituationswhereitmatters.
Forthispurpose,weintroducetheconceptoftheprincipalvalueArgz in(5),p.614,andillustrateitin
Example1,Probs.9and13.
Thetriangleinequalitydefinedin(6),p.614,andillustratedinExample2,p.615,isveryimportant
sinceitwillbeusedfrequentlyinestablishingboundssuchasinChap.15.
Oftenitwillbeusedinitsgeneralizedform(6*),p.615,whichcanbeunderstoodbythefollowing
geometricreasoning.Drawseveralcomplexnumbersaslittlearrowsandleteachtailcoincidewiththe
precedinghead.Thisgivesyouazigzaginglineofnparts,andtheleftsideof(6*)equalsthedistance
fromthetailofz totheheadofz .Canyou“see”it?Nowtakeyourzigzaglineandpullittaut;thenyou
1 n
havetherightsideasthelengthofthezigzaglinestraightenedout.
Inalmostallcaseswhenweuse(6*)inestablishingbounds,itwillnotmatterwhetherornottheright
sideof(6*)ismuchlargerthantheleft.However,itwillbeessentialthatwehavesuchanupperboundfor
theabsolutevalueofthesumontheleft,sothatinalimitprocess,thelattercannotgotoinfinity.
Thelasttopicisrootsofcomplexnumbers,illustratedinFigs.327–329,p.617,andProb.21.Lookat
thesefiguresandseehow,fordifferentn,therootsofunity(16),p.617,liesymmetricallyontheunit
circle.
ProblemSet13.2.Page618
1. Polarform.Sketchz D 1Ci tounderstandwhatisgoingon.Pointz istheppoint.1;1/inthe
complexplane.Fromthisweseethatthedistanceofz fromtheoriginisjzj D 2.Thisisthe
262 ComplexAnalysis PartD
absolutevalueofz.Furthermore,z liesonthebisectinglineofthefirstquadrant,sothatitsargument
(theanglebetweenthepositiverayofthex-axisandthesegmentfrom0toz)is45ı or(cid:25)=4.
Now weshowhowtheresultsfollowfrom(3)and(4),p.613.Inthenotationof(3)and(4)we
havez D xCiy D 1Ci:Hencetherealpartofz isx D 1andtheimaginarypartofz isy D 1.
From(3)weobtain
p
p
jzj D 12C12 D 2;
asbefore.From(4)weobtain
y (cid:25)
tan(cid:18) D D 1; (cid:18) D 45ı or :
x 4
Hencethepolarform(2),p.613,is
p
(cid:16) (cid:25) (cid:25)(cid:17)
z D 2 cos Ci sin :
4 4
NotethatherewehaveexplainedthefirstpartofExample1,p.614,ingreatdetail.
y
1 1 + i
45o
x
1
Sec.13.2 Prob.1. Graphofz D 1Ci inthecomplexplane
5. Polarform.Weuse(7),p.610,inSec.13.1,toobtain
p p p
2C 1i 2C 1i (cid:0) 8C 2i
(A) p 3 D p 3 (cid:1) p 3 :
(cid:0) 8(cid:0) 2i (cid:0) 8(cid:0) 2i (cid:0) 8C 2i
3 3 3
Thenumeratorof(A)simplifiesto
p p p p p p p
(cid:16) (cid:17)(cid:16) (cid:17) (cid:16) (cid:17) (cid:16) (cid:17)
2C 1i (cid:0) 8C 2i D (cid:0) 16C 2 2(cid:0) 1 8 i (cid:0) 2 D 38 C 2 2(cid:0) 12 2 i D (cid:0)38:
3 3 3 3 9 9 3 3 9
Thedenominatorof(A)is
p
(cid:16)(cid:0) 8(cid:17)2C(cid:0)2(cid:1)2 D 8C 4 D 72 C 4 D 76:
3 9 9 9 9
Puttingthemtogethergivesthesimplificationof(A),thatis,
p
2C 1i (cid:0)38 (cid:18) 38(cid:19)(cid:18) 9 (cid:19) 38 1
p 3 D 9 D (cid:0) D (cid:0) D (cid:0) :
(cid:0) 8(cid:0) 2i 76 9 76 76 2
3 9
Hencez D (cid:0)1 correspondsto(cid:0)(cid:0)1;0(cid:1)inthecomplexplane.Furthermore,by(3),p.613,
2 2
q
jzj D r D px2Cy2 D (cid:0)(cid:0)1(cid:1)2C02 D 1
2 2
Chap.13 ComplexNumbersandFunctions.ComplexDifferentiation 263
andby(4),p.613,
y 0
tan(cid:18) D D D 0I (cid:18) D 180ı D (cid:25):
x (cid:0)1
2
Henceby(2),p.613,thedesiredpolarformis
z D r.cos(cid:18) Cisin(cid:18)/ D 1 .cos(cid:25) Cisin(cid:25)/:
2
y
180o
x
1
–
2
Sec.13.2 Prob.5. Graphofz D (cid:0)1 inthecomplexplane
2
7. Polarform.Forthegivenz wehave
q q
jzj D 12C(cid:0)1(cid:25)(cid:1)2 D 1C 1(cid:25)2;
2 4
y 1(cid:25) 1 (cid:18)1 (cid:19)
tan(cid:18) D D 2 D (cid:25)I (cid:18) D arctan (cid:25) :
x 1 2 2
Thedesiredpolarformofz is
q
z D jzj.cos(cid:18) Cisin(cid:18)/ D 1C 1(cid:25)2(cid:2)cos(cid:0)arctan 1(cid:25)(cid:1)Cisin(cid:0)arctan 1(cid:25)(cid:1)(cid:3):
4 2 2
9. Principalargument.Thefirstandsecondquadrantscorrespondto0 (cid:20)Argz (cid:20) (cid:25).Thethirdand
fourthquadrantscorrespondto(cid:0)(cid:25) <Arg z (cid:20) 0.NotethatArg z iscontinuousonthepositivereal
semiaxisandhasajumpof2(cid:25) onthenegativerealsemiaxis.Thisisaconvenientconvention.Points
onthenegativerealsemiaxis,e.g.,(cid:0)4:7,havetheprincipalargumentArgz D (cid:25):
Tofindtheprincipalargumentofz D (cid:0)1Ci,weconvertz topolarform:
p
p
jzj D .(cid:0)1/2C12 D 2;
y 1
tan(cid:18) D D D (cid:0)1:
x (cid:0)1
Hence
(cid:18) D 3(cid:25) D 135ı:
4
Hencez,inpolarform,is
p
z D 2(cid:0)cos 3(cid:25) Cisin 3(cid:25)(cid:1):
4 4
Asexplainedneartheendofp.613,(cid:18) iscalledtheargumentofz anddenotedbyargz: Thus(cid:18) is
(cid:18) D argz D 3(cid:25) ˙2n(cid:25); n D 0;1;2;(cid:1)(cid:1)(cid:1) :
4
264 ComplexAnalysis PartD
Thereasonisthatsineandcosineareperiodicwith2(cid:25),so135ı looksthesameas135ıC360ı,etc.
Toavoidthisconcern,wedefinetheprincipalargumentArgz [see(5),p.614].Wehave
Argz D 3(cid:25):
4
Youshouldsketchtheprincipalargument.
13. Principalargument.Thecomplexnumber1Ci inpolarformis
p (cid:18) (cid:19)
(cid:25) (cid:25)
1Ci D 2 cos Cisin byProb.1.
4 4
p
Then,usingDeMoivre’sformula(13),p.616,withr D 2andn D 20,
p (cid:20) (cid:21)
(cid:16) (cid:17)20 (cid:16) (cid:25)(cid:17) (cid:16) (cid:25)(cid:17)
.1Ci/20 D 2 cos 20(cid:1) Cisin 20(cid:1) byProb.1:
4 4
D 210.cos5(cid:25) Cisin5(cid:25)/
D 210.cos(cid:25) Cisin(cid:25)/:
Hence
argz D (cid:25) ˙2n(cid:25); n D 0;1;2;(cid:1)(cid:1)(cid:1) I Argz D (cid:25):
Furthermore,notethat
.1Ci/20 D 210.cos(cid:25) Cisin(cid:25)/ D 210.(cid:0)1Ci (cid:1)0/ D (cid:0)210 D (cid:0)1024:
Graphtheprinicipalargument.
17. Conversiontox+iy.ToconvertfrompolarformtotheformxCiy,wehavetoevaluate
sin(cid:18) andcos(cid:18) forthegiven(cid:18):Here
p p p p
p (cid:18) (cid:19) p !
1 1 2 2 16 16
8 cos (cid:25) Ci sin (cid:25) D 8 Ci D C i D 2C2i:
4 4 2 2 2 2
21. Roots.FromProb.1andExample1,p.614inthissection,weknowthat1Ci inpolarformis
p
1Ci D 2(cid:0)cos 1(cid:25) Cicos 1(cid:25)(cid:1):
4 4
Henceby(15),p.617,
p (cid:16)p (cid:17)1=3(cid:18) 1(cid:25) C2k(cid:25) 1(cid:25) C2k(cid:25)(cid:19)
3 1Ci D .1Ci/1=3 D 2 cos 4 Cicos 4 :
3 3
Nowwecansimplify
p
(cid:16) (cid:17)1=3 (cid:16) (cid:17)1=3
2 D 21=2 D 21=6
and
1(cid:25) C2k(cid:25) (cid:25)=4 2k(cid:25) (cid:25) 8k(cid:25) (cid:25).1C8k/
4 D C D C D :
3 3 3 12 12 12
Chap.13 ComplexNumbersandFunctions.ComplexDifferentiation 265
Hence
p (cid:20) (cid:25).1C8k/ (cid:25).1C8k/(cid:21)
3 1Ci D 21=6 cos Cisin ;
12 12
where k D 0;1;2 (3roots;thus3valuesofk).Writtenoutweget
(cid:16) (cid:25) (cid:25) (cid:17)
Fork D 0 z D 21=6 cos Cisin :
0
12 12
(cid:18) (cid:19)
9(cid:25) 9(cid:25)
Fork D 1 z D 21=6 cos Cisin :
1
12 12
(cid:18) (cid:19)
17(cid:25) 17(cid:25)
Fork D 2 z D 21=6 cos Cisin :
2
12 12
Thethreerootsareregularlyspacedaroundacircleofradius21=6 D 1:1225withcenter0.
y
120o
~1.12
~1.12
15o
x
120o
~1.12
p
Sec.13.2. Prob.21. Thethreerootsz ;z ;z ofz D 3 1Ci inthecomplexplane
0 1 2
29. Equationsinvolvingrootsofcomplexnumbers.Applyingtheusualformulaforthesolutions
ofaquadraticequation
p
(cid:0)b˙ b2(cid:0)4ac
z D
2a
to
(Eq) z2(cid:0)zC1(cid:0)i D 0;
wefirsthave
p
(cid:0)1˙p12(cid:0)4(cid:1)1(cid:1).1(cid:0)i/ (cid:0)1˙ (cid:0)3C4i
(A) z D D :
2(cid:1)1 2
p
Now,in(A),wehavetosimplify (cid:0)3C4i:Letz D pCqi beacomplexnumberwherep;q are
real.Then
z2 D .pCqi/2 D p2(cid:0)q2C2pqi D (cid:0)3C4i:
266 ComplexAnalysis PartD
Weknowthatfortwocomplexnumberstobeequal,theirrealpartsandimaginarypartsmustbe
equal,respectively.Hence,fromtheimaginarypart
2pq D 4;
(B) pq D 2;
2
p D :
q
Thiscanthenbeused,intherealpart,
p2(cid:0)q2 D (cid:0)3;
4
p2C D (cid:0)3;
p2
p4C4 D (cid:0)3p2;
p4(cid:0)3p2C4 D 0:
Tosolvethisquarticequation,weseth D p2 andgetthequadraticequation
h2C3h(cid:0)4 D 0;
whichfactorsinto
.h(cid:0)1/.hC4/ D 0 sothat h D 1 and h D (cid:0)4:
Hence
p2 D 1 and p2 D (cid:0)4:
Sincep mustbereal, p2 D (cid:0)4isofnointerest.Weareleftwithp2 D 1so
(C) (a) p D 1; (b) p D (cid:0)1:
Substituting[C(a)]into(B)gives
pq D 1(cid:1)q D 2 so q D 2:
Similarly,substituting[C(b)]into(B)gives
pq D .(cid:0)1/(cid:1)q D 2 so q D (cid:0)2:
Wehavep D 1;q D 2 andp D (cid:0)1,q D (cid:0)2.Thus,forz D pCqi (seeabove),weget
1C2i and (cid:0)1(cid:0)2i D (cid:0).1C2i/:
Hence(A)simplifiesto
p
(cid:0)1˙ (cid:0)3C4i (cid:0)1˙p.1C2i/2 (cid:0)1˙.1C2i/
z D D D :
2 2 2
Thisgivesusthedesiredsolutionsto(Eq),thatis,
(cid:0)1C.1C2i/ 2i
z D D D i
1
2 2
and
(cid:0)1(cid:0).1C2i/ (cid:0)2(cid:0)2i
z D D D (cid:0)1(cid:0)i:
2
2 2
Verifytheresultbypluggingthetwovaluesintoequation(Eq)andseethatyougetzero.