asmar pde solution 外文学习材料

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1、Section1.1ComplexNumbers1SolutionstoExercises1.11.Wehave−i=0+(−1)i.Soa=0andb=−1.5.Wehave(2−i)2=(2+i)2(because2−i=2−(−i)=2+i)=−1!"#$=4+4i+(i)2=3+4i.Soa=3andb=4.9.Wehave=17%&%&%&!"#$1i31331i252+−i=+i−−i=−i272221427287Soa=25andb=−2.28713.Multiplyinganddividingbytheconjugateofthedenom

2、inator,i.e.by2−i=2+iweget14+13i(14+13i)(2+i)14·2+14·i+13i·2+13i2==2−i(2−i)(2+i)4+128+14i+26i−131540==+i=3+8i555Soa=3andb=8.17.Applyingthequadraticformulaweget√√0±−24−24√x==0±=0±i62221.Putu=x2.Thenwegetu2+2u+1=0.Bythequadraticformula√−2±4−4u==−1.2Thereforeweget−1=x2andthesolutionis

3、x=±i.2Chapter1APreviewofApplicationsandTechniques25.Supposewcanbewrittenintheformw=c+id.Thenfromtheidentityz+w=0wehavea+ib+c+id=0,or(a+c)+i(b+d)=0.Itfollowsthata+c=0andb+d=0.Hencec=−aandd=−b.And,thismeansw=−a−ib.(b)Letesrepresentacomplexnumbersuchthatz+es=zforallcomplexz.Showthate

4、s=0;thatis,Re(es)=0andIm(es)=0.Thuses=0istheuniqueadditiveidentityforcomplexnumbers.Solution.Letusputz=0intoz+es=z.Thisgives0+es=0,orifes=a+ibwegeta+ib=0+i0.Sincetwonumbersareequalifandonlyiftheirrealandimaginarypartsarethesamewehavea=0andb=0.So,es=0.29.(a)Wehaveanzn+an−1zn−1+···+

5、a1z+a0=anzn+an−1zn−1+···+a1z+a0=an·zn+an−1·zn−1+···+a1·z+a0[Sincea0,a1,···,an−1,anareallreal]=anzn+an−1zn−1+···+a1z+a0[byusingthepropertythatzn=(z)n]=a(z)n+a(z)n−1+···+a(z)+ann−110(b)Supposethatz0isarootofthepolynomialp(z)=anzn+an−1zn−1+···+a1z+a0.Thismeansthatazn+azn−1+···+az+a=0

6、.n0n−10100Byusingthepart(a)wehavea(z)n+a(z)n−1+···+az+a=azn+azn−1+···+az+a=0.n0n−10100n0n−10100Hence,thenumberz0isalsoarootofthesamepolynomial.33.ProjectProblem:Thecubicequation.(a)Usingthegivenchangeofvariablesx=y−aweget3'a(3'a(2'a(x3+ax2+bx+c=y−+ay−+by−+c33332a'a(2'a(32a'a(2=y−3

7、y+3y−+ay−2ay+a33333a+by−b+c332a22a2a3a3ab=y+y(−a+a)+y(−+b)+(−+−+c)332793a22a3ab=y3+y(b−)+(−+c)3273=y3+py+q=0Section1.1ComplexNumbers3ifweputp=b−a2andq=2a3−ab+c.3273(b)Nowwesubstitutey=u+vintotheequationwereceivedinthepreviousstep:y3+py+q=(u+v)3+p(u+v)+q=u3+3u2v+3uv2+v3+p(u+v)+q=u3

8、+v3+3uv(u+v)+p(u+v)+q=u3+v3+(3uv+